Plus Two

Kerala State Board New Syllabus Plus Two Zoology Previous Year Question Papers and Answers.

Kerala Plus Two Zoology Previous Year Question Paper March 2017 with Answers

Board	SCERT
Class	Plus Two
Subject	Zoology
Category	Plus Two Previous Year Question Papers

Time: 2½ Hours
Cool off time : 15 Minutes

General Instructions to Candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2½ hrs.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and the only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Question 1.
The following table shows the F₂ generation of a dihybrid cross. Identify the ‘Phenotype’ with homozygous re­cessive genotype. Find out AB: C:D.

No	Phenotype	No.of offspring (F2 gen.)
1	A</	21
2	B	7
3	e	63
4	D	21

Question 2.
values of a frugivorous bat species are given below. Which value is not applicable to continents?
1. 0.6
2. 65
3. 20
4. 0.68

Question 3.
Distinguish in-situ conservation from ex-situ conservation with one example each.

Question 4.
Which of the following pairs of STDs is completely curable?
1. HIV, Hepatitis-B
2. Hepatitis-6, Gonorrhoea
3. Syphilis, Gonorrhoea
4. Chlamydomonasgenital herpes

Question 5.
Which of the following do not have similar sex chromosomes? (Homogam­etic)
1. Human female
2. Drosophila female
3. Bird female
4. Bird male

Question 6.
Feeding…………. in the first few days is essential for preventing infections in a newly born baby.

Question 7.
L.H. and FSH are gonadotrophins. Distinguish their roles in males and females.

Question 8.
Examine the following fragment of the beta-globin chain in human hemoglobin and identify the hereditary disease with reason.

Question 9.
A population of 208 people of MN blood group was sampled and it was found that 119 were MM group, 76 MN group, and 13 NN groups. Answer the following questions:
a. Determine the gene frequencies of M and N alleles in the population.
b. How do the above frequencies af­fect evolution?
OR
Examine the pictures of Darwin’s finches given below and answer the following:
a. What phenomenon in evolution is represented in the picture?
b. Explain the phenomenon with the help of an additional example.

Question 10.
What is the advantage of   biofertiliz­ers over chemical fertilizers / Give an example for biofertilizers?

Question 11.
What is ART? Categorize the fol­lowing ARTs based on their applica­tions in male sterility and female ste­rility: GIFT, Al.

Question 12.
Which of the following sets of gases were used in Miller’s experiment?
1. CH₄, no₂, H₂0, C0₂
2. NH₃, CH₃, H₂0, H₂
3. H₂, CH₄, NH₃ , H₂O
4. H₂0, N, CH₄, H₂

Question 13.
Which of the following combinations do not apply to DNA?
a. Deoxyribose, Guanine
b. Ribose, Adenine
c. Deoxyribose, Uracil
d. Guanine, Thymine
1. (a) and (b)
2. (b) and (c)
3. (c) and (d)
4. (a) and (d)

Question 14.
Examine the diagram of mRNA given below. Mark the 5’ and 3’ end of the mRNA by giving reasons.

Question 15.
A small fragment of the skin of a differ­ent person was extracted from the nails of a murdered person. This fragment of skin led the crime investigators to the murderer. Based on this incident answer the following questions:
a. What technique was used by the investigators?
b. What is the procedure involved in this technique?
OR
In an E. coli culture, lactose is used as food instead of glucose. If so, answer the following questions:
a. How do the bacteria respond to the above situation at the genetic level?
b. If lactose is removed from the medium what will happen?

Question 16.
Morphine is said to be an abused drug. Discriminate the terms ‘use’ and ‘abuse’ of drugs based on this example.

Question 17.
Differentiate Active immunity from Passive immunity. Give an example of Passive immunity.

Kerala State Board New Syllabus Plus Two Zoology Previous Year Question Papers and Answers.

Kerala Plus Two Zoology Previous Year Question Paper March 2018 with Answers

 

Board	SCERT
Class	Plus Two
Subject	Zoology
Category	Plus Two Previous Year Question Papers

Time: 1 Hour
Cool off time : 10 Minutes
Maximum : 30 Score

General Instructions to Candidates :

• There is a ‘cool off time’ of 10 minutes each for Botany and Zoology in addition to the writing time of 1 hour each. Further, there is ‘5 minutes’ ‘Preparatory Time’ at the end of the Botany Examination and before the commencement of Zoology Examination.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’ and ‘Preparatory time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before answering.
• All questions are compulsory and the only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Answer all questions from 1 – 3. Each question carries 1 Score. (Scores: 3 × 1 = 3)

Question 1.
Name the cells in testes which synthesize and secrete the androgens.
Answer:
Interstitial cells or Leydig cells

Question 2.
Different contraceptive methods are given below. Pick out the odd one.
a) CuT
b) Saheli
c) Multiload 375
d) Lippes loop
Answer:
b) Saheli

Question 3.
Expressed sequences in the gene are called
a) Introns
b) Mutons
c) Exons
d) Cistrons
Answer:
c) Exons

II. Answer any nine questions from 4 – 14. Each question carries 2 Scores. (Scores: 9 × 2 = 18)

Question 4.
In a classroom discussion, a student said that the sex of the baby is determined by father. Analyse the statement and give reason for it.
Answer:
Chromosome pattern of human female is XX and male is xy
Haploid gametes produced by female contain only X chromosome while male produces either x or y.
Hence father determines sex of a child

Question 5.
Different contraceptive methods are used to control population explosion. Summarise the natural method and barrier method of contraception.
Answer:
a) Natural methods:
Periodic abstinence, Coitus interruptus, etc

b) Barrier method:
Use of condoms, diaphragms, etc

Question 6.
DNA is tightly packed structure and is found as units called nucleosomes.
a) Explain the concept of nucleosomes.
b) Differentiate between euchromatin and heterochromatin
Answer:
a) Nucleosomes are histone octamer, in this negatively charged DNA wrapped around it.

b) Euchromatin:

• Here, chromatin region is loosely arranged and they are transcriptionally active Heterochromatin:
• Here, Hchromatin region is densely packed and transcriptionally inactive

Question 7.

a) Observe the above cross and name the phenomenon.
b) Write down the theoretically given explanations of the phenomenon.
Answer:
a) Incomplete dominance
b) It is the blending of characters, that lead to the production of intermediate flower colour-pink.
In this less efficient enzyme is produced.

Question 8.
Complete the boxes with the suitable words given below:
[Analogus, Homologus, Convergent evolution, Divergent evolution]

Answer:
A – Homologous organ
B – Divergent evolution
C – Analogous organs
D – Convergent evolution

Question 9.
Explain the factors affecting the Hardy-Weinberg equilibrium.
Answer:

• Gene flow
• Genetic drift
• Mutation and
• Natural selection

Question 10.
Innate immunity is a non-specific type of defense and consists of four types of barriers. Categorize the barriers and give one example for each.
Answer:
i) Physical barriers:
It includes skin or mucous coating of respiratory system

ii) Physiological barriers:
Acid in stomach or saliva

iii) Cellular barriers:
It includes natural killer cells and monocytes

iv) Cytokines barriers:
It includes interferon

Question 11.
Complete the table given below:

Answer:
A – Salmonella
B – common cold
C – plasmodium falciparum
D – elephantiasis

Question 12.
Consumption of drug and alcohol affects the persons mental and physical health very badly. List the warning signs of alcohol or drug abuse.
Answer:

• Drop in academic performance
• Lack of interest in personal hygiene
• Depression
• Withdrawal symptoms

Question 13.
Identify the disadvantages of RNA over DNA as a genetic material and explain it.
Answer:

• Hydroxyl group of RNA is reactive and easily degradable
• RNA is catalytic and highly reactive
• Mutation takes place at a faster rate
• Presence of uracil makes RNA less stable

Question 14.
Complete the table with appropriate terms.

Answer:
A – Aspergillus Niger
B – bacterium
C – cyclosporine A
D – Monascus purpureus

III. Answer any three from 15 – 18 questions. Each question carries 3 scores. (Scores: 3 × 3 = 9)

Question 15.
Observe the graph and answer the following questions:

a) Name S, A, Z and C in the graph.
b) Name the scientists who explained Species – Area relationship.
Answer:
a) S – Species richness
A – Area
Z – Regression coefficient
C – Y Intercept

b) Alexander Von Humboldt

Question 16.
a) In lac Operon lactose act as inducer molecule. Evaluate the statement and explain it.
b) Observe the diagram of lac Operon and identify labelled parts A, B, C and D.

Answer:
a) Lactose that binds with repressor and make structural modification hence it cannot bind with operator therefore lactose switch off Lac Operon act as inducer.

b) A – Repressor mRNA
B – Bete galactosidase
C – Permease
D – Transacetylase

Question 17.
Sexually transmitted disease (STD) are mainly transmitted through sexual contact.
a) Name any two example of sexually transmitted disease.
b) Explain any two methods adopted to prevent STDs.
Answer:
a) Gonorrhea, syphilis
b)

1. Avoid sex with multiple partners
2. Always use condoms during coitus

Question 18.
Hemophilia, Sickle cell anemia, and Phenyl Ketonurea are Mendelian disorders.
a) What do you mean by Mendelian disorder?
b) Which one of the above is an example of inborn error of metabolism? Mention the cause of the disorder.
Answer:
a) Mendelian disorder is determined by a mutation in a single gene
b) Phenylketonuria.
It is an autosomal recessive trait in which lack of enzyme for the conversion of phenylalanine to tyrosine.

Kerala Plus Two Botany Model Question Paper 5

Time: 1 Hours
Cool off time : 10 Minutes

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Your are not allowed to write your answers nor to discuss anything with others during the ‘cool off,time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question’itself.                  ,
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer all the questions. Each question carries 1 score.

Question 1.
Insects feeding on plant sap and other parts are known to be?
.

Question 2.
Name the first transgenic cow.

Question 3.
The hard outer layer of pollengrain is com-posed of
a. cellulose
b. Pectin
c. Suberin
d. Sporopollenin

Question 4.
MOET is a programme for herd improvement. Name the hormone used in it

Answer any 10 questions. Each question carries 2 score.

Question 5.
Artificial hybridisation is one of the major approaches of crop improvement program-me. Suggest the techniques used to protect the stigma of bisexual flowers during hybridisation.

Question 6.

Observe the above figure
a. Identify i and ii
b. Distinguish i and ii

Question 7.
Bee keeping though relatively easy does require some specialised knowledge. List out the important points for its success.

Question 8.

Question 9.
List out any four changes that occur in flower after fertilisation

Question 10.
Isolation of DNA from plant cell involves many steps. Explain the different steps.

Question 11.
Flowering plants have developed many devices to discourage self pollination and to encourage cross pollination. List out such features found in plants.

Question 12.
Succession takes place on bare rock is called Xerarch succession. List out its sequential stages

Question 13.
In angiosperms during fertilization two types of fusion occurs in the embryo sac.
a. Name the types of fusion
b. Which are the nuclei involved in each fusion.

Question 14.
Match the column A with B

Question 15.
The first clinical gene therapy was given to a four year old girl child,
a. What was her genetic disorder?
b. Briefly describe the clinical procedure adopted in this case.

Question 16.
High level of noise is considered as pollution.
a. What are its effects?
b. How it can be controlled?

Answer any 2 questions. Each question carries 3

Question 17.
Identify the figure given below and explain the principles behind it.

Question 18.
For reducing the air pollution in delhi the entire fleet of public transport buses were converted to CNG
a. Exand CNG
b. List out its advantages.

Question 19.
Identify the following interactions
a. Barnacles on whale
b. Wasp on fig.
c. Ticks on dog
d. Abigdon tortoise and goats
e. Tiger and deer
f. Mycorrhiza

Answers

Answer 1.
Phytophagous

Answer 2.
Rosie

Answer 3.
Sporopollenin

Answer 4.
FSH (Follicle Stimulating Hormones)

Answer 5.
This is achieved by emasculation and bagging techniques. If the female parent bears bisexual flowers, removal of anthers from the flower bud before the anther dehisces using a pair of forceps is necessary. This step is referred to as emasculation. Emasculated flowers have to be covered with a bag of suitable size, generally made up of butter paper, to prevent contamination of its stigma with unwanted pollen. This process is called bagging. When the stigma of bagged flower attains receptivity, mature pollen grains collected from anthers of the male parent are dusted on the stigma, and the flowers are rebagged, and the fruits allowed to develop.

Answer 6.
a. i. Proinsulin
ii. Insulin
b. Insulin used for diabetes was earlier extracted from pancreas of slaughtered cattle and pigs. Insulin from an animal source, though caused some patient to develop allergy or other types of reactions to the foreign protein. Insulin consists of two short polypeptide chains. Chain A and chain B, that are linked together by disulphide bridges. In mammals, including humans, insulin is synthesised as a pro-hormone which contains an extra stretch called the C peptide.

Answer 7.

• Knowledge of the nature and habits of bees.
• Selection of suitable location for keeping rtthebeehives.
• Catching and hiving of swarms (group of bees).
• Management of beehives during different seasons.
• Handling and collection of honey and of beewax.

Answer 8.
Inverted pyramid of biomass. Pyramid of biomass shows relationship between producers and consumers in an ecosystem terms of biomass. If can be inverted in case of pond ecosystem. If shows small standing crop of phytoplankton supports large standing crop of zooplankton

Answer 9.

1. Zygote formed changes into embryo
2. Ovule changes into seed
3. Ovary develops into fruit and
4. Sepals and petals neither away

Answer 10.
The separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation (you cannot see pure DNA fragments in the visible light and without staining). You can see bright orange coloured bands of DNA ! in a ethidium bromide stained gel exposed to UV light. The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece. This step is known as elution. The DNA fragments purified in this way are used in constructing recombinant DNA by joining them with cloning vectors.

Answer 11.

1. Release of pollen much before or much after the stigma is ready to accept it.
2. Placement of anther & stigma at such positions so that the pollens do not reach the stigma of the same plant or the same flower.
3. It is a chemical method in which the stigma inhibits the growth of the pollen tubes into the pistil. This process 15 known as self-incompatibility.
4. Production of flowers or plants that are uni-sexuel.

Answer 12.

• Pioneer species are the first ones to in vade a bare area, i.e., on rocks.
• These are usually lichens which secrete acids to dissolve rock, helping in weathering and soil formation.
• This paves way for some very small plants like bryophytes, which are able to grow in the small amount of soil.
• Over the time, they are succeeded by bigger plants and after several more stages, ultimately a stable climax forest community is formed.
• The climax community remains stable if the environment remains unchanged.
• With time, the xerophytic habitat gets converted into a mesophytic one.

Answer 13.
a. Syngamy & Triple fusion
b. Syngamy: The pollen tube releases two male gametes into a cytoplasm of a synergid out of which one male gamete fuses with the egg cell to form a diploid zygote. This process is called syngamy.
Triple Fusion : The other male gamete fuses with the two polar nuclei in the central cell to form triploid PEN (Primary Endosperm Nucleus). This process is called triple fusion.

Answer 14.
1 – d
2 – a
3 – c
4 – b

Answer 15.
a. Adenosine deaminase (ADA) deficiency
b. In some children ADA deficiency can be cured by bone marrow transplantation; in others it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. But the problem with both of these approaches that they are not completely curative. As a first step towards gene therapy, lymphocytes 18. from the blood of the patient are grown in a culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. However, as 19. these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes. However, if the gene isolate from marrow cells producing ADA is introduced into cells at early embryonic stages, it could be a permanent cure.

Answer 16.
a.

• It may cause sleeplessness, increased heart rate, breathing problem, feeling of stress.
• Discomfort, permanent hearing loss and other psychological and physiological disorders,

b.

1. Turn off appliances at Home & Offices
2. Use ear plugs
3. Follow the limits of noise level
4. Go Green by planning trees
5. Regularly check noise level
6. Hotify authorities about disobedience of noise rules.

Answer 17.
Gel electrophoresis The cutting of DNA by restriction endonucleases results in the fragments of DNA. These fragments can be separated by a technique known as gel electrophoresis. Since DNA fragments are negatively charged molecules they can be separated by forcing them to move towards the anode under an electric field through a medium/ matrix. Nowadays the most commonly used matrix is agarose which is a natural polymer extracted from sea weeds. The DNA fragments separate (resolve) according to their size through sieving effect provided by the agarose gel. Hence, the smaller the fragment size, the farther it moves.

Answer 18.
a. CNG – Compressed Natural Gas
b.

1. CNG burns most efficiently, unlike petrol, petrol or diesel
2. CNG is cheaper than petrol or diesel
3. It cannot be siphoned off by thieves.

Answer 19.
a. Commensalism
b. Mutualism
c. Parasitism
d. Competition
e. Predation
f. Mutualism

Plus Two Botany Previous Year Question Papers and Answers

Kerala Plus Two Botany Model Question Paper 4

Time: 1 Hours
Cool off time : 10 Minutes

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Your are not allowed to write your answers nor to discuss anything with others during the ‘cool off,time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question’itself.                  ,
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

(Questions 1 to 2): Answer all questions. Each question carries 1 score.

