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Match List-I with List-II.
Column I
- A. $[Fe(CN)_6]^{3-}$
- B. $[Fe(H_2O)_6]^{3+}$
- C. $[Fe(CN)_6]^{4-}$
- D. $[Fe(H_2O)_6]^{2+}$
Column II
- I. 5.92 BM
- II. 0 BM
- III. 4.90 BM
- IV. 1.73 BM
Correct answer: A → IV, B → I, C → II, D → III
Detailed Solution
$[Fe(CN)_6]^{3-}$: $Fe^{3+}$ ($3d^5$), strong field – 1 unpaired electron, $\mu = 1.73$ BM
$[Fe(H_2O)_6]^{3+}$: $Fe^{3+}$ ($3d^5$), weak field – 5 unpaired electrons, $\mu = 5.92$ BM
$[Fe(CN)_6]^{4-}$: $Fe^{2+}$ ($3d^6$), strong field – 0 unpaired electrons, $\mu = 0$ BM
$[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ ($3d^6$), weak field – 4 unpaired electrons, $\mu = 4.90$ BM
Correct match: A-IV, B-I, C-II, D-III.