Paramagnetic Character and Unpaired Electrons of Brown Ring Complex

To evaluate the given statements, let us analyze the brown ring complex, $[\text{Fe}(\text{H}_2\text{O})_5\text{NO}]\text{SO}_4$:

1. Oxidation state and electronic configuration of $\text{Fe}$:
   The complex consists of the coordination cation $[\text{Fe}(\text{H}_2\text{O})_5\text{NO}]^{2+}$ and the sulfate anion $\text{SO}_4^{2-}$.
   In this complex, the nitric oxide ligand acts as a nitrosonium ion ($\text{NO}^+$), having transferred one electron to the iron center:
   $\text{Fe}^{2+} + \text{NO} \longrightarrow \text{Fe}^+ + \text{NO}^+$
   The water molecules ($\text{H}_2\text{O}$) are neutral ligands. Therefore, the oxidation state of iron is $+1$:
   $\text{Oxidation state of Fe} = +1$
   The ground-state electronic configuration of neutral iron ($Z = 26$) is:
   $\text{Fe}: [\text{Ar}]\,3d^6 4s^2$
   Hence, the electronic configuration of $\text{Fe}^+$ is:
   $\text{Fe}^+: [\text{Ar}]\,3d^7$

2. Crystal Field Splitting and Unpaired Electrons:
   In an octahedral crystal field with weak-field ligands like $\text{H}_2\text{O}$, the $d$-orbitals split into the lower energy $t_{2g}$ set and the higher energy $e_g$ set. The $3d^7$ configuration follows high-spin distribution:
   $t_{2g}^5 \, e_g^2$

   • The $t_{2g}$ orbitals contain 5 electrons: two pairs and one unpaired electron ($\uparrow\downarrow, \uparrow\downarrow, \uparrow$).
   • The $e_g$ orbitals contain 2 electrons: each unpaired ($\uparrow, \uparrow$).

   Thus, the total number of unpaired electrons is:
   $n = 1 + 2 = 3$
   Therefore, STATEMENT-2 is True.

3. Paramagnetic Character:
   A substance is defined as paramagnetic if it contains one or more unpaired electrons. Since $\text{Fe}^+$ has $n = 3$ unpaired electrons, the complex possesses a non-zero net spin magnetic moment:
   $\mu_{\text{eff}} = \sqrt{n(n+2)}\,\text{BM} = \sqrt{3(3+2)}\,\text{BM} = \sqrt{15}\,\text{BM} \approx 3.87\,\text{BM}$
   This confirms that the complex is paramagnetic. Therefore, STATEMENT-1 is True.

Since the paramagnetism of $[\text{Fe}(\text{H}_2\text{O})_5\text{NO}]\text{SO}_4$ directly arises from the presence of three unpaired electrons in the $\text{Fe}$ center, STATEMENT-2 is the correct explanation for STATEMENT-1.

Hence, the correct option is (A).