Question

The value of ‘spin only’ magnetic moment for one of the following configurations is $ 2.84BM. $ The correct one is:
(A) $ {d^4} $ (in strong ligand field)
(B) $ {d^4} $ (in weak ligand field)
(C) $ {d^3} $ (in strong as well as weak ligand field)
(D) $ {d^5} $ (in strong ligand field)

Answer 1

Hint: Strong ligand: Ligands that produce large splitting, are known as strong fields. Examples are $ CO,C{N^ - } $ .
Weak ligands: Ligands that produce small splitting, are known as weak fields. Examples are $ {H_2}O $ , $ C{l^ - } $ .

Complete step by step solution:
Let us first talk about strong ligands and weak ligands.
Strong ligand: Ligands that produce large splitting, are known as strong fields. Examples are $ CO,C{N^ - } $ . They cause forcible pairing of electrons within the $ 3d $ orbitals. So they form low spin complexes.
Weak ligand: Ligands that produce small splitting, are known as weak fields. Examples are $ {H_2}O $ , $ C{l^ - } $ . They cannot cause forcible pairing of electrons within the $ 3d $ orbitals. So they form high spin complexes.
The spin only magnetic moment is calculated as $ \sqrt {n(n + 2)} $ , where $ n $ is the number of unpaired electrons.
Here we are given a spin only magnetic moment as $ 2.84 $ so $ \sqrt {n(n + 2)} = 2.84 $ . If we try to find the value of $ n $ from this equation then it comes out to be $ 2 $ (first squaring both sides and then find the roots of the quadratic equation by factorisation). Hence the number of unpaired electrons will be two. Now we know that in the strong field $ {d^4} $ configuration the number of unpaired electrons is two and one paired electron is there.
So option A is the correct answer.

Note:
We know that there are five d-orbitals. And the splitting of these orbitals are in the ratio of $ 2:3 $ . In octahedral complexes two orbitals are in higher energy levels and three are in lower energy levels and the opposite happens in the case of tetrahedral complexes.