Question

The compounds which uses $s{{p}^{3}}$- hybrid orbitals for bond formation are-
(A)- $HCOOH$
(B)- ${{({{H}_{2}}N)}_{2}}CO$
(C)- ${{(C{{H}_{3}})}_{3}}COH$
(D)- $C{{H}_{3}}CHO$

Answer 1

Hint: A $s{{p}^{3}}$- hybridized atom is linked to four other atoms by single bonds (sigma). A $s{{p}^{3}}$- hybrid orbital contains 4 hybrid orbitals. Carbon has tetra valency.

Complete step by step solution:
Let’s look at the given options:
-In $HCOOH$ the carbon atom is linked to 3 atoms only by single bonds. Its hybridization is $s{{p}^{2}}$. So OPTION (A) is incorrect.
- In ${{({{H}_{2}}N)}_{2}}CO$the carbon atom is again linked to 3 atoms only by single bonds. Its hybridization is $s{{p}^{2}}$. Also the 2 N atoms are also linked to 3 atoms only. So this compound does not contain any $s{{p}^{3}}$ hybrid orbital. So OPTION (B) is incorrect.
-In ${{(C{{H}_{3}})}_{3}}COH$all the carbon atoms present are linked to 4 other atoms by single bonds. Its hybridization is $s{{p}^{3}}$. So, OPTION (C) is correct.
-In $C{{H}_{3}}CHO$ one carbon atom is linked to 4 other atoms by single bonds. Its hybridization is $s{{p}^{3}}$. So, OPTION (D) is also correct.

So, the correct answer is “Option B and D”.

Note: The geometry of a $s{{p}^{3}}$ hybrid orbital is tetrahedral. The shape of a $s{{p}^{3}}$ hybrid orbital is also the same that is tetrahedral. But if a lone pair is present then the geometry remains the same but the shape changes. Only count single bonds (sigma bonds) while checking hybridization.