Spin Only Magnetic Moment of Cr(CO)6

Spin only magnetic moment is the magnetic moment of an atom or ion in the absence of any magnetic field. It is calculated based on the number of unpaired electrons in the ground state of the ion or atom.

Formula to calculate spin only magnetic moment:

μspin = √n(n+2)BM

Where, n = number of unpaired electrons, BM = Bohr Magneton

Calculation of Spin Only Magnetic Moment of Cr(CO)6

To calculate the spin only magnetic moment of Cr(CO)6, we need to find the number of unpaired electrons in the ground state of Cr(CO)6.

Step 1: Identify the oxidation state of Cr

Cr(CO)6 is a coordination complex in which Cr is bonded to six CO (carbon monoxide) ligands. Each CO ligand donates one electron to the central Cr atom, making the oxidation state of Cr as 0.

Step 2: Determine the electronic configuration of Cr

The atomic number of Cr is 24. In the ground state, the electronic configuration of Cr is [Ar] 3d4 4s2.

Step 3: Calculate the number of unpaired electrons

In the ground state, there are four unpaired electrons in the 3d sublevel of Cr.

Step 4: Calculate the spin only magnetic moment

Using the formula, we get:

μspin = √n(n+2)BM
μspin = √4(4+2)BM
μspin = 2.83 BM

Therefore, the spin only magnetic moment of Cr(CO)6 is 2.83 BM.

However, the correct answer given in the options is 0, which indicates that Cr(CO)6 has no unpaired electrons in the ground state. This is because the complex undergoes backbonding, where the CO ligands donate their lone pair of electrons to the empty 3d orbitals of Cr. This results in the pairing of the four unpaired electrons, making the complex diamagnetic and having a spin only magnetic moment of 0.