Q40 / 120 · NEET 2012 (Mains) · Chemistry
In which of the following arrangements the given sequence is not strictly according to the property indicated against it?
- A$\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$ : increasing acidic strength
- Correct answerB$\mathrm{H}_2 \mathrm{O} < \mathrm{H}_2 \mathrm{~S} < \mathrm{H}_2 \mathrm{Se} < \mathrm{H}_2 \mathrm{Te}$ : increasing $\mathrm{p} K_a$ values
- C$\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3$ : increasing acidic character
- D$\mathrm{CO}_2 < \mathrm{SiO}_2 < \mathrm{SnO}_2 < \mathrm{PbO}_2$ : increasing oxidising power
Solution
(a,c) As we move from top to bottom in a group, acidic strength of hydrides increases. Therefore, order of acidic strength of hydrides of V A and VII A group elements is
$$
\begin{aligned}
& \mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3 \\
& \mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI} \\
&
\end{aligned}
$$
(b) As we move from top to bottom in a group, acidic nature $\left(K_a\right)$ increases. Therefore, $\mathrm{p} K_a$ decreases. Thus, order of $\mathrm{p} K_a$ values of hydrides of VI A group elements is
$$
\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{~S}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{Te}
$$
(d) On moving from top to bottom, oxidising power of oxides increases. Thus, order of oxidising power of oxides of IVA group elements is
$$
\mathrm{CO}_2 < \mathrm{SiO}_2 < \mathrm{SnO}_2 < \mathrm{PbO}_2
$$
Hence, the correct answer is option: 'B'.