JEE Chemistry 2025 Question with Solution

Given: The complexes are $[Co(NH_3)_6]^{3+}, [Co(C_2O_4)_3]^{3-}, [MnCl_6]^{3-}, [Mn(CN)_6]^{3-}, [CoF_6]^{3-}, [Fe(CN)_6]^{3-}$ and $[FeF_6]^{3-}$.

Find: The number of paramagnetic species having the same number of unpaired electrons.

Use crystal field theory by identifying the oxidation state, $d$-electron count, and whether the ligand is strong field or weak field.

1. $[Co(NH_3)_6]^{3+}$: cobalt is $Co^{3+}$, so it is $d^6$. With $NH_3$, it is taken as low spin here, so it has 0 unpaired electrons and is diamagnetic.
2. $[Co(C_2O_4)_3]^{3-}$: cobalt is again $Co^{3+}$, so it is $d^6$. From the extracted solution, this complex is treated as having 4 unpaired electrons.
3. $[MnCl_6]^{3-}$: manganese is $Mn^{3+}$, so it is $d^4$. With weak-field $Cl^-$, it is high spin and has 4 unpaired electrons.
4. $[Mn(CN)_6]^{3-}$: manganese is $Mn^{3+}$, so it is $d^4$. With strong-field $CN^-$, it is low spin and has 2 unpaired electrons.
5. $[CoF_6]^{3-}$: cobalt is $Co^{3+}$, so it is $d^6$. With weak-field $F^-$, it is high spin and has 4 unpaired electrons.
6. $[Fe(CN)_6]^{3-}$: iron is $Fe^{3+}$, so it is $d^5$. With strong-field $CN^-$, it is low spin and has 1 unpaired electron.
7. $[FeF_6]^{3-}$: iron is $Fe^{3+}$, so it is $d^5$. With weak-field $F^-$, it is high spin and has 5 unpaired electrons.

Now count the paramagnetic species grouped by equal unpaired electrons:

• 1 unpaired electron: $[Fe(CN)_6]^{3-}$
• 2 unpaired electrons: $[Mn(CN)_6]^{3-}$
• 4 unpaired electrons: $[Co(C_2O_4)_3]^{3-}, [MnCl_6]^{3-}, [CoF_6]^{3-}$
• 5 unpaired electrons: $[FeF_6]^{3-}$

The largest group of paramagnetic species with the same number of unpaired electrons contains 3 complexes.

Therefore, the required number is $3$.

Given: A list of coordination complexes is provided.

Find: How many paramagnetic species have an identical number of unpaired electrons.

The key idea is to calculate the unpaired electrons for each complex and then look for the repeated value.

From the solution:

• $[Co(NH_3)_6]^{3+}$ has $0$ unpaired electrons.
• $[Co(C_2O_4)_3]^{3-}$ has $4$ unpaired electrons.
• $[MnCl_6]^{3-}$ has $4$ unpaired electrons.
• $[Mn(CN)_6]^{3-}$ has $2$ unpaired electrons.
• $[CoF_6]^{3-}$ has $4$ unpaired electrons.
• $[Fe(CN)_6]^{3-}$ has $1$ unpaired electron.
• $[FeF_6]^{3-}$ has $5$ unpaired electrons.

Ignoring the diamagnetic species, the paramagnetic complexes are grouped as:

The repeated unpaired-electron count occurring most often is 4, and it appears for 3 paramagnetic complexes.

Therefore, the answer is $3$.

Note: One extracted solution block contains inconsistent intermediate statements, but the second solution block and the displayed correct answer both support $3$.