Question 1.
A unisexual flower having no andreorecium is called
a. Dithecous
b. Dioecious
c. Monoecious
d. Pistillate

Question 2.
In honey bees and some lizards female gamete undergoes development to form new oroanisms without fertilization This
phenomenon is called…………

(Questions 3 to 15): Answer any eleven quesions from the following. Each questions carries 2 scores.

Question 3.
In human beings, certain diseases are caused due to genetic disorders.
a. Name the method that allows the correction of a aene defect that has been diagnosed in a child or embryo.
b. How this method has been used for treatina ADA (Adenosine deaminase) deficiency?

Question 4.
A novel strategy to prevent nematode infestation is based on ‘ RNA interference’.
a. Explain RNA interference.
b. Can you suggest, how it can be used for producing nematode resistant plant?

Question 5.
Species diversity decreases from equator towards poles. Ecologists proposed various hypothesis. Suggest two hypothesis for this phenomenon.

Question 6.
Match the following

A                         B
Zoospores      Hydra
Conidia          Sponges
Bud                Chlamydomonas
Gemmules    Penicillium

Question 7.
The gradual and fairly predictable changes in the species composition in an area is called ecological succession.
a. Name the pioneer species in the primary succession in water.
b. Give the sequence of events and climax community in the hydrarch succession.

Question 8.
Infection by nematodes cause threat to t cultivation and yield loss of tobacco plants. A strateav has been developed at RNA level to control this infestation.
a. Name the process.
b. Explain how this process works at the molecular level.

Question 9.
In maize, the chromosome number present in the meiocvte is 20. Give the number of chromosomes present in the following:
a. Maize pollen
b. Maize endosperm

Question 10.
The gradual change in the species composition of a given area leading to the formation of climax community is called ecological succession. In a rocky area,
a. What is the expected type of pioneer species?
b. How this pioneer species leads to the establishment of a stable climax community?

Question 11.
During a .study tour teacher showed the primary colonisers on the banks of the river ‘Nila’.
a. Identify the succession and justify your answer.
b. List the different stages of the identified succession.

Question 12.
Jaya read in a biotechnology book that alien DNA can be introduced into host cell by micro injection and biolistics. Explain these methods.

Question 13.
Sophie was born with a genetic disorder – ADA deficiency.
a. What is ADA deficiency?
b. Can you suggest methods to treat this ADA deficiency?

Question 14.
An article in the newspaper reports that ‘Refrigerants like Chlorofluorocarbons (CFCs) pose threat to the environment’. How CFCs are harmful to the environment?

Question 15.
Genetically Modified Organism (GMO) is always a debatable topic among scientists, academicians and public. State any four usefulness of GMOs.

Question 16.
Interspecific interaction from the interaction of populations of two different species. If we assign + for beneficial, – for detrimental and 0 for neutral interactions, copy and complete the following chart.

Question 17.
A newspaper report read like this. “Conventional agricultural products like cereals, pulses and other seeds may not be able to meet the demand of food according to the increase in population. So focus has to be shifted to alternate food sources like SCPs.
a. What are SCPs? ”
b. Give one example of SCPs.
c. What are the advantages of SCPs?

Question 18.
Restriction endonucleases are the enzymes used to cut the DNA molecules.
a. Give the general term of the specific sequences where these enzymes cut the DNA.
b. Name the enzyme that joints the foreign DNA and vector DNA.
c. Give any two procedures to introduce the recombinant DNA into the host cell.

Answers

Answer 1.
d. Pistillate

Answer 2.
Parthenogenesis.

Answer 3.
a. Gene therapy.
b. ADA deficiency can be treated using bone marrow transplantation or by enzyme therapy. In gene therapy, lymphocytes from the blood of the patient are grown in a culture outside the body. A functional ADA cDNA is then introduced into lymphocytes and return back to patient.

Answer 4.
a. RNA interference (RNAi) is the phenomenon of inhibiting activity of a gene through production of sense RNA and anti-sense RNA
b. The nematode specific genes introduced into the host plant using vectors. The introduction of DNA, it produce both sense and anti sense RNA in the host cells. These two DNA’s it form double stranded RNA that initiated RNA which silenced the specific mRNA of the nematode. Thus parasite could not survive in the host.

Answer 5.

1. Tropical regions have relatively more constant and predictable climate.
2. Tropical regions remained undisturbed for millions of years which results in greater diversification.
3. Availability of more solar energy con-tribute much productivity at tropical regions.

Answer 6.
A                     B
Zoospores     Chlamydomonas
Conidia          Penicillium
Bud                 Penicillium
Gemmules     Sponges

Answer 7.
a. Phytoplanktons
b.

1. Phytoplanktons
2. Submerged stage
3. Floating stage
4. Reed swamp stage
5. Sedge meadow stage
6. Woodland stage
7. Climax forest stage.

Answer 8.
a. RNA interference.
b. RNA interference (RNA) is the phenomenon of inhibiting activity of a gene through production of sense RNA and anti-sense RNA.

Answer 9.
a. 10
b. 30

Answer 10.
a. Lichen, ie., crustose lichen is the pioneer species.
b. Crustose lichen stage → Foliose lichen stage → moss stage → herb stage → shrub stage → forest stage.
Forest stage is a stable climax community.

Answer 11.
a. Hydarch succession.
b.

1. Phytoplanktons
2. Submerged stage
3. Floating stage
4. Reed swamp,stage
5. Sedge meadow stage
6. Woodland stage
7. Climax forest stage.

Answer 12.
Microinjection: Recombinant DNA is directly injected into the nucleus of animals cell. ,
Biolistics: Cells are bombarded with high velocity microparticles of gold or tungsten coated with DNA.

Answer 13.
a. Adenosine deaminase (ADA) enzyme is crucial for the immune system function. Absense or decrease of this enzyme cause disorder. It is due to the deletion of the gene for adenosine deaminase enzyme.’ This is called ADA deficiency,
b. Gene therapy, Bine marrow transplantation

Answer 14.
CFC released from refrigerant to the atmosphere, is split in to chlorofluro components and combined with ozone in the ozone layer and induce its decomposition. So ozone readily become O₂ molecule. This decreases the concentration of ozone in ozone layer which cause holes in the ozone layer. Through the holes, UV rays and other harmful rays can be easily fall up on earth which cause other diseases like cancer.

Answer 15.

1. The GM props have enhanced nutritional quality and yield.
2. Transgenic crops grow well in saline soil and show salt tolerance.
3. They increase productivity of crops generally by showing resistance against plant pathogens and to some herbicides.
4. Transgenic pig organs like heart, kidney, pancreas can be transplanted into humans.

Answer 16.
Species Species Name of interaction

Answer 17.
a. Single Cell Proteins.
b. Spirullina.
c.

1. It can be grown easily on materials like waste water, straw, sewage etc.
2. It can be serve on food rich in protein, minerals,fats, carbohydrate and vitamins.
3. It reduce environmental pollution.

Answer 18.
a. Palindrome sequences
b. DNA Ligases
c. Micro-injection : Recombinant DNA is directly injected into the nucleus of animal cells.
Biolistics or gene gun: Cells are bom barded with high velocity micro par tides of gold or tungsten coated with DNA

Plus Two Botany Previous Year Question Papers and Answers

Kerala Plus Two Botany Model Question Paper 3

Time: 1 Hours
Cool off time : 10 Minutes

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Your are not allowed to write your answers nor to discuss anything with others during the ‘cool off,time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question’itself.                  ,
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer all questions. Each question carried 1 score.

Question 1.
Bv observinq the relationship of the first, fill in the blanks.
Unisexual male flower : Staminate
Unisexual female flower : ………………

Question 2.
Johny, a plus two student is from a tribal colony with lower level of vitamins, minerals and protein deficiency. He wishes to be a plant breeder to help the public by producing new crops with high levels of vitamins, minerals, protein etc. Identify the phenomenon.

Question 3.
After syngamy and triple fusion in embryosac, embryo will be diploid and endosperm will be…………….

Question 4.
In some seeds the nucellus may be persistent. Such nucellus is called
a. Endosperm
b. Scutellum
c. Plumule
d. Perisperm

Answer any 10 questions. Each question j carried 2 score.

Question 5.
Raju is searching blight resistant plants for his crossing experiments. Unfortunately no such varieties are available.
a. Suggest a process to create such plants.
b. Justify your suggested method.

Question 6.
Given below is a table which shows the interspecific interaction. ‘ + ’ sign indicates beneficial, sign indicates detrimental and ‘0’ indicates neutral.
a. Fill in the blanks.
Species A        Species B     Interaction
–                             –               Competition
0                            –                …………………
+                            –                …………………
b. Name the interactions where one species is benefited and the other is detrimental.

Question 7.
Given below are 3 gametes a, b and c. a and b gametes undergoes fusion.

a. Identify the fusion.
b. Give the explanation for the identification.

Question 8.
Industrial effluents and domestic sewage seriously affect fresh water bodies. For protecting aquatic life Govt, of India recently declared an animal as national aquatic animal.
a. Identify the animal.
b. Distinguish biomagnification from eutrophication.

Question 9.
Copy the picture given below and mark the following:

a. Connective tissue
b. Endothecium
c. Tapetum
d. Sporogenous tissue

Question 10.
Ram singh is a conventional wheat breeder. One of the promising wheat varieties is found to be susceptible to leaf rust. , What breeding steps will be adopt to make his original promising variety resistant to leaf rust?

Question 11.
Consider pond as an ecosystem showing the number of individuals in the following ‘ categories.
Carnivores – 2500,
Producers – 15000,
Herbivores – 5000
a. Draw the pyramid of numbers in this ecosystem.
b. Comment on the energy flow in the ecosystem.

Question 12.
Meena, an environmental activist, noticed a gradual decline in the population of birds in the open agricultural fields near her place. She has heard of the excessive use of pesticides like DDT around that area.
a. What might have led to the decline of bird population in that area?
b. Name the process that has caused this phenomenon.

Question 13.
The regeneration of whole plants from any part of the plant grown under sterile conditions is called tissue culture.
a. The general term of the part of the plant taken out for tissue culture is …………
b. The capacity to generate a whole plant from any plant cell is………….

Question 14.
In summer we use air conditioners and in winter we use heaters. Here homeostasis is accomplished by artificial means. Explain four ways by which other living organisms cope with the situation.

Question 15.
Match the Column A with B:
A                        B
1. Bulbil             a. Bulbophyllum
2. Offset            b. Sponge
3. Gemmules   c. Water hyacinth
4. Leaf buds     d. Agave

Question 16.
In a debate one of the speaker reported like this. “Continuous inbreeding leads to inbreeding depression”. If so define the following:
a. Outcross
b. Cross breeding

Answer any 2 questions. Each question carried 3 score.

Question 17.
Three different flowers are given to you in the practical class.
i. Maize
ii. Vallisneria
iii. Rose
You are asked to group them based on pollinating agents. Describe the adaptations of each flower related with the agents of pollination.

Question 18.
Population interactions may be beneficial or not. Write any three interactions in detail.

Question 19.
Resistance is the ability to prevent the pathogen from causing disease.
1. Elucidate the steps in breeding for disease resistance.
2. Cite two examples for virus resistant plants.

Answers

Answer 1.
Pistillate.

Answer 2.
Biofortification: Breeding of crops for higher levels of nutritional quality.

Answer 3.
Triploid.

Answer 4.
d. Perisperm

Answer 5.
a. Mutation breeding
b. A new character can be created by mutation through the changes in the genotype of plants.

Answer 6.
a.
–              –                  – Competition
–             0                  – Ammensalism
+            0                  – Commensalism
b. Parasitism

Answer 7.
a. Isogamy
b. Two identical gametes take place in sexual reproduction.

Answer 8.
a. Gangetic Dolphin.
b. Eutrophication : Natural ageing of a lake by nutrient enrichment of its water.
Biomagnification : To the increases in concentration of the toxicant at successive trophic level.

Answer 9.

Answer 10.

1. Collection of variability.
2. Evaluation & selection of parents.
3. Hybridisation among the selected pa-rents.
4. Selection and testing of superior combinants.
5. Testing, releasing and commercialization of new characters.

Answer 11.
a.

b. Energy flow is unidirectional. The . amount of energy flow decreases with successive trophic level.

Answer 12.
a. High concentration of DDT disturbs calcium metabolism in bird which causes thinning of egg shell and their . premature breaking eventually causing decline in bird population,
b. Biomagnification.

Answer 13.
a. Explant
b. Totipotency

Answer 14.

1. Migration: It is the movement of organism from one place to another region and return when the stressful period is over.
2. Suspend: Organism like bacteria and fungi produce thick walled spores and tide over unfavourable condition.
3. Hibernation (winter sleep): It is the period of dormacy during winter, eg., cold blooded animals like frog.
4. Aestivation (summer sleep): It is the period of dormancy during summer.

Answer 15.

1. Bulbil           d. Agave
2. Offset           c. Water hyacinth
3. Gemmules  b. Sponge
4. Leaf buds    a. Bulbophyllum

Answer 16.
a. The breeding of the unrelated animals of same breed is called out-crossing. The offspring of such a mating is known as an out-cross. It is the best breeding method for animals that are below average in productivity.
b. Out-breeding is the breeding of the unrelated animals, which may be between individuals of the same breed but having no common ancestors for 4-6 generations (out¬crossing) or between different breeds (cross-breeding) or different species (interspecific hybrid i-sation).

Answer 17.
i. Maize – Anemophily – Wind pollination. Numerous pollen grains, pollen is dry and light, stigma is feathery.
ii. Vallisneria – Hydrophily – Water pollination.In vallisneria pollination by water, male flower occur in clusters when mature it detached and reach to the surface of water.
iii. Rose – Entamophily – Insect pollination. Flowers large, colourful, fragrant and rich in nactar.

Answer 18.
a. Predation : It is a kind of interaction between species in which one individual kills another for food, eg., prey is deer and predator is tiger.
Amensalism : In this, one population inhibits the arowth and development of other population.
Mutualism : In which both the species benefit, eg., Lichens (between a fungus and algae), mycorrhizae (between funai and the mots of hiaher plants).

Answer 19.
a.

1. The various sequential steps are:
   • Screening germplasm for resistance sources.
   • Hybridisation of selected parents
   • Selection and evaluation of the hybrids.
   • Testing and release of new varieties.
2. Crop – Chilli
   • Variety – Pusa Sadabahar
     Disease – Chilly mosaic virus, leaf curl
   • Crop – Tomato
     Variety – Bolseno
     Disease – Tomato mosaic virus

Plus Two Botany Previous Year Question Papers and Answers

Kerala Plus Two Botany Model Question Paper 2

Time: 1 Hours
Cool off time : 10 Minutes

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Your are not allowed to write your answers nor to discuss anything with others during the ‘cool off,time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question’itself.                  ,
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

(Questions 1 to 2): Answer all questions. Each questions carries 1 score.

Question 1.
From the following, select the two having haploid chromosome number.
a. egg
b. endosperm
c. zygote
d. pollen

Question 2.
Bt Cotton is regarded as an important achievment of genetic engineering. What does Bt stands for?

(Questions 3 to 15) : Answer any eleven questions from the following. Each questions carries 2 score.

Question 3.
Ecological pyramids are usually upright. Meanwhile some, pyramid of biomass is inverted. Explain the reason.

Question 4.
Figure representing the reactions associated with Polymer Chain Reaction (PCR). Name the steps A, B, C in the process.

Question 5.
Recombinant DNA technology can be accomplished only if we have the following key tools ie., Restriction enzymes, Polymerase enzyme, Ligases and Vectors. State the functions of
a. Ligases
b. Restriction enzymes

Question 6.
Match the following

Question 7.
Given below are the components related to simplified model of mineral cycling in a terrestrial ecosystem. Construct a flow chart.
(Hint : weathering of rock)

Question 8.
Primary succession of rocks is known as Xerosere. Answer the following related with Xerosere.
a. Name the pioneer community.
b. Organic acids have important roles in this succession. Justify.

Question 9.
An aquatic ecosystem having luxurious growth of cyanobacteria (Algal bloom) leads to eutrophication.
a. What kind of pollutants cause algal bloom to colonize the aquatic ecosystem?
b. What are the consequences of eutrophication ?

Question 10.
A list of different organisms in an ecosystem are given below. Arrange them in 1st, 2nd, 3rd and 4th trophic level.
i. Phytoplankton
ii. Man
iii. Fish
iv. Zooplankton

Question 11.
Use of thermostable DNA polymerase from the bacterium, Thermus aquaticus, made it possible to generate billion copies of DNA in a very short time using a process.
a. Name the process.
b. Name the three important steps in volved in this process.

