Question

Which of the following are isoelectronic species?
(i) $N{H_3}$ (ii) $C{H_3}^ +$ (iii) $\;N{H_2}^ -$ ; (iv) $\;N{H_4}^ +$
Choose the correct answer from the codes given below:
A (i), (ii), (iii)
B (i), (iii), (iv)
C (i), (ii), (iv)
D (ii), (iii)

Answer 1

If any molecules contain the same number of the total electron in that molecule, then those molecules are called isoelectronic to each other. To solve this question, you have to calculate the total number of electrons of each molecule.
Formula used:
Total number of electron=( electron in central atom+ total electron in surrounding atoms)

Complete step by step answer:
In the case of $N{H_3}$ the total number of the electron is $(7 + 3) = 10$
In the case of $C{H_3}^ +$ the total number of the electron is $(6 + 3 - 1) = 8$
In the case of $\;N{H_2}^ -$ the total number of the electron is $(7 + 2 + 1) = 10$
In the case of $\;N{H_4}^ +$ the total number of the electron is $(7 + 4 - 1) = 10$
Therefore, from the calculation of the total number of the electron, it is clear that the isoelectronic species are, $N{H_3}$ , $\;N{H_2}^ -$ , $\;N{H_4}^ +$ as they have same number of electrons.

Therefore the correct answer is, B.

Note:
The hybridization can be calculated by using the formula shown below. $H = \dfrac{1}{2}\left[ {V + X - C + A} \right]$
where V is the number of valence electrons of the central atom, X is the number of monovalent atoms attached to the central atom, C is the total cationic charge and A is the total anionic charge.
For example the hybridization of the given species,
For, $N{H_3}$

$$H = \dfrac{1}{2}\left[ {V + X - C + A} \right] \\\ = \dfrac{1}{2}\left[ {5 + 3 + 0 + 0} \right] \\\ = \dfrac{1}{2}\left[ 8 \right] \\\ = 4 \\\$$

For H=4 the hybridization is $s{p^3}$