Question

Which of the following complex compounds will exhibit highest paramagnetic behaviour?

(At. No: Ti=22, Cr=24,Co=27 and Zn=30)

(a) ${{[Ti{{(N{{H}_{3}})}_{6}}]}^{3+}}$

(b) ${{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}$

(c) ${{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}$

(d) ${{[Zn{{(N{{H}_{3}})}_{6}}]}^{2+}}$

Answer 1

The presence of unpaired electrons indicates that the complex is paramagnetic and they are strongly attracted by the magnet. First find the number of unpaired electrons in the above complexes and then, you can easily identify the complex having the highest number of unpaired electrons exhibiting highest paramagnetic behaviour.

Complete step by step solution:

By paramagnetic we mean the presence of the unpaired electrons in the complexes. Paramagnetic complexes have the permanent magnetic moment and such complexes are strongly attracted by the magnetic field due to the presence of the unpaired electrons in them. These paramagnetic complexes lose their magnetism in the absence of the magnetic field.

Now, considering the given complexes one by one as;

1. ${{[Ti{{(N{{H}_{3}})}_{6}}]}^{3+}}$

Atomic of Ti=22 (given)

Electronic configuration of Ti=${{[Ar]}^{18}}3{{d}^{2}}4{{s}^{2}}$

Electronic configuration of $T{{i}^{3+}}={{[Ar]}^{18}}3{{d}^{1}}$

Since, $N{{H}_{3}}$ is a weak ligand, so there will be no pairing in ${{[Ti{{(N{{H}_{3}})}_{6}}]}^{3+}}$

So, in ${{[Ti{{(N{{H}_{3}})}_{6}}]}^{3+}}$, there is one unpaired electron.

2. ${{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}$

Atomic of $Cr =24$ (given)

Electronic configuration of Cr =${{[Ar]}^{18}}3{{d}^{5}}4{{s}^{1}}$

Electronic configuration of $C{{r}^{3+}}={{[Ar]}^{18}}3{{d}^{3}}$

Since, $N{{H}_{3}}$ is a weak ligand, so there will be no pairing in ${{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}$

So, in${{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}$,there are three unpaired electrons.

3. ${{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}$

Atomic of $Co =27$ (given)

Electronic configuration of $Co ={{[Ar]}^{18}}3{{d}^{7}}4{{s}^{2}}$

Electronic configuration of $C{{o}^{3+}}={{[Ar]}^{18}}3{{d}^{6}}$

Since, $N{{H}_{3}}$ is a weak ligand, but in case of $Co$ metal it becomes strong ligand and thus, results into the pairing of the unpaired electrons in${{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}$.

So, in${{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}$, there are no unpaired electrons and is, thus, diamagnetic complex.

4. ${{[Zn{{(N{{H}_{3}})}_{6}}]}^{2+}}$

Atomic of Zn =30 (given)

Electronic configuration of Zn =${{[Ar]}^{18}}3{{d}^{10}}4{{s}^{2}}$

Electronic configuration of $Z{{n}^{3+}}={{[Ar]}^{18}}3{{d}^{10}}$

So, in ${{[Zn{{(N{{H}_{3}})}_{6}}]}^{2+}}$, there is no unpaired electron ,so it is a diamagnetic complex.

Hence, from the above, it is clear that the complex having the highest number of unpaired electrons is ${{[Cr{{(N{{H}_{3}})}_{6}}]}^{3+}}$ and thus, exhibit highest paramagnetic behaviour.

Thus, option (b) is correct.

Note: The complexes which do not have any unpaired electrons in the complexes are called the diamagnetic complexes. Diamagnetic complexes do not have the permanent magnet moment and such complexes are also not strongly attracted by the magnetic but are weakly repelled by the magnet due to the absence of the unpaired electrons in them. Since they do not have any unpaired electrons, they have no effect on them in the absence of the magnetic field.