JEE Main · 2009 · 12 Apr 2009 · Offline · Classification of Elements and Periodicity

In which of the following arrangements, the sequence is not strictly according to the property written against it ?

(A) $CO_{2} < SiO_{2} < SnO_{2} < PbO_{2}$ : increasing oxidising power
(B) $HF < HCl < HBr < HI$ : increasing acid strength
(C) $NH_{3} < PH_{3} < AsH_{3} < SbH_{3}$ : increasing basic strength✓ Correct
(D) $B < C < O < N$ : increasing first ionization enthalpy.

Answer: (C)

Solution

Basic strength of the group-15 hydrides falls as the central atom grows larger and its lone pair spreads out, so the true order is

N H_{3} > P H_{3} > As H_{3} > Sb H_{3}

. The sequence written as increasing basic strength therefore runs the wrong way. The other three orders are genuine: the oxidising power of the group-14 dioxides rises down the group, the acid strength of the hydrogen halides rises from HF to HI, and the first ionisation enthalpies increase across the period with nitrogen standing above oxygen because of its stable half-filled shell.

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option

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JEE Main · 2009 · 12 Apr 2009 · Offline · Classification of Elements and Periodicity

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