Hexaamminenickel(II) hexanitrocobaltate(III) can be written as?

[N i (N H_{3})_{6}]_{2} [C o (N O_{2})_{6}]_{3}

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[N i (N H_{3})_{6}]_{3} [C o (N O_{2})_{6}]_{2}

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[N i (N H_{3})_{6}] [C o (N O_{2})_{6}]

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[N i (N H_{3})_{6} (N O_{2})_{6}] C o

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Solution

The correct option is D $[N i (N H_{3})_{6}]_{3} [C o (N O_{2})_{6}]_{2}$
Hexaamminenickel(II) hexanitrocobaltate(III) can be written as $[N i (N H_{3})_{6}]_{3} [C o (N O_{2})_{6}]_{2}$ . The oxidation number of Ni is +2. Hence, $[N i (N H_{3})_{6}]$ has +2 charge and can be written as $[N i (N H_{3})_{6}]^{2 +}$ . Cobalt has +3 oxidation state. nitro group has -1 oxidation state. Hence, $[C o (N O_{2})_{6}]$ has -3 oxidation state and can be written as $[C o (N O_{2})_{6}]^{3 -}$ . When three $[N i (N H_{3})_{6}]^{2 +}$ complex cations combine with two $[C o (N O_{2})_{6}]^{3 -}$ complex anions, electrical neutrality is maintained.

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