# Plus One Chemistry Previous Year Question Paper 2017

## Kerala Plus One Chemistry Previous Year Question Paper 2017

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.

Question 1:
a. Determine the number of moles present in 0.55 mg of electrons.
imgg
i. 1 mole
ii. 2 moles
iii. 1.5 moles
iv. 0.5 mole [1]

b. Give the empirical formula of the following. [2]
C6H12O6, C6H6. CH3COOH, C6H6C16

c. Two elements, carbon and hydrogen combine to form C2H6, C2H4 and C2H2. Identify the law illustrated here. [1]

Question 2:
a. i. Write the electronic configuration of chromium (Z=24).
ii. Find the number of electrons in the subshells with azimuthal quantam number l=2.
iii. Represent the orbital with quantum numbers n = 1 and 1 = 0.
b. Give the mathematical representation of Heisenberg’s uncertainty principle and its one important significance.

Question 3:
Electron gain enthalpy is one of the important periodic property.
a. Define electron gain enthalpy. [1]
b. Explain any two factors affecting electron gain enthalpy. [2]
c. Write the oxidation state and covalency of Al in (A/ F6)3- [1]

Question 4:
The geometry of the molecule is decided by type of hybridization.
a. Discuss the shape of PCl5 molecule using hybridization. [2]
b. Give the reason for the high reactivity of PCl5 [2]
c. Isoelectronic species have the samebond order. Among the following, choose the pair having same bond order. Among the following, choose the pair having same bond order.
CN, O2 ,NO+, CN+

Question 5:
a. Give the reason behind the following.

i. The glass window pannels of old buildings are thicker at the bottom than at the top. [1]
ii. Sharp glass edges are heated for making them smooth. [1]

b. Maxwell and Boltzmann have shown that actual distribution of molecular speeds depends, on temperature and molecular mass.
i. What do you mean by most probable velocity? [1]
ii. At the same temperature which will move faster, N2 or Cl2 ? [1]

Question 6:
a. Some macroscopic properties are given below. Help Reena to classify them into two groups under suitable titles.
[Heat capacity, Entropy, Refractive in¬dex, Surface tension.] [2]
b. For the reaction
2A(g) + 2B(g) → 2D(g)
∆U°= -10.5 kJ/mol
∆S°= – 44.1J/k/mol at 298 K.
Calculate ∆G° for the reaction. [2]

Question 7:
a. Classify the following solutions into acidic, basic and neutral
NaCl, NH4NO3, NaCN, NaNO2 [2]
b. pH of blood remains constant in spite of the variety of goods and spices we eat. Give a reason. [1]
c. The solubility ofMg(OH)2 at 298K is 1.5 x 10-4.Calculate the solubility product.[2]

Question 8:
Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced equa¬tion for the reaction using oxidation number method.
Skeletal equation is
MnO4 + Br → MnO2 + BrO3 [3]

Question 9:
Hydrogen is the most abundant element in the universe, But in free state it is almost not found in earth’s atmosphere.
a. Suggest any three methods for the preparation of H2 gas by selecting suitable substance given below.

b. Do you expect carbohydrides of the type CnH2n+2to act as Lewis acid or base? Why?

Question 10:
The s-block of periodic table constitutes alkali metals and alkaline earth metals.
a. The hydroxides and carbonates of sodium and pottassium are more soluble than that of corresponding salts of Magnesium and Calcium. Explain. [2]
b. Write the chemical name of the following:
i. Caustic soda
ii. Baking soda
iii. Slaked lime
iv. Milk of lime [2]

Question 11:
Borax is an important compound of Boron.
a. The solution of borax is alkaline. Give a reason. [2]
b. Give any two uses of borax. [1]
c. Diamond has covalent bonding. Yet it has high melting point. Give a reason. [1 ]

Question 12:
a. Give the structural formula of the following compounds:
i. 2, 4, 7 – Trimethyloctane
ii. 2 – chloro – 4 – methyl pentane [2]
b. CH3C2 or (CH3)2CH -which is more stable? Explain. [2]
c. Explain the chemistry behind crystallization. [2]
Or
a. Give the IUPAC names of the following:

b. Which is more stable (CH3)3 C+ or CH3C+H2? Give a reason. [2]
c. Give the chemistry behind distillation under reduced pressure. [2]

