Question

Octet rule is not followed in:
(a)- $CC{{l}_{4}}$, ${{N}_{2}}{{O}_{4}}$, and ${{N}_{2}}{{O}_{5}}$
(b)- $B{{F}_{3}}$, $BeC{{l}_{2}}$, and $N{{O}_{2}}$
(c)- NaCl, $MgC{{l}_{2}}$, and MgO
(d)- $PC{{l}_{3}}$, $N{{H}_{3}}$, and ${{H}_{2}}O$

Answer 1

Hint: Octet rule means when the total number of eight. If the compound is made up of two elements then add the valence electrons, if the total electron is either less than eight or greater than eight then the Octet rule is not followed.

Complete step-by-step answer:
Octet rule means when the total number of eight. If the compound is made up of two elements then add the valence electrons, if the total electron is either less than eight or greater than eight then the Octet rule is not followed.
(b)- $B{{F}_{3}}$, $BeC{{l}_{2}}$, and $N{{O}_{2}}$
In $B{{F}_{3}}$, the valence electron of the boron atom is 3, and the valence electrons of the fluorine atom is 1. There are 3 fluorine atoms in $B{{F}_{3}}$. The total valence electrons is:
$\text{3 + 1 + 1 + 1 = 6}$
Which means octet rule is not followed.
In $BeC{{l}_{2}}$, the valence electron of the beryllium atom is 2, and the valence electrons of the chlorine atom is 1. There are 2 chlorine atoms in $BeC{{l}_{2}}$. The total valence electrons is:
$\text{2 + 1 + 1 = 4}$
Which means octet rule is not followed.
In $N{{O}_{2}}$, the valence electron of the nitrogen atom is 3, and the valence electrons of the oxygen atom are 2. There are 2 oxygen atoms in $N{{O}_{2}}$. The total valence electrons are:
$\text{3 + 2 + 2 = 7}$
This means the octet rule is not followed.
So, in all three compounds, the Octet rule is not followed.

Therefore, the correct answer is an option (b).

Note: In some compounds, the total valence electrons come out to be 16, 24, 32, etc and these numbers are divisible by 8, so these compounds will also follow the Octet rule.