Question

Odd electrons bond is present in:
(a)- $NO$
(b)- $N{{O}_{2}}$
(c)- $Cl{{O}_{2}}$
(d)- All of the above

Answer 1

Hint: Odd electron bond means there is an odd number of the electron in the overall molecule. Or we can say that the molecule has unpaired electrons. If any of the atoms has an odd number of valence electrons then the molecule will have odd electrons bond.

Complete step by step answer:
An odd electron bond means there is an odd number of the electron in the overall molecule. Or we can say that the molecule has unpaired electrons.
So, in $NO$, the valence electrons in nitrogen are 5, and valence electrons in oxygen are 6. So, by adding we get, $6+5=11$
Therefore, the $NO$ molecule has an odd number of electrons. There are 2 bonds between nitrogen and oxygen. The nitrogen has one unpaired electron. So, it has an odd electron bond.

In $N{{O}_{2}}$ the valence electrons in nitrogen are 5 and the valence electrons in oxygen are 6. So, by adding we get, $6+6+5=17$
Therefore, $N{{O}_{2}}$ molecules have an odd number of electrons. There are 2 bonds between nitrogen and oxygen. The nitrogen has one unpaired electron. So, it has an odd electron bond.

In $Cl{{O}_{2}}$, the valence electrons in chlorine are 7, and valence electrons in oxygen are 6. So, by adding we get, $6+6+7=19$
Therefore, $Cl{{O}_{2}}$ molecule has an odd number of electrons. There are 2 bonds between chlorine and oxygen. The chlorine has one unpaired electron. So, it has an odd electron bond.

Hence all the options are odd electron bonds.
So, the correct answer is “Option D”.

Note: In a molecule having two atoms and both of them have unpaired electrons then the unpaired electrons will combine to form a bond. For example, in ${{N}_{2}}$, a total of 10 electrons are there.