Question

Which‌ ‌one‌ ‌of‌ ‌the‌ ‌following‌ ‌does‌ ‌not‌ ‌have‌ ‌‌$s{p^2}$‌‌ ‌hybridised‌ ‌carbon?‌ ‌
A.Acetone‌ ‌
B.Acetic‌ ‌Acid‌ ‌
C.Acetonitrile‌ ‌
D.Acetamide‌ ‌

Answer 1

In this question, we need to determine that of the given options which compound does not have $s{p^2}$ hybridised carbon. For this, we will first draw the structure of each of the given options and then back bonding form. Hybridisation can easily be recognized by seeing the structure of the compound.

Complete step-by-step answer:
Let’s see the (a) acetone structure,

In $C = O$, the C is $s{p^2}$ hybridised.
In both $C{H_3}$, the$C - H$ single bound therefore the C is $s{p^3}$ hybridised.

Let’s see the (b) acetic acid structure,

In $C = O$, the C is $s{p^2}$ hybridised.
In both $C{H_3}$, the $C - H$ single bound therefore the C is $s{p^3}$ hybridised.

Let’s see the (c) acetonitrile structure,

In $C \equiv N$, the C is $sp$ hybridised .
In both $C{H_3}$, the $C - H$ single bound therefore the C is $s{p^3}$ hybridised.

Let’s see the (d) acetamide structure,

In $C = O$, the C is $s{p^2}$ hybridised .
In both $C{H_3}$, the $C - H$ single bound therefore the C is $s{p^3}$ hybridised .

Therefore, (C) is the correct answer. Because the (c) option doesn’t have $s{p^2}$ hybridised Carbon.

Note: Remember to draw structures and make sure that common names and IUPAC names should be known. Nitrogen can form triple and in ionic form it forms four bonds. In this question the acetonitrile was not showing any $s{p^2}$ hybridised Carbon because of the structure. $s{p^2}$ carbon shows $one\;\sigma - two\;\pi$ bonds, whereas in acetonitrile’s case we are able to observe that it shows $sp$ hybridised $one\;\sigma - two\;\pi$ bonds.