Question

Hybridisation in $C{H}_3,\stackrel{+}{C}{H}_3$ and $\overline{C}{H}_3$ are respectively

• A. $s{p}^2,s{p}^2,s{p}^3$
• B. $s{p}^2,s{p}^3,s{p}^3$
• C. $s{p}^3,s{p}^3,s{p}^3$
• D. $s{p}^3,s{p}^2,s{p}^2$

Answer 1

Answer: (A)

$C{H}_3$ (methyl free radical) has planar structure with $s{p}_2$ hybridisation of ' $C$ ' atom. The odd electron is present in unhybridised $2p_z$ orbital.
${\stackrel{\oplus }{\mathrm{CH}}}_3$ (methyl carbonium ion) has also trigonal planar structure ( $s{p}_2$ )
${\stackrel{\ominus }{\mathrm{CH}}}_3$ (methyl carbanion ion) has tetrahedral structure $\left(s{p}_3\right)$ and one of the hybrid orbital contains the lone pair of electrons.