Match of the following column.
(metal ion of 3d - series ) (characteristics )

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${Ni}^{2 +}$

${Cr}^{2 +}$

$V^{2 +}$

${Ti}^{4 +}$

Produce blue aq. solution

Half filled $t_{2 g}$ orbitals 10 octahedral complex

Diamagnetic ion

Calculated µ = 2.84 B.M. (spin only) "

answer is D, A, B, C.

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Detailed Solution

The picture for ${Ni}^{2 +}$ Determined = 2.84 B.M. (spin only) The magnetic moment of ${Ni}^{2 +}$ in aqueous solution with respect to spin alone is n(n+2) =2(2+2) =8 BM.
When combined with aqueous potassium fluoride, aqueous copper sulphate solution (blue in colour) produces two distinct results: I a vivid green solution; and (ii) a green precipitate.
In complex crystal fields, degenerate atomic orbitals are translated into $t_{2 g}$ and eg in octahedral geometry, but in tetrahedral geometry, e and $t_{2}$ .
${Ti}^{4 +}$ is diamagnetic and contains no unpaired electrons.