Question 12.
Read the statements below and identify the mode of interaction between the species.
a. Tiger eating deer.
b. Butterfly feeding pollen.
c. Human liver fluke feeding on snail.
d. Lice on humans.
e. Orchid attached to a tree.
f. Mycorrhizal association of fungi and roots of higher plants.
g. Sparrow eating seed.
h. Egrets foraging close to cattle.

Question 13.
Environmentalists usually says: “There are may causes for biodiversity losses”. Illustrate four major causes of biodiversity loss.

Question 14.
In a marine ecosystem, a population of phytoplankton (1,50,000) supports a standing crop of fishes (40,000).
a. Draw the pyramid of biomass.
b. Draw the pyramid of numbers in this ecosystem.

Question 15.
Geetha resides in a city nearby a lake. Water from this lake was used for various domestic purposes earlier. Now-a- days, this water has become turbid and is with an unpleasant odour.
a. What can be the reason for this?
b. Name the scientific term that explains this effect.

(Questions 16 to 18): Write any two from the following questions. Each questions carries 3 score.

Question 16.
Raju is a diabetic patient who takes insulin injections regularly. The insulin used by such patients is produced by genetically engineered organism. Write the different steps involved in the production of insulin by genetic engineering.

Question 17.
Raman is learning the post-fertilization changes of an angiosperm embryosac with the help of slides. He identified the egg nucleus and polar nuclei with the help of his teacher.
a. Name the other nuclei present in the embryosac.
b. Help Raman by giving the changes that takes place with egg nucleus and polar nuclei after fertilization.

Question 18.
Given below is the bar diagram showing age structure of three different populations. Observe the diagram carefully and answer the following questions.

Answers

Answer 1.
egg, pollen

Answer 2.
B – Bacillus
T – thuringiensis

Answer 3.
In most ecosystems, all the pyramids, of number, of energy and biomass are upright, i.e., producers are more in number and biomass than the herbivores, and herbivores are more in number and biomass than the carnivores. Also energy at a lower trophic level is always more than at a higher level.
The pyramid of biomass in sea is also generally inverted because the biomass of fishes far exceeds that of phytoplankton. Pyramid of energy is always upright, can never be inverted, because when energy flows from a particular trophic level to the next trophic level.

Answer 4.
A – Denaturation
B – Annealing,
C – Primer extension

Answer 5.
a. Ligases are enzymes that can rejoin
cut fragments of DNA by sealing the sugar-phosphate back bone,
b. These are the molecular scissors that cut long DNA at specific points, into smaller fragments.
Restriction enzymes belong to a larger class of enzymes called nucleases.

Answer 6.
i. 4
ii. 1
iii. 2
iv. 3

Answer 7.

Answer 8.
a. Lichens are the pioneer community.
b. Lichens secrete acids to dissolve the rock, helping in weathering and soil formation.

Answer 9.
a. Due to the addition of domestic
sewage, water bodies become rich in nutrient content. Addition of nutrients stimulates profuse growth of planktonic algae (free-floating algae).
b. Luxuriant growth of algal bloom totally cover the water surface. This causes reduction in DO (dissolved oxygen content). Decomposition of algal blooms causes deficiency of 02 in water. Algal blooms also release some toxins in water. Toxins inhibit the growth of other algae in bloom infected water body. So, in this poor oxygenated water with high level of CO₂, fishes and other aquatic animals begins to die due to the toxicity.

Answer 10.

Answer 11.
a. PCR (Polymerase chain reaction).
b. Step I – Denaturation,
Step II – Annealing,
Step III – Primer extension.

Answer 12.
Predation:

1. Tiger eating deer
2. Sparrow eating seed

Commensalism:

1. Orchid attached to a tree
2. Egrets foraging close to cattle

Parasitism:

1. Human liver fluke feeds on snail
2. Lice on humans

Mutualism:

1. Butterfly feeding pollen.
2. Mycorrhizal association of fungi and roots of higher plants.

Answer 13.
Major cause of biodiversity loss are:

1. Habitat loss and fragmentation: Due to the loss of natural habitat or use of natural habitat of animals or plants for other purposes will decrease the biodiversity in that area.
2. Co-extinction: Extinction of an organism or tree may cause the extinction of other organisms or plants that depends on it.
3. Over exploitation: With the over use of natural resources the biodiversity decreases in earth.
4. Alien species invasion: The species that enter in a new area when compete with the indigenous species results in the decrease in the species or extinction of indegeneous species

Answer 14.

Answer 15.
a. Domestic sewage and fertilizers add large quantities of nutrients in the lake. The rich supply of these nutrient increased productivity of water bodies i.e., algal blooming or nutrient enrichment. The decomposed water has become turbid and is with an unpleasant odour.
b. Eutrophication.

Answer 16.

1. 1st two DNA sequences corresponding to A and B chains of human insulin are prepared.
2. They are introduced into plasmids of E-coli to produce insulin chain.
3. Chains A and B were produced separately.
4. Extracted and combined by creating disulphide bonds to form human insulin.

Answer 17.
a. Synergids and antipodals.
b. One of the two male gametes fuses with the egg to form zygote (2n).
The other male gametes fuses with the secondary nucleus to form endosperm which is triploid.

Answer 18.
a. B is stable.
b. A- Pre-reproductive: High percentage of young individual positive growth and board based age pyramid.
B- Reproductive: Growth rate is slow and stable, ie., zero growth. It is belt shaped polygon.
C- Post reproductive: Growth rate is decline, negative growth phase. An urn-shaped figure.

Plus Two Botany Previous Year Question Papers and Answers

Kerala State Board New Syllabus Plus Two Botany Previous Year Question Papers and Answers.

Kerala Plus Two Botany Model Question Paper 1 with Answers

Board	SCERT
Class	Plus Two
Subject	Botany
Category	Plus Two Previous Year Question Papers

Time: 1 Hour
Cool off time : 10 Minutes
Maximum : 30 Score

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.                  ,
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

I. Answer All questions. Each question carries 1 score. (4 × 1 = 4)

Question 1.
Insects feeding on plant sap and other parts are known to be?
Answer:
Pest

Question 2.
Name the first transgenic cow ………..
Answer:
Rosie

Question 3.
The hard outer layer of pollen grain is composed of ………….
a) cellulose
b) pectin
c) suberin
d) sporopollenin
Answer:
d) Sporopollenin

Question 4.
MOET is a programme for herd improvement. Name the hormone used in it.
Answer:
FSH

II. Answer any 10 questions. Each question carries 2 Scores. (10 × 2 = 20)

Question 5.
Artificial hybridisation is one of the major approaches of crop improvement programme. Suggest the techniques used to protect the stigma of bisexual flowers during hybridisation.
Answer:
In artificial hybridisation bagging is used to prevent the entry of unwanted pollen grains, then pollen grains from desired plant is dusted on the stigma, after this stigma again bagged, this is called rebagging.

Question 6.

Observe the above figure
a) Identify i and ii
b) Distinguish i and ii
Answer:
a) i) it contains polypeptide chains A B and C connected together
ii) it contains polypeptide chains A and B connected together but the c-peptide chain is separated

b) i) non functional insulin
ii) functional insulin

Question 7.
Beekeeping though relatively easy does require some specialised knowledge. List out the important points for its success.
Answer:

• Knowledge of the natural habitat of bees
• Selection of suitable location for keeping beehives
• Catching and having of swarms
• Handling and collection of honey and bee wax

Question 8.

Identify the above pyramid and justify your answer.
Answer:
It is called pyramid of biomass, here biomass increases in successive trophic levels.

Question 9.
List out any four changes that occur in flower after fertilisation
Answer:

• PEN develops into endosperm
• Zygote develops into embryo
• Ovary develops into fruit
• Ovule develops into seed

Question 10.
Isolation of DNA from plant cell involves many steps. Explain the different steps.
Answer:
At first plant cell wall is removed by using cellulose and pectinase enzyme, after this other components of cell can be removed by protease, lipase, carbohydrates and RNAase.

Then fine DNA Threads are suspended in chilled alcohol, it is obtained for slicing by spooling technique. Thus genetic material is isolated.

Question 11.
Flowering plants have developed many devices to discourage self-pollination and to encourage cross-pollination. List out such features found in plants.
Answer:

• Pollen release and stigma receptivity are not at the same time.
• The anther and stigma are placed at different positions.
• Self-incompatibility
• Production of unisexual flowers.

Question 12.
Succession takes place on bare rock is called Xerarch succession. List out its sequential stages.
Answer:
Lichen communities are first appear in the rock, after death and decay more amount of organic matter is added to the soil, in addition to this weathering process takes place and all this helps in the formation of next plant communities like bryophytes. Later, Herbs shrubs and finally climax vegetation (forest) appears in each stage one plant the community is replaced by other takes place.

Question 13.
In angiosperms during fertilization two types of fusion occurs in the embryo sac.
a) Name the types of fusion
b) Which are the nuclei involved in each fusion.
Answer:
a) Syngamy and triple fusion
b) In syngamy one male garnets and one egg cell is involved while in triple fusion one male gamete and two polar nuclei are involved.

Question 14.
Match the column A and B

A	B
1. Catalytic converter	a. Thermal Power plant
2. Electrostatic precipitator	b. Ozone depletion
3.Montreal protocol	c. Deforestation
4.Jhum cultivation	d. Platinum-Palladium
	e. Joint Forest Management

Answer:
1. Catalytic converter – Platinum Palladium
2. Electrostatic precipitator – thermal power plants
3. Montreal protocol – ozone depletion
4. Jhum cultivation – deforestation

Question 15.
The first clinical gene therapy was given to a four-year-old girl child.
a) What was her genetic disorder?
b) Briefly describe the clinical procedure adopted in this case
Answer:
a) Denosine deaminase deficiency

b) Before gene therapy the diseases was cured by bone marrow transplantation and in enzyme replacement therapy, after genetic engineering, Lymphocytes taken from the patient’s blood, it is cultured outside and into this functional ADA cDNA are introduced, then it is taken back into patients body or functional ADA cDNA are introduced into embryonic stage

Question 16.
A high level of noise is considered as pollution
a) What are its effects?
b) How it can be controlled?
Answer:
a) Noise causes sleeplessness, increased heart beating, and altered breathing patterns in humans.

b) Use sound-absorbing materials in industries. Govt, laws are there to reduce noise pollution around hospitals and schools by avoiding the use of air – horns, restricting the use of loudspeakers, etc.

III. Answer any 2 questions. Each question carries 3 scores. (2 × 3 = 6)

Question 17.
Identify the figure given below and explain the principles behind it.

Answer:
Gel electroforesis:
It is used to separate DNA fragments under electric field, At first the DNA fragments are loaded in well, it it moves through the gel and move towards anode according to their size and charge.

For this, DNA are stained by using ethidium bromide and it is visualised by using UV light DNA fragments are appeared as orange coloured bands. It is separated and cut out from the gel by using elusion.

Question 18.
For reducing the air pollution in Delhi the entire fleet of public transport buses were converted to CNG
a) Expand CNG
b) List out its advantages
Answer:
a) Compressed natural gas
b)

• It burns most efficiently.
• It is cheaper than petrol or diesel and
• It cannot be adulterated like petrol or diesel

Question 19.
Identify the following interactions
a) Barnacles on whale
b) Wasp on fig.
c) Ticks on dog
d) Abingdon tortoise and goats
e) Tiger and deer
f) Mycorrhizae
Answer:
a) Barnacles on whale – Commensalism
b) Wasp on fig – Mutualism
c) Ticks on dog – Parasitism
d) Abingdon tortoise and goats – Competition
e) Tiger and deer – Predation
f) Mycorrhizae – Mutualism

Kerala State Board New Syllabus Plus Two Botany Previous Year Question Papers and Answers.

Kerala Plus Two Botany Previous Year Question Paper March 2017 with Answers

Board	SCERT
Class	Plus Two
Subject	Botany
Category	Plus Two Previous Year Question Papers

Time: 2½ Hours
Cool off time : 15 Minutes

General Instructions to Candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2½ hrs.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and the only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Question 1.
A date palm seed discovered during archeological investigation retained viability even after 10000 years. The retention of viability is due to the state of inactivity of the embryo called…………

Question 2.
The plant in which adventitious buds along the margin of leaves give rise to new plants is
a. Water Hyacinth
b. Agave
c. Bryophyllum
d. Dahlia

Question 3.
Match the following varieties wit their respective crops:

Question 4.
Sequences of base pairs in DNA that reads the same on both the strands when the Orientation of reading is kept the same are called…………

Question 5.
When the pollen is transferred from anther to the stigma of the same flower,the pollination is called autogamy.
a. Cleistogamous flowers are Kinvara bly autogamous. Explain.
b. Geitonogamy is functionally cross-pollination, but genetically similar to autogamy. Justify the statement.

Question 6.
The thick protective covering of the fruit is known…………

Question 7.
Match the following

Question 8.
Insulin getting assembled into a mature form was the major challenge in commercial insulin production by rDNA technology. How did Eli Nilly Company found a solution to this problem?

Question 9.
In a given habitat, the maximum number possible for a species is called ………… of that species in that habitat.

Question 10.
A common cause of deforestation is slash and bum agriculture.
a. What is the common name attributed to such type of cultivation?
b. Explain how this type of cultivation is practiced?

Question 11.
A. Different types of population interaction have been observed in a population.
Write the type of interaction observed among the following species:

OR
B. Organisms other than human beings manage or adapt to stressful conditions by adopting different mechanisms. Explain any three mechanisms adopted by them to maintain the internal environment.

Question 12.
Breeding crops with the objective of increased nutritional quality is called…………

Question 13.
A. The following photograph shows the result of a technique showing the separation of DNA.

a. Name the technique.
b. How the separated DNA is visualized?
c. DNA fragments of size 500 bp, 1600 bp and 2000 bp are separated by this process. Which fragment will migrate fast. Why?
B. Different methods have been suggested to introduce alien DNA into host cells. Give and explain any three methods adopted for this purpose.

Question 14.
The different stages of primary succession in water are represented below. Fill the gaps that are unfilled.
a. Phytoplankton
b. ……………………
c. Submerged free-floating plant stage
d. …………………….
e………………………
f. Shrub stage
g………………………

Question 15.
Particulate matter in polluted air is removed by the application of an electrostatic precipitator. Explain the working principle of the electrostatic precipitator.

Question 16.
Nature has mechanisms to promote outbreeding in plants. Explain any two mechanisms existing in plants to promote outbreeding.

Question 17.
An ecosystem consists of the following population: Phytoplankton, Man, Fish, Zooplankton Draw a food chain denoting each trophic level.

Kerala State Board New Syllabus Plus Two Botany Previous Year Question Papers and Answers.

Kerala Plus Two Botany Previous Year Question Paper March 2018 with Answers

Board	SCERT
Class	Plus Two
Subject	Botany
Category	Plus Two Previous Year Question Papers

Time: 1 Hour
Cool off time : 10 Minutes
Maximum : 30 Score

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.                  ,
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

I. (Questions 1 to 2): Answer Alt questions. Each question carries 1 score. (2 × 1 = 2)

Question 1.
Select the disease resistant variety of Bhindi produced by rotation breeding.
a) Himgiri
b) Parbhani Kranti
c) Pusa Gaurav
d) Pusa Komal
Answer:
b) Parbhani Kranti

Question 2.
Under unfavourable conditions many zooplankton species in lakes and ponds are known to enter a stage of suspended development. Name that stage.
Answer:
Diapause

II. (Questions 3 to 15) : Answer any eleven questions from the following. Each question carries 2 Scores. (Scores : 11 × 2 = 22)

Question 3.
Ozone acts as a shield for absorbing ultraviolet radiation from the sun.
a) Name the region of atmosphere where ozone is seen.
b) Write the unit used to measure ozone layer
Answer:
a) Troposphere or Stratosphere
b) Dobson

Question 4.
Match the following organisms to their life span:

A		B
1	Butterfly	a.	140 years
2.	Crow	b.	100 – 150 years
3.	Parrot	c.	1 – 2 weeks
4.	Tortoise	d.	15 years
		e.	60 years

Answer:

A	B
Butterfly	1 – 2 weeks
Crow	15 years
Parrot	140 years
Tortoise	100 – 150 years

Question 5.
The DNA fragments can be separated using get electrophoresis.
a) Name the gel used in this technique.
b) Write the name of technique used to remove the DNA from the gel.
Answer:
a) Agarose
b) Elution

Question 6.
Observe the following figure:

a) Identify the figure
b) What does (i) tet^R (ii) rop (iii) ori denote?
Answer:
a) Cloning vector – pBr322
b) i) tet^R – tetracycline resistant gene
ii) rop – Gene codes for proteins involved in the replication of plasmid
iii) ori – Orgin of replication

Question 7.
Synergids have special cellular thickening at micropylar tip. Write the name and function of this structure.
Answer:

• Filiform apparatus
• It guides pollen tube into synergid

Question 8.
Controlled breeding experiments can be carried out using artificial insemination. What are the advantages of this process?
Answer:
The semen is collected from the male partner and injected into the reproductive tract of selected female. The collected semen can be stored in freezing state and used at later date.

