In nitroprusside ion the iron and $N O$ exist as $F e^{I I}$ and $N O^{+}$ rather than $F e^{I I I}$ and $N O$ . These forms can be differentiated by :

estimating the concentration of iron.

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measuring the concentration of

C N^{-}

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measuring the solid state magnetic moment.

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thermally decomposing the compound

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Solution

The correct option is C measuring the solid state magnetic moment.
As we know,
$μ = \sqrt{n (n + 2)}$ wher n is no. of unpaired electron.
Fe(II) have 4 unpaired electron so it magnetic moment will be less than Fe (III)(5 unpaired electron).

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Similar questions

F e^{+ 2}

N O^{+}

F e^{+ 3}

N O

N a_{2} (C N)_{5} N O]

N a_{2} [F e (N O) {(C N)}_{5}]

N a_{4} [F e {(C N)}_{5} (N O S)]

0.1 M

A g {N O}_{3}

0.2 M

N a C l

{N O}_{3}^{-}

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