Oxidation state of Fe in F eSO 4 N H 42 SO 4- 6 H 2 O isA- 2B- 1C- -2D- -1

The correct option is A 2Here, O. S. for H_2O = 2(1) + ( 2) = 0 O. S. for NH^+_4 = 3 + 4(1) = + 1 O. S. for SO^2 _4 = 2 Sum of O. S. for (NH_4)_2SO_4 = ( ...

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Standard XII

Chemistry

Redox Reactions and Couple

Oxidation sta...

Question

Oxidation state of Fe in $F e S O_{4} (N H_{4})_{2} S O_{4} .6 H_{2} O$ is

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Solution

The correct option is A 2
Here,
O. S. for $H_{2} O$ = 2(1) + (-2) = 0
O. S. for $N H_{4}^{+}$ = -3 + 4(1) = + 1
O. S. for $S O_{4}^{2 -}$ = -2
Sum of O. S. for $(N H_{4})_{2} S O_{4}$ = (+1)2 + (-2) = 0
Therefore,
$F e S O_{4} (N H_{4})_{2} .6 H_{2} 0$ $\Rightarrow$ n+(-2) + 0 + 0 = 0
$\Rightarrow$ Fe = +2 Ans

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Oxidation state of Fe in FeSO4(NH4)2SO4.6H2O is

The correct option is A 2Here, O. S. for H2O = 2(1) + (-2) = 0 O. S. for NH+4 = -3 + 4(1) = + 1 O. S. for SO2−4 = -2 Sum of O. S. for (NH4)2SO4 = (+1)2 + (-2) = 0 Therefore, FeSO4(NH4)2.6H20 ⇒ n+(-2) + 0 + 0 = 0 ⇒ Fe = +2 Ans

Oxidation state of Fe in $F e S O_{4} (N H_{4})_{2} S O_{4} .6 H_{2} O$ is