Question

Some of the properties of the two species, $NO_{3}^{-}$ and ${{H}_{3}}{{O}^{+}}$ described below. Which one of them is correct?
(a) dissimilar in hybridization for the central atom with different structures
(b) isostructural with same hybridization for the central atom
(c) isostructural with different hybridization for the central atom
(d) similar in hybridization for the central atom with different structures

Answer 1

Hint: By calculating the by hybridization of $NO_{3}^{-}$ and ${{H}_{3}}{{O}^{+}}$ (it is process of inter-mixing of the orbitals to form new orbitals of equivalent energy),we can know whether they have same hybridization or not. The hybridization can be calculated by the formula:
           \[H=\dfrac{1}{2}~\left( V+M-C+A \right)\]
Here, V represents the number of electrons in the valence shell of the atom, M represents the monovalent atoms attached to that atom and C and A represents the charges of the cations and anions and both $NO_{3}^{-}$ and ${{H}_{3}}{{O}^{+}}$are not isostructural with one another. Now, Identify the correct statement.

Complete answer:
First of all, let’s discuss what hybridization is. By the term hybridization we mean the phenomenon of inter-mixing of the orbitals of slightly different energies so as to redistribute their energies and to give a new set of orbitals of equivalent energies and shape.
First, we have to fi8nd the hybridization of $\text{P}{{\text{F}}_{5}}$molecule by the formula as:
          \[H=\dfrac{1}{2}~\left( V+M-C+A \right)\]---------(1)
Here, V= number of the valence electrons in the atom, M= number of the monovalent atoms bonded to the central atom, C=the charge on the cation and A= the charge on the anion.
Now, calculating the hybridization of both $NO_{3}^{-}$ and ${{H}_{3}}{{O}^{+}}$by using this formula as;
In $NO_{3}^{-}$ ;
Number of valence electrons in \[N=5\]
Charge on the $NO_{3}^{-}$\[=1\]
Number of monovalent atoms=0
Then, we get;
Hybridization of $NO_{3}^{-}$ is
\[\begin{align}
  & =\dfrac{1}{2}\left( 5+0+0+1 \right) \\
 & =3 \\
\end{align}\]
Since, its hybridization is $s{{p}^{2}}$, therefore, it has trigonal planar geometry as;
                   

On the other hand, in${{H}_{3}}{{O}^{+}}$;
Number of valence electrons in oxygen =$6$
Number of monovalent electrons=$3$
Charge on the cation =\[1\]
Then, we get;
Hybridization of ${{H}_{3}}{{O}^{+}}$is
\[\begin{align}
  & =\dfrac{1}{2}\left( 6+3-1+0 \right)~ \\
 & =\text{ }4 \\
\end{align}\]
Since, its hybridization is $s{{p}^{3}}$, therefore, it has pyramidal geometry due to the presence of one lone pairs of electrons instead of the tetrahedral geometry as;
                        

So, thus, both $NO_{3}^{-}$ and ${{H}_{3}}{{O}^{+}}$ neither have same hybridization nor are isostructural with one another. So, from above all the given statements ,only statement a is correct.

Hence, option(a) is correct.

Note:
Hybridization helps us to predict the shape, geometry and structure of the molecules and isostructural are those structures which have the same number of atoms and they are arranged in the same way in the similar structures.