Question

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

• A. $H_{2}O<H_{2}S<H_{2}Se<H_{2}Te$ : increasing $p{K}_a$ values
• B. $N{H}_3<P{H}_3<As{H}_3<Sb{H}_3$ : increasing acidic character
• C. $C{O}_2<Si{O}_2<Sn{O}_2<Pb{O}_2$ : increasing oxidising power
• D. $HF<HCl<HBr<HI$ : increasing acidic strength

Answer 1

The correct option is A $H_{2}O<H_{2}S<H_{2}Se<H_{2}Te$ : increasing $p{K}_a$ values

If acidic strength of a particular compound is high, its $p{K}_{\alpha }$ value is low. In option A, the acidic strength of $H_{2}O$ is lowest and that of $H_{2}Te$ is highest. So, the $p{K}_{\alpha }$ value of $H_{2}O$ should be the highest and should decrease till $H_{2}Te$. The bond between $H$ and $Te$ is the longest in length because $Te$ is the largest atom among the given options. Therefore, this bond is the easiest to break to release protons. Thus it has highest acidic strength and lowest $p{K}_{\alpha }$ value.