Question 9.
Bio-fortification is a practical approach to improve the public health. Name any four such crops released by IARI.
Answer:
Vitamin A enriched carrot and spinach
Protien enriched beans – lablab

Question 10.
The products of ecosystem processes are named as ecosystem services. List out any four such services.
Answer:

1. Ecosystems purify air and water
2. Mitigate droughts and floods
3. Cycle nutrients
4. Provide wildlife habitat

Question 11.
In 1990’s Delhi ranked fourth among the most polluted cities of the world. What are the parallel steps taken to reduce vehicular pollution?
Answer:

1. Avoid the use of old vehicles
2. use of unleaded petrol
3. use of low-sulphur petrol and diesel
4. use of catalytic converters in vehicles

Question 12.
Increase in the level of green house gases leads to global warming. How can it be controlled?
Answer:
Global warming is controlled by

1. cutting down use of fossil fuel
2. improving efficiency of energy usage
3. reducing deforestation
4. planting trees

Question 13.
Genetically modified plants have been used in many ways. Give any four advantages of such plants.
Answer:

1. crops are tolerant to abiotic stresses
2. reduced reliance on chemical pesticides
3. helped to reduce post harvest losses.
4. increased efficiency of mineral usage by plants

Question 14.
In 1943 American Company Eli lilly produced human insulin artificially. Write down the method followed in this technique.
Answer:
American company Eli Lilly prepared Genetically engineered insulin. Steps are given below

1. Isolation of desired genes.
2. Insertion of desired Genes into plasmids of E. coli.
3. Introduction of plasmids into E coli cells.
4. Culture of E coli cells.
5. After this, a polypeptide chains A and B are separated and connected together by disulphide linkages.

Question 15.
Pyramid of energy is never been inverted, why?
Answer:
When energy flows from one trophic level to the next higher trophic level, energy is lost as heat. This is based on 10 % law.

III. (Questions 16 to 18) : Write any two from the following questions. Each question carries 3 Scores. (Scores : 2 × 3 = 6)

Question 16.
Meloidogyne incognitia is a nematode parasite infects the root of tobacco plants. It’s infection can be prevented by biotechnological methods.
a) Name the strategy.
b) Explain the principle behind this strategy.
Answer:
a) RNA interference

b) Nematode Meloidegyne incognitia infects the roots of tobacco plants and reduces the yield. Nematode-specific genes are introduced into the host plant by Agrobacterium vectors, it produce both sense and anti-sense RNA in the host cells that initiated RNAi and prevent the translation of the specific mRNA of the nematode.

Question 17.
The following graph shows two types of population growth curves:

a) Name the growth curves.
b) What does ‘K’ stand for?
Answer:
a) Exponential growth (J shaped growth curve)
Logistic growth (S shaped growth curve)

b) Carrying capacity

Question 18.
In Angiosperms female gametophyte is known as embryo sac. Explain it’s development.
Answer:
A single megaspore mother cell differentiates in the micropylar region of the nucellus. It undergoes meiotic division to form four megaspores.

Out of four, three degenerates and one megaspore is functional. The functional megaspore undergoes three repeated mitotic division to form 8 nucleate and 7 celled embryosac (female gametophyte).

Kerala Plus Two Chemistry Model Question Paper 5

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Scores

General Instructions to candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Questions from 1 to 7 carry 1 score. Answer any 6 questions.

Question 1:
The edge length and axial angles of a unit cell are a = b = c and α = β = γ = 90° respectively. Identify the type of Crystal System.

Question 2:
Name the polymer used for non-stick surface coated utensils

Question 3:
Which of the following forms colored compounds.

i. Sc³⁺
ii. Cr²⁺
iii. Zn³⁺
iv. Cu⁺
[Atomic number Sc – 21, Cr – 24, Zn – 30, Cu – 29]

Question 4:
The rate expression of a reaction is Rate = K [A]^{\(\frac { 1 }{ 2 }\)} [B]^{\(\frac { 1 }{ 2 }\)}. Calculate its overall order.

Question 5:
Among the following amines, identify the one that does not reach with Hinsberg reagent

i. C₂H₅NH₂
ii. (CH₃)₂NH
iii. (C₂H₅)₂NH
iv. (CH₃)₃N

Question 6:
The Carbohydrate, that is known as ‘animals starch’ is ………

Question 7:
Name the class of drug used for the treatment of stress and mental diseases.

Questions from 8 to 20 carry 2 score. Answer any 10 questions.

Question 8:
Write the names and structure of the monomers of the polymer Buna – S

Question 9:
Calculate the packing efficiency in body centred cubic structure.
[Given that r = \(\frac { \sqrt { 3 } }{ 2 }\)a, Where ‘r’ is the radius of sphere and ‘a’ is the edge length of unit cell]

Question 10:
Write one method for the preparation of sulphur dioxide. How is its presence detected?

Question 11:
Write the IUPAC names of following com-pounds.

(a) K₃[Fe(CN)₆]
(b) CO[Cl₂(en)₂Cl

Question 12:
Define Van’t Hoff factor ‘i’. What would be the value of ‘i’ for a dilute solution of K₂SO₄ in water?

Question 13:
By taking ethyl bromide as starting material, prepare ethyl iodide and ethyl fluoride.

Question 14:
Write the complete reaction representing the disproportionation of benzaldehyde.

Question 15:
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Question 16:
Name the products of hydrolysis of sucrose is not a reducing sugar.

Question 17:
Tetiary buty bromide react with aqueous NaOH to give tertiary butyl alcohol proceeds via S_N1 reaction. Write the mechanism of the reaction.

Question 18:
Give the structures of A and B in the following reaction.

Question 19:
Match the following

Question 20:
Write a simple chemical test to distinguish ethanal from propanone.

Questions from 21 to 29 carry 3 score. Answer any 6 questions.

Question 21:
The resistance of a 0.5 M solution of an electrolyte enclosed between two platinum electrode 1.5 cm apart and having an area of 2.0 cm² was found to be 30Ω. Calculate the molar conductivity of the solution.

Question 22:
write notes on different allotropic forms of phosphorous.

Question 23:
Integrated rate equation for a first order reaction.

a. Derive an expression for half life of first order reaction

b. Find the half life of first order reaction having rate constant K = 5.5 x 10^-14S^-1

Question 24:
What are adsorption isotherms? Write the Freundlich absorption isotherm equation and plot agraph based on it.

Question 25:
Describes the steps involved in the manufacture of sulphuric acid by contact process.

Question 26:
Differentiate ferrimagnetism from paramagnetism and predicts the change that takes place to ferrimagnetic substances on heating.

Question 27:
a. Draw the structure of dichromate ion.
b. How can dichromate ion be converted to chromate ion?
c. Write examples for the oxidising action of potassium dichromate in acidic medium.

Question 28:
Explain the following
a. Hydraulic washings
b. zone refining
c. Roasting of ores

Question 29:
Describe the following reactions.
a. Aldol condensation
b. Etard reaction
c. Rosenmund reduction

Questions from 30 to 33 carry 4 score. Answer any 3 questions.

Question 30:
Write chemical equations for the conversion of phenol to acetyl salicylic acid (aspirin)

Question 31:
With the help of diagram describe the construction and working of a fuel cell using H₂, and O₂. What are the advantages of fuel cell over conventional cells.

Question 32:
Sketch the Crystal field splittng of d-orbit- als of the Central metal ion in [CoF₆]^3-. Also write the electronic configuration of central metal ion and predict its magnetic behaviour based on crystal field theory.

Question 33:
Give the structure and IUPAC names of the products formed on the following reactions
a. Hydration of propene in the presence of dilute sulphuric acid.
b. Reaction of methanal with methyl mag-nesium bromide followed by hydrolysis.

Answers

Answer 1:
Cubic

Answer 2:
Teflon

Answer 3:
Cr²⁺(4 unpaired d electrons)

Answer 4:
Order of the reaction = \(\frac { 1 }{ 2 }\) + \(\frac { 3 }{ 2 }\) = \(\frac { 4 }{ 2 }\) = 2

Answer 5:
iv

Answer 6:
Glycogen

Answer 7:
Tranquilizers

Answer 8:

Answer 9:
In bcc

Answer 10:
Sulphur dioxide is formed together with a little (6 – 8%) sulphur trioxide when sul¬phur is burnt in air or oxygen.

S(s) + O₂(g) → SO₂(g)

SO₂ is a colourless and pungent smell gas It can detected by using potassium permag- nate solution when SO₂ is passed through an acidified potassium permanganate solu¬tion, then it decolonises the solution as it reduces MnO₄^– ions to Mn²⁺ ions. SO₂ give out suffocating smell which indicates its presence

Answer 11:
a. Potassium hexacyanoferrate (III).
b. Dichloridobis (ethane -1,2- diamine) cobalt (III) chloride.

Answer 12:
Vant Hoff factor is the ratio of the observed college five property to the normal colligative property
i = \(\frac { observed collegative property }{ Theoretical collegative property }\)
Dilute solution of K₂SO₄( would dissociate into two K⁺ ions and a sulphate ion.
K₂SO₄ → 2K⁺ + SO₄^2-
The Van’t hoff factor for a dilute solution of K₂SO₄ is 3.

Answer 13:

Answer 14:
Cannizzaro reaction which do not have an a-hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on heating with concetrated alkali. In this reaction, one molecule of the aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt.

Answer 15:

Answer 16:
Hydrolysis of sucrose gives equimolar mixture of D – (+) – glucose and D – (-) fractose.

Sucrose ¡s non-reducing sugar because aldehyde group of glucose and keto group of fractose are not fuce and involved in glycosidic bond formation.

Answer 17:
The reaction between tert-butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first of order kinetics, i.e, the rate of reaction depends upon the concetration of only one reactant, which is tert butyl bromide.
(CH₃)₃ CBr + NaOH → (CH₃)₃COH + NaBr
If occurs in two steps
Step I : The polarised C-Br bond under¬goes slow clearage to produce a carbocation and a bromide ion. The carbocation thus formed is then attacked by nucleophile in step II to complete the substitution reaction.

Answer 18:

Answer 19:
a. Antiseptic                                  –            bithional
b. Artificial sweetners                 –            Saccharin
c. Soap                                            –            Sodium stearate
d. Vitamin C                                  –           ascorbic acid

Answer 20:
Ethanal give silver mirror with tollens reagent where as propanone will not react

Answer 21:

Answer 22:
There are 3 allotropic forms of phosphorus.

White phosphrous:

• Translucent white waxy solid
•  Garlic smell
• Poisonous
•  Less stable due to angle strain. (60° PPP bond angle)
• Readily catches fire in air.
  

Red phosphorus:

• Iron grey lustrous powder
• Odourless
• Non-poisonous
•  More stable than white phosphorus
  

Black phosphorus:
It has two forms α black phosphorus and β black phosphorus. When white phosphorus is heated under high pressure at 473 K β black phosphorus is formed.

Answer 23:

Answer 24:
The relationship between extent of adsorption and pressure of the gas at constant temperature is called adsorption isotherm and is given by
\(\frac { x }{ m }\) = kp ^{\(\frac { 1 }{ n }\)} (n> 1)
Freundlich Adsorption Isotherm
where x = mass of gas adsorbed on ‘m’ mass of the adsorbent at p pressure, k, n are constants.

Observations:

• At low pressure \(\frac { x }{ m }\) ∝ p (graph is linear).
•  At high pressure \(\frac { x }{ m }\) is independent of p (graph – straight line).
• At intermediate pressure \(\frac { x }{ m }\) = kp^{\(\frac { 1 }{ n }\)}

Taking logarithm on both sides. We get

In case of a solution, the isotherm takes the form, x = k(C)^{\(\frac { 1 }{ n }\)}

Answer 25:
H₂SO₄ is manufactural by the contact process which involves three steps
(1) Burning of sulphur or sulphide uses in air to generate SO₂
S + O₂ → SO₂

(2) Conversion of SO₂ to SO₃ by the reaction with oxygen in the presence of a catalyst
2SO₂(g) + O₂(g) \(\underrightarrow { { V }_{ 2 }{ O }_{ 5 } }\) 2SO₃(g)
Absorption of SO₃in H₂SO₄ to give oleum (H₂S₂O₇)
SO₃ + H₂SO₄(dil) → H₂S₂O₇ (Oleum)
Oleum is diluted with water to form H2S04
H₂S₂O₇+ H₂O → 2H₂SO₄

Answer 26:
Ferrimagnetism is observed when the magnetic moments of the domains in the sub-stances are aligned in parallel and antiparallel directions in unusual numbers. They are weakly attracted by magnetic field as compared to ferromagnetic substances.
Paramagnetic substances are weakly attracted by a magnetic field. They are magnetised in a magnetic field in the same direction. They lose their magnetism in they absence of magnetic field. Paramagnetism is due to presence of one or more unpaired electrone which are attracted by the magnetic field.
Ferrimagnetic substance lose ferrimagnetism on heating and become paramagnetic.

Answer 27:

b. The chromate and dichromateions are interconvertible in aqueous solution depending upon the pH of the solution

Acidified potassium dichromate will oxidise iodides to iodine sulphides to sulphur.
tin (II) to tin (iv) and iron (ii) salt to iron (iii)

Answer 28:
a. It is also called gravity separation. It is based on the difference in densities (gravities) of ore and the gangue. Ore is washed with stream of water so that light impurities are washed away, eg., haematite.
b. It is based on the principle that the im-purities are more soluble in the melt than in the solid state of metal.

The impure metal is heated with the help of a circular mobile heater at one end. This results in the formation of a molten zone. As the heater is moved along the rod, pure metal crystallises out of the melt and impurities pass onto the adjacent molten zone. This is reimpurities at one end of the rod is cut off and discarded.

•  Metals like Si, Ge, Ga, B etc are refined by this method (semi-conductors and metals of high purity).

c. It is a process in which an ore is heated in regular supply of air at a temperature below the melting point of the metal so as to convert the give ore into oxide ore.
2ZnS + 3O₂ → 2ZnO + 2SO₂

Answer 29:
a. Aldehydes and ketones having at least one α hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β – hydroxy aldehydes (aldol) or β hydroxy ketones (ketol), respectively. This is known as aldol reaction. The aldol and ketol readily lose water to give α, β – unsaturated carbonyl compounds which are aldol condensation products and the reaction is called aldol condensation

b. Strong oxidising agents oxidise tolerene and derivatives to benzoic acids. How-ever, it is possible to stop the oxidation at the aldehyde stage with suitable reagents that convert the muthyl group to an inter¬mediate. Chromy chloride (CrO₂Cl₂) oxidise methyl group to a chromium complex, which on hydrolysis gives corresponding benzaldehyde.

This reaction is called Etard reaction

c. Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on burium sulphate. This reaction is called Rosenmund reduction.

Answer 30:
Kolbe’s reaction : Phenoxide ion generated by treating phenol with sodium hydroxide is even more reactive than phenol towards electrophilic aromatic substitution. Hence if undergoes electrophilic substitution with CO₂, a weak electrophite. Ortho hydroxybenzoic acid is formed as the main reaction product.

Answer 31:
Fuel Cells: They produce electrical energy directly from the combustion of fuels such as hydrogen. carbon monoxide or methane. It is a Galvanic cell.

The electrode reactions of the most popular fuel cell H₂– O₂ cell is given below:

Advantages:

• PoHution-free.
•  High efficiency (70%).
• Continuous source of energy if gas supply ‘s maintained.
•  The electrodes used in fuel cells are not affected during its working.

Answer 32:
[C0F₆]³Here cobalt is in the +3 oxidation state.

But F^– is a weak ligand. So pairing does not take place. Hence in metal ion under goes sp³d² hybridisation. There are four unpaired electrons present, therefore the complex is paramagnetic.

Answer 33:
a. When hydration of propene in the presence of dilute sulphuric acid. If gives


The grignard reagents (R-Mg -X) react with compounds containing \(\begin{matrix} | \\ C \\ | \end{matrix}\) = o to form addition compounds, which on hydrolysis with dil acid give the corresponding alcohols.Methanal react with grignai reagent to give a primary alcohol.

Plus Two Chemistry Previous Year Question Papers and Answers

Kerala Plus Two Chemistry Model Question Paper 4

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Scores

General Instructions to candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

(Questions 1 to 7): Carry one score each. Answer all questions.

Question 1.
Saline water is a solution.

a. Isotonic
b. Hypertonic
c. Hypotonic
d. Heavy water

Question 2.
Name the law which states that partial pressure of a gas is directly proportional to its mole fraction.

Question 3.
A crystalline solid

a. changes abruptly from solid to liquid
b. has no definite melting point
c. Undergoes deforestation of its geometry easily
d. has an irregular 3dimensional arrangement.

