Identify the correct order of bond angle in the following species.

${C H_{3}}^{+}, C H_{4}, {C H_{3}}^{-}$

{C H_{3}}^{-} > C H_{4} > {C H_{3}}^{+}

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C H_{4} > {C H_{3}}^{-} > {C H_{3}}^{+}

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{C H_{3}}^{+} > C H_{4} > {C H_{3}}^{-}

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{C H_{3}}^{+} > {C H_{3}}^{-} > C H_{4}

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Solution

The correct option is C ${C H_{3}}^{+} > C H_{4} > {C H_{3}}^{-}$
The correct order of the bond angle is ${C H_{3}}^{+} > {C H}_{4} > {C H_{3}}^{-}$

In ${C H_{3}}^{+}$ , three bond pairs and zero lone pairs are present which result in $s p^{2}$ hybridization of C atom with bond angle of $120^{0}$ .
In methane, $s p^{3}$ hybridization results in tetrahedral geometry with bond angle of $109^{0}$ .
In ${C H_{3}}^{-}$ , the lone pair of elections repels the bond pair of electrons.
Hence, the bond angle is smaller than the tetrahedral angle.

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Similar questions

C H_{4}, C H_{3} - O H, C H_{3} - F, C H_{3} - N H_{2}

(1) C l_{2} \to ∙ 2 C l (2) ∙ C H_{3} + ∙ C H_{3} \to C H_{3} - C H_{3} (3) ∙ C l + C H_{4} \to C H_{3} C l + ∙ C l (4) ∙ C H_{3} + ∙ C l \to C H_{3} - C l (5) ∙ C H_{3} + C l_{2} \to C H_{3} - C l + ∙ C l

C l_{2}

C H_{4}

1. C H_{3} - C \equiv C - C H_{3}

2. C H_{3} - C H_{2} - C H_{2} - C H_{3}

3. C H_{3} - C H_{2} C \equiv C H

4. C H_{3} - C H = C H_{2}

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