Match List - I with List - II and choose the correct answer.

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

To match each iron complex with its magnetic moment, we determine three things for every species:

1. The oxidation state of iron → gives the d‑electron count.
2. Whether the ligands are strong‑field (low‑spin) or weak‑field (high‑spin).
3. The number of unpaired electrons n, and then compute the spin‑only magnetic moment using the formula

$\mu \; =\; \surd (n · (n + 2)) BM$

Now we analyse each complex.

(a) [Fe(CN)₆]^3‑

CN⁻ is a strong‑field ligand. The overall charge is –3, and six CN⁻ give –6, so Fe must be +3 (Fe³⁺). Fe³⁺ is d⁵. In a strong field, the d⁵ ion adopts a low‑spin configuration (t₂g⁵), leaving only one unpaired electron (n = 1).
Applying the formula:

$\mu \; =\; \surd (1·(1 + 2))\; =\; \surd 3\; \approx \; 1.73 BM$

Hence (a) corresponds to value (iv) 1.73 BM.

(b) [Fe(H₂O)₆]³⁺

H₂O is a weak‑field ligand. The complex charge is +3, so Fe is also +3 (d⁵). With weak field, the ion remains high‑spin, giving five unpaired electrons (n = 5).
Magnetic moment:

$\mu \; =\; \surd (5·(5 + 2))\; =\; \surd 35\; \approx \; 5.92 BM$

Thus (b) matches value (i) 5.92 BM.

(c) [Fe(CN)₆]^4‑

Again CN⁻ is strong‑field. Charge –4 with six CN⁻ (‑6) makes Fe +2 (Fe²⁺), which is d⁶. In a strong field the d⁶ ion is low‑spin (t₂g⁶), fully paired, so n = 0.
Magnetic moment:

$\mu \; =\; \surd (0·(0 + 2))\; =\; 0 BM$

Therefore (c) pairs with value (ii) 0 BM.

(d) [Fe(H₂O)₆]²⁺

H₂O is weak‑field. The complex charge is +2, giving Fe²⁺ (d⁶). High‑spin d⁶ distributes as t₂g⁴ e_g², leaving four unpaired electrons (n = 4).
Magnetic moment:

$\mu \; =\; \surd (4·(4 + 2))\; =\; \surd 24\; \approx \; 4.90 BM$

Consequently (d) corresponds to value (iii) 4.90 BM.

Putting the matches together we obtain the correct list:

(a) → (iv), (b) → (i), (c) → (ii), (d) → (iii).