Question 4.
Galvanization of iron denotes coating with

a. Al
b. Sn
c. Cd
d. Zn

Question 5.
Write an example of an emulsion.

Question 6.
The formation of micelles takes place on above

a. Inversion temperature
b. Boyle temperature
c. Critical temperature
d. Kraft temperature

Question 7.
Complete the reaction.

(Questions 8 to 20) : Answer any ten. Each questions carries two scores.

Question 8.
Vapour pressure of a solution is different from that of the pure solvent.
i. Name the law which helps us to determine partial vapour pressure of a volatile component in solution.
ii. State the above law?

Question 9.
Zero-order reaction means that the rate of a reaction is independent of the concentration of reactants.
i. Write an example of a zero order reaction.
ii. Write the integral rate expression for the zero order reaction, R → P

Question 10.
a. How will you distinguish primary and secondary alcohols using Luca’s test?
b. Write the correct pair of reactants for the preparation of butyl ethyl ether by Williamson synthesis.

Question 11.
Two chemical reactions are given below:
1. Identify the products of each reaction.
2. Give the name of each reaction.

Question 12.
200 cm³ of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of a solution of 300 K is found to be 8.3 × 10^-2 bar. Calculate the molar mass of protein. (R= 0.083 L bar K^-1 mol^-1)

Question 13.
i. The number of moles of the solute per kilogram of the solvent is
a. Mole fraction
b. Molality
c. Molarity
d. Molar mass
ii. ‘The extent to which a solute is dissociated or associated can be expressed by Can’t Hoff factor.’ Substantiate the statement.

Question 14.
Presence of excess sodium makes NaCl crystal coloured. Explain on the basis of crystal defects.

Question 15.
Explain the following:
i. Esterification
ii. Williamson Synthesis

Question 16.
Aldehydes, Ketones and Acids contain

a. Name the product obtained by the reaction between Acetic acid and Ethanol.
b. Give any two tests to distinguish between aldehydes and ketones.

Question 17.
a. Among the following which is not a colligative property?
i. Osmotic pressure
ii. Bevation of boiling point
iii. Vapour pressure
iv. Depression of freezing point
b. What is the significance of Can’t Hoff factor?

Question 18.
One of the fuel cells uses the reaction of hydrogen and oxygen to form water. Write down the cell reaction taking place in the anode and cathode of that fuel cell.

Question 19.
The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, nonelectrolyte solid weighing 0.5g when added to 39 g of benzene (molar mass 78 g/mol), vapour pressure becomes 0.845 bar. What is the molar mass of the solid substance?

Question 20.
a. Write down the ionization isomer of [CO(NH₃)₅Cl] SO₄
b. Write the IUPAC name of the above compound.

(Questions 21 to 29): Answer any seven. Each question carries three scores.

Question 21.
i. Which of the following oxidation state is ! not shown by Manganese?
a. + 1
b. +2
c. +4
d. +7
ii. Represent the structure of dichromate ion.
iii. Potassium permanganate (KMnO₄) is a strong oxidizing agent. Write any two oxidizing reactions of KMnO₄.

Question 22.
Account for the following:
i. H₂O is a liquid while H2S is a gas.
ii. Noble gases have very low boiling points.
iii. NO₂ dimerises to N₂O₄.

Question 23.
How will you bring below the following conversions? (Write the chemical equations)
i. Ethanol → Ethanoic acid
ii. Benzamide → benzoic acid
iii. Benzaldehyde → meta nitro benzaldehyde

Question 24.
Amines are classified as primary, secondary and tertiary.
a. Write the IUPAC name of the following compound: NH₂-(CH₂)₆ NH₂
b. Which is stronger base CH₃NH₂ or C₆H₅NH₂? Why?

Question 25.
a. Which named reaction is used to reduce CH₃COCl to CH₃CHO?
b. Aldehydes and Ketones undergo reactions due to the presence of hydrogen I atom.
i. Write the name of the reaction of aldehydes which take place only because of the presence of a hydrogen atom.
ii. How will you bring about the above reaction?

Question 26.
Amines are classified as a primary, secondary and tertiary amine.
a. How will you convert nitrobenzene to aniline?
b. Represent the structure of secondary and tertiary amine.
c. Aniline does not undergo Friedel Craft’s reaction. Why?

Question 27.
Nucleophilic substitution reactions are of two types of SN reactions and SN reactions,
a. Write any 2 differences between \({ S }_{ N }^{ 1 }\) and \({ S }_{ N }^{ 2 }\) reactions.
b. Write any 2 reasons for the less reactivity of aryl halides towards nucleophilic substitution reactions.

Question 28.
Primary, secondary and tertiary amines can be distinguished using Hinsberg’s reagent.
i. What is Hinsberg’s reagent?
ii. How will you distinguish primary, secondary and tertiary amines using Hinsberg’s reagent?

Question 29.
Schottky defects and Frenkel defects are two stoichiometric defects shown by crystals.
i. Classify the following crystals into those showing Schottky defects and Frenkel defects: NaCl, AgCl, CsCl, CdCl₂
ii. Name a crystal showing both Schottky defect and Frenkel defect.

(Questions 30 to 33): Answer any three. Each question carries four scores.

Question 30.
a. Account for the following:
i. NH₃ acts as a Lewis base.
ii. PCl₃ fumes in moist air.
b.
i. Suggest any two fluorides of Xenon.
ii. Write a method to prepare any one of the above-mentioned Xenon fluorides.

Question 31.
Some elements in block show allotropy.
a. What are the allotropic forms of sulphur?
b.
i. How will you manufacture sulphuric acid by contact process?
ii. What is an interhalogen compound?

Question 32.
Potassium dichromate is an orange coloured crystal and is an important compound used as an oxidant in many reactions.
a. How do you prepare K₂Cr₂O₇ from chromite ore?
b. How will you account for the colour of ! potassium dichromate crystals?

Question 33.
a. Most important chemical reactions of haloalkanes are their substitution reactions.
i. What is \({ S }_{ N }^{ 1 }\) reaction?
ii. Arrange the four isomeric Bromo butanes in the increasing order of their reactivity towards SN1 reaction.
b. How will you prepare chlorobenzene from benzene diazonium chloride?

Answers

Answer 1.
hypertonic

Answer 2.
Henry’s law

Answer 3.
a

Answer 4.
d

Answer 5.
MIlk

Answer 6.
d

Answer 7.
Bakelite

Answer 8.
i. Raoult’s law.
ii. For a solution of volatile liquids, the partial vapour pressure of each component ¡n the solution is directly proportional to its mole fraction of component in the solution.

Answer 9.
i. Decomposition of ammonia gas on a hot platinum surface.
ii. K = [R₀]-[R]; where R₀ is the initial concentration of the reactants, R is the concentration at a time and K is the rate constant.

Answer 10.
a. Alcohols are soluble in Lucas reagent and their halides are immiscible and form turbidity 3° alcohols produce turbidity immediately 2° alcohols produce turbidity within a few minutes whereas 1° alcohol does not produce turbidity at all at room temperature.

Answer 11.

2. Clemmensen reduction, Gattermann Koch reaction.

Answer 12.

Answer 13.
i. b. Molality
ii. In the case of dissociation, the observed molecular weight has a lower value than the normal molecular weight while in the case of an association, the observed molecular weight is higher than the normal molecular weight. In order to explain such j abnormal cases, Can’t Hoff factor is used. This factor is defined as

Answer 14.
Presence of excess sodium makes NaCl crystal coloured, this is caused by metal excess defect due to anionic vacancies. When NaCl crystal is heated in an atmosphere of sodium vapour, the sodium j atoms are deposited on the surface of the crystal and combine with Na atoms to give NaCl. The released electrons diffuse into the crystal and occupy an anionic site. As a result of the crystal now has an excess of sodium. The anionic sites occupied unpaired electrons are F centres. They impart a yellow colour to the crystals of NaCl.

Answer 15.
i. Alcohols and phenols react with carboxylic acids, acid chlorides and acid anhydrides to form esters.

ii. It is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.

Answer 16.
a. Acetic acid and ethanol react in the presence of H₂SO₄/H₂O to form ethyl acetate. This is called esterification.

b. Aldehydes answer Tollen’s test. On warming an aldehyde with Tollen’s reagent, a bright silver mirror is produced due to the formation of silver metal. Ketones do not answer this test, in presence of HCl.

Fehling’s test: Aldehydes on heating with Fehling’s solution give a reddish brown precipitate. Ketones do not answer this.

Answer 17.
a. Vapour pressure.
b. In the case of dissociation, the observed molecular weight has a lower value than the normal molecular weight while in the case of an association, the observed molecular weight is higher than the normal molecular weight. In order to explain such abnormal cases, Can’t Hoff factor is used.

Answer 18.
The electrode reactions are:

Answer 19.

Answer 20.
a. [Co(NH₃)₅SO₄]Cl
b. Pentaamminesulphatocobalt (II) chloride

Answer 21.

Answer 22.
i. Due to the presence of H-bonding in between H₂O molecule they exist as a liquid. But no such H bonding is there In H₂S due to less electronegative nature and large size compared to oxygen.
ii. Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low melting points.
iii. NO₂ contains an odd number of valence electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N.O. molecule with an even number of electrons.

Answer 23.

Answer 24.
a. Hexane 1,6 diamine.
b. The stronger base is CH₃NH₂. Presence of + 1 group (CH₃) increases the electron density on Natom and thus basicity increases.

Answer 25.
a. Rosenmund reduction.
b.
i. Aldol condensation.
ii. Aldehydes and ketones having at least one a hydrogen atom in presence of dilute alkali given hydroxy aldehydes.

Answer 26.


c. The catalyst in FriedalCraft’s reaction is any. AlCl₃, it, is a Lewis acid so get coordinated with electron pair of N atom in aniline. So an electrophile cannot attack the ring.

Answer 27.
a.

\({ S }_{ N }^{ 1 }\)	\({ S }_{ N }^{ 2 }\)
The rate of the reaction depends only on the molar concentration of the substrate.	Rate depends on the molar concentration of the alkyl halide and the OFF ions.
Takes place in two steps.	Takes place in a sin­gle step.

b.
1) Aryl halides are electron rich.
2) In aryl halides, the halogen is attached to an sp2 hybridised C, which is highly stable.

Answer 28.
i. Benzene sulphonyl chloride.
ii. 1° amine react with Flinsberg’s reagent to form sulphonamides which is soluble in alkali. 2° amines react with Hinsberg’s reagent forming sulphonamides insoluble in alkali. 3° amines do not react with Flinsberg’s reagent.

Answer 29.
i. Schottky defect : NaCl, CsCl Frenkel defect: AgCl, CdCl₂
ii. AgBr.

Answer 30.
a. i. The nitrogen atom in NFH₃ has one lone pair of electrons which is available for donation. So it acts as a Lewis base,
ii. PCl₃ undergo hydrolysis in presence of moisture giving fumes of HCl.

b.

Answer 31.
a. Rhombic sulphur or sulphur and monoclinic sulphur or p sulphur,
b.
i. Contact process.
Burning of sulphur in the air to generate SO₂. Conversion of SO₂ to SO₃ by there action with oxygen in the presence of a catalyst (V₂O₅).
Absorption of SO₃ in H₂S0₄ to give Oleum. Oleum with water gives H₂SO₄.
ii. Halogens can react themselves to form binary compounds known as interhalogen compounds. They are more reactive than halogens because in the difference in electronegativity of the combining halogens.

Answer 32.
a. K₂Cr₂ O₇ is prepared from chromite ore by heating with Na₂CO₃ in the presence of air when sodium chromate is formed.

The soluble Na₂CrO₄ on acidification with H₂SO₄ produces sodium dichromate which on treating with KCl gives potassium dichromate.

b. Coloured property of potassium dichromate is due to the presence of unpair electron when light falls on the compound the unpair electron undergoes dd electron transmission. This creates the colour for the potassium dichromate.

Answer 33.
a. i. First, the carbocation is formed (slow, rate determining) and then nucleophile attacks (fast). Order of reactivity of alkyl halides is 3°> 2° > 1 “(due to decreasing stability of carbocation). It results in racemization. ie., retention as well as inversion.

Plus Two Chemistry Previous Year Question Papers and Answers

Kerala Plus Two Chemistry Model Question Paper 3

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Scores

General Instructions to candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

(Questions 1 to 7): Carry one score each. Answer all questions.

Question 1.
The polymer used in orthopaedic devices j and in controlled drug release is

Question 2.
Which one among the following is not an analgesic?

a. Ibuprofen
b. Naproxen
c. Aspirin
d. Valium

Question 3.
Addition of Grignard reagent to an aldehyde j or ketone gives an ;

a. anomer
b. ether
c. ester
d. alcohol

Question 4.
Following graph is a plot of the rate of a reaction vs concentration of the reactant, What is the order of the reaction?

Question 5.
How will the pH of brine ( aq. NaCl solution) effect when it is electrolysis?

Question 6.
Which stoichiometric defect in crystal increased the density of a solid?

Question 7.
Given an example of a material used for making semipermeable membrane for carrying out reverse osmosis.

(Questions 8 to 20) : Answer any ten. Each questions carries two scores.

Question 8.
a. Identify an analgesic from the following:
i. Equanil
ii. aspirin
iii. serotonin
iv. cimetidine
b. Differentiate between antiseptics and antibiotics.

Question 9.
a. Among the following which one is chlorine containing insecticide?
(i) DDT
(ii) Freon
(iii) Phosgene
(iv) Iodoform
b. Haloarenes undergo WurtzFitting reaction. What is WurtzFitting reaction?

Question 10.

Write the formulae of A and B in the above reaction.

Question 11.
Every substance has some magnetic properties associated with it. How will you account for the following magnetic properties?
i. Paramagnetic property
ii. Ferromagnetic property

Question 12.
Osmotic pressure is a colligative property and it is proportional to the molarity of the solution.
a. What is osmotic pressure?
b. The molecular mass of NaCI determined by osmotic pressure measurement is found to be half of the actual value. Account for it.

Question 13.
Aldol condensation reaction is a special reaction of aldehydes.
i. What is aldol condensation reaction?
ii. Write the structural formula of aldol formed from ethanol.

Question 14.
Synthetic rubber is a Vulcanisable rubber-like polymer.
i. Wite one example for synthetic rubber.
ii. Write the method of preparation of the above synthetic rubber.

Question 15.
A compound is formed by two elements P and Q. Atoms of Q (as anions) make hep lattice and those of the element P (as cations) occupy all the tetrahedral voids. What is the formula of the compound?

Question 16.
Write simple chemical tests and observations used to distinguish between the following compounds.
i. Propanal and propanone
ii. Phenol and benzoic acid

Question 17.
Calculate the osmotic pressure exerted to a solution prepared by dissolving 1.5 g of a polymer of molar mass 185000 in 500 ml of water at 37°C. [R = 0.0821 L atm K^-1 mol^-1]

Question 18.
Colligative properties can be used to determine the molecular mass of solutes in solutions.
a. What do you mean by coffigative property’?
b. For determining the molecular mass of polymers, osmotic pressure is preferred to other properties. Why?

Question 19.
The limiting molar conductivity of an electrolyte is obtained by adding the limiting molar conductivities of cation and anion of the electrolyte.
a. Name the above law.
b. What is meant by limiting molar conductivity?

Question 20.
Which are the monomers of Nylon6 and Nylon66?

(Questions 21 to 29): Answer any seven. Each questions carries three scores.

Question 21.
a. Explain how conductivity measurements help to determine the ionisation con start a weak electrolyte like acetic acid.
b. Explain the change of conductivity and molar conductivity of a solution with dilution?

Question 22.
a. Atomic sizes increase as we come down
a group, but in 4^th group of the periodic table Zr, Hf has almost the same atomic sizes. Why?
b. E° (Standard electrode potential) values generally become less negative as we move across a transition series, but E° values of Ni²⁺/Ni and Zn²⁺/Zn values are exceptions.

Question 23.
Phenols are more acidic than alcohols.
a. Name the product obtained when phenol is treated with chloroform in presence of NaOH.
b. Name the above reaction.
c. What is the product obtained when phenol is treated with conc.HNO₃?

Question 24.
a. Amines are versatile functional group useful in the preparation of many organic compounds. How can you convert?

b. A student tried to prepare nitro aniline by nitrating aniline with a cone. HNO₃ cone. H₂SO₄ mixture, but he got only m nitroaniline. Why?
c. Explain how to prepare nitro aniline from aniline?

Question 25.
Colloids exhibit certain special properties,
a. Name the property of colloid involved in j the construction of ultramicroscope?
b. Explain the above property.
c. What are the conditions to be satisfied to exhibit Tyndall effect?

Question 26.
Aromatic amines are important synthetic intermediates.
i. What are the products obtained when aniline is treated with Bromine water?
ii. How will you convert nitrobenzene to aniline?
iii. Write down the isocyanide test for the primary amines.

