Kerala Plus One Chemistry Board Model Paper 2021 with Answers

Reviewing Kerala Syllabus Plus One Chemistry Previous Year Question Papers and Answers Pdf Model 2021 helps in understanding answer patterns.

Kerala Plus One Chemistry Board Model Paper 2021

Answer any 6 questions from 1 to 12. Each carries 2 scores. (6 × 2 = 12)

Question 1.
State Hund’s rule of maximum multiplicity.
Answer:
Pairing occurs only after all the orbitals are completely half filled.

Question 2.
Calculate the de-Broglie wavelength associated with an electron with velocity 2.05 × 10⁷ ms^-1.
Answer:
de-Broglie wavelength = \(\frac{h}{m v}\)
= \(\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.05 \times 10^7}\)
= 3.55 × 10^-11 = 35.5

Question 3.
Based on VSEPR theory predict the shape of H₂O and NH₃.
Answer:
H₂O – bent or V Shape
NH₃ – Tetra hedral

Question 4.
Real gases do not follow gas laws perfectly under all conditions. Why?
Answer:
Due to two faulty assumptions in kinetic theory

• There is no force of attraction between the gas molecules.
• The volume of a gas molecule is negligible compare to the volume of a gas.

Question 5.
State the first law of thermodynamics.
Answer:
Energy can neither be created nor be destroyed.

Question 6.
Give the relation between k_p and k_c for the reaction given below.
2NOCl(g) → 2NO(g) + Cl₂(g)
Answer:
K_p = K_c . (RT)An
(An = no of moles of gaseous pdt – no of moles of gaseous reactant)
= 3 – 2 = 1
K_p = K_c. (RT)
K_p = K_c.RT

Question 7.
Classify the following into Lewis acids and Lewis bases.
(i) H₂O
(ii) NH₃
(iii) BCl₃
(iv) H⁺
Answer:
Lewis acids : – BCl,, H⁺
Lewis bases : NH₃, H₂O.

Question 8.
Write any two anomalous behaviuor of Li.
Answer:
(i) Li is very hard
(ii) Li forms monoxide

Question 9.
Givethe IUPAC name of the following
(A) CH₃ – CH = CH – CH₂ – CH₂ – COOH
(B)

Answer:
(A) Hex-4-en-1-oicacid
(B) Cyclohex-2-en-1-ol

Question 10.
Write any two chain isomers of a molecule having molecular formula C₅H₁₂.
Answer:

Question 11.
Name the following reaction

Answer:

Question 12.
Cyclopenta dienyl anion is aromatic. Why?
Answer:
Because it is planar, cyclic, conjugated and contains huckels no. of electrons.

Answer any 8 questions from 13 to 28. Each carries 3 scores. (8 × 3 = 24)

Question 13.
(i) State the law of definite proportion. (2)
Answer:
The same compound always contains same elements in a fixed ratio by mass.

(ii) Define limiting reagent of a reaction. (1)
Answer:
The reactant which is completely used up in the reaction.

Question 14.
(i) What is atomic mass unit? (1)
Answer:
amu = \(\frac{1}{12}\) th of a C – 12 atom

(ii) Calculate the number of atoms in each of the following:
(A) 52 mole of Ar
(B) 52 g of He (2)
Answer:
(A) 52 mole of Ar.
= 52 × 6.022 × 10²³ no.of Ar atoms 52

(B) 52 g of He = \(\frac{52}{4}\) × 6.022 × 10²³ no. of atoms.

Question 15.
(i) Which of the following ion is not isoelectronic with Ne atom.
(a) O^–
(b) Na⁺
(c) Al³⁺
(d) Mg²⁺ (1)
Answer:
(a) O^–

(ii) Explain why Na⁺ ion is smaller than Na atom.(2)
Answer:
In Na+ ion 11 proton attracts 10 electron so the size contracts. In Na 11 proton attracts 10 electron. So the size is large.

Question 16.
(i) Consider the process.

Which is greate IE₂ or IE₁? Justify. (2)
Answer:
IE₂ is greater.
IE₁ is the energy to remove one electron and attain stable electronic configuration. So it is low.
IE₂ is the energy needed to remove one electron from stable electronic configuration. So it is high.

(ii) Electron gain enthalpy of F is less than that of Cl. Why? (1)
Answer:
In F the electron enters to 2p orbital. But in Cl the enters to 3p orbital 2p is smaller compared to 3p and the electron repulsion is more in F than Cl.