Question 13:
Benzene and benzenoid compounds show aromatic character.
a. Select the aromatic compounds from the following: [1]

b. Suggest a method to convert Ethyne to benzene. [2]
c. Give the products formed when benzene reacts with
i. CH3 Cl/AlCl3
ii. Cl2 /hυ [2]

Question 14:
Environmental pollution is the effect of undesirable change in surroundings that have harmful effect on plants, animals and human beings.
a. Explain the adverse effect of global warming. [2]
b. Choose the one which is not a component of photochemical smog.
i. NO2
ii. O3
iii. SO2

Answers

Answer 1:
1. a.i. 1 mole
Mass of one electron = 9.1 x 10-28 g
= 9.1 x 10-25 mg
Mass of 1 mol of electron
= 9.1 x 10-25 x 6.022 x 10-23= 0.55 mg
So 0.55 mg of electron = 1 mol of electron

b. Empirical formula of C6 H12O6 is CH2O
Empirical formula of C6H6 is CH
Empirical formula of CH3COOH is CH2O
Enpiric formula of C6H6Cl6 is CHCl
c. Law of multiple proportion.

Answer 2:
a. i. 24Cr – 1s2 , 2s2, 2p6, 3s2, 3p6, 4s1, 3dor [Ar] 4s1, 3d5
ii. 10 electrons in d subshell.
iii. 1s

b. Heisenberg’s uncertainty states that it is impossible to determine simultaneously and precisely both the exact position and exact momentum or velocity of a subatomic particle like an electron.
Mathematically, ∆x ∆p ≥ $$\frac { h }{ 4\pi }$$

where ∆p is the uncertainty in the momentum of the particle, ∆x is the uncertainty in position.
From this equation, it is seen that if ∆p increases, the ∆x decreases, and vice versa. It is also can be written as follows:
∆x ∆p ≥ $$\frac { h }{ 4\pi m }$$

Answer 3:
a. Electron gain enthalpy is the enthalpy change when an electron is added to the outer most shell of an isolated gaseous atom in the ground state.
b. The factors affecting electron gain enthalpy are atomic size, nuclear charge, shielding effect, electronic configuration etc. As the atomic size increases. ∆eg H decreases. When nuclear charge increases, electron gain enthalpy increases.
c. Oxidation state is +3
Covalency is 6.

Answer 4:
a. In PCl5, five sp3d orbitals of phosphorus overlap with singly occupied p-orbitals of chlorine atoms to form five P-Cl bonds. Three P-Cl bond lie in one plane and make bond angle of 120°. These bonds are equitorial bonds. The remaining two P-Cl bond lye above and below the equitorial plane making an angle of 90° with the plane. These bonds are called axial bonds.

b. Two P-Cl axial bonds are slightly elongaed to minimise repulsion from equitorial bond pairs. So axial bonds are weaker than equitorial bonds in PCl5.
c. CN and NO+

Answer 5:
a.i. Glass is a very viscous liquid. So it has a tendency to flow very slightly. Hence the window panes of old buildings are thicker at the bottom than at the top.
ii. On heating, the glass melts and the surface of the liquid tend to take the rounded shape at the edges, which makes the edges smooth.
b. i. Most probable velocity is the velocity possessed by a maximum number of gas molecules.
ii. N2.

b. Given All0 = -10.5 kJ/mol = -10.5 x 103 J/mol
AS0 = -44.1 J/K/ mol, T= 298 K We know that AH° = AU° + AnRT Here AH° = -10500 + (-1)x 8.314 x 298 = -12978 J/ mol
AG° = AH0 – T AS0 = -12978 – (298 X44.1) = -164 J/ mol = -0.164 kJ/ mol.

Answer 6:
a.

 Extensive properties Intensive properties Heat capacity Refractive Index Entropy Surface tension

b. Given ∆U° = -10.5 kJ/mol = -10.5 x 103 J/mol
∆S° = -44.1 JIKI mol, T = 298 K
We know that ∆H° = ∆U° + ∆nRT
Here ∆H° = -10500 + (-1)x 8.314 x 298
= -12978 J/ mol
∆G° = ∆H° – T∆S° = -12978- (298 X 44.1)
= -164 J/ mol = -0.164 kJ/ mol.

