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JEE Chemistry Solved Question Paper 2008

Multiple Choice Questions

The percentage of an element M is 53 in its oxide of molecular formula M₂O₃. Its atomic mass is about

The electronic configuration of the element with maximum electron affinity is

1s², 2s²,2p³
1s², 2s²,2p⁵
1s², 2s²,2p⁶, 3s², 3p⁵
1s², 2s²,2p⁶, 3s², 3p³

3.
The incorrect statement/s among the following is/are
I. NCl₅, does not exist while PCl₅, does.
II. Lead prefers to form tetravalent compounds.
III. The three C-O bonds are not equal in the carbonate ion.
IV. Both O⁺₂and NO are paramagentic

I, III and IV

I and IV

II and III

I and III

II and III

(I) In nitrogen d-orbitals are absent, so it does not form NCl₅. Thus, NCl₅ does not exist but PCl, does.

(II) Pb²⁺ is more stable than Pb⁴⁺, due to inert pair effect.

(III) In carbonate ion (CO^2-₃) all the three C-O bonds are identical due to resonance.

$(IV) O_{2}^{+} (8 + 8 - 1 = 15) = σ 1 s^{2}, σ^{*} 1 s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2}, σ 2 p_{x}^{2}, π 2 p_{y}^{2} \approx 2 p_{z}^{2}, \overset{*}{π} 2 p_{y}^{1}$

NO (7 + 8 = 15)
Hence, both O⁺₂ and NO contains one unpaired electron, so paramagnetic.

The first ionisation energy of oxygen is less than that of nitrogen. Which of the following is the correct reason for this observation?

Lesser effective nuclear charge of oxygen than nitrogen.
Lesser atomic size of oxygen than nitrogen
Greater interelectron repulsion between two electrons in the same p-orbital counter balances the increase in effective nuclear charge on moving from nitrogen to oxygen.
Greater effective nuclear charge of oxygen than nitrogen.

The correct match of contents in Column I with those in Column II is

	Column I		Column II
(A)	He	(i)	High electron affinity
(B)	Cl	(ii)	Most electropositive element
(C)	Ca	(iii)	Strongest reducing agent
(D)	Li	(iv)	Highest ionisation energy

A-iv, B-i, C-ii, D-iii
A-iii, B-i, C-ii, D-iv
A-iv, B-iii, C-ii, D-i
A-ii, B-iv, C-i, D-iii

Identify the nuclear reaction that differs from the rest.

γ- decay
κ- capture
Positron emission
β-decay

Using the following thermochemical equations

(i) S(rh) + 3/2O₂(g) → SO₃ (g) ΔH = -2x kJ mol^-1

(ii) SO₂(g) + 1/2O₂ (g) → SO₃ (g) ΔH = -y kJ mol^-1

Find out the heat of formation of SO₂ (g) in kJ mol^-1.

(2x + y)
(x + y)
(2x/y)
(y- 2x)

The lattice enthalpy and hydration enthalpy of four compounds are given below

Compound Lattice enthalpy (in kJ mol^-1) Hydration
enthalpy
(in kJ mol^-1)

P +780 -920

Q +1012 -812

R +828 -878

S +632 -600

The pair of compound which is soluble in water is

P and R
Q and R
P and Q
Rand S

1.6 moles of PCl₅(g) is placed in 4 dm closed vessel. When the temperature is raised to 500 K, it decomposes and at equilibrium 1.2 moles of PCl₅(g) remains. What is the K, value for the decomposition of PCl₅(g) to PCl₃(g) and Cl₅(g) at 500 K?