Match List - I with List - II

List - I	List - II
(a) [Fe(CN)6]3-	(i) 5.92 BM
(b) [Fe(H2O)6]3+	(ii) 0 BM
(c) [Fe(CN)6]4-	(iii) 4.90 BM
(d) [Fe(H2O)6]2+	(iv) 1.73 BM

Choose the correct answer from the options given below.

The correct option is (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).

To determine the spin-only magnetic moment ($\mu$) of each complex, we can use the spin-only formula:
$\mu =\sqrt{n(n+2)}$ Bohr Magnetons (BM)
where $n$ is the number of unpaired electrons in the central metal ion.

Let us analyze each complex one by one:

(a) ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$:
Here, iron is in the +3 oxidation state (${\mathrm{Fe}}^{3+}$). The electronic configuration of ${\mathrm{Fe}}^{3+}$ is $\left[\mathrm{Ar}\right]3d^5$.
Since ${\mathrm{CN}}^{-}$ is a strong field ligand, it causes pairing of the $3d$ electrons. Thus, the configuration becomes low-spin:
$t_{2g}^5{e}_g^0$
This gives the number of unpaired electrons as $n=1$.
Calculating the magnetic moment:
$\mu =\sqrt{1(1+2)}=\sqrt{3}\approx 1.73$ BM
Therefore, (a) matches with (iv).

(b) ${\left[\mathrm{Fe}{\left(H_{2}O\right)}_6\right]}^{3+}$:
Here, iron is in the +3 oxidation state (${\mathrm{Fe}}^{3+}$) with a $3d^5$ configuration.
Since $H_{2}O$ is a weak field ligand, it does not cause pairing of electrons. Thus, the configuration remains high-spin:
$t_{2g}^3{e}_g^2$
This gives the number of unpaired electrons as $n=5$.
Calculating the magnetic moment:
$\mu =\sqrt{5(5+2)}=\sqrt{35}\approx 5.92$ BM
Therefore, (b) matches with (i).

(c) ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}$:
Here, iron is in the +2 oxidation state (${\mathrm{Fe}}^{2+}$). The electronic configuration of ${\mathrm{Fe}}^{2+}$ is $\left[\mathrm{Ar}\right]3d^6$.
Since ${\mathrm{CN}}^{-}$ is a strong field ligand, it causes complete pairing of the $3d$ electrons. Thus, the configuration becomes low-spin:
$t_{2g}^6{e}_g^0$
This gives the number of unpaired electrons as $n=0$.
Calculating the magnetic moment:
$\mu =\sqrt{0(0+2)}=0$ BM
Therefore, (c) matches with (ii).

(d) ${\left[\mathrm{Fe}{\left(H_{2}O\right)}_6\right]}^{2+}$:
Here, iron is in the +2 oxidation state (${\mathrm{Fe}}^{2+}$) with a $3d^6$ configuration.
Since $H_{2}O$ is a weak field ligand, it does not pair the electrons. Thus, the configuration remains high-spin:
$t_{2g}^4{e}_g^2$
This gives the number of unpaired electrons as $n=4$.
Calculating the magnetic moment:
$\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.90$ BM
Therefore, (d) matches with (iii).

Combining all the matched pairs:
(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)