Question 17.
(i) Explain the covalent character of LiCl using Fajan’srule. (1)
Answer:
Smaller the cation, large the anion more is the covalent character.

(ii) Write any two differences between sigma (σ) bond and pi(π) bond. (2)
Answer:

σ bond	π bond
Strong bond	Weak bond
By axial overlap	By sidewise overlap

Question 18.

(i) Name of the gas low shown by the above graph.
Answer:
Boyle’s law

(ii) A vessel 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transffed to another vessel of volume 180 mL at 35°C. What would be its pressure?
Answer:
From Boyle’s law,
P₁V₁ = P₂V₂
Here P₁ = 1.2 bar, V₁ = 120 mL, V₂ = 180 mL,
P₂ = ?
1.2 × 120 = P₂ × 180
P₂ = 1.2 × 120/180 = 0.8 bar

Question 19.
(i) Write the van der Waal’s equation for ‘n’ mole of a gas. (1)
Answer:
(P + \(\frac{n^2 a}{V^2}\)) (V – nb) = nRT

(ii) State any four postulates of kinetic theory of gases. (2)
Answer:
(1) Gas consists of large number of minute particles called molecules.
(2) Molecules are always in random motion.
(3) During their motion they collide with each other this creates pressure.
(4) As the temperature increase K.E. also increases.

Question 20.
(i) Which among the following is an intensive property:
(a) Mass
(b) Volume
(c) Enthalpy
(d) Temperature (1)
Answer:
(d) Temperature

(ii) State and explain Hess’s law of constant heat summation. (2)
Answer:
The total enthalpy change for a process is same whether it lakes place in single step or several step.

Question 21.
(i) Define the Lattice enthalpy. (1)
Answer:
It is the enthalpy change when 1 mole of a compound is formed from its constituent ions.

(ii) Draw the Born-Haber cycle for the calculation of lattice enthalpy of NaCl. (2)
Answer:
Born-Haber cycle for the formation of NaCl

S – Sublimation energy
D – Dissociation energy
IE – Ionisation energy
EA- Electron gain enthalpy /electron affinity
U – Lattice energy
∆H_f – Enthalpy of formation

Question 22.
(i) The oxidation state of Mn in KMnO₄ is …….. (1)
Answer:

(ii) Identify the element undergoing disproportionation reaction in the following. (2)
2H₂O₂(aq) → 2H₂O(6) + O₂(g)
Answer:
Assign the oxidation number of each element

The element undergoing disproportionation reaction is oxygen in H₂O₂.

Question 23.
Balance the following equation by the half reaction method in acidic medium. (3)
Fe²⁺_(aq) + CrO^2-_7(aq) → Fe³⁺_(aq) + Cr³⁺_(aq) (3)
Answer:
Balance the equation by half reaction
Fe²⁺ + Cr₂O^2-₇ → Fe³⁺ + Cr³⁺
Step I : Fe²⁺ → Fe³⁺
Cr₂O^2-₇ → 2Cr³⁺

Step II : Balancing atoms other than oxygene and hydrogen
Fe²⁺ → Fe³⁺
Cr₂O^2-₇ → 2Cr³⁺

Step III : Balancing oxygen and hydrogen by adding H₂O and H⁺.
Fe²⁺ → Fe³⁺
Cr₂O^2-₇ + 14H⁺ → 2Cr³⁺ + 7H₂O

Step IV: Adding electrons to balance charges
Fe²⁺ → Fe³⁺ + e^–
Cr₂O^2-₇ + 14H⁺ + 6e → 2Cr³⁺ + 7H₂O

Step V: Equalising the no. of electrons in both
6Fe²⁺ → 6Fe³⁺ + 6e^–
Cr₂O^2-₇ + 14H⁺ + 6e → 2Cr³⁺ + 7H₂O

Step VI: Adding the two half reactions
6Fe²⁺ Cr₂O^2-₇ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

Question 24.
(i) Write the name of one salt responsible for the permanent hardness of water. (1)
Answer:
CaCl₂, CaSO₄, MgCl₂, MgSO₄

(ii) Explain any one method used for the removal of permanent hardness of water. (2)
Answer:
Add washing soda to hard water.

Question 25.
(i) Write the name of any two electron rich hydrides. (1)
Answer:
NH₃, PH₃,HCl

(ii) Explain with suitable chemical equation, why hydrogen peroxide is stored in waxlined glass or plastic vessel in dark. (2)
Answer:
H₂O₂ reacts with silica and decomposes easily in presence of light.
H₂O₂ → H₂O + O₂