Plus One Chemistry Model Question Paper 2

Kerala Plus One Chemistry Model Question Paper 2

Time Allowed: 2 hours
Cool off time: 15 Minutes
Maximum Marks: 60

General Instructions to candidates

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

Answer all questions from question numbers 1 to 7. Each carry one score.

Question 1:
The ozone in the stratosphere is destroyed by

Question 2:
The geometry of SF₆ molecule is …………
i. tetrahedral
ii. planar
iii. octahedral
iv. triagonal bipyramidal

Question 3:
Syn gas is a mixture of …….
i. CO and H₂O
ii. CO and H₂
iii. CO₂ and H₂
iv. CH₄ and CO

Question 4:
Representation of orbital with quantum numbers n = 3, l = 1 is ………….

Question 5:
The metal present in the chlorophyll of plants ¡s …………

Question 6:
Thermodynamically, the most stable allot rope of carbon is ………..

Question 7:
The atomic number of the element with IUPAC nane Ununbium is ……….

Answer any ten from question numbers 8 to 20. Each carries two scores.

Question 8:
Explain sp³ hybridisation taking methane (CH₄) as an example.

Question 9:
Write the IUPAC name of the following compounds:

Question 10:
The Taj Mahal in India has been affected by ‘acid rain’. Explain the causes and harmful effects of acid rain.

Question 11:
Calculate the number of molecules in each of the following:
i. 1 g N₂
ii. 1 g CO₂
(Given that N_A is 6.02 x 10²³, molecular mass of N₂ is 28 and CO₂is 44).

Question 12:
a. How many moles of dioxygen are present in 64 g of dioxygen (Molecular mass of dioxygen is 32).

b. The following data were obtained when dinitrogen (N₂) and dioxygen (O₂) react together to form different compounds.

Question 13:
a. The number of electrons, protons and neutrons in a species are equal to 18,16 and 16 respectively. Assign the proper symbol to the species.

b. Among the following electronic configurations, which one is correct?

Question 14:
Give reasons:
i. KO₂ is paramagnetic.
ii. Solutions of alkali metals in liquid ammonia are blue in colour.

Question 15:
An alkene ‘A’ on ozonolysis gave two molecules of formaldehyde. Write the name of ‘A’ and the chemical equation of ozonolysis.

Question 16:
Photoelectric effect was first observed by Hertz.

1.  The number of electrons ejected in the photoelectric effect is proportional to ……….. of light used. [Frequency, intensity]
2. Select the correct statement related to the photoelectric effect:
   • Threshold frequency is the maximum frequency required to cause photoelectric emission from a particular metal.
   • The kinetic energy of the photoelectrons is directly proportional to the frequency of incident light.
   • Work function is the same for all metals.

Question 17:
Le Chatelier’s principle helps to explain the effect of change in conditions on equilibrium.
Discuss the effect of pressure in the following equilibrium on the basis of Le Chatelier’s principle:

Question 18:
Analyse the following graph between ionization enthalpy and atomic number.

What do you observe from the graph? Give justification for your observation.

Question 19:
Carbocations are formed by the heterolytic cleavage of a covalent bond.
a. What is heterolytic bond fission?
b. Arrange the following carbocations in the increasing order of stability:

Question 20:
Balance the following equation by the half-reaction method.

Answer any seven from question numbers 21 to 29. Each carries three scores.

Question 21:
Two important oxides of carbon are carbon monoxide and carbon dioxide.
a. Why is CO called a poisonous gas?
b. How is CO₂ responsible for global warming?
c. What are producer gas and water gas? Mention their uses.

Question 22:
Complete the following reactions:

Question 23:
Hybridization influences the bond length and bond enthalpy in organic compounds:
a. Compare the bond length and bond strength of C-H bonds formed by sp and sp₃ hybridized carbon atoms. Give reason.
b. How many s and p bonds are present in the following molecules?
i. CH₃ – CH₂ – CH₃
ii. CH₃ – C = CH

Question 24:
Particulate pollutants are the minute solid particles or liquid droplets in air.
a. Suggest two examples for non-viable particulate pollutants.
b. Write any two adverse effects of photochemical smog.

Question 25:
The behaviour of acids and bases can be explained using different concepts.
a. Select the Lewis acid from the following:
(NH₃, OH^–, BCl₃, Cl^–)
b. What are conjugate acid-base pairs? illustrate using a suitable example.

Question 26:
Valence Bond Theory(VBT) and Mole-cular Orbital Theory (MOT) are the two important theories of chemical bonding.
a. Out of the following, which is the hybridization of phosphorus in PCl₅?
(sp², sp³, dsp², sp²d)
b. Explain the geometry of the PCl₅ molecule and account for its high reactivity.

