The electronic configuration of the given elements is

$S{c^{3 + }}(18)\, = \,1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,3{d^0}\,4{s^0}$   no unpaired $e^-$

$T{i^{4 + }}(18)\, = \,1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,3{d^0}\,4{s^0}$   no unpaired $e^-$

${V^{3 + }}(18)\, = \,1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,3{d^2}\,4{s^0}$  - Two unpaired $e^-$.

$Z{n^{2 + }}(28)\, = \,1{s^2}\,2{s^2}\,2{p^6}\,3{s^2}\,3{p^6}\,3{d^{10}}\,4{s^0}$   no unpaired $e^-$  

hence $[V(NH_3)_6]^{3+}$ will moost likely absorb visible light.