The correct decreasing order of spin only magnetic moment values (BM) of $\mathrm{Cu}^{+}, \mathrm{Cu}^{2+}, \mathrm{Cr}^{2+}$ and $\mathrm{Cr}^{3+}$ ions is :

(C) $\mathrm{Cr}^{2+}>\mathrm{Cr}^{3+}>\mathrm{Cu}^{2+}>\mathrm{Cu}^{+}$

(A) $\mathrm{Cu}^{+}>\mathrm{Cu}^{2+}>\mathrm{Cr}^{3+}>\mathrm{Cr}^{2+}$

(B) $\mathrm{Cu}^{2+}>\mathrm{Cu}^{+}>\mathrm{Cr}^{2+}>\mathrm{Cr}^{3+}$

(D) $\mathrm{Cr}^{3+}>\mathrm{Cr}^{2+}>\mathrm{Cu}^{+}>\mathrm{Cu}^{2+}$

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Coordination Compounds

The correct decreasing order of spin only magnetic moment values (BM) of

Cu^{+}, Cu^{2 +}, Cr^{2 +}

and

Cr^{3 +}

ions is :

(A) $Cu^{+} > Cu^{2 +} > Cr^{3 +} > Cr^{2 +}$
(B) $Cu^{2 +} > Cu^{+} > Cr^{2 +} > Cr^{3 +}$
(C) $Cr^{2 +} > Cr^{3 +} > Cu^{2 +} > Cu^{+}$ ✓ Correct
(D) $Cr^{3 +} > Cr^{2 +} > Cu^{+} > Cu^{2 +}$

Answer: (C)

Solution

The spin-only moment rises with the number of unpaired electrons,

μ = n (n + 2) BM

C u^{+}

3 d^{10}

with no unpaired electron,

μ = 0

;

C u^{2 +}

3 d^{9}

with one,

μ = 3 \approx 1.73

;

C r^{3 +}

3 d^{3}

with three,

μ = 15 \approx 3.87

; and

C r^{2 +}

3 d^{4}

with four,

μ = 24 \approx 4.90

. Arranged in decreasing order of magnetic moment, the ions run

C r^{2 +} > C r^{3 +} > C u^{2 +} > C u^{+}

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option

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