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${SO}_{3} + H_{2} O \to A,$ hybridisation of anion of central atom in A is sp³

${SCl}_{4} \to_{- 4 HCl}^{4 H_{2} O} X \overset{- H_{2} O}{\to} Y,$ Y cannot act like reducing agent

A is more acidic than Y

A has tetrahedral geometry

answer is B.

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Detailed Solution

Option-(1) -- true

A is H₂SO₄ ;

A has tetrahedral geometry where Sulphur atom is sp³ hybridized

option-(4) is also true

option(2) - false

begin{gathered} S{(OH)_4}xrightarrow{{ - {H_2}O}}{H_2}S{O_3} \ ,,(X),,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,(Y) \ end{gathered}

Oxidation state of Sulphur in H₂SO₃ is '+4'

Highest ox.state of Sulphur is +6 and least Ox.state is -2

$∴$ H₂SO₃can act like both Oxidant and reductant

option(3) is true

Acidic strength of H₂SO₄(A) > H₂SO₃(Y)

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