Question

The type of hybrid orbitals used by Chlorine atom in \[ClO_{2}^{-}\] ion is:
a.) \[s{{p}^{3}}\]
b.) \[s{{p}^{2}}\]
c.) \[sp\]
d.) 1 & 2

Answer 1

Hint: We can calculate the hybridization of the central atom in the molecule by using the steric number. “The steric number is the number of atoms bonded to a central atom of a molecule and the number of lone pairs attached to the central atom”.

Complete step by step answer:
The given compound is \[ClO_{2}^{-}\].
The hybridization of chlorine in \[ClO_{2}^{-}\] can be calculated as follows by using steric numbers.
Steric number : \[\dfrac{1}{2}\][V+M−C+A]
V= number of valence electrons
M= number of monovalent atoms surrounding the atom
C= Charge on cation.
A= Charge on anion.
Number of valence electrons of chlorine in \[ClO_{2}^{-}\] , V= 7
Number of monovalent atoms surrounding the chlorine atom, M= 0
Charge on cation, C= 0
Charge on anion, A=1
Steric number = \[\dfrac{1}{2}\] [7+0+0+1] = 4
Steric number = 4 means the hybridization of chlorine atom in \[ClO_{2}^{-}\] is \[s{{p}^{3}}\].
So, the correct option is A.

Note: Don’t be confused with the terms cation and anion.
Cation: it is an ion containing positive charge and formed by the loss or donation of electrons to other atoms.
Anion: it is an ion containing negative charge and formed by the gain or accepting electrons from other atoms.
Valence electrons: the electrons which are present in the outermost orbital of an atom are called valence electrons.