Question

Which species do not have \[s{p^3}\] hybridisation?
A) Ammonia
B) Methane
C) Water
D) Carbon dioxide

Answer 1

Hint: An \[s{p^3}\] hybridised atom has four groups surrounding it. The surrounding groups may be bonded atoms or lone pairs present on the atom. Here, we will use a formula to calculate hybridisation.

Formula used: \[H = \dfrac{{V + X + A - C}}{2}\]; Here, H means hybridisation, X is the count of monovalent atoms, A stand for anionic charge, and C stands for cationic charge

Complete Step by Step Answer:
Let’s discuss all the options one by one.
Option A is ammonia. The formula for ammonia is \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\] .
Valence electrons of N atom = 5, Count of monovalent atoms = 3
So, \[H = \dfrac{{5 + 3}}{2} = 4\]
The value of H 4 means the hybridisation is \[s{p^3}\].

Option B is methane. The formula for methane is \[{\rm{C}}{{\rm{H}}_4}\] .
Valence electrons of C atom = 4, Count of monovalent atoms=4
So, \[H = \dfrac{{4 + 4}}{2} = 4\]
The value of H 4 means the hybridisation is \[s{p^3}\].

Option C is water. The formula for water is \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\] .
Valence electrons of O atom = 6, Count of monovalent atoms = 2
So, \[H = \dfrac{{6 + 2}}{2} = 4\]
The value of H 4 means the hybridisation is \[s{p^3}\].

Option D is carbon dioxide. The formula for carbon dioxide is \[{\rm{C}}{{\rm{O}}_2}\] .
Valence electrons of C atom = 4, Count of monovalent atoms = 0
So, \[H = \dfrac{4}{2} = 2\]
The value of H 2 means the hybridisation is \[sp\].
Therefore, among the given molecules, only \[{\rm{C}}{{\rm{O}}_2}\]molecule does not have\[s{p^3}\] hybridisation.
Hence, option D is right.

Note: Another method to find out whether the hybridisation is \[s{p^3}\] or not is to count the surrounding groups of the atom. If the count of the surrounding groups is four, the hybridisation is \[s{p^3}\]. Here, \[{\rm{O}} = {\rm{C}} = {\rm{O}}\], the C atom is surrounded by two O atom, so its hybridisation is \[sp\].