Question

Match the following lists: A) $[MnCl_6]^{3-}$, B) $[Mn(CN)_6]^{3-}$, C) $[Fe(CN)_6]^{4-}$, D) $[CoF_6]^{3-}$ with their d-electronic configurations.

• A. A-III, B-IV, C-I, D-II
• B. A-III, B-IV, C-II, D-I
• C. A-II, B-III, C-IV, D-I
• D. A-II, B-I, C-IV, D-III

Hint:

Strong field ligands ($CN^-$) cause pairing; weak field ligands ($Cl^-$, $F^-$) do not!

Answer 1

The Correct Option is B

Step 1: Get the metal d-count for each complex.
In $[MnCl_6]^{3-}$ and $[Mn(CN)_6]^{3-}$ the metal is $Mn^{3+}$, which is $d^4$. In $[Fe(CN)_6]^{4-}$ the metal is $Fe^{2+}$, which is $d^6$. In $[CoF_6]^{3-}$ the metal is $Co^{3+}$, which is $d^6$.

Step 2: Sort ligands as strong or weak field.
$CN^-$ is a strong field ligand and forces electrons to pair up (low spin). $Cl^-$ and $F^-$ are weak field ligands and let electrons spread out (high spin).

Step 3: Fill $[MnCl_6]^{3-}$ ($d^4$, weak).
Weak field high spin gives $t_{2g}^3 e_g^1$. That matches configuration III. So A is III.

Step 4: Fill $[Mn(CN)_6]^{3-}$ ($d^4$, strong).
Strong field low spin pairs them up as $t_{2g}^4 e_g^0$. That matches IV. So B is IV.

Step 5: Fill $[Fe(CN)_6]^{4-}$ ($d^6$, strong).
Strong field low spin gives $t_{2g}^6 e_g^0$. That matches II. So C is II.

Step 6: Fill $[CoF_6]^{3-}$ ($d^6$, weak).
Weak field high spin gives $t_{2g}^4 e_g^2$. That matches I. So D is I.

Step 7: Conclusion.
So the answer is A-III, B-IV, C-II, D-I. \[ \boxed{\text{A-III, B-IV, C-II, D-I}} \]