Question 27:
Free rotation is possible with respect to a C – C bond in the case of alkanes.
a. The repulsive interaction between the adjacent bonds in a conformation is called ………
b. Draw Newman’s projections of the two conformers of ethane. Which among these is more stable? Justify.

Question 28:
Equilibrium is possible only in a closed system at a given temperature.
a. Write the expression for equilibrium constant, K_c^– for the reaction.
4NH_3(g) +5O_2(g) ⇌ 4NO_(g) + 6H₂O_(g)
b. What happens to the value of the equilibrium constant (Kc) when the above reaction is reversed?

Question 29:
a. What is the number of hydrogen atoms in 1 mole of methane (CH₄)?
b. Calculate the amount of carbon dioxide formed by the complete combustion of 80g of methane as per the reactions
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂ O_(g)
(Atomic masses: C= 12.01u, H= 1.008u, O = 16u).

Answer any three from question numbers 30 to 33. Each carries four scores.

Question 30:
The Valence Shell Electron Pair Repulsion (VSEPR) theory helps in predicting the shapes of covalent molecules.
a. Arrange the bond pair electron and lone pair electron in the decreasing order of the repulsive interactions among them.
b. A molecule of the type. AB₃E₂ has three bond pairs and two lone pairs of electrons. Predict the most stable arrangement of electronpairs in this molecule.
c. The bond order value is an important property of a molecule. How is bond order related to bond length?
d. Write the electronic configuration of an oxygen molecule and justify its magnetic nature.

Question 31:
Most of the naturally occuring processes are spontaneous.
a. Give the criteria for spontaneity of a process in terms of free energy change (∆G).
b. Exothermic reactions associated with a decrease in entropy are spontaneous at lower temperatures. Justify on the basis of Gibbs equation.
c. Find the temperature above which the reaction.
MgO_s + C_s → Mg_s + CO_(g)
becomes spontaneous. (Given ∆H° = 490 KJ mol^-1 and ∆S° = 198JK mol^-1)

Question 32:
What do you mean by the following terms?
a. Homolytic fission
b. Heterolytic fission
c. Nucleophiles
d. Electrophiles

Question 33:
The gases which obey Gas Laws at all temperatures and pressures are called ideal gases.
a. Give reasons for the deviation of real gases from the ideal gas behaviour.
b. Calculate the minimum pressure required to compress 500 ml of air at 1 atm to 300 ml at the same temperature.

Answers

Answer 1:
Cl
Answer 2:
Octahedral

Answer 3:
CO and H₂

Answer 4:
3p

Answer 5:
Magnesium

Answer 6:
graphite

Answer 7:
112

Answer 8:
i. It is defined as the phenomenon of inter-mixing of pure atomic orbitals of slightly different energies and shapes to form new hybrid orbitals of equivalent energies and identical shapes.
ii. Electronic configuration of C (6) is 1s² 2s² 2p_x¹ 2p_y¹ in ground state. It has electronic configuration 1s2 2s1 2px1 2py1 2pz1 in excited state (since C is tetravalent only in excited state).

Answer 9:
a. pent – 4 – ene – 2 – ol
b. 2, 5 – dimethylheptane

Answer 10:
Due to acid rain caused from exhaust of industries the marble layers of Taj Mahal get destroyed. Acid rain is caused by of acid ox-ides such as oxides of sulphur and nitrogen in the air.
Acid rain destroys crops, aquatic lives and buildings. The acid rain will carrode build ing material like marble, limestone etc. and dissolve heavy metals like Cu,Pb,Hg, etc. in water and make it unfit for drinking. It fades the colour of fabrics, leather and paper.

Answer 11:
i. Number of molecules present in 1g of
N² = 6.02 x 10²³/28 = 0.215 x 10²³
ii. Number of molecules present in 1 g of CO₂ = 6.02 x 10²³/44 = 0.136 x 10²³

Answer 12:
a. No. of moles = 64/32 = 2 mole
b. Law of multiple proportions.

Answer 13:
\(_{ 16 }^{ 32 }{ { S }^{ 2- } }\)
b. fig (iii) is the correct one. fig(i) and fig(ii) do not obey Hund’s rule.

Answer 14:
i. Presence of one unpaired e’.
ii. Due to ammoniated electrons.

Answer 15:

Answer 16:
a. Intensity.
b. Kinetic energy of the photoelectrons is directly proportional to the frequency of incident light.

Answer 17:

The forward reaction involves a decrease in number of molecules. Hence a high pressure favours forward reaction (formation of products).

