In an octahedral complex of ${\mathrm{Fe}}^{+2}$in a high spin state, what is the magnetic moment (spin only)?

The correct option is A

$4.89\mathrm{BM}$

The correct option is option A.

The explanation for the correct option:

1. A magnetic field is created when an electron moves freely, this magnetic field is what is referred to as the magnetic moment. A spin-only magnetic moment occurs when an electron rotates around its own axis and produces a magnetic field.
2. Spin-only magnetic moment is given by ${\mu }_s=\sqrt{\mathrm{n}(\mathrm{n}+2)}$
3. When the complex is in low spin the pairing of electrons is preferred over the transition of electrons and if a high spin octahedral complex is formed, electron transition is preferred over pairing.
4. The given ${\mathrm{Fe}}^{+2}$ is in a high spin state i.e. electron transition is preferred over pairing.
5. Electronic configuration of ${\mathrm{Fe}}^{+2}$is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^6$so, number of electrons in d-orbital is six.
6. Arrangement of six electron in d-orbital is given by

1. Spin magnetic moment$=\sqrt{4(4+2)}\Rightarrow \sqrt{24}\Rightarrow 4.89\mathrm{BM}$

Final answer: The magnetic moment (spin only) for ${\mathrm{Fe}}^{+2}$ is $4.89\mathrm{BM}$