Question

Some of the properties of the two species, $NO_3^ -$ and ${H_3}{O^ + }$ are described below. Which one of them is correct ?
a.) Dissimilar in hybridisation for the central atom with different structures
b.) Isostructural with the same hybridisation for the central atom
c.) Isostructural with different hybridisation for the central atom
d.) Similar in hybridisation for the central atom with different structures

Answer 1

According to the VSEPR theory, the number of bond pair and lone pair electrons determine the shape of a molecule. The lone pair-lone pair repulsion is greater than the lone pair-bond pair repulsion and the repulsion between the bond pair-bond pair is the lowest.

Complete step by step answer:
Let us first study the molecules and then we can get to know their similarities and dissimilarities.
- The first molecule is $NO_3^ -$.
The hybridisation is as -
Hybridisation = $\dfrac{1}{2}$[ 5 + 0 - 0 + 1 ]
Hybridisation = 3
So, the hybridisation of $NO_3^ -$ = $s{p^2}$
Thus, its geometry is trigonal planar.

- The second molecule is ${H_3}{O^ + }$.
The hybridisation is as -
Hybridisation = $\dfrac{1}{2}$[ 6 + 3 - 1 + 0 ]
Hybridisation = 4
So, the hybridisation of ${H_3}{O^ + }$ = $s{p^3}$
So, its geometry is tetrahedral. But the two lone pairs are at one place.
Thus, its shape is pyramidal.
So, both the molecules have dissimilar hybridisations for the central atoms and different geometries. The correct answer is option “A” .
Additional Information :

Note: It must be noted that the molecules with $s{p^3}$ hybridisation can have tetrahedral or square planar geometries. But here, the ${H_3}{O^ + }$ has tetrahedral geometry. The Oxygen has donated one lone pair to a hydrogen atom and it has one lone pair free. So, it has a pyramidal shape. Thus, the shape is derived from geometry.