Statement $\mathbf{1}\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4$ is paramagnetic.
Statement 2 The $\mathrm{Fe}$ in $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4$ has three unpaired electrons.
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
Statement 1 is true, Statement 2 is false.
Statement 1 is false, Statement 2 is true
Solution
The oxidation number of $\mathrm{Fe}$ in the complex, $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4$ is $+1$ [ $\mathrm{NO}$ has $+1$ charge ]
$
\mathrm{Fe}^{+}=[\mathrm{Ar}] 3 d^6 4 s^1
$
$\mathrm{NO}^{+}$causes pairing of 4 s electron inside .
Thus, the configuration is $3 d^7$ and number of unpaired electrons $=3$