Question

The compound in which $\underset{–}{C}$ uses its $s{p}^3$ hybrid orbitals in bonding is:

• A. $H\underset{–}{C}OOH$
• B. $(H_{2}N{)}_2\underset{–}{C}O$
• C. $(C{H}_3{)}_3\underset{–}{C}OH$
• D. $C{H}_3\underset{–}{C}HO$

Answer 1

The correct option is C $(C{H}_3{)}_3\underset{–}{C}OH$

In the given options, only the C in compound of option (C) uses the $s{p}^3$ hybrid orbitals as C in this compound is attached to -OH and three $-(C{H}_3{)}_3$ with the sigma bonds which implies $s{p}^3$ hybridization while others have $s{p}^2$ hybridization.