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 answer is: (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

This problem requires us to determine the number of unpaired electrons in each iron complex and then calculate the spin-only magnetic moment using the formula:

$\mu =\sqrt{n(n+2)}$ BM, where n = number of unpaired electrons.

The key concept is that CN⁻ is a strong field ligand (causes electron pairing → low spin) and H₂O is a weak field ligand (does not cause pairing → high spin) in octahedral complexes.

Step 1: Determine the oxidation state and d-electron configuration of Fe in each complex.

• Fe³⁺ → [Ar] 3d⁵ (5 d-electrons)
• Fe²⁺ → [Ar] 3d⁶ (6 d-electrons)

(a) [Fe(CN)₆]³⁻ — Fe³⁺ (d⁵) with CN⁻ (strong field ligand → low spin)

In a low-spin octahedral field, the 5 d-electrons are arranged as:
t_2g⁵ e_g⁰ → The t_2g orbitals hold 5 electrons (↑↓, ↑↓, ↑), giving n = 1 unpaired electron.

$\mu =\sqrt{1(1+2)}=\sqrt{3}=1.73$ BM → matches (iv) 1.73 BM

(b) [Fe(H₂O)₆]³⁺ — Fe³⁺ (d⁵) with H₂O (weak field ligand → high spin)

In a high-spin octahedral field, the 5 d-electrons occupy all five orbitals singly (maximum multiplicity):
t_2g³ e_g² → All electrons are unpaired, giving n = 5 unpaired electrons.

$\mu =\sqrt{5(5+2)}=\sqrt{35}=5.92$ BM → matches (i) 5.92 BM

(c) [Fe(CN)₆]⁴⁻ — Fe²⁺ (d⁶) with CN⁻ (strong field ligand → low spin)

In a low-spin octahedral field, all 6 d-electrons pair up in the three t_2g orbitals:
t_2g��� e_g⁰ → All orbitals are fully paired, giving n = 0 unpaired electrons.

$\mu =\sqrt{0(0+2)}=0$ BM → matches (ii) 0 BM

(d) [Fe(H₂O)₆]²⁺ — Fe²⁺ (d⁶) with H₂O (weak field ligand → high spin)

In a high-spin octahedral field, the 6 d-electrons are distributed as:
t_2g⁴ e_g² → (↑↓, ↑, ↑) in t_2g and (↑, ↑) in e_g, giving n = 4 unpaired electrons.

$\mu =\sqrt{4(4+2)}=\sqrt{24}=4.90$ BM → matches (iii) 4.90 BM

Summary of matching:

(a) [Fe(CN)₆]³⁻ → (iv) 1.73 BM (1 unpaired electron, low spin d⁵)
(b) [Fe(H₂O)₆]³⁺ → (i) 5.92 BM (5 unpaired electrons, high spin d⁵)
(c) [Fe(CN)₆]⁴⁻ → (ii) 0 BM (0 unpaired electrons, low spin d⁶)
(d) [Fe(H₂O)₆]²��� → (iii) 4.90 BM (4 unpaired electrons, high spin d⁶)