NEB Grade 12 Chemistry Model Paper Solution

NEB Grade 12 Chemistry Model Paper Solution of 2078 of New Syllabus

Check out NEB Grade 12 Chemistry Model Paper solution of new curriculum 2078. NEB grade 12/XII Chemistry model paper solution of new syllabus 2078. Download the paper here. Also, check out our chemistry notes here.

Group A: Multiple Choice Questions (11×1 = 11)

1. What is the equivalent weight of H₃PO₃ in the reaction; 2NaOH + H₃PO₃ → Na₂HPO₃ + 2H₂O
A) 2M
B) M/1
C) M/2
D) M/3
Ans: C is the correct answer.

2. The solubility product of chalk is 9.3 × 10^-8. What is its solubility in gram per liter?
A) 3.04 × 10-1
B) 3.04 × 10^-2
C) 3.04 × 10^-3
D) 3.04 × 10^-4
Ans: B is the correct answer.

3. What is the concentration of N₂O5 in the following first-order reaction in which the rate is 2.4 × 10^-5 mol/L and rate constant is 3.0 × 10^-5S^-1?
2N2O5 → 4NO2 + O2
A) 0.04
B) 0.8
C) 1.2
D) 1.4
Ans: B is the correct answer.

4. What happens when the lead storage battery is discharged?
A) SO₂ is evolved
B) PbSO₄ is consumed
C) Lead is formed
D) H₂SO₄ is consumed
Ans: D is the correct answer.

5. What is the general electronic configuration of transition metal?
A) (n-1)s²p⁶d^1-10ns^0-2
B) (n-1)s²p⁶ns²np¹
C) (n-1)s²p⁶d⁵ns¹
D) (n-1)s²p⁶ns¹
Ans: A is the correct answer.

6. Which of the following ore is concentrated by forth-flotation process?
A) Hematite
B) Siderite
C) Galena
D) Malachite
Ans: C is the correct answer.

7. Which of the following products is obtained when nitrobenzene is electrolytically reduced?
A) P-aminophenol
B) azobenzene
C) azoxybenzene
D) hydrazobenzene
Ans: A is the correct answer.

8. Which of the following compounds is a pi-bonded organo-metallic compound that has ethene as one of its components and is the first synthesized organometallic compound?
A) Zeise‟s salt
B) Ferrocene
C) Dibenzene chromium
D) Tetraethyl tin
Ans: A is the correct answer.

9. What effect does calcium sulphate have on cement?
A) Retards setting action
B) Acts as flux
C) Imparts color
D) Reduces strength
Ans: A is the correct answer.

10. Removal of which of the following leads to higher fiber-fiber bonding strength in the paper?
A) Softwood
B) Hardwood
C) Lignin
D) Pulp
Ans: C is the correct answer.

11. In the figure given below which one is correct?

A) Alpha rays deviate towards A, beta rays deviate towards C and gamma rays direct towards B.
B) Alpha rays direct towards B, beta rays deviate towards C and gamma rays towards A.
C) Alpha rays deviate towards C, beta rays direct towards B and gamma rays towards A.
D Alpha rays deviate towards C, beta rays deviate towards A and gamma rays direct towards B.
Ans: D is the correct answer.

Group B: Short Answer Questions (8×5 = 40)

1. Standard solution of Na₂CO₃ is used to determine the strength of H₂SO₄ during Titration.

A) How is the completion of the reaction in this titration detected? Is the solution prepared from Na₂CO₃ primary standard? Why? [1+1]
Ans: Na₂CO₃ is a weak base and H₂SO₄ is a strong acid. So, methyl orange is used as an indicator to complete the reaction since the pH range is 3.1 to 4.4.
Yes, the solution prepared from Na₂CO₃ is the primary standard because Na₂CO₃ is soluble in water and it is available in pure form.