Answer 7:
a. Neutral – NaCl
Acidic – NH4NO3
Basic – NaCN, NaNO2
b. Because blood is a buffer solution.
c. Mg (OH)2 → Mg2+ + 2 OH
Solubility: S S 2S
Ksp = [Mg2+][OH]2 = S X (2S)2 = 4S3
= 4 x (1.5 x 10-4 )3 = 1.35 x 10-11

Answer 8:
Step 1 : The skeletal equation is :
MnO4 +Br → MnO2 + BrO3

Step 2 : Assign oxidation number each element and identify the elements under-going change in oxidation number.
Mn+1O4-2 + Br-1 → Mn+4O-22 + BC+5O3-2
Here the oxidation number of Mn and Br are changed.

Step 3 : Calculate the increase and decrease in oxidation number and make them equal by multiplying with suitable number.
2MnO4 + Br → 2 MnO2+ BrO3

Step 4 : Since the reaction takes place in the basic medium, 20H ions on the RHS to balance the ionic charges on both sides.
2MnO+4 + Br → 2 MnO2 + BrO3 + 2OH

Step 5 : Now balance hydrogen atom by adding a sufficient number of H2O molecules. Here add one H20 molecule on LHS.
2MnO+4+ Br +H2O → 2MnO2+ BrO3+ 2OH
This is the balanced equation.

Answer 9:
a. i. 2Na + 2H2O → 2 NaOH + H2
or
2Na + 2 HCl → 2NaCl + H2
ii. Zn + 2 HCl → ZnCl2 + H2
OR
Zn + 2 NaOH → Na2ZnO2 + H2
iii. CaH2 + 2H2O → Ca(OH)2 + 2H2
iv. 2Al + 6 HCl → 2AlCl3 + 3H2 (Any three)

b. Cn H2n+1 is neither a Lewis acid nor a Lewis base due to the absence of vacant orbitals or lone pairs of electrons on carbon.

Answer 10:
a. This is due to greater atomic radii (size) or lesser lattice enthalpy of sodium and potassium ions.
b. i. Caustic soda : Sodium hydroxide (NaOH)
ii. Baking soda : Sodium bicarbonate (NaHCO3)
iii. Slaked lime : Calcium hydroxide (Ca(OH)2)
iv. Milk of lime : Magnesium hydroxide (Mg(0H)2)

Answer 11:
a. Borax dissolves in water to give NaOH and orthoboric acid. Since NaOH is a strong alkali and orthoboric acid is a weak acid, the solution is basic in nature.
Na2B4O7 + 7 H2O → 2NaOH + 4H3BO3

b. Borax is used for the detection of transition elements by Borax bead test. It is used for the manufacture of heat-resistant glasses, glass-wool, and fiberglass. It is also used as a flux for soldering metals, for heat, scratch and stain resistant, glazed coating to earthenwares and as a  constituent of medicinal soaps. :
c. Diamond has a three-dimensional network structure with extended covalent bonding.
Since it is very difficult to break these covalent bonds, it has high m.p Or, due to the tetrahedral (sp3) hybridization of carbon atoms in diamond.

Answer 12:
a. i.

b. CH3 – CH2 is more stable than (CH3)2CH since for carbanions the stability order 1° > 2° > 3° due to inductive effect.

c. Crystallization is based on the difference in the solubilities of the compound which is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature. The solution is concentrated to get a nearly saturated solution at a higher temperature. On cooling the solution , pure compound crystallizes out and is removed by filtration.

OR

a. i. 3- Bromo-3-chloroheptane
ii. 4-Ethyl-2-methylaniline

b. (CH3)3C+ is more stable than CH3 – CH2+ since the stability of carbocations follow the order 3° > 2° > 1° due to inductive effect and hyperconjugation.

c. This method is used to purify liquids having very high boiling points and those, which decompose at or below their boiling points. Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface. The pressure is reduced with the
help of a water pump or vacuum pump.

Answer 13:
a. Aromatic compounds are :

b. Ethylene on passing through a red-hot iron tube at 873 K, we get benzene (C6H6).
c. i. Toluene (C6H5 – CH3)

Answer 14:
a. Due to global warming, the average global temperature will increase. This will lead to the melting of polar ice caps and flooding of low lying areas all over the earth. Increase in global temperature results in infectious diseases like dengue, malaria, yellow fever, sleeping, sickness etc.
b. iii. SO2