Question 27.
Carbohydrates can be divided into three major classes monosaccharides, oligosaccharides and polysaccharides.
a. What are polysaccharides?
b. Give two examples for polysaccharides.
c. What is invert sugar?

Question 28.
Antibiotics, antiseptics and disinfectants are antimicrobial drugs.
a. Give 2 examples of antibiotics.
b. Write one similarity between antiseptics and disinfectants.
c. Write one difference between antiseptics and disinfectants.

Question 29.
[Cr(NH₃)₄CI₂]Br is a coordination compound.
a. Identify the central metal ion of the above compound.
b. Name the ligands present in it.
c. What is the coordination number?

(Questions 30 to 33): Answer any three. Each questions carries four scores

Question 30.
Colligative properties are properties of solutions which depend on the number of solute particles irrespective of their nature.
a. Name the four important colligative properties.
b. What happens to the colligative properties when ethanoic acid is treated with benzene? Give reason.

Question 31.
The order of a chemical reaction can be zero and even a fraction, but molecularity cannot be zero or a non-integer.
i. What do you mean by the order of a reaction?
ii. What is molecularity of a reaction?
iii. The conversion of molecules ‘A’ to‘B’ follows second order kinetics. If concern traction of ‘A’ is increased to three times, how will it affect the rate of formation of ‘B’?

Question 32.
The geometry and magnetic properties of complexes can be explained by VB Theory. The octahedral complex [CO(NH₃)₆]³⁺ is diamagnetic, while the octahedral complex [COF₆]^3-is paramagnetic. Explain using VB theory?

Question 33.

a. What is its IUPAC name?
b. Explain the conversion of the above acid into the following.

Answers

Answer 1.
d

Answer 2.
Valium

Answer 3.
d

Answer 4.
Zero-order reaction

Answer 5.
It will increase due to the formation of NaOH.

Answer 6.
Interstitial defect

Answer 7.
Cellulose acetate

Answer 8.
a. ii. aspirin
b. Antiseptics are substances which prevent the growth of microorganisms or kill them but are not harmful to the living human tissues. Antibiotics are drugs used to treat infections in our body. it inhibits the growth of microorganisms.

Answer 9.
a. ii.DDT
b. When aryl halide is heated with an alkyl halide in the presence of sodium in dry ether, a halogen atom is replaced by an alkyl group and alkylarene is formed. This process is called Wurtz Fitting reaction.

Answer 10.

Answer 11.
i. Paramagnetic property is the property formed by the substance as it is early attracted by the magnetic field m.field). Magnetic alignments are in same direction, eg., O₂, Fe³⁺.
ii. ferromagnetic property is the property formed by the substance due to strongly attracted by the magnetic field. Magnetic alignments are in same direction, eg., CO, Ni.

Answer 12.
a. osmotic pressure is the external pressure squired to stop the flow of solvent mole ule from higher concentration to lower concentration through a semipermeable membrane. It is a colligative property.
b. dilution no. of molecules increases, therefore mass decreases. On dilution, laCI dissociates to two molecules? So a solar mass decreases to half.

Answer 13.
i. Idol condensation reaction. dehyde and Ketones having one hydrogen react in the presence of dilute skill to form aldol and Ketol, which no further reaction and release water molecule to form α, β unsaturated compound.

Answer 14.
i. Neoprene.
ii. By the polymerisation of chlorine

Answer 15.
No. of atoms of Q = N No. of atoms of P = 2N
∴ The ratio of atoms of P & Q = 21
∴ The formula = P₂Q

Answer 16.
i. Tollen’s test
ii. Iodoform test

Answer 17.

Answer 18.
a. Properties which depend on the number of particles and not on it nature are called colligative property
b. This is because in the osmotic pressure method, the pressure is measured at room temperature and molarity the solution is used instead of mole

Answer 19.
a. Kohlrausch’s law.
b. It is the molar conductivity of electrolytes at zero concentration or at infinite dilution.

Answer 20.
Nylon-6 Caprolactam
Nylon-66 Hexamethylenediamine and adipic acid

Answer 21.
a. The degree of dissociation can be determined by the equation, α = \(\frac { { \lambda }_{ m } }{ { \lambda }_{ 0 }^{ m } }\) To obtain ionisation constant, K_a, sub stitute a in the following equation, \({ K }_{ a }=\frac { { Ca }^{ 2 } }{ \left( 1-\alpha \right) }\) ;where C is the concentration a of acetic acid in moles/litre.
b. Conductivity of a solution decreases with a decrease in concentration as the number of ions present per unit volume of the solution actually decreases. The molar conductivity of electrolytes increases with a decrease in concentration.

Answer 22.
a. Due to lanthanoid contraction, size of 2nd series Zr and 3^rd series Hf elements are almost similar.
b. The exceptional behaviour of Ni is due to its high negative enthalpy of hydration. Since Zn has completely filled orbitals, it E? value is more negative than expected.

Answer 23.
a. Salicylaldehyde.
b. Reimer-Tiemann reaction.
c. Picric acid is formed.

Answer 24.

b. Due to the formation of anilinium ion, since -N⁺H₃ is a meta directing group, meta nitro aniline is obtained.
c. NH₂ group is protected from protonation by acetylation.

Answer 25.
a. Tyndall effect.
b. The phenomenon of scattering of light by colloidal particles when a beam of light is passed through it is called the Tyndall effect.
c.
i. The size of the colloidal particles and the wavelength of the light used sho could be comparable.
ii. There should be a very large difference between the dispersed phase arid I dispersion medium.

Answer 26.

iii. Carbylamine reaction.
1° amine on heating with chloroform j and ethanolic KOH form isocyanides.

Answer 27.
a. Polysaccharides give a large number of monosaccharide units on hydrolysis.
b. Starch and cellulose.
c. Sucrose solution is dextrorotatory. When it is subjected to hydrolysis, the resultant sugar-solution becomes levorotatory. This sugar is called invert sugar.

Answer 28.
a. Penicillin, ofloxacin.
b. Antiseptics and disinfectants are chemicals which prevent or kill the growth of microorganisms.
c. Antiseptics are applied to living tissues such as wounds, cuts, ulcers etc to kill microorganisms. Disinfectants are applied on floors, drainage system like nonliving materials to kill microorganisms.

Answer 29.
a. Chromium.
b. Ammonia, Chloride,
c. 6

Answer 30.
a.
1. Lowering of vapour pressure
2. The salvation of boiling point
3. Depression in freezing point
4. Osmotic pressure.
b. Due to an association of molecules, the ‘ number of particle decreases and colligative property decreases. Molecules of ethanoic acid dimerise in benzene due to hydrogen bonding.

Answer 31.
i. Order of a reaction is the sum of the powers of the concentration terms of the reactants as expressed in the rate equation.
ii. It is the no. of molecules of the reactants that collide simultaneously to bring about the chemical reaction.
iii. r = K. [A]²
r, = K. [3A]² = 9. K [A]²
∴ Rate increases by 9 times.

Answer 32.
The outer electronic configuration of cobalt is 3d⁷ 4s². In this complex, cobalt is in the +3 oxidation state. So electronic configuration of CO³⁺ is 3d⁶ 4s° is

Ammonia is a strong ligand. Therefore pairing of electrons takes place,

There are six ammonia molecules acting as ligands. So there should be six vacant orbitals for Co³⁺. Hence the metal ion undergoes d2sp3 hybridisation. Since there are no unpaired electrons, the complex is diamagnetic. [COF₆]^3- Here also cobalt is in the +3 oxidation state.

But P is a weak ligand. So pairing does not take place. Hence in metal ion undergoes sp³d² hybridisation. There are four unpaired electrons present, .. the complex is paramagnetic.

Answer 33.
a. 3,4dinitrobenzoic acid,

Plus Two Chemistry Previous Year Question Papers and Answers

Kerala Plus Two Chemistry Model Question Paper 2

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Scores

General Instructions to candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

(Questions 1 to 7): Carry one score each. Answer all questions.

Question 1.
Which type of stoichiometric defect am I shown by AgCI?

Question 2.
What is the two-dimensional coordination; a number of a molecule in a square close-packed later?

Question 3.
Mercury is a ……… cell.

Question 4.
For the reaction ; Cl₂ (g) + 2NO(g) → 2NOCI(g) ; the rate law is expressed as rate = K[CI₂][NO]² ! What is the overall order of this reaction?

Question 5.
Name the method used for removing gangue I from sulfide ore.

Question 6.
Which one of the following compound reacts, with chlorobenzene to produce DDT?

Question 7.
Biotin is an organic compound present in yeast deficiency in the diet causes dermatitis 1 and paralysis. It is also known as

a. Vitamin H
b. Vitamin B₃
c. Vitamin B₁₂
d. VitaminD

(Questions 8 to 20) : Answer any ten. Each question carries two scores.

Question 8.
i. Which of the following is a molecular solid?
a. Diamond
b. Graphite
c. Ice
d. Quartz
ii. Unit cells can be classified into primitive and centered unit cells. Differentiate between primitive and centered unit cells

Question 9.
i. Which of the following is a secondary cell?
a. Dry cell
b. Leclanche cell
c. Mercury cell
d. None of these
ii. What is the relationship between resistance and conductance?

Question 10.
Complete the following.

Question 11.
Match the items of Column I with items of Column II.


Question 12.
How will you convert propanoic acid into the following compounds?
i. Ethane 0
ii. Butane

Question 13.
Haloalkanes and haloarenes react with metals to give hydrocarbons or products from which hydrocarbons are obtained easily.

Identify the product and name the reaction.

Identify the product and name the reaction.

Question 14.
Most of the organic chlorides, bromides and iodides react with certain metals to give compounds containing carbon-metal bonds.
i. Give one example of such a compound.
ii. How will you prepare the above compound?

Question 15.
Schematic alignment of magnetic moments of ferromagnetic, antiferromagnetic and ferrimagnetic substances are given below. Identify each of them.

Question 16.
Aldehydes and ketones are organic compounds containing a carbonyl group.
i. Write a chemical reaction to distinguish between aldehydes and ketones.
ii. Aldehydes and ketones can be subjected to Clemmensen reduction and wolfishness reduction. Name the reagents used in both cases.

Question 17.
Write any two electrophilic substitution reactions of chlorobenzene.

Question 18.
a. Which of the ones mentioned in the above table can be concentrated by magnetic separation. justify your answer.
b. Identify the ones that can be concentrated by leaching.

Question 19.
The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation. With the help of the Arrhenius equation calculate the rate constant for the first order reaction, C₂ H₅ I_(g) → C₂ H_4(g) + HI_(g) at 700 K. Energy of activation (E_a) for the reaction is 209 KJmof1 and rate constant at 600 K is 1.60 x 105 S’1. [Universal gas constant R= 8.314K1mol’1].

Question 20.
Chemotherapy is a term found in medical terminology. What is chemotherapy?

(Questions 21 to 29): Answer any seven. Each questions carries three scores.

Question 21.
a. The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is termed adsorption.
i. What is adsorption isotherm?
ii. Write the mathematical expression of Freundlich adsorption isotherm.
b. Enzymes are known as biochemical catalysts. Write any two important characteristics of enzyme catalysis.

Question 22.
Biomolecules are formed by certain specific linkages between simple monomeric units. Write the names of linkages and monomeric units in the following class of biomolecules.
i. Starch
ii. Protein
iii. Nucleic acid

Question 23.
Polymers are macromolecules formed by a union of monomers.
a. Name natural polymer and synthetic polymer.
b. Distinguish between thermoplastic and thermosetting polymers with an example.

Question 24.
Carbohydrates are broadly divided into monosaccharides, oligosaccharides, and polysaccharides.
a. Write one example each of monosaccharide and oligosaccharide.
b. i. Write any one method for the preparation of glucose.
ii. What is peptide linkage?

Question 25.
a. Amines are basic in nature. Arrange the following compounds in the increasing order of their basic strength.
NH₂, C₆H₅NH₂, CH₃ NH₂, (CH₃) 2NH, (CH₃)₃N.
b. How will you convert Toluene (C₆H₅CH₃) to Benzaldehyde

Question 26.
Consider the coordination compound [CO(NH₃)₅ SO₄]Br.
a. Write the IUPAC name of the above compound.
b. What is the primary valence and secondary valence of the central metal, Cobalt, in the above coordination compound?
c. Which type of structural isomerism is exhibited by the above coordination compound?

Question 27.
Different drugs have different therapeutic action in our body. Write the therapeutic action of the following drugs in our body.
a. Tranquilizers
b. Analgesics
c. Antibiotics

Question 28.
The concept of AG° of coupled reactions f is used to explain reductions in metallurgy.
a. Explain the above statement?
b. In the blast furnace for manufacturing iron, most of the reduction is carried out by CO rather than C (coke). How can you account for this?

Question 29.
a. LDPE is a homopolymer, while Nylon66 is a copolymer. Explain?
b. Classify the following into homopolymer or copolymers. Nylon-66 and HDPE

(Questions 30 to 33): Answer any three. Each questions carries four scores.

Question 30.
Detergents are used to remove oil and dirt from surfaces.
a. What are synthetic detergents? Give an example.
b. What are anionic, cationic and nonionic detergents? Illustrate with examples?
c. A synthetic detergent is used as a dishwasher. To which of the above type would it belong?

Question 31.
Crystal defects give rise to certain special properties in the solids.
a. What is meant by Frenkel defect?
b. VWiy does Li Cl not exhibit Frenkel defect?
c. Explain the pink color of LiCI when heated in the vapors of Li.

Question 32.
The hydrolysis of an ester in acid medium is a first order reaction.
a. What do you mean by an Ist order reaction?
b. What is the relation between the rate of the instant and half-life period of a reaction?
c. Half-life period of a first-order reaction is 20 seconds. How much time will it take to complete 90% of the reaction?

Question 33.
The limiting molar conductivity of an electrolyte is obtained by adding the limiting molar conductivities of cation and anion of the. electrolyte.
a. Name the above law.
b. What is meant by limiting molar conductivity?
c. Explain how conductivity measurements help to determine the ionization constant of a weak electrolyte like acetic add.
d. Explain the change of conductivity and molar conductivity of a solution with dilution?

Answers

Answer 1.
Frenkel defect

Answer 2.
4

Answer 3.
Primary cell

Answer 4.
3

Answer 5.
Froth flotation process

Answer 6.
d

Answer 7.
a

Answer 8.
i. c. Ice I
ii. In primitive unit cells, constituent particles are present only on the corner position I of unit cell whereas in centered unit cell, one or more constituent particles are present at a position other than corners in addition to those at corners.

Answer 9.
i. d. None of these
ii. Conductance (C) is the reciprocal of the resistance (R). C = 1/R

Answer 10.

Answer 11.

Answer 12.

ii. Kolbe’s electrolysis: By the electrolysis of aqueous solution of sodium salt of the propanoic acid.

Butane is formed at the anode.

Answer 13.

Answer 14.
i. Grignard reagent -R MgX
ii. The alkyl halide is treated with magnesium in dry ether to get Grignard reagents.

Answer 15.
i. Antiferromagnetic.
ii. Ferrimagnetic.
iii. Ferromagnetic.

Answer 16.
i. Oxidation: Aldehydes on oxidation give carboxylic acid containing same no. of C atoms.

Ketones on oxidation give a mixture of carboxylic acids containing lesser no.of carbon atoms.

ii. Zn amalgam and HCI for Clemmensen reduction. Hydrazine heated with KOH in ethylene glycol for WolfKishner reduction.

Answer 17.

Answer 18.
a. Haematite, magnetite , Iron pyrites,
b. Bauxite.

Answer 19.
K = ln A – \(\frac { { E }_{ a } }{ RT }\);
where K = Rate constant,
A = Arrhenius factor, E_a = Activation energy, T= Temperature, R = Gas constant

Answer 20.
i. Use of chemicals for therapeutic effect is called chemotherapy.

Answer 21.
a. i. The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature is expressed using a curve called adsorption isotherm.
ii. \(\frac { X }{ m } =k.{ p }^{ 1/2 },n>1;\)
where x/m is the extent of adsorption, P is the pressure, K and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.
b. i. Most highly efficient. One molecule of an enzyme may transform one million molecules of the reactant per minute.
ii. Highly specific in nature. Each enzyme is specific for a given reaction.

Answer 22.
i. α – D- glucose, the bond is glycosidic linkage.
ii. An amino acid, a peptide bond is a linkage.
iii. Nucleotide, Bond is a hydrogen bond.

Answer 23.
a. Natural polymer Cellulose, Starch Synthetic polymer PVC, Nylon,
b. Thermoplastics are linear or slightly branched long chain molecules, which can be remolded again and again by heating and cooling. The intermolecular force is intermediate between elastomers and fibers.

Thermosetting are cross-linked or heavily branched molecules. On heating, they undergo extensive cross-links and becomes infusible. They cannot be remolded.

