Question

Using the previous question.
Predict the magnetic nature of A and B:

[IIT-JEE, 2006]

• A. Both are diamagnetic.
• B. A is diamagnetic and B is paramagnetic with one unpaired electron.
• C. A is diamagnetic and B is paramagnetic with two unpaired electrons.
• D. Both are paramagnetic.

Answer 1

The correct option is C A is diamagnetic and B is paramagnetic with two unpaired electrons.

To predict the magnetic nature, we will need to use VBT and figure out if the ligand is a strong field or weak field ligand. Once we have this, we can then analyse the pairing of electrons. If all electrons are paired then the compound will be diamagnetic, otherwise paramagnetic. Simple enough? Let's do this!

i. $\left[Ni\right(CN{)}_4{]}_2,Ni(Z=28)\Rightarrow 3d^{8}4s^{2}N{i}^{2+}=3d^8$

Cyano is a strong field ligand, it forces pairing of electrons. As you can see, the hybridisaiton can also be worked our easily to be $ds{p}^2$- which would have a square planar geometry.
There are no unpaired electrons, so the complex is diamagnetic.

ii. $[NiC{l}_4{]}^2$
$N{i}^{2+}=\left[Ar\right]3d^8$

Chlorido is a weak field ligand, so it does not force pairing of electrons.
$s{p}^3$ - hybridisation (tetrahedral)

Unpaired electrons exist, so the complex is paramagnetic.
If needed, we can also calculate the spin magnetic moment using the formula below:
Remember, there are 2 electrons which are unpaired here.
$\mu =\sqrt{n(n+2)}BM=\sqrt{2(2+2)}BM=\sqrt{8}BM$