Q8 / 160 · AP EAMCET 2021 (23 Aug Shift 1) · Chemistry

The hybridisations of $\mathrm{N}$-orbitals in $\mathrm{NO}_3^{-}, \mathrm{NO}_2^{-}$ and $\mathrm{NH}_4^{+}$respectively are

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

Solution

As we know, $$ \text { Hybridisation }=\frac{1}{2}[V+M-C+A] $$ $$ \begin{aligned} \mathrm{H}_{\mathrm{NO}_3^{-}} & =\frac{1}{2}[5+0-0+1] \\ & =3 \text { i.e. } s p^2 \\ \mathrm{H}_{\mathrm{NO}_2^{-}} & =\frac{1}{2}[5+0-0+1] \\ & =3 \text { i.e. } s p^2 \\ \mathrm{H}_{\mathrm{NH}_4^{+}} & =\frac{1}{2}[5+4-1+0] \\ & =4 \text { i.e. } s p^3 \end{aligned} $$ The hybridisation of $\mathrm{N}$ in $\mathrm{NO}_3^{-}, \mathrm{NO}_2^{-}$and $\mathrm{NH}_4^{+}$are $s p^2, s p^2$ and $s p^3$ respectively.

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

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