Question

The pair in which both the species have the same magnetic moment (spin only) is :

• A. \([\mathrm{Cr(H_2O)_6}]^{2+}\) and \([\mathrm{CoCl_4}]^{2-}\)
• B. \([\mathrm{Cr(H_2O)_6}]^{2+}\) and \([\mathrm{Fe(H_2O)_6}]^{2+}\)
• C. \([\mathrm{Mn(H_2O)_6}]^{2+}\) and \([\mathrm{Cr(H_2O)}]^{2+}\)
• D. \([\mathrm{Co(OH)_4}]^{2-}\) and \([\mathrm{Fe(NH_3)_6}]^{2+}\)

Hint:

Spin-only magnetic moment is \(\mu = \sqrt{n(n+2)}\) BM, so count the unpaired electrons in each complex (water is a weak-field ligand, so high spin).

Answer 1

The Correct Option is B

Step 1: \(\mu = \sqrt{n(n+2)}\) BM, so equal moments mean equal numbers of unpaired electrons n.
Step 2: \([\mathrm{Cr(H_2O)_6}]^{2+}\): \(\mathrm{Cr^{2+}}\) is \(d^4\); water is a weak-field ligand, so high spin \(t_{2g}^3 e_g^1\), n = 4, \(\mu = \sqrt{24} = 4.90\) BM.
Step 3: \([\mathrm{Fe(H_2O)_6}]^{2+}\): \(\mathrm{Fe^{2+}}\) is \(d^6\), high spin \(t_{2g}^4 e_g^2\), n = 4, \(\mu = 4.90\) BM.
Step 4: \([\mathrm{CoCl_4}]^{2-}\): \(\mathrm{Co^{2+}}\) is \(d^7\), tetrahedral, \(e^4 t_2^3\), n = 3, \(\mu = \sqrt{15} = 3.87\) BM; \([\mathrm{Co(OH)_4}]^{2-}\) likewise n = 3; \([\mathrm{Fe(NH_3)_6}]^{2+}\) is high spin \(d^6\) with n = 4; \([\mathrm{Mn(H_2O)_6}]^{2+}\) is \(d^5\) with n = 5 (\(5.92\) BM).
Step 5: Option (1): 4 vs 3, option (3): 5 vs 4, option (4): 3 vs 4 are unequal. Option (2): 4 and 4 are equal.
Answer: (2) \([\mathrm{Cr(H_2O)_6}]^{2+}\) and \([\mathrm{Fe(H_2O)_6}]^{2+}\)