Question

\[NH_2^ - \] is isoelectronic and isostructural with:
A. $B{H_3}$
B. $C{H_4}$
C. ${H_2}O$
D. ${H_3}{O^ + }$

Answer 1

Hint: We must know that isoelectronic species are those, which have the identical number of electrons. Isostructural species are those that contain the similar shape and hybridization.

Formula used:
We can predict the number of orbitals involved in hybridisation by using formula which is given below as,
$H = \dfrac{1}{2}\left[ {V + M - C + A} \right]$
Where,
H-Number of orbitals involved in hybridisation
V-Valence electron of central atom
M-Number of monovalent atoms linked to central atom
C-Charge of cation
A-Charge of anion

Complete answer:
Let us remember that if two species have the identical number of electrons, they are known to be isoelectronic with each other.
Now let us calculate the number of electrons for those species given in question.
We shall also discuss the isostructural species. These species have identical shape and hybridization.
In case of \[NH_2^ - \],
The valence electron of central atom$ = 5$
No. of monovalent atom linked to central atom$ = 2$
Charge of anion$ = 1$
Substitute the values in formula we get,
$H = \dfrac{1}{2}\left( {5 + 2 + 1} \right)$
$ \Rightarrow H = \dfrac{1}{2} \times 8$
On simplifying we get,
$H = 4$
$\therefore $The hybridization of $NH_2^ - $ is $s{p^3}$ and its bent V shape with 10 electrons. We can draw the structure as below,

In case of \[B{H_3}\],
The valence electron of central atom$ = 3$
No. of monovalent atom linked to central atom$ = 3$
There is no charge on this molecule. Substitute the values in formula we get,
$H = \dfrac{1}{2}\left( {3 + 3} \right)$
$ \Rightarrow H = \dfrac{1}{2} \times 6$
On simplifying we get,
$ \Rightarrow H = 3$
$\therefore $The hybridisation of $B{H_3}$is $s{p^2}$and it has a trigonal planar shape with eight electrons. We can draw the structure as below,

In case of \[C{H_4}\],
The valence electron of central atom$ = 4$
No. of monovalent atom linked to central atom$ = 4$
There is no charge on this molecule. Substitute the values in formula we get,
$H = \dfrac{1}{2}\left( {4 + 4} \right)$
$ \Rightarrow H = \dfrac{1}{2} \times 8$
On simplifying we get,
$ \Rightarrow H = 4$
$\therefore $ The hybridization of $C{H_4}$ is $s{p^3}$ and its tetrahedral shape with 10 electrons. We can draw the structure as below,

In case of \[{H_2}O\],
The valence electron of central atom$ = 6$
No. of monovalent atom linked to central atom$ = 2$
There is no charge on this molecule. Substitute the values in formula we get,
\[H = \dfrac{1}{2}\left( {6 + 2} \right)\]
$ \Rightarrow H = \dfrac{1}{2} \times 8$
On simplifying we get,
$ \Rightarrow H = 4$
The hybridization of ${H_2}O$ is $s{p^3}$ and its bent V shape with 10 electrons. We can draw the structure as below,

In case of \[{H_3}{O^ + }\],
the valence electron of central atom$ = 6$
No. of monovalent atom linked to central atom$ = 3$
Charge of cation$ = 1$
Substitute the values in formula we get,
$H = \dfrac{1}{2}\left( {6 + 3 - 1} \right)$
$ \Rightarrow H = \dfrac{1}{2} \times 8$
On simplifying we get,
$ \Rightarrow H = 4$
The hybridization of ${H_3}{O^ + }$ is $s{p^3}$ and its trigonal pyramidal shape with 10 electrons. We can draw the structure as below,

We can see that $NH_2^ - $ is isoelectronic with $C{H_4}$,${H_2}O$,${H_3}{O^ + }$. Among them we can see $NH_2^ - $ is isostructural with ${H_2}O$.
Therefore, the option (C) is correct.

Note: We can calculate the hybridization of a molecule using the bond pair and lone pair present in it. Also we must know Isostructural species contains equal numbers of valence electrons.