Match the complexes given in List-I with the number of unpaired electrons given in List-II
|
List-I Complexes |
List-II Number of unpaired electrons |
|
(A) \( [Ti(H_2O)_6]^{3+} \) |
(I) 1 |
|
(B) \( [NiF_6]^{4-} \) |
(II) 5 |
|
(C) \( [Fe(CN)_6]^{4-} \) |
(III) 2 |
|
(D) \( [MnF_6]^{4-} \) |
(IV) 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
To count unpaired electrons, find the oxidation state of the metal from the charges, derive its d electron count from Z, and then fill the octahedral t2g and eg levels. A strong field ligand (\( \ce{CN^{-}} \)) pairs the electrons in t2g first; a weak field ligand (\( \ce{F^{-}} \), \( \ce{H2O} \)) lets them spread into eg before pairing.
(A) \( \ce{[Ti(H2O)6]^3+} \) — This matches (I) 1. Ti is +3. Ti (Z = 22) is \( \mathrm{[Ar]\,3d^{2}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives 3d1: one unpaired electron.
(B) \( \ce{[NiF6]^4-} \) — This matches (III) 2. x + 6(−1) = −4, so Ni is +2. Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives 3d8, \( t_{2g}^{6}e_{g}^{2} \): two unpaired electrons.
(C) \( \ce{[Fe(CN)6]^4-} \) — This matches (IV) 0. x + 6(−1) = −4, so Fe is +2. Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); losing the two 4s electrons gives 3d6. Strong field \( \ce{CN^{-}} \) gives \( t_{2g}^{6}e_{g}^{0} \): no unpaired electron.
(D) \( \ce{[MnF6]^4-} \) — This matches (II) 5. x + 6(−1) = −4, so Mn is +2. Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing the two 4s electrons gives 3d5. Weak field \( \ce{F^{-}} \) gives \( t_{2g}^{3}e_{g}^{2} \): five unpaired electrons.
Shortcut: (A)-(I) appears in Options (1) and (2), and (C)-(IV) then leaves only Option (1).