Sulphur reacts with chlorine in $1 : 2$ ratio and forms (X). (X) on hydrolysis gives a sulphur compound (Y). What is the hybridisation of the 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 $s p^{3}$
$S + 2 C l_{2} \to S C l_{4} (X)$

$S C l_{4} 4 H_{2} O - -- \to S (O H)_{4} (unstable) + 4 H C l$

$S (O H)_{4} (unstable) \to H_{2} S O_{3} (Y) + H_{2} O$

Hybridisation of $S$ atom in $S O_{3}^{2 -}$ 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?

Q. Sulphur reacts with chlorine in 1 : 2 ratio and forms X. Hydrolysis of X gives a sulphur compound Y. What is the structure and hybridisation of 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 hybridisation of the central atom in the anion of (Y)?

The correct option is B sp3S+2Cl2→SCl4(X) SCl44H2O−−−→S(OH)4(unstable)+4HCl S(OH)4(unstable)→H2SO3(Y)+H2O Hybridisation of S atom in SO2−3 is sp3

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

The correct option is B $s p^{3}$
$S + 2 C l_{2} \to S C l_{4} (X)$

$S C l_{4} 4 H_{2} O - -- \to S (O H)_{4} (unstable) + 4 H C l$

$S (O H)_{4} (unstable) \to H_{2} S O_{3} (Y) + H_{2} O$

Hybridisation of $S$ atom in $S O_{3}^{2 -}$ is $s p^{3}$