Match the LIST-I with LIST-II.
| List-I (Complex) | List-II (Magnetic moment) |
| A. [CoF6]3− | I. 4.89 BM |
| B. [Ni(NH3)6]2+ | II. 5.20 BM |
| C. [FeF6]3− | III. Zero |
| D. [Zn(NH3)6]2+ | IV. 2.82 BM |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-I, B-IV, C-II, D-III
The spin-only magnetic moment depends only on the number of unpaired electrons (n):
\( \displaystyle \mu=\sqrt{n(n+2)}\ \mathrm{BM} \)
n = 0, 1, 2, 3, 4, 5 gives μ = 0, 1.73, 2.83, 3.87, 4.90, 5.92 BM.
To find n, write the dn configuration of the metal ion and fill the t2g and eg sets. A weak field ligand such as \( \ce{F^{-}} \) gives a small splitting, so the electrons stay unpaired (high spin). For d8 and d10 ions the ligand does not change the count.
A. \( \ce{[CoF6]^{3-}} \) — This matches I. 4.89 BM. Co (Z = 27) is [Ar]3d74s2; the charge balance x + 6(−1) = −3 gives x = +3, and losing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): 3d6. With \( \ce{F^{-}} \) it is high spin, t2g4eg2, with 4 unpaired electrons.
\( \displaystyle \mu=\sqrt{4\times6}=\sqrt{24}\approx4.90\ \mathrm{BM} \)
B. \( \ce{[Ni(NH3)6]^{2+}} \) — This matches IV. 2.82 BM. Ni (Z = 28) is [Ar]3d84s2; \( \ce{NH3} \) is neutral, so nickel is +2, and losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d8, t2g6eg2, so it has 2 unpaired electrons with any ligand.
\( \displaystyle \mu=\sqrt{2\times4}=\sqrt{8}\approx2.83\ \mathrm{BM} \)
C. \( \ce{[FeF6]^{3-}} \) — This matches II. 5.20 BM, by elimination: 5 unpaired electrons give 5.92 BM, not 5.20 BM. Fe (Z = 26) is [Ar]3d64s2; x + 6(−1) = −3 gives x = +3, and losing the two 4s electrons and one 3d electron gives \( \ce{Fe^{3+}} \): 3d5. With \( \ce{F^{-}} \) it is high spin, t2g3eg2, with 5 unpaired electrons, the largest number in the list.
\( \displaystyle \mu=\sqrt{5\times7}=\sqrt{35}\approx5.92\ \mathrm{BM} \)
D. \( \ce{[Zn(NH3)6]^{2+}} \) — This matches III. Zero. Zn (Z = 30) is [Ar]3d104s2; \( \ce{NH3} \) is neutral, so zinc is +2, and losing the two 4s electrons gives \( \ce{Zn^{2+}} \): 3d10, so every electron is paired and the complex is diamagnetic.
Shortcut: The pair A–I appears in Options (2) and (3), and D–III then leaves only Option (2).
Note: It seems that there is a mistake in the original NTA paper, which prints 5.20 BM in item II when it should read 5.92 BM.