Explanation

(d)
V(23$) = \lbrack$Ar$\rbrack\ 3d^{3},\ 4s^{2}$
$V^{3 +}$=[ Ar] $3d^{2}\ ,4s^{0}$ (two unpaired electrons)
Cr(24)=[ Ar] $3d^{5}\ ,4s^{1}$
${Cr}^{3 +}$=[ Ar] $3d^{3}\ ,4s^{0}$ (three unpaired electrons)
Co(27)= [ Ar] $3d^{7}\ ,4s^{2}$
${CO}^{3 +}$=[ Ar] $3d^{7}\ ,4s^{0}$ (three unpaired electrons)
Sc(21)= [ Ar] $3d^{1}\ ,4s^{2}$
${Sc}^{3 +}$=[ Ar] $3d^{0}\ ,4s^{2}$ (no unpaired electrons)
Thus, in ${Sc}^{3 +}$ no unpaired $d$-electron is present.
Hence, no $d - d$ transition is possible and it is colourless ion.