An inorganic compound (A) is a strong reducing agent. Its hydrolysis in water gives a white turbidity (B). Aqueous solution of (A) gives a white precipitate (C) with NaOH solution which is soluble in excess of NaOH. (A) reduces auric chloride to produce purple of cassius. (A) also reduces iodine and gives chromyl chloride test. What are (A) to (C) and give the chemical reactions.

$A = (S n O_{2} B = S n (O H)_{2} C = S n C l_{2}$

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$A = S n (N O_{3})_{2} B = S n C l_{2} C = S n (O H)_{2}$

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$A = S n C l_{2} B = S n (O H) C l C = S n (O H)_{2}$

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$A = S n C l_{2} B = S n H_{4} C = S n (O H)_{2}$

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Solution

The correct option is C
$A = S n C l_{2} B = S n (O H) C l C = S n (O H)_{2}$

$(A) = S n C l_{2}; (B) = S n (O H) C l : (C) = S n (O H)_{2}$

$S n C l_{2} + H_{2} O (\to) S n (O H) C l + H C l$

$(A) (B)$

$S n C l_{2} + N a O H (\to) S n (O H)_{2} + 2 N a C l, S n (O H)_{2} + 2 N a O H (\to) N a_{2} S n O_{2} + 2 H_{2} O$

$3 S n C l_{2} + 2 A u C l_{3} (\to) 3 S n C l_{4} + 2 A u$

$S n C l_{2} + 2 H C l + I_{2} (\to) S n C l_{4} + 2 H I]$

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An inorganic compound (A) is a strong reducing agent. Its hydrolysis in water gives a white turbidity (B). Aqueous solution of (A) gives a white precipitate (C) with NaOH solution which is soluble in excess of NaOH. (A) reduces auric chloride to produce purple of cassius. (A) also reduces iodine and gives chromyl chloride test. What is 'B' in the chemical reactions.

Q. An inorganic compound (A) is a strong reducing agent. Its hydrolysis in water gives a white turbidity (B). An aqueous solution of, (A) gives a white precipitate (C) with