Question

Assertion :Statement - I : [Fe(H${}_2$O)${}_5$NO]SO${}_4$ is paramagnetic. Reason: Statement - II : The Fe in [Fe(H${}_2$O)${}_5$NO]SO${}_4$ has three unpaired electrons.

• A. Statement - I is true; Statement - II is also true; Statement - II is a correct explanation for Statement - I.
• B. Statement - I is true; Statement - II is true; Statement - II is not the correct explanation for Statement - I.
• C. Statement - I is true; Statement - II is false.
• D. Statement - I is false; Statement - II is true,

Answer 1

The correct option is A Statement - I is true; Statement - II is also true; Statement - II is a correct explanation for Statement - I.

NO has charge equal to +1 in this complex. Therefore, the oxidation number of Fe in the complex becomes +1.
$Fe=\left[Ar\right]3d^{6}4s^2$
$F{e}^{+}=\left[Ar\right]3d^{6}4s^1$

$N{O}^{+}$ causes pairing of 4s electron inside. Thus, the configuration is $3d^7$ and the number of unpaired electrons is 3.

Hence, it is paramagnetic.

So Statement - I is true and Statement - II is also true but Statement - II is a correct explanation for Statement - I.

Hence option A is correct.