Answer 24.
a. Monosaccharide Glucose, Fructose-Oligosaccharide Maltose, Lactose, Sucrose Polysaccharide Cellulose
b.
i. Starch.

ii. The linkage between units of amino acids in a polypeptide linkage is called peptide linkage.

Answer 25.

Answer 26.
a. Pentaamminesulphatocobait (III) bromide.
b. Primary valence is 3 and secondary valence is 6.
c. Ionization isomerism.

Answer 27.
a. Tranquilizers are used for the treatment of stress and mental diseases.
b. Analgesics reduce pain without causing any disturbance to the nervous system.
c. Antibiotics are drugs used to treat infections in our body. It inhibits the growth of microorganisms.

Answer 28.
a. For the feasibility of a reaction, ΔrG° must be negative. The positive value of ArG° means that the reaction is not spontaneous. This reaction can be made spontaneous by coupling with a reaction having very large ve Gibb’s energy value so that Gibb’s energy for the two combined reactions become ve.
b. A better reducing agent should have a larger ve value for AG°. If coke is used ArG° = 13 KJ/mol and for CO, ArG° = 56.2 KJ/mol Comparing the two values, it is clear that CO is a better reducing agent than C.

Answer 29.
a. If the polymer is formed only one type of monomer units, then it is a homopolymer. LDPE is low-density polyethylene whose monomer is ethylene. Hence it is a homopolymer.
Nylon66 is the polymer containing two different monomer units hexamethylene diamine and adipic acid. Hence it is a copolymer.
b. Nylon6: Homopolymer HDPE: Homopolymer

Answer 30.
a. Synthetic detergents are cleansing agents which have all the properties of soap, but actually, do not contain any soap. They can be used both in soft and hard water.
b.
i. Cationic detergents are quarternary ammonium salts of amines with acetates, Cl or Bras anions. Cationic part posses a long hydrocarbon chain. They are expensive and have germicidal properties used in hair conditioners, eg., Cetyl trimethyl ammonium bromide.
ii. Anionic detergents are sodium salts of sulfonated long chain alcohols or hydrocarbons. The anionic part of the molecule is involved in cleaning action. Used in toothpaste, household work, eg., sodium dodecyl benzene sulphonate.
iii. Nonionic detergents do not contain any ion. eg., Detergent formed by stearic acid and polyethylene glycol. Liquid dishwashing detergents are nonionic.
c. It belongs to a nonionic type.

Answer 31.
a. Frenkel defect arises due to the dislocation of a cation from its normal site to an interstitial site.
b. Frenkel defect is shown by crystals with a large difference in size between the anion and cation. In LiCI, the cation and anion have comparable (similar) size. So it will not exhibit Frenkel defect.
c. This is due to metal excess defect due to anionic vacancies. When LiCI is heated in vapors of Li, the Cf ions diffuse to the surface of the crystal creating vacant sites. Instead, the electrons diffuse into the crystal and occupy these anionic sites, forming F centers. These electrons absorb energy from white light and show a pink color.

Answer 32.
a. If the sum of powers of the concentration terms in the rate expression is one, then the reaction is a first order reaction.

Answer 33.
a. Kohlrausch’s law.
b. It is the molar conductivity of electrolytes at zero concentration or at infinite dilution.
c. The degree of dissociation can be determined by the equation, \(\alpha =\frac { { \lambda }_{ m } }{ { \lambda }_{ 0 }^{ m } }\) To obtain ionisation constant, K_a substitute α in the following equation, \({ K }_{ a }=\frac { { C\alpha }^{ 2 } }{ (1-\alpha ) }\) where C is the concentration a of acetic acid in moles/liter.
d. The conductivity of a solution decreases with a decrease in concentration as the number of ions presents per unit volume of the solution actually decreases. The molar conductivity of electrolytes increases with a decrease in concentration.

Plus Two Chemistry Previous Year Question Papers and Answers

Kerala State Board New Syllabus Plus Two Chemistry Previous Year Question Papers and Answers.

Kerala Plus Two Chemistry Model Question Paper 1 with Answers

Board	SCERT
Class	Plus Two
Subject	Chemistry
Category	Plus Two Previous Year Question Papers

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Score

General Instructions to candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Questions from 1 – 7 Carry 1 Score each. Answer all the questions. (7 × 1 = 7)

Question 1.
The edge length and axial angles of a unit cell are a = b = c and α = β = γ = 90° respectively. Identify the type of Crystal System.
Answer:
Cubic

Question 2.
Name the polymer used for non-stick surface coated utensils.
Answer:
Polytetrafluoroethene/Teflon

Question 3.
Which of the following forms coloured compounds
i) Sc³⁺
ii) Cr²⁺
iii) Zn²⁺
iv) Cu⁺
[Atomic number Sc-21, Cr-24, Zn-30, Cu-29]
Answer:
ii) Cr²⁺

Question 4.
The rate expression of a reaction is Rate = K [A]^1/2 [B]^1/2. Calculate its overall order.
Answer:
Overall order = \(\frac{1}{2}+\frac{3}{2}\) = 2

Question 5.
Among the following amines, identify the one that does not reach with Hinsberg reagent.
i) C₂H₅NH₂
ii) (CH₃)₂NH
iii) (C₂H₅)₂NH
iv) (CH₃)₃N
Answer:
iv) (CH₃)₃N

Question 6.
The Carbohydrate, that is known as ‘animal starch’ is …………
Answer:
Glycogen

Question 7.
Name the class of drug used for the treatment of stress and mental diseases.
Answer:
Tranquilizers

Questions from 8 – 20 Carry 2 Score each. Answer any 10 questions. (10 × 2 = 20)

Question 8.
Write the names and structures of the monomers of the polymer Buna-S.
Answer:
1, 3-Butadiene CH₂=CH-CH=CH₂
Styrene/Vinyl benzene C₆H₅-CH=CH₂

Question 9.
Calculate the packing efficiency in body centred cubic structure.
[Given that r = \(\frac{\sqrt{3}}{4}\)a, where ‘r’ is the radius of sphere and ‘a’ is the edge length of unit cell]
Answer:
In body centred cubic (bcc) structure, total number of atoms per unit cell is 2.
Volume of atoms in the unit cell = 2 × (\(\frac{4}{3}\))πr³
Volume of the cube = a³ = (\(\frac{4}{\sqrt{3}}\)r)³

Question 10.
Write one method for the preparation of sulphur dioxide. How is its presence detected?
Answer:
Sulphur dioxide is obtained when sulphur is burnt in air or oxygen.
S(s) + O₂(g) → SO₂(g)
OR
In the laboratory, it is prepared by treating a sulphite with dilute sulphuric acid.
SO₃^2-(aq) + 2H⁺(aq) → H₂O(l) + SO₂(g)
OR
Industrially, it is produced as a by-product of the roasting of sulphide ores.
4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(g)
The sulphur dioxide gas decolourises acidified potassium permanganate solution. This is a convenient test for the gas.
5SO₂ + 3MnO₄^– + 2H₂O → 5SO₄^2- + 4H⁺ + 2Mn²⁺

Question 11.
Write the IUPAC names of following compounds.
a) K₃[Fe(CN)₆]
b) Co[Cl₂(en)₂]Cl
Answer:
a) Potassium hexacyanoferrate(III)
b) Dichloridobis(ethane-1, 2-diamine)cobalt(III) chloride

Question 12.
Define van’t Hoff factor ‘i’. What would be the value of ‘i’ for a dilute solution of K₂SO₄ in water?
Answer:
The van’t Hoff factor, ‘i’ is defined as:


K₂SO₄ ionises in water as,
K₂SO₄ → 2K⁺ + SO₄^2-
Therefore, for a dilute solution of K2S04 in water the value of i = 2 + 1 = 3

Question 13.
By taking ethyl bromide as starting material, prepare ethyl iodide and ethyl fluoride.
Answer:
Ethyl bromide is treated with Nal in dry acetone to get ethyl iodide. (Finkelstein reaction)
CH₃CH₂Br + Nal → CH₃CH₂l + NaBr
Ethyl bromide is heated with metallic fluorides such as AgF, Hg₂F₂, CoF₂ or SbF₃ to get ethyl fluoride. (Swarts reaction)
CH₃CH₂Br + AgF → CH₃CH₂F + AgBr

Question 14.
Write the complete reaction representing the disproportionation of benzaldehyde.
Answer:

This reaction is called Cannizzaro reaction.

Question 15.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25g of glucose is dissolved in 450 g of water.
Answer:
\(p_{1}^{0}\) = 17.535 mm Hg, w₂ = 25g, M₂ = 180g mol^-1, w₁ = 450 g, M₁ = 18 g mol^-1, p, = ?

Therefore, vapour pressure of glucose solution,
p₁ = \(p_{1}^{0}\) – 0.09742
= 17.535 – 0.09742 = 17.4376 mm Hg

Question 16.
Name the products of hydrolysis of sucrose and explain why sucrose is not a reducing sugar.
Answer:
Sucrose on hydrolysis gives equimolar mixture of D-(+)-glucose and D-(-)-fructose.

In sucrose, the two monosaccharide units (glucose and fructose) are held together by a glycosidic linkage between C1 of α – glucose and C2 of β – fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non-reducing sugar.

Question 17.
Tertiary butyl bromide react with aqueous NaOH to give tertiary butyl alcohol proceeds via S_N¹ reaction. Write the mechanism of the reaction.
Answer:

Mechanism: This reaction occurs in two steps:
Step I: The polarised C-Br bond undergoes slow cleavage to produce tert-butyl carbocation and a bromide ion.

Step II: The carbocation is attacked by the nucleophile OH^– to form the product, tert-butyl alcohol.

Question 18.
Give the structures of A and B in the following reaction.

Answer:

Question 19.
Match the following

Answer:

Question 20.
Write a simple chemical test to distinguish ethanal from propanone.
Answer:
Tollens’ Test – On warming with Tollens’ reagent ethanal gives a bright silver mirror while propanone will not give silver mirror.
OR
Fehling’s Test – On heating with Fehling’s reagent ethanal gives a reddish brown precipitate while propanone do not respond to this test.

Questions from 21 – 29 Carry 3 Score each. Answer any 7 questions. (7 × 3 = 21)

Question 21.
The resistance of a 0.5 M solution of an electrolyte enclosed between two platinum electrodes 1.5 cm apart and having an area of 2.0 cm² was found to be 30Ω. Calculate the molar conductivity of the solution.
Answer:
M = 0.5 M, l = 1.5 cm, A = 2 cm², R = 30Ω, \(\Lambda_{m}\) = ?

Question 22.
Write notes on different allotropic forms of phosphorous.
Answer:
The important allotropic forms of phosphorus are white, red and black phosphorus.
White phosphorus: white waxy solid, poisonous, insoluble in water but soluble in CS₂, glows in dark, dissolves in boiling NaOH solution in an inert atmosphere giving PH₃.
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
It readily catches fire in air to give dense white fumes Of P₄O₁₀.
P₄ + 5O₂ → P₄O₁₀
It consists of discrete tetrahedral P₄ molecules. It is less stable because of angular strain in P₄ molecule where the angles are only 60°.

Red phosphorus: obtained by heating white P at 573K in an inert atmosphere for several days. This when heated under high pressure, a series of phases of black P is formed. It possesses iron grey lustre. It is odourless, non-poisonous, insoluble in water as well as in CS₂, much less reactive than white P, does not glow in dark. It is polymeric, consisting of chains of P₄ tetrahedra linked together as shown:

Black phosphorus: It has two forms, α -Black P and β -Black P.
α -Black P: It is formed when red P is heated in a sealed tube at 803 K, can be sublimed in air, has opaque monoclinic or rhombohedral crystals, does not oxidise in air.
β -Black P: It is prepared by heating white P at 473K under high pressure. It does not burn in air upto 673 K.

Question 23.
Integrated rate equation for a first order reaction.
K = \(\frac{2.303}{t} \log \frac{\left[R_{0}\right]}{[R]}\)
a) Derive an expression for half life of first order reaction.
b) Find the half life of a first order reaction having rate constant K = 5.5 × 10^-14 S^-1.
Answer:
a) For a first order reaction R → P, the rate constant

Question 24.
What are adsorption isotherms? Write the Freundlich adsorption isotherm equation and plot a graph based on it.
Answer:
Adsorption isotherms are curves used to express the variation of the amount of gas adsorbed by the adsorbent with pressure at constant temperature.

The Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. The relationship can be expressed by the equation:
\(\frac{x}{m}=k \cdot P^{1 / n}\)
where x is the mass of the gas adsorbed on mass m of the adsorbent at pressure P, k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.

The relationship can be represented in the form of a curve where mass of the gas adsorbed per gram of the adsorbent (x/m ) is plotted against pressure(P).

Question 25.
Describe the steps involved in the manufacture of sulphuric acid by contact process.
Answer:
The Contact process for the manufacture of sulphuric acid involves three steps:
Step 1: Burning of sulphur or sulphide ores in air to generate SO₂.
S(s) + O₂(g) → SO₂(g)
Step 2: Conversion of SO₂ to SO₃ by the reaction with oxygen in the presence of V₂O₅ catalyst at a pressure of 2 bar and a temperature of 720 K.

Step 3: Absorption of SO₃ in H₂SO₄ to give oleum (H₂S₂O₇).
SO₃ + H₂SO₄ → H₂S₂O₇
Dilution of oleum with water gives H₂SO₄ of the desired concentration.

Question 26.
Differentiate ferrimagnetism from paramagnetism and predict the change that takes place to ferrimagnetic substances on heating.
Answer:
Paramagnetism: Magnetic property which arises due the presence of one or more unpaired electrons. Such substances are weakly attracted by the magnetic field. They are magnetised in a magnetic field in the same direction and lose their magnetism in the absence of magnetic field.
e.g. O₂, Cu²⁺, Fe³⁺, Cr³⁺.

Ferrimagnetism: Magnetic property observed when the magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers as shown:

They are weakly attracted by magnetic field as compared to ferromagnetic substances.
e.g. Fe₃O₄ (magnetite) and ferrites like MgFe₂O₄ and ZnFe₂O₄.
Ferrimagnetic substances on heating lose ferrimagnetism and become paramagnetic.

Question 27.
a) Draw the structure of dichromate ion.
b) How can dichromate ion be converted to chromate ion?
c) Write examples for the oxidising action of potassium dichromate in acidic medium.
Answer:
a)

b) When an aqueous solution of dichromate is treated with alkali (i.e., by increasing the pH of the medium) it changes to chromate.
Cr₂O₇^2- + 2OH^– → 2CrO₄^2- + H₂O

c) Acidified potassium dichromate oxides iodides to iodine, sulphides to sulphur, tin(II) to tin(IV) and iron(II) to iron(III).

Question 28.
Explain the following
a) Hydraulic washing
b) Zone refining
c) Roasting of ores
Answer:
a) Hydraulic washing: This is a method used for the concentration of ores. It is based on the differences in gravities of the ore and the gangue particles. It is type of gravity separation. Here an upward stream of running water is used to wash the powdered ore. The lighter gangue particles are washed away and the heavier ores are left behind.

b) Zone refining: It is a method used for obtaining extra pure germanium, silicon, boron, gallium and indium. It is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal. Here a circular mobile heater is fixed at one end of a rod of the impure metal.

The molten zone moves along with the heater which is moved forward, the pure metal crystallises out of the melt and the impurities pass on into the adjacent molten zone. This process is repeated several times and the heater is moved in the same direction. At one end, impurities get concentrated. This end is cut off.

c) Roasting of ores: Roasting is used for converting ore to its oxide. It is mainly applied to sulphide ores. Here the ore is heated in a regular supply of air in a furnace at a temperature below the melting point of the metal, e.g.
2ZnS + 3O₂ → 2ZnO + 2SO₂ OR
2PbS + 3O₂ → 2PbO + 2SO₂ OR
2CU₂S + 3O₂ → 2Cu₂O + 2SO₂

Question 29.
Describe the following reactions.
a) Aldol condensation
b) Etard reaction
c) Rosenmund reduction
Answer:
a) Aldol condensation: Aldehydes and ketones having at least one α – hydrogen atom undergo a reaction in the presence of dilute alkali as catalyst to form β – hydroxy aldehydes (aldol) or β – hydroxy ketones (ketol) respectively. The aldol and ketol on heating readily lose water to give α, β – unsaturated aldehydes or ketones respectively.

e.g. Ethanal on treating with dil NaOH gives 3-hydroxy butanal which on heating loses water to give but-2-enal.

b) Etard reaction: When toluene is treated with chromyl chloride (CrO₂Cl₂) in CS₂ a chromium complex is formed which on hydrolysis gives benzaldehyde.

C) Rosenmund reduction: When an acyl chloride(acid chloride) is hydrogenated over catalyst, palladium on barium sulphate the corresponding aldehyde is formed.
e.g. Benzoyl chloride on hydrogenation over catalyst, palladium on barium sulphate gives
benzaldehyde.

