Question

The spin only magnetic moment value (in Bohr magneton units) of $Cr{\left( {CO} \right)_6}$ is?

Answer 1

The spin only magnetic moment value can be calculated from the number of unpaired electrons in the central metal atom of the complex compound, in some complex compounds the unpaired electrons will pair up due to the presence of a strong ligand like carbonyl or cyanide. In the given complex the six unpaired electrons of chromium will pair up due to the presence of carbonyl ligand and does not show any spin magnetic moment.
$\mu = \sqrt {n\left( {n + 2} \right)}$
$n$ is number of unpaired electrons
$\mu$ is a magnetic moment.

Complete Step By Step Answer:
Chromium is a chemical element with atomic number $24$ and has six valence electrons. Generally, chromium exhibits half-filled electronic configuration to attain stability.
In the given complex, $Cr{\left( {CO} \right)_6}$ chromium is a central metal atom and carbonyl is a strong ligand. The oxidation state of chromium in given complex is zero. Thus, chromium has six electrons in $4s$ and $3d$ orbitals.
But these six electrons will pair up due to the presence of a strong field carbonyl ligand. Thus, the number of unpaired electrons is zero. Substitute the number of electrons as zero in the above formula.
$\mu = \sqrt {0\left( {0 + 2} \right)} BM = 0BM$
The unit of magnetic moment is Bohr magneton.
The spin only magnetic moment value (in Bohr magneton units) of $Cr{\left( {CO} \right)_6}$ is zero.

Note:
In the absence of a strong field ligand chromium has six electrons, out of which two were paired up in s-orbital and has only four valence electrons, which shows the spin only magnetic moment value. Carbonyl is a strong ligand which is responsible for the pairing of electrons in the given molecule.