Answer 18:
The graph shows variation of ionisation enthalpy along a group.
Moving down the group, atomic size increases and the outer electrons are far away from the nucleus. There is also an increase in the screening of nuclear charge by electron in inner shells. As a result the removal of electrons becomes easier down the group and ionization enthalpy decreases.

Answer 19:
a. When a bond between 2 atoms breaks in such a way that the shared pair of electrons stays on any one of the atoms. The process is called hetrolytic fission.

Answer 20:

Answer 21:
a. Carbon monoxide reacts with haemo globin to form carboxy haemoglobin which is highly stable complex. This prevent haemoglobin in red blood cells from carrying oxygen and ultimately resulting in death. Hence CO is considered as poisonous gas.
b. CO absorbs IR radiations. This lead to increase in green house effect and thereby raising the temperature of the atmosphere.
c. The mixture of CO and H₂ is called water gas. The mixture of CO and N₂ is called producer gas. Both water gas and pro ducer gas are very important industrial fuels.

Answer 22:

Answer 23:
a. Bond length of C-H bond for sp hybridised carbon is less and that of sp³ hybridised carbon is high. Bond strength is greater for C-H bond is sp hybridised carbon than that for sp³ hybridised carbon.
b. 1.10 σ bonds.
2. 6 σ bonds and 2 π bonds.

Answer 24:
a. Mist, smoke, dust
b. *Causes irritation to eyes.
*Causes health problems like head ache, chest pain, dryness of throat, cough and difficulty in breathing.
*PAN adversely affects green plants.

Answer 25:
a. BCl₃
b. The acid-base pair which is formed from each other by loss or gain of proton is called as conjugate acid-base pairs. Consider the reaction,

Water act as a base since it accepts proton. HCl act as an acid since it donate proton.
Cl^– is the conjugate base of acid HCl.
H₃O⁺ is the conjugate acid of base H₂O.

Answer 26:

In PCl₅, 5 sp³ d orbitals of phosphorus over¬lap with singly occupied P orbitals of chlorine atoms to form 5 P-Cl bonds. Three P- C/ bond lie in one plane and make bond angle of 120°. These bonds are equitorial bonds. The remaining 2 P-Cl bond lie above and below the equitorial plane making an angle of 90° with the plane. These bonds are called axial bonds. Axial bonds are longer and weaker than equitorial bonds since they suffer more repulsive interactions from equitorial bond pairs. As a result PCl₅ is highly reactive.

Answer 27:
a. Torsional strain.
b. Newman’s projection

Staggered conformation is more stable than eclipsed conformation. This is because in staggered conformation, the hydrogen atoms are at maximum distance apart and there is minimum repulsion and it is more stable. But in eclipsed conformation, H atoms of one carbon atom are directly behind those of other and thus repulsion is maximum and hence least stable.

Answer 28:

Answer 29:
a. 4 moles of hydrogen or 4 x 6.023 x 10²³ atoms of hydrogen.
b. According to the equation, 1 mole CH₄ gives 1 mole CO₂ number of moles of methane = \(\frac { 80 }{ 16 }\)
= 5 moles \ 5 moles of CO₂ is formed, ie., 5 x 44 = 220g CO₂

Answer 30:
a. Lone pair – lone pair > lone pair – bond pair > bond pair – bond pair.

Answer 31:
a. ∆G = ∆H – T∆S
When ∆G is negative (∆G < 0), the pro-cess is spontaneous, b. For exothermic reactions, ∆H = -ve and ∆S = -ve When ∆H = -ve, ∆S = -ve, ∆G become -ve only. When ∆H >T∆S. Such reactions are spontaneous only at low temperature.

The reaction is spontaneous above this temperature.

Answer 32:
a. When the bond between two atoms A-B breaks in such a way that each fragment carries one unpaired electron, the process is called homolytic fission,
b. When the bond between two atoms A-B breaks in such a way that the shared pair of electrons stays on any one of the atoms, the process is called heterolytic fission.
c. Electron rich species are called nucleo-philes, (nucleus loving). They has attraction for a positively charge centre.
eg:, OH^–, CN^–, I^–
d. These are the reagents which attacks electron rich centres in the substrate.

Answer 33:
a. Two postulates of kinetic theory of gases are defective:

i. Volume occupied by gas molecule is negligible as compared to the total volume of the gas.
ii. The force of attraction and repulsion among the gas molecule are negligible.

At low temperature and high pressure, gas is in compressed states. But molecules are incompressible. As a result the volume of the molecules in the gas cannot be ignored in comparison to the volume occupied by gas.
At low temperature and high pressure, the volume of the gas becomes small, and molecules come closer to each other. Under these conditions, the at¬tractive force between the molecules become operative. Hence under these conditions real gas deviate from ideal behaviour.

Plus One Chemistry Previous Year Question Papers and Answers