Spin-only magnetic moment can be defined as the magnetic moment of an atom in the absence of any orbital angular momentum. It is calculated by using the formula μ = √n(n+2)BM, where n is the number of unpaired electrons and BM is the Bohr magneton.



Out of the given options, the configuration with 2.84BM can only be possible for d4 in strong ligand field. Let's understand why.

- d4 (in strong ligand field): In this case, all the four electrons are paired in the lower energy orbitals and there are no unpaired electrons. Therefore, the spin-only magnetic moment will be zero (μ = √0(0+2)BM = 0BM). This is not possible as the question states that the value of spin-only magnetic moment is 2.84BM.

- d4 (in weak ligand field): In this case, there will be two unpaired electrons in the higher energy orbitals. Therefore, the spin-only magnetic moment will be μ = √2(2+2)BM = 2.83BM, which is close to 2.84BM but not the exact value.

- d3 (in weak as well as strong ligand field): In this case, there will be one unpaired electron in the higher energy orbitals. Therefore, the spin-only magnetic moment will be μ = √1(1+2)BM = 1.73BM, which is not even close to 2.84BM.

- d5 (in strong ligand field): In this case, there will be five unpaired electrons in the higher energy orbitals. Therefore, the spin-only magnetic moment will be μ = √5(5+2)BM = 3.87BM, which is much higher than 2.84BM.

Therefore, the only possible option is d4 in strong ligand field.



The value of spin-only magnetic moment for the given configuration of 2.84BM can only be possible for d4 in strong ligand field where there are two unpaired electrons in the higher energy orbitals.

Spin only magnetic moment = √(n(n+2)) B.M.

Where n = number of unpaired electron

Given, √(n(n+2)) = 2.84

n(n + 2) = 8.0656

n = 2

In an octahedral complex, for a d4configuration in a strong field ligand, number of unpaired electrons = 2