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

s p

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s p^{3}

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s p^{2}

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s p^{3} d

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Solution

The correct option is B $s p^{3}$
$S + 2 C l_{2} \to S C l_{4} (A) 4 H_{2} O - -- \to S (O H)_{4} + 4 H C l ↓ H_{2} S O_{3} + H_{2} O (Y)$
The oxidation state of S in $H_{2} S O_{3}$ is $s p_{3}$

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

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

The correct option is B $s p^{3}$
$S + 2 C l_{2} \to S C l_{4} (A) 4 H_{2} O - -- \to S (O H)_{4} + 4 H C l ↓ H_{2} S O_{3} + H_{2} O (Y)$
The oxidation state of S in $H_{2} S O_{3}$ is $s p_{3}$