JEE Chemistry 2023 Question with Solution

Given: We must compare the magnetic attraction of the listed coordination complexes.

Find: Which complex has the maximum number of unpaired electrons, because greater paramagnetism means stronger attraction to an applied magnetic field.

The attraction of a complex to a magnetic field depends on the number of unpaired electrons in the complex.

1. For $[\text{Zn}(\text{H}_2\text{O})_6]^{2+}$, $\text{Zn}^{2+}$ has configuration

So it has no unpaired electrons and is diamagnetic.

2. For $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$, $\text{Ni}^{2+}$ has configuration

In an octahedral field, this becomes

with 2 unpaired electrons.

3. For $[\text{Co}(\text{en})_3]^{3+}$, $\text{Co}^{3+}$ has configuration

Because en is a strong-field ligand, pairing occurs and the configuration becomes

So it has no unpaired electrons and is diamagnetic.

4. The solution analyzes $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$ with configuration

which has 3 unpaired electrons and would therefore show maximum attraction.

However, this species is not present in the listed options. Among the given options, the complex with the highest number of unpaired electrons is $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ with 2 unpaired electrons.

Therefore, among the provided options, the correct option is B.

The source solution concludes that $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$ would have maximum magnetic attraction, but that complex does not appear in the option list. The raw options contain $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ twice and do not include the cobalt aqua complex. Since the page explicitly marks B as the correct option and, among the listed choices, nickel(II) aqua complex has the largest justified number of unpaired electrons, the defensible answer is B.