In sodium nitroprusside Na2[Fe(CN)5NO], treat NO as the nitrosonium ligand NO+. With sulphide ions in alkaline solution, a purple complex Na4[Fe(CN)5NOS] forms. The oxidation state of Fe in the purple complex, the number of unpaired electrons in it (strong-field low-spin) and the type of the colour are respectively

JEE Advanced levelSingle correct optionDifficulty 3 of 3qa033

In sodium nitroprusside $Na X_{2} [Fe (CN) X_{5} NO]$ , treat NO as the nitrosonium ligand $NO X^{+}$ . With sulphide ions in alkaline solution, a purple complex $Na X_{4} [Fe (CN) X_{5} NOS]$ forms. The oxidation state of Fe in the purple complex, the number of unpaired electrons in it (strong-field low-spin) and the type of the colour are respectively

In the product the ligand

NOS X^{-}

is formed from

NO X^{+}

and

S X^{2 -}

(charge -1). Charge balance:

x + 5 (- 1) + (- 1) = - 4

gives

x = + 2

. Fe(II) is

d^{6}

and with strong-field CN- ligands is low-spin with 0 unpaired electrons, so the complex is diamagnetic. The colour is purple.