Question

Structure anions of acids, $HN{O}_3,H_{3}P{O}_4$ and $H_{2}S{O}_4$ are, respectively

• A. tetrahedral, tetrahedral and trigonal bipyramidal
• B. angular, tetrahedral and trigonal bipyramidal
• C. tetrahedral, tetrahedral and angular
• D. planar, tetrahedral and tetrahedral

Answer 1

Answer: (D)

Anions of the acids $HN{O}_3,H_{3}P{O}_4$ and $H_{2}S{O}_4$ are $N{O}_3^{-},P{O}_4^{3-}$ and $S{O}_4^{2-}$ respectively
Their structures can be explained as follows:
$N{O}_3^{-}$ has planar structure It has nitrogen as a central atom without lone pair, which is bonded to three oxygen atoms. Its hybridisation can be predicted by $\frac{1}{2}[V+M-C+A]$
Where $V$ = number of valence electrons
$M$ = number of monovalent ion
$C$= cationic charge
$A$ - anionic charge
$\therefore$ Hybridisation $=\frac{1}{2}[5+0-0+1]=3$

$P{O}_4^{3-}$ has tetrahedral structure. It has phosphorus as a central atom without lone pair which is bonded to four oxygen atoms. Its hybridisation can be predicted as $\frac{1}{2}[5+0-0+3]=3$

$S{O}_4^{2-}$ has tetrahedral structure. It has phosphorus as a central atom without lone pair which is bonded to four oxygen atoms. Its hybridisation can be predicted as
$\frac{1}{2}[6+0-0+2]=4$