Question

Among the following which one is a wrong statement?

• A. $P{H}_5$ and $BiC{l}_5$ do not exist
• B. $p\pi -d\pi$ bonds are present in $S{O}_2$
• C. $Se{F}_4$ and $C{H}_4$ have same shape
• D. $I_3^{+}$ has bent shape

Answer 1

The correct option is C $Se{F}_4$ and $C{H}_4$ have same shape

$Se{F}_4$ with $s{p}^{3}d$ hybridization is see-saw whereas $C{H}_4$ with $s{p}^3$ is tetrahedral.

$I_3^{+}-s{p}^3$, lone pair = 2; bond pair= 2, thus, shape is bent.


The extent of overlapping of d-orbital of phosphorus and 1s-orbital of hydrogen is very less because the energy difference between them is too high for favourble bond formation so that the most of 1s-orbital of hydrogen interacts with p-orbitals to form $P{H}_3$.

$P{H}_5$ does not exist as d-s orbital overlapping is less, it will dissociate into $P{H}_3$ and $H_2$.

In case of $BiC{l}_5$, it is due to the inert pair, +5 oxidation state is not stable.

In $S{O}_2$, both $p\pi -p\pi$ and $p\pi -d\pi$ bonds are present because S is a third period element and has favourable energy difference for $p\pi -d\pi$ overlap between $S$ and $O$.

Hence, option (c) is the correct answer.