Single Choice

Decreasing order of bond angle is:

A$$BeCl_{2} > NO_{2} > SO_{2}$$
Correct Answer
B$$BeCl_{2} > SO_{2} > NO_{2}$$
C$$SO_{2} > BeCl_{2} > NO_{2}$$
D$$SO_{2} > NO_{2} > BeCl_{2}$$

Solution

Compound: $$BeCl_{2} > NO_{2} > SO_{2}$$
$$sp$$ $$sp^2$$ $$sp^2$$ ---- Hybridisation

Bond angle: $$180^{\circ} > 132^{\circ} > 119.5^{\circ}$$


SIMILAR QUESTIONS

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Molecule obtained by $$sp^{3}d^{2}$$ hybridisation has bond angle of:

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The bond angles of $$NH_{3}, NH_{4}^{+}$$ and $${NH_{2}}^{-}$$ are in the order:

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In which of the following is the angle between the two covalent bonds greatest ?

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