Questions from 30 – 33 Carry 4 Score each. Answer any 3 questions. (3 × 4 = 12)

Question 30.
Write chemical equations for the conversion of phenol to acetylsalicylic acid (aspirin).
Answer:

Question 31.
With the help of diagram describe the construction and working of a fuel cell using H₂ and O₂. What are the advantages of fuel cell over conventional cells.
Answer:
In the H₂ – O₂ fuel cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous NaOH solution. Catalysts like finely divided platinum or palladium metal are incorporated into the electrodes for increasing the rate of electrode reactions.

This cell uses the reaction of hydrogen with oxygen to form water. The electrode reactions are:
Anode reaction: 2H₂(g) + 4OH^–(aq) → 4H₂O(l) + 4e^–
Cathode reaction: O₂(g) + 2H₂O(l) + 4e^– → 40H^–(aq)
The overall reaction is, 2H₂(g) + O₂(g) → 2H₂O(l)

Advantages:

• The cell runs continuously as long as the reactants are supplied.
• Fuel cells produce electricity with an efficiency of about 70%, which is much higher than that of conventional cells.
• Fuel cells are pollution free.

Question 32.
Sketch the Crystal field splitting of d-orbitals of the Central metal ion in [CoF₆]^3-. Also write the electronic configuration of central metal ion and predict its magnetic behaviour based on Crystal field theory.
Answer:

The electronic configuration of the central metal ion. Co³⁺ is [Ar]3d⁶ 4s⁰.
Here F- is a weak field ligand. Therefore, Δ₀ < P and no pairing occurs (\(t_{2 g}^{4} e_{g}^{2}\)). Thus, there are four unpaired electrons and hence [CoF₆]^3- is an outer orbital/high spin/spin free paramagnetic octahedral complex.

Question 33.
Give the structures and IUPAC names of the products formed on the following reactions.
a) Hydration of propene in the presence of dilute sulphuric acid.
b) Reaction of methanal with methyl magnesium bromide followed by hydrolysis.
Answer:

Kerala Plus Two Chemistry Previous Year Question Paper 2017

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Scores

General Instructions to Candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• You are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and the only internal choice is allowed.
• When you select a question, all the subquestions must be answered from the same question itself.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except nonprogrammable calculators are not allowed in the Examination Hall.

Question 1.
a. Identify the nonstoichiometric defect
i. Schottky defect
ii. Frenkel defect
iii. interstitial defect
iv. Metal deficiency defect
b. What type of substance could make better permanent magnets ferromagnetic or ferrimagnetic? Justify your answer.
c. In terms of Band theory write the differences between conductors and insulators.

Question 2.
a. Henry’slaw is related to the solubility of a gas in a liquid.
i. State Henry’s law
ii. Write any two applications of Henry’s law.
b. 1000 cm³ of an aqueous solution of a protein contains 1.26 gm of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10^-3 bar. Calculate the molar mass of the protein. (R = 0.083 L bar mol¹K¹)

Question 3.
a. Represent the galvanic cell based on the cell reaction given below:

b. Write the half-cell reactions of the above cell.
c. \(\wedge _{ m }^{ 0 }\) for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2 mol^-1 respectively. Calculate \(\wedge _{ m }^{ 0 }\) for HAc.

Question 4.
a. Plot a graph showing variation in the concentration of reactants against time for a zero order reaction.
b. What do you mean by zero order reaction?
c. The initial concentration of the first order reaction, N₂O_5(g) → 2NO_2(g) + 1/2O_2(g) was 1.24 × 10^-2 mol L^-1 at 300 K. The concentration of N₂O₅ after ‘ 1 ’ hour was 0.20 × 10^-10 mol L^-1. Calculate the rate constant of the reaction of 300 k.

Question 5.
There are mainly two types of adsorption. They are physisorption and chemisorption.
a. Differentiate between physisorption and chemisorption.
b. Write any two applications of adsorption.

Question 6.
Leaching is a process of concentration of ores. Explain the leaching of alumina from bauxite.

Question 7.
Nitrogen forms a number of oxides and oxoacids.
a. Which of the following is a neutral oxide of nitrogen?
i. N₂O
ii. N₂O₅
iii. NO₂
iv.N₂O₄
b. Prepare a short write-up on Nitric acid highlighting its laboratory preparation, chemical properties and uses.
OR
Phosphorous forms a number of compounds, a. The gas liberated when calcium phosphide is treated with dil.HC/ is
i. Cl₂
ii.H₂
iii. PH₃
iv. All of the above
b. Prepare a short write up on PCl₃ and PCl₅ highlighting the preparation and chemical properties of PCl₃ and structure of PCl_r

Question 8.
a. Transition elements are d block elements.
i. Write any four characteristic properties of transition elements.
ii Cr²⁺ and Mn³⁺ have d⁴ configuration. But Cr²⁺ is reducing and Mn³ is oxidizing. Why?
b. Which of the following is not a lanthanoid element?
i. Cerium
ii. Europium
iii Lutetium
iv. Thorium

Question 9.
[CO(NH₃)₅S0₄]Cl and [Co(NH₃)₅ Cl]SO₄ are coordination compounds.
a. Identify the isomerism shown by the above compounds.
b. Write the IUPAC names of the above compounds.
c. Identify the ligands in each of the above compounds.

Question 10.
a. An ambident nucleophile is
i. Ammonia
ii Ammonium ion
iii Chloride ion
iv. Nitrite ion
b.Haloalkanes and Haloarenes are organohalogen compounds.
i. Suggest a method for the preparation of alkyl chloride.
ii. Aryl halides are less reactive towards Nucleophilic substitution reactions. Give reasons.

Question 11.
a. Arrange the following compounds in the order of increasing boiling points: Ethanol, Propan1 o 1, Butan1 o 1, Butan2ol
b. In the lab, students were asked to carry out the reaction between phenol and cone. HN03. But one student, ‘A’ carried out the reaction between phenol and dil. HNOr Do you think that the student ‘A’ got the same result as others. Substantiate with suitable explanations. [Also write the chemical equations wherever necessary]

Question 12.
a. The product obtained when benzene is treated with carbon monoxide and hydrogen chloride in presence of anhydrous AlCl₃ is
i. Chlorobenzene
ii. Phenol
iii. Benzaldehyde
iv. Benzoic acid
b. How will you carry out the following conversions?

OR

b. Explain the following:
i. Esterification
ii. Tollen’s test
iii. HVZ reaction
iv. Dicarboxylic acid

Question 13.
a. Classify the following amines as primary, secondary and tertiary

Identify the products B and C and write their formulae.

Question 14.
a. Which of the following is a polysaccharide?
i. Maltose
ii. Sucrose
iii. Fructose
iv. Cellulose
b. Explain the amphoteric behavior of amino acid.

Question 15.
a. Which of the following is not applicable to Nylon 6, 6?
i. Synthetic polymer
ii. Fiber
iii. Addition polymer
iv. Condensation polymer.
b. Differentiate between thermoplastics and thermosetting plastics. Write one example for each.

Question 16.
‘Antibiotics, antiseptics, and disinfectants are antimicrobial drugs. Explain any one of the above-mentioned drugs with examples.

Answers

Answer 1:
a. Metal deficiency defect
b. Ferromagnetic substances make better permanent magnets because, in the solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains. Thus, each domain acts as a tiny magnet. In an unmagnetized piece of a ferromagnetic substance, the domains are randomly oriented and their magnetic moments get canceled. When the substance is placed in the magnetic field, a strong magnetic effect is produced. This ordering of domains persists even when the magnetic field is removed and the ferromagnetic substance becomes a permanent magnet. However, this is not possible in ferrimagnetic substances,

c. In conductors, the valence band and conduction band overlap. Hence electrons can easily flow. In insulators the energy gap between the valence band and the conduction band is large, hence electrons cannot jump from one band to the other.

Answer 2.
a. i. It states that “the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution”. The most commonly used form of Henry’s law states that “the partial pressure(p) of gas in the vapor phase is proportional to the mole fraction of the gas(x) in the solution” and it is expressed as, P = K_H.X, where KH is Henry’s law constant and ‘x’, is mole fraction of gas in solution.
ii.
1. To increase the solubility of CO2 in soda water and soft drinks, the bottle is sealed under high pressure.
2. To avoid the toxic effects of a high concentration of nitrogen in blood, the tanks used by scuba divers are filled with air diluted with helium.
b. p = 2.57 × 10^-3 bar
V = 1000 cm³ = 1L
T = 300k
R = 0.083 L bar mol^-1K^-1

Answer 3.

Answer 4.
a.

b. In zero order reactions, the rate remains constant throughout the course of reaction i.e., the rate does not change with a change in concentration of reactants.

Answer 5.
a. Physical adsorption (Physisorption):

• Force of attraction between adsorbate and adsorbent are Van der Waals forces.
• Lacks specificity.
• Reversible.
• Enthalpy of ads 40kJmol^-1)
• Occurs at low temperature.

Chemical adsorption (chemisorption):

• Force of attraction I and adsorbent I bonds.
• Highly specific.
• Irreversible.
• Enthalpy of advisor 240kJ/mol).
• Occurs at high terr

b.

• Production of high
• Gas masks
• Heterogenous cat?
• Froth floatation pr
• Control of humidity;
• Chromatographic analysis

Answer 6.
Bauxite ore is treated with aqueous NaOH at 473-523K and 35-36 bar pressure. This way Al₂O₃ is leached out as sodium aluminate leaving the impurities behind.

The aluminate in solution is neutralized by passing C0₂ gas and hydrated Al₂O₃ is precipitated.

The sodium silicate remains in the solution and hydrated alumina is filtered, dried and heated to give back pure Al₂O₃

Another example: Gold or silver ore is leached with a dilute solution of NaCN or KCN in the presence of air from which the metal is obtained by replacement (using Zn as a reducing agent).

Answer 7.
a. N₂O
b. In the laboratory, nitric acid is prepared by heating KNO₃ or NaNO₃ and concentrated H₂SO₄ in a glass retort.

On large scale, it is prepared mainly by Ostwald’s process. This method is based upon catalytic oxidation of NH₃ by atmospheric oxygen.

Nitric oxide thus formed combines with oxygen giving NO₂,

Nitrogen dioxide so formed dissolves in water to give HN03.

NO thus formed is recycled and the aqueous HNO₃ can be concentrated by distillation up to 68% by mass. Further concentration to 98% can be achieved by dehydration with concentrated H₂SO₄. Chemical properties
1. Acidic nature

2. Oxidizing agent behaves as a strong oxidizing agent because of its tendency to give nascent oxygen.

Uses :
Used in the manufacture of various fertilizers, explosives, plastics, fibers, dyestuffs etc. It is used as a laboratory reagent and as an oxidizing agent. It is also used in the purification of silver an gold.
OR
a. (iii) pH₃
b. Phosphorus forms two types of halides, PX₃ (X = F, Cl, Br, I) and PX₅ (X = F, Cl, Br). Preparation
It is obtained by passing dry chlorine overheated white phosphorus. P₄ + 6CI₂ → 4PCI₃ It is also obtained by the action of thionyl chloride with white phosphorus. P₄ + 8SOCI₂ → 4PCI₃ + 4SO₂ + 2S₂CI₂ Properties It is a colorless oily liquid and hydrolyzes in the presence of moisture.
PCI₃ + 3H₂O → H₃PO₃ + 3HCI It reacts with organic compounds containing OH group such as CH₃COOH, C₂H₅OH.
3CH₃COOH + PCI₃ → 3CH₃COCI + H₃PO₃ 3C₂H₅OH + PCI₃ → 3C₂H₅CI + H₃PO₃ It has a pyramidal shape as shown, in which phosphorus is sp3 hybridized. Preparation
Phosphorus pentachloride is prepared by the reaction of white phosphorus with an excess of dry chlorine. P₄ + 10Cl₂ → 4PCL₅
It can also be prepared by the action of SO₂CI₂ on phosphorus.
P₄ + 10SO₂CI₂ → 4PCI₃ + 10SO₅Properties PCI₅ is a yellowish white powder and in moist air, it hydrolyzes to POCI₃ and finally gets converted to phosphoric acid. PCI₅ + H₂O → POCI₃ + 2HCI POCI₃ + 3HO → HPO₄ + 3HCI In gaseous and liquid phases, it has a trigonal bipyramidal structure as shown. The three equatorial P_CI bonds are equivalent, while the two axial bonds are longer than equatorial bonds.

This is due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs. In the solid state, it exists as an ionic solid, [PCI₄]⁺ [PCI₆]^– in which the cation, [PCI₄]⁺ is tetrahedral and the anion, [PCI₈]^– cathedral.

Answer 8.
a. i. The transition elements are those elements which have partially filled subshells in their elementary form or in any one of its oxidation states. They have low ionization energy. They are malleable and ductile. They are good conductors of heat and electricity. High melting and boiling point.
ii. Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level. On the other hand, the change from Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability,
b. Thorium

Answer 9.
a. Ionization isomerism
b. [C0(NH₃)₅S0₄]CI- Pentaamminesulphatocobalt (III) chloride [CO(NH₃)₅Cl]SO₄ – Pentamminechloridocobalt (III) sulfate
c. Counterion in a complex salt itself a potential ligand. ligands are Ct and SO₄^2-

Answer 10.
a.iv. Nitrite ion
b.i. The preparation of alkyl chloride is carried out either by passing dry hydrogen chloride gas through a solution of alcohol or by heating a solution of alcohol in concentrated aqueous acid.

ii.
1. Aryl halides are electron rich.
2. In aryl halides, the halogen is attached to an sp² hybridized C, which is highly stable.
3. Because of the possible repulsion, it is less likely for the electron-rich arenes.
4. In the case of haloarenes, the phenyl cation formed as a result of self-ionization will not be stabilized by resonance and therefore, SN, a mechanism is ruled out.

Answer 11.
a. Ethanol < Propane -l-ol < Butan 2-ol < Butan -1-ol.
b. They don’t get the same result. Phenol with dilute nitric acid at low temperature (298 K) phenol yields a mixture of ortho and para nitrophenols.

Phenol with concentrated nitric acid, phenol is converted to 2, 4, 6 trinitrophenol. The product is commonly known as picric acid. The yield of the reaction product is poor.

Answer 12.
a. (iii) Benzaldehyde


OR
b.
i. Carboxylic acids react with alcohols in the presence of a strong acid catalyst like H₂SO₄ to form esters. The reaction is reversible and the forward reaction is called esterification,

ii. On warming an aldehyde with freshly prepared ammoniacal silver nitrate solution (Tollen’s reagent) a bright silver mirror is produced due to the formation of silver metal. The aldehydes are oxidized to corresponding carboxylate anion. The reaction occurs in alkaline medium.

iii. HVZ reaction
Carboxylic acids having an α – hydrogen is halogenated at the α – position on treatment with chlorine or bromine in the presence of small amount of red phosphorous to give α – halocarboxylic acids. The reaction is known as Hell Volhard Zelinsky reaction.
iv. Carboxylic acids lose carbon dioxide to form hydrocarbons when their

Sodium salts are heated with soda lime (NaOH and CaO in the ratio of 3:1). The reaction is known as decarboxylation.

Answer 13.

Answer 14.
a.iv. Cellulose
b. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as twitter ion. This is neutral but contains both positive and negative charges. In winter ionic form amino acids show amphoteric behavior as they react both with acids and bases.

Answer 15.
a.iii. Addition polymer
b. Thermoplastic polymers These are the linear or slightly branched long chain molecules capable of repeatedly softening on heating and hardening on cooling. These polymers possess intermolecular forces of attraction intermediate between elastomers and fibers. Some common thermoplastics are polyethylene, polystyrene, polyvinyls, etc.

Thermosetting polymers These polymers are cross-linked or heavily branched molecules, which on heating undergo extensive cross-linking in molds and again become infusible. These cannot be reused. Some common examples are bakelite, urea formaldehyde resins, etc

Answer 16.
Antibiotics: These are chemical substances which are produced wholly or partly by chemical synthesis, which in low concentration inhibits the growth or destroys microorganisms by intervening in their metabolic processes, eg., penicillin, streptomycin, ampicillin, tetracycline. Antibiotics are of two types:
Bactericidal antibiotics: They have killing (cidal) effect on microbes, eg., pencil in, ofloxacin, etc.
Bacteriostatic antibiotics: They have an inhibitory (static) effect on microbes, eg., Tetracycline, chloramphenicol, etc.
Antibiotic spectrum: The range of microorganisms attacked by an antibiotic is called its spectrum.

• Antibiotics which kill or inhibit a wide range of Gram-positive and Gram-negative bacteria are called broad-spectrum antibiotics.
  eg., Ampicillin, tetracycline, ofloxacin.
• Those which are effective mainly against Gram-positive or Gram-negative bacteria are narrow spectrum antibiotics.
  eg., penicillin G.

Plus Two Chemistry Previous Year Question Papers and Answers