Sodium nitroprusside, Na2[Fe(CN)5NO], gives a purple colour with sulphide. Treating NO as NO+, the oxidation state of Fe in nitroprusside and the number of unpaired electrons (cyanide is strong field) are:

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Sodium nitroprusside, $Na X_{2} [Fe (CN) X_{5} NO]$ , gives a purple colour with sulphide. Treating NO as $NO X^{+}$ , the oxidation state of Fe in nitroprusside and the number of unpaired electrons (cyanide is strong field) are:

The complex ion is

[Fe (CN) X_{5} NO] X^{2 -}

x - 5 + 1 = - 2

gives

x = + 2

. Fe(II) is

d^{6}

and with strong-field ligands is low spin

t_{2 g}^{6}

, with 0 unpaired electrons, so the ion is diamagnetic. The other values come from taking NO as neutral or ignoring the strong field.