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The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the complex $\left[ {{\text{M}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]{\text{C}}{{\text{l}}_{\text{2}}}$, is:
(A)${\text{C}}{{\text{r}}^{{\text{2 + }}}}$and ${\text{M}}{{\text{n}}^{2 + }}$
(B)${{\text{V}}^{2 + }}$and ${\text{C}}{{\text{o}}^{2 + }}$
(C)${{\text{V}}^{2 + }}$and ${\text{F}}{{\text{e}}^{2 + }}$
(D)${\text{C}}{{\text{o}}^{2 + }}$and ${\text{F}}{{\text{e}}^{2 + }}$

Last updated date: 13th Jun 2024
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Hint: The formula for calculating spin only magnetic moment of coordination compounds is $\mu = \sqrt {n\left( {n + 2} \right)} $ BM. The value 3.9BM is obtained by complexes with a number of unpaired electrons are 3.

Complete Step-by-step solution:
Magnetic moment is the measure of magnetic property of compounds. To know the magnetic moment value let us use the formula for spin only magnetic moment. Let us find out which metal ions pair has three unpaired electrons each. The number of unpaired electrons in a complex depends on the ligands (both outer and inner sphere). The formula for magnetic moment is \[\mu = \sqrt {n\left( {n + 2} \right)} \] where, $\mu $ -Magnetic moment, n- Number of unpaired electrons in the valence shell of the metal ion
The option (A) chromium (II) ion (Z=22) has 4 unpaired electrons as two electrons are removed from neutral electronic configuration $\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^5}{\text{4}}{{\text{s}}^1}$ and manganese (II) ion (Z= 23) gets 5unpaired electrons upon removal of two electrons from z=25. In option (B) Vanadium (II) ion (Z=21) has three unpaired d-electrons and Cobalt (II) has 7 electrons in the d-orbital in which three electrons remain unpaired. In the option Vanadium (II) ion has three electrons and Iron (II) gets four unpaired d-electrons upon removal of two s-electrons. In option (D) cobalt (II) has 7 electrons in the d-orbital where four are paired ones and three are unpaired.

So, the answer for the above question is option (B) ${{\text{V}}^{{\text{2 + }}}}$ and ${\text{C}}{{\text{o}}^{2 + }}$

Note: As the ligands are both weak field ligands they are high spin and pairing of electrons does not take place. The electrons are filled in d-orbital with one electron each in ${{\text{t}}_{{\text{2g,}}}}{{\text{e}}_{\text{g}}}$ orbitals and then the pairing of the extra electrons takes place.