Q23 / 180 · KCET 2018 · Chemistry

The state of hybrid orbitals of carbon in \( \mathrm{CO}_{2}, \mathrm{CH}_{4} \) and \( \mathrm{CO}_{3}{ }^{2-} \) respectively is

  • A
    \( \mathrm{sp}^{3}, \mathrm{sp}^{2} \) and \( \mathrm{sp} \)
  • B
    \( s p^{3}, s p \) and \( s p^{2} \)
  • Correct answer
    C
    \( \mathrm{sp}, \mathrm{sp}^{3} \) and \( \mathrm{sp}^{2} \)
  • D
    \( \mathrm{sp}^{2}, \mathrm{sp}^{3} \) and \( \mathrm{sp} \)

Solution

Using the formula State of hybridisation $=\frac{(V+M-C+A)}{2}$ Where, $\mathrm{V}=$ No. of valance electrons. $\mathrm{M}=$ No. of monovalent atoms. $\mathrm{C}=$ Cationic charge. $\mathrm{A}=$ Anionic charge. $\mathrm{CO}_{2}=\frac{4+0}{2}=\frac{4}{2}=2 \Rightarrow s p h y b r i d i z e d$ $\mathrm{CH}_{4}=\frac{4+4}{2}=\frac{8}{2}=4 \Rightarrow s p^{3}$ hybridized. $\mathrm{CO}_{3}{ }^{2-}=\frac{4+0+2}{2}=\frac{6}{2} \Rightarrow 3 \Rightarrow s p^{2}$ hybridized.

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

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