The hybridization of the central atoms of compounds $A, B, C$ and $D$ are $s p^{3} d, s p^{3}, s p^{2}$ and $s p$ respectively. If the compounds $A$ and $D$ have the same shape like $I_{3}^{-}$ and the compounds $B$ and $C$ have the same shape like water molecule, then the value of $P + Q + R + S$ is:(Given that $P, Q, R$ and $S$ are the number of lone pairs of electrons on central atoms of compounds $A, B, C$ and $D$ respectively.)

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Solution

The given data is summarized in the following table:

Compound:
$A$
$B$
$C$
$D$
Hybridization:
$s p^{3} d$
$s p^{3}$ $s p^{2}$ $s p$
Shape like:
$I_{3}^{-}$ $H_{2} O$ $H_{2} O$ $I_{3}^{-}$
Number of bonding domains:
$2$
$2$
$2$
$2$
Number of lone pairs:
$3$ $2$ $1$ $0$

Since the compound $A$ is $s p^{3} d^{2}$ hybridised and has only $2$ bonding domians, it must have $3$ lone pair of electrons. Similarly, $B, C$ and $D$ have $2, 1$ and $0$ lone pairs of electrons respectively.

Hence, the sum $P + Q + R + S = 3 + 2 + 1 + 0 = 6$

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Compound:	$A$	$B$	$C$	$D$
Hybridization:	$s p^{3} d$	$s p^{3}$	$s p^{2}$	$s p$
Shape like:	$I_{3}^{-}$	$H_{2} O$	$H_{2} O$	$I_{3}^{-}$
Number of bonding domains:	$2$	$2$	$2$	$2$
Number of lone pairs:	$3$	$2$	$1$	$0$