Chemistry Q73 – JEE-MAIN 2025

Home Chemistry Coordination CompoundsQuestion #39

The number of paramagnetic metal complex species among

[C o (N H_{3})_{6}]^{3 +}

[C o (C_{2} O_{4})_{3}]^{3 -}

[M n C l_{6}]^{3 -}

[M n (C N)_{6}]^{3 -}

[C o F_{6}]^{3 -}

[F e (C N)_{6}]^{3 -}

and

[F e F_{6}]^{3 -}

with same number of unpaired electrons is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Determine Oxidation State

First, find the oxidation state of the central metal atom in each complex.

🧠 Full Solution Path

Step 1: Determine oxidation state and d-electron configuration✦ Active

For all given complexes, the central metal ion is in the +3 oxidation state. Let's determine their d-electron configurations:

{Co}^{3 +} : [A r] 3 d^{6}

{Mn}^{3 +} : [A r] 3 d^{4}

{Fe}^{3 +} : [A r] 3 d^{5}

Step 2: Determine spin state and calculate unpaired electrons for each complex○ Expand

We classify ligands as strong field (SF) or weak field (WF) to determine the spin state (low spin (LS) or high spin (HS)) and then count unpaired electrons:

1. $[C o (N H_{3})_{6}]^{3 +}$ : $C o^{3 +}$ ( $d^{6}$ ), $N H_{3}$ (SF) $⟹$ LS. $t_{2 g}^{6} e_{g}^{0}$ . Unpaired electrons = 0 (Diamagnetic).

2. $[C o (C_{2} O_{4})_{3}]^{3 -}$ : $C o^{3 +}$ ( $d^{6}$ ), $C_{2} O_{4}^{2 -}$ (SF) $⟹$ LS. $t_{2 g}^{6} e_{g}^{0}$ . Unpaired electrons = 0 (Diamagnetic).

3. $[M n C l_{6}]^{3 -}$ : $M n^{3 +}$ ( $d^{4}$ ), $C l^{-}$ (WF) $⟹$ HS. $t_{2 g}^{3} e_{g}^{1}$ . Unpaired electrons = 4 (Paramagnetic).

4. $[M n (C N)_{6}]^{3 -}$ : $M n^{3 +}$ ( $d^{4}$ ), $C N^{-}$ (SF) $⟹$ LS. $t_{2 g}^{4} e_{g}^{0}$ . Unpaired electrons = 2 (Paramagnetic).

5. $[C o F_{6}]^{3 -}$ : $C o^{3 +}$ ( $d^{6}$ ), $F^{-}$ (WF) $⟹$ HS. $t_{2 g}^{4} e_{g}^{2}$ . Unpaired electrons = 4 (Paramagnetic).

6. $[F e (C N)_{6}]^{3 -}$ : $F e^{3 +}$ ( $d^{5}$ ), $C N^{-}$ (SF) $⟹$ LS. $t_{2 g}^{5} e_{g}^{0}$ . Unpaired electrons = 1 (Paramagnetic).

7. $[F e F_{6}]^{3 -}$ : $F e^{3 +}$ ( $d^{5}$ ), $F^{-}$ (WF) $⟹$ HS. $t_{2 g}^{3} e_{g}^{2}$ . Unpaired electrons = 5 (Paramagnetic).

💡 Teacher's Secret Hint

Remember the spectrochemical series to correctly identify strong and weak field ligands.

Step 3: Count paramagnetic complexes with the same number of unpaired electrons○ Expand

Listing the paramagnetic complexes and their unpaired electrons:

- $[M n C l_{6}]^{3 -}$ : 4 unpaired electrons

- $[M n (C N)_{6}]^{3 -}$ : 2 unpaired electrons

- $[C o F_{6}]^{3 -}$ : 4 unpaired electrons

- $[F e (C N)_{6}]^{3 -}$ : 1 unpaired electron

- $[F e F_{6}]^{3 -}$ : 5 unpaired electrons

The complexes $[M n C l_{6}]^{3 -}$ and $[C o F_{6}]^{3 -}$ both have 4 unpaired electrons. These are the only two paramagnetic species that share the same number of unpaired electrons. Therefore, the number of such species is 2.

💡 Teacher's Secret Hint

Carefully compare the number of unpaired electrons for all paramagnetic species.

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Chemistry Question 73 – JEE-MAIN 2025

Clue 1 of 3: Determine Oxidation State

🧠 Full Solution Path