Sulphur reacts with chlorine in 1 : 2 ratio and forms X. Hydrolysis of X gives a sulphur compound Y. What is the hybridization state of central atom in the anion of Y?

$s p^{3}$

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$s p$

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$s p^{2}$

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$s p^{2} d$

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Solution

The correct option is A
$s p^{3}$

$S + 2 C l_{2} \to S C l_{4} (X); S C l_{4} + 3 H_{2} O \to H_{2} S O_{3} (Y) + 4 H C l$

In $S O_{3}^{2 -}$ , sulphur has $s p^{3}$ hybridization.

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Sulphur reacts with chlorine in 1 : 2 ratio and forms X. Hydrolysis of X gives a sulphur compound Y. What is the hybridization state of central atom in the anion of Y?

The correct option is A sp3 S+2Cl2→SCl4(X);SCl4+3H2O→H2SO3(Y)+4HCl In SO2−3, sulphur has sp3 hybridization.

The correct option is A
$s p^{3}$

$S + 2 C l_{2} \to S C l_{4} (X); S C l_{4} + 3 H_{2} O \to H_{2} S O_{3} (Y) + 4 H C l$

In $S O_{3}^{2 -}$ , sulphur has $s p^{3}$ hybridization.