The correct option is C [Ti(H2O)6]3+
A) Ti has the electronic configuration [Ar]3d2. In this complex, it is present as Ti3+ which has one unpaired electron in 3d orbital. It undergoes d2sp3 (octahedral) hybridization to accommodate 6 pairs of water.
So, the complex is octahedral in shape, will be coloured and paramagnetic due to the presence of one unpaired electron.
B) Sc has the electronic configuration [Ar]3d1. Since, it is present in the complex as Sc3+, it does not have any unpaired electron. It undergoes d2sp3 hybridization to accommodate 6 pairs of electrons from water. So, it is octahedral in shape, colourless and diamagnetic due to the absence of unpaired electrons.
C) In this complex, Zn is present as Zn2+ and there is no unpaired electron in the complex. So, it is colourless.
D) In this complex, Ti is present as