B) 2.16 g of pure Na₂CO₃ is added to 400 ml deci-normal solution of H₂SO₄. How many grams of H₂SO₄ is further required to neutralize the resultant solution completely? [3]
Solution:

No. of gm.eq. of Na₂CO₃ No. of gm. eq. of H₂SO₄ + x gm.eq. of required H₂SO₄

\begin{align*}V_{1}N_{1}&=V_{2}N_{2}+V_{3}N_{3}\\or,\ \not V_{1} \times\frac{w}{\not V_{1}}\times\frac{1000}{eq.\ wt}&=400\times\frac{N}{10}+\not V_{3} \times\frac{w}{\not V_{3}}\times\frac{1000}{eq.\ wt}\\2.16\times\frac{1000}{53}&=40+x\times\frac{1000}{49}\\\therefore x&=0.037\ gm\end{align*} 

OR

A) Derive the relation k =log(2.303/t) log(a/a-x) . Show that for the first-order reaction the time required for half the change (half-life period) is independent of the initial concentration. (2+1)
Ans: The reaction in which the rate of reaction depends upon one concentration term is called a first-order reaction.
Let us consider a general first-order reaction. Let the initial concentration of the reactant be a mol L^-1 and after time t, x mol of reactant changes into product.

The rate of the above reaction is given by

\begin{align*}Rate &= K_{a}[A]^{1}\\\frac{dx}{dt}&=K_{1}(a-x)^{1}\end{align*}

It is a differential rate law equation for a first-order reaction.

\frac{dx}{(a-x)}=K_{1}dt

Integrating it, we get

\int\frac{dx}{(a-x)}=K_{1}dt\\or,\ -ln(a-x) = K_{1}t+I

where I is the integration constant. To know its value, let us apply the initial condition i.e. when time = 0, the amount of product is also 0 i.e. x=0. Substituting this value in the above equation, we get

ln(a-0) = K₁ x 0 + 1
so, – ln a = I
Hence,

\begin{align*}-ln(a-x)&=k_{1}t-lna\\or,\ K_{1}t=&lna-ln(a-x)\\or,\ k_{1}t&=ln\frac{a}{(a-x)}\\K_{1}=\frac{2.303}{t}&log\frac{a}{a-x}---(i)\end{align*}

This is the integrated rate law for the first-order reaction.

Half-life period: It is the time during which half of the initial concentration of reactant is converted into product. It is denoted by t1/2.
According to the definition, at the half-life period (t₁/2), x = a/2, where a is the initial concentration of reaction.
Substituting this value in eqn (I),

\begin{align*}K_{1}&=\frac{2.303}{t_{1/2}}\ log\ \frac{a}{a-a/2}\\K_{1}&=\frac{2.303}{t_{1/2}}log2\\t_{1/2}&= \frac{0.693}{K_{1}}\end{align*}

It shows that the half-life of the first-order reaction is independent of initial concentration.

B) A first-order reaction is 50% completed in 1.26×10¹⁴⁵. How much time would it take for 90% completion? (2)
Solution:

Note: This question seems to have a data error. We have used time as 20 minutes for half-life reaction.

\begin{align*}K_{1}&=\frac{2.303}{t}log\frac{a}{(a-x)}---(i)\\&=\frac{2.303}{20}\ log2\\&=0.035\\so,\\0.035&=\frac{2.303}{t}\ log\frac{a}{(a-x)}\\or,\ 0.035&=\frac{2.303}{t}\times log10\\ \therefore t&= 65.7\ minutes\end{align*}

2. Study the following data for the thermodynamic process H₂O (l) → H₂O (s) at different temperatures and at 1 atmospheric pressure.

a. Calculate the total entropy of the universe at given condition 3. (1)
Ans: Total entropy = Entropy of system + Entropy of surrounding
= -27.62 + 27.42
= -0.2

b. Can we predict the spontaneity of the given reaction at 0°C? (1)
Ans: If total entropy is positive, the process is spontaneous.
Total entropy = Entropy of system + Entropy of surrounding
= -26.55 + 26.88
= 0.33
It is a positive value. So, in the given reaction at O°C, the process is spontaneous.

c. Calculate the equilibrium constant for the fusion of ice at 1°C. What is the effect of temperature on the entropy change of reaction? (2+1)
Solution:

\begin{align*}\Delta G_{sys}&=-T\Delta S_{universe}\\-R\not TlnK&=-\not T\times 0.2\\lnK&=\frac{0.2}{8.314}\\K&=10^{0.02}=1.05\end{align*}

Entropy increases as temperature increases. An increase in temperature means that the particle of the substance has greater kinetic energy. The faster-moving particles have more disorder than particles that are moving slowly at a lower temperature.

