Q42 / 180 · AIIMS 2001 · Chemistry

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

  • Correct answer
    A
    $s p^2, s p^2, s p^3$
  • B
    $s p^2, s p^3, s p^3$
  • C
    $s p^3, s p^3, s p^2$
  • D
    $s p^3, s p^2, s p^2$

Solution

$\dot{\mathrm{C}} \mathrm{H}_3$ (methyl free radical) has planar structure with $s p^2$ hybridisation of ' $\mathrm{C}$ ' atom. The odd electron is present in unhybridised $2 p_z$ orbital. $\oplus$ $\mathrm{CH}_3$ (methyl carbonium ion) has also trigonal planar structure $\left(s p^2\right)$. $\stackrel{\ominus}{\mathrm{CH}_3}$ (methyl carbanion) has tetrahedral structure $\left(s p^3\right)$ and one of the hybrid orbitals contains the lone pair of electrons.

Hence, the correct answer is option: 'A'.

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