What is the oxidation state of central atom in $[C o (N H_{3})_{4} (O H_{2}) I] S O_{4}$ ?

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Solution

The correct option is C $3$
$[C O {(N H_{3})}_{4} (H_{2} O) I] S O_{4}$
$⇓$
${[C O {(N H_{3})}_{4} (H_{2} O) I]}^{2 +} {S O_{4}}^{2 -}$
Central atom $= C O$
Oxidation state of $N H_{3}$ in general $= 0, H 2 O = 0$ and $I = - 1$
Applying formula, $Sum of total oxidation state of all atoms = Overall charge on the compound .$
Let oxidation state of $C O = x x + 4 (0) + 1 \times (0) + 1 \times (- 1) = + 2, x = 3$
$∴$ Oxidation state of $C O = + 3.$

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