Ncert Exercises & Solutions

Which of the following orders are correct as per the properties mentioned against each?

Column I		Column II
(a)	As₂O₃ < SiO₂ < P₂O₃ < SO₂	(i)	Acid strength.
(b)	AsH₃ < PH₃ < NH₃	(ii)	Enthalpy of vaporisation.
(c)	S < O < Cl < F	(iii)	More negative electron gain enthalpy.
(d)	H₂O > H₂S > H₂Se > H₂Te	(iv)	Thermal stability.

1. (a, b)
2. (b, c)
3. (c, d)
4. (a, d)

Hint: Enthalpy of vaporization increases down the group for the 15th group hydride.

(a)

As the electronegativity of the central atom increases, the acidic character also increases.

Increases order is As₂O₃ < SiO₂ < P₂O₃ < SO₂

Hence, first is the correct statement.

(b) The enthalpy of vaporization in correct order is as follows:

AsH₃ > PH₃ > NH₃

Hence, statement second is incorrect.

(c)

The negative electron gain enthalpy value is as follows:

S = -200 kJ mol^-1
O = – 141 kJ mol^-1
Cl = – 349 kJ mol^-1
F = – 328 kJ mol^-1
The correct order of negative electron gain enthalpy is Cl > F > S > O.

Hence, third statement is incorrect.

(d)

Thermal stability decreases on moving from top to bottom due to an increase in its bond length.

H₂O>H₂S>H₂Se>H₂Te

This is related to the bond dissociation enthalpies of the E-H bonds where E stands for the element.
E-H bond: O−H, S−H, Se−H, and Te−H,
Bond dissociation enthalpy : (kJ mol−1) 463, 347, 276, 238

Based on bond dissociation enthalpy, H₂O is maximum stable thermally while H₂Te is the least stable. Hence, statement fourth is correct.

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