Sulphur has highest oxidation state in:

{S O}_{2}

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H_{2} {S O}_{4}

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{N a}_{2} S_{2} O_{3}

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H_{2} S

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Solution

The correct option is C $H_{2} {S O}_{4}$
${S O}_{2} ↓ x + (- 2 \times 2) = 0 x = + 4 H_{2} {S O}_{4} ↓ (1 \times 2) + x + (2 \times 4) = 0 2 + x - 8 x = + 6 (h i g h e s t) {N a}_{2} S_{2} O_{3} ↓ (1 \times 3) + 2 x + (- 2 \times 3) 2 + 2 x - 6 = 0 x = + 2 H_{2} S ↓ (1 \times 3) + x = 0 x = - 2$

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Similar questions

S F_{6}

{N a}_{2} S_{2} O_{3}

K_{2} {C r}_{2} O_{7} + K I + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {C r}_{2} {({S O}_{4})}_{3} + I_{2} + H_{2} O

K M n O_{4} + {N a}_{2} {S O}_{3} \to M n O_{2} + {N a}_{2} {S O}_{4} + K O H

C u + H {N O}_{3} \to C u {({N O}_{3})}_{2} + {N O}_{2} + H_{2} O

K M {n O}_{4} + H_{2} C_{2} O_{4} + H_{2} {S O}_{4} \to K_{2} {S O}_{4} + {M n S O}_{4} + {C O}_{2} + H_{2} O

H_{2} {S O}_{4}

$H_{2} S$ acts only as a reducing agent while $S O_{2}$ can act both as a reducing and oxidizing agent because

H_{2} {S O}_{4}

(a) H_{2} {S O}_{4} + N a O H \to N a H {S O}_{4} + H_{2} O

(b) H_{2} {S O}_{4} + 2 N a O H \to {N a}_{2} {S O}_{4} + 2 H_{2} O

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