Treat NO as NO+ in the brown-ring complex [Fe(H2O)5NO]^2+, which is high spin. Which statements are correct?

The oxidation state of Fe in the complex is +1.

The complex has three unpaired electrons.

Nitrite also gives a brown ring-type colour with FeSO4 and dilute acetic acid.

The oxidation state of Fe in the complex is +1.

The complex has three unpaired electrons.

Nitrite also gives a brown ring-type colour with FeSO4 and dilute acetic acid.

The brown ring is formed with dilute H2SO4 alone, no conc. acid layer being needed for nitrate.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3qa049

Treat NO as $NO X^{+}$ in the brown-ring complex $[Fe (H X_{2} O) X_{5} NO] X^{2 +}$ , which is high spin. Which statements are correct?

AThe oxidation state of Fe in the complex is +1.
BThe complex has three unpaired electrons.
CNitrite also gives a brown ring-type colour with $FeSO X_{4}$ and dilute acetic acid.
DThe brown ring is formed with dilute $H X_{2} SO X_{4}$ alone, no conc. acid layer being needed for nitrate.

Charge balance with neutral

H X_{2} O

and

NO X^{+}

gives

x + 1 = + 2

, so

x = + 1

, a

d^{7}

ion. In a high-spin

d^{7}

arrangement there are three unpaired electrons. Nitrite is reduced to NO by

Fe X^{2 +}

even in acetic acid, so 3 is true. Nitrate needs conc.

H X_{2} SO X_{4}

to be reduced, so 4 is false.