3. The figure shows the octahedral distortion of the d-block orbital in the presence of a ligand.

a. Why does octahedral distortion occur in the presence of ligand? Explain on the basis of CFT. (2)
Ans: When the ligands approach the central metal ion, the degeneracy of electronic orbital state usually d-orbitals is broken due to the static electric field produced by the surrounding charge distribution. Because electrons repel each other, the d-electrons closer to the ligands will have higher energy than those further away, resulting in the d-orbitals splitting. In most cases, the d-orbitals are degenerate, but sometimes they can split with the e_g and t_2g subsets having different energy. The d_{x² – y²} and d_z² all point directly along the x,y and z axes. They form an eg set. On the other hand, the lobes of the d_xy, d_xz and d_yz all line up in the quadrants, with no electron density on the axes. These three orbitals form the t_2g set. CFT is the bonding model that explains many important properties of transition metal complexes, including their colours, magnetism, structure, stability and reactivity. The central assumption of CFT is that metal-ligand interactions are purely electrostatic. According to this theory, in a free isolated gaseous ion, the five d orbitals are degenerate (have equal energy) but in the solution or compound state, the energy of the d-orbitals is changed which cause the splitting into different energy d-orbitals. This is called crystal field splitting.

b. On the basis of the given distortion, how can you explain [Cu(H₂O)₆]⁺⁺ is blue coloured complex. (1)
Ans: The colour of transition metal ions arises from the excitation of electrons from the d-orbitals of lower energy to the d-orbitals of higher energy. Light radiations corresponding to such small amounts of energy which are required for the d-d transition are available in the visible region. It is for this reason that transition metal ions have the property to absorb certain radiations from the visible regions and exhibit complementary colours.
An unpaired electron is present in Cu⁺⁺ (d⁹) which makes electronic transition feasible and imparts blue colour.

c. Out of Fe⁺⁺ and Fe⁺⁺⁺ which one is more stable? Explain on the basis of distortion seen in the above figure. (1)
Ans: Fe+++ is more stable due to a half-filled stable electronic configuration.

d. Why do such elements which give such splitting show good catalytic properties? (1)
Ans: Due to the presence of vacant d-orbitals and providing a large surface area due to their finely divided nature, they have the ability to exhibit variable valencies and show good catalytic properties.

4. X is an ore of a metal M. X on calcination gives black precipitate (W) of metal oxide which belongs to group II of basic radical in qualitative analysis. X on roasting gives the metal (M) and gas as major byproducts. The gas when passed through an acidified K₂Cr₂O₇ solution turns green.

a. Identify the metal X. (1)
Ans: Copper (Note: Copper pyrite on calcination gives black sulphide. It does not give black oxide on calcination.)

b. Write the reaction involved during calcination of X. (1)
Ans: 2CuFeS₂ → Cu₂S + 2FeS + SO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂

c. Write the action of the gas on acidified K₂Cr₂O₇. (1)
K₂Cr₂O₇ + H₂SO₄ + 3SO₂ → K₂SO₄ + Cr₂(SO₄)₃ +H₂O

d. Convert metal X into it’s vitriol. (2)
Ans:

Cu+O_{2}\xrightarrow{upto\ 110^{\circ}}CuO\\CuO+H_{2}SO_{4}\rightarrow CuSO_{4}(aq.)+H_{2}O\\CuSO_{4}(aq.)\xrightarrow{crystallization}CuSO_{4}.5H_{2}O 

5. The given table shows the compounds and their molecular formula. How can you convert P to Q, where Q is a compound in which two methyl groups are substituted at adjacent carbons? How is P obtained from T, where T is secondary alcohol? Write the reactions involved in the conversion of P into R and S? [5×1=5]

Ans:
i. Conversion of P to Q

ii. Conversion of T to P

iii. Conversion of P into R and S

OR

An aromatic compound [A] in which one chlorine atom is substituted at a benzene ring. When the compound [A] is heated with 2, 2, 2-trichloro ethanal in presence of conc. H₂SO₄ gives an insecticide [B]. The compound [A] when treated with an acid chloride-containing two carbon atoms in the presence of anhydrous AlCl₃ gives [C].

a. Identify B and C. (1 +1)
Ans:

b. Reaction of aq. NaOH on the compound [A] is more difficult than with chloroethane, justify with a suitable explanation.(2)
Ans:

Due to partial double bond development in the C-Cl bond of haloarene caused by resonance, the C-Cl bond becomes rigid and replacement of halogen by nucleophile becomes difficult. So, haloarenes are less reactive than haloalkanes.

c. How would you obtain compound A from benzene diazonium chloride? (1)