JEE Main: Count Unpaired Electrons in These Complexes — Quick Trick!

 

❓ Question

The number of paramagnetic metal complex species among the following — that have the same number of unpaired electrons — is asked for:

$$[{\text{Co(NH}}_3{)}_6{]}^{3+},[{\text{Co(C}}_2{\text{O}}_4{)}_3{]}^{3-},[{\text{MnCl}}_6{]}^{3-},[{\text{Mn(CN)}}_6{]}^{3-},[{\text{CoF}}_6{]}^{3-},[{\text{Fe(CN)}}_6{]}^{3-},[{\text{FeF}}_6{]}^{3-}$$

We must find which of these are paramagnetic and then see how many of them share the same number of unpaired electrons.

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🖼️ Question Image

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✍️ Short Solution (stepwise)

We analyse each complex:

General steps:

1. Determine metal oxidation state from complex charge.

2. Write the d-electron count.

3. Decide whether ligand is strong-field (low-spin) or weak-field (high-spin).

4. Count unpaired electrons in octahedral splitting.

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1. $[ \text{Co(NH}_3)_6 ]^{3+}$

• Co oxidation: $+3$ → $d^6$.

• NH₃ is a borderline/relatively strong ligand for Co³⁺ → typically low-spin.

• Low-spin $d^6$ → $t_{2g}^6 e_g^0$ → 0 unpaired → diamagnetic.

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2. $[ \text{Co(C}_2\text{O}_4)_3 ]^{3-}$

• Co: $+3$ (oxalate is a moderate/chelate ligand, behaves strong enough for low-spin in Co³⁺) → $d^6$ low-spin → 0 unpaired → diamagnetic.

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3. $[ \text{MnCl}_6 ]^{3-}$

• Mn oxidation: $+3$ → $d^4$.

• Cl⁻ = weak field → high-spin d⁴ → configuration $t_{2g}^3 e_g^1$ → 4 unpaired → paramagnetic.

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4. $[ \text{Mn(CN)}_6 ]^{3-}$

• Mn: $+3$ → $d^4$.

• CN⁻ = strong field → low-spin d⁴ → $t_{2g}^4 e_g^0$ → 2 unpaired → paramagnetic.

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5. $[ \text{CoF}_6 ]^{3-}$

• Co: $+3$ → $d^6$.

• F⁻ = weak field → high-spin d⁶ → $t_{2g}^4 e_g^2$ → 4 unpaired → paramagnetic.

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6. $[ \text{Fe(CN)}_6 ]^{3-}$

• Fe: $+3$ → $d^5$.

• CN⁻ strong → low-spin d⁵ → $t_{2g}^5 e_g^0$ → 1 unpaired → paramagnetic.

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7. $[{\text{FeF}}_6{]}^{3-}$

• Fe: $+3$ → $d^5$.

• F⁻ weak → high-spin d⁵ → $t_{2g}^3 e_g^2$ → 5 unpaired → paramagnetic.

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🧮 Tabulated summary

Complex	Metal Ox. State	d-count	Ligand field	Unpaired e⁻	Paramagnetic?
Co(NH₃)₆³⁺	+3	d⁶	NH₃ (strongish)	0	No
Co(C₂O₄)₃³⁻	+3	d⁶	oxalate (moderate)	0	No
MnCl₆³⁻	+3	d⁴	Cl⁻ (weak)	4	Yes
Mn(CN)₆³⁻	+3	d⁴	CN⁻ (strong)	2	Yes
CoF₆³⁻	+3	d⁶	F⁻ (weak)	4	Yes
Fe(CN)₆³⁻	+3	d⁵	CN⁻ (strong)	1	Yes
FeF₆³⁻	+3	d⁵	F⁻ (weak)	5	Yes

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✅ Final reasoning & answer

We look for paramagnetic species that have the same number of unpaired electrons. From the table:

• $[ \text{MnCl}_6 ]^{3-}$ → 4 unpaired

• $[ \text{CoF}_6 ]^{3-}$ → 4 unpaired

These two complexes are paramagnetic and share the same number (4) of unpaired electrons. The other paramagnetic species have unpaired counts 2, 1, and 5 (each unique).

$$\boxed{\text{Answer: }2}$$