Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 27.

$ 3.87\text{ BM} $

The correct answer is Option (4) → $ 3.87\text{ BM} $

For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the 'spin-only' formula, i.e.

$ \mu=\sqrt{n(n+2)} $

in Bohr magneton (BM)

where n is the number of unpaired electrons and

In question it is given atomic number is 27, so the divalent ion in aqueous solution will have $d^7$ configuration (3 unpaired electrons). The magnetic moment is

$ \mu=\sqrt{3(3+2)}={3.87}\text{ BM} $