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

• 45

• 9

• 27

• 36

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

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⁺₂ and NO are paramagentic

• I, III and IV

• I and IV

• II and III

• I and III

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

• 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

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?

• 0.013

• 0.050

• 0.033

• 0.067

For a concentrated solution of a weak electrolyte A_xB_y of concentration 'C', the degree of dissociation 'a' is given as

• α = √ K_eq/ C ( x + y)

• α = √ K_eq/ C ( xy)

• α =( K_eq/ Cxy)

• α =( K_eq/ C^xy)