Question

Match the colum:

List - A	List - B
(Molecule)	(Pure and hybrid orbitals involved)
A) $C_2{H}_6$	1) 12, 18
B) $C_2{H}_4$	2) 6, 4
C) $C_2{H}_2$	3) 6, 6
D) $C_6{H}_6$	4) 6, 8

• A. A-1, B- 2, C-3, D-4
• B. A-2, B-3, C-4, D-1
• C. A-3, B-4, C-2, D-1
• D. A-4, B-3, C-2, D-1

Answer 1

The correct option is D A-4, B-3, C-2, D-1

In $C_2{H}_6$, every carbon is $s{p}^3$ hybridize so it has 8 hybridize orbitals and 6 pure orbitals.
In $C_2{H}_4$, every carbon is $s{p}^2$ hybridize so it has 6 hybridize orbitals and 6 pure orbitals.
In $C_2{H}_2$, every carbon is $sp$ hybridize so it has 4 hybridize orbitals and 6 pure orbitals.
In $C_6{H}_6$, every carbon is $s{p}^2$ hybridize so it has 18 hybridize orbitals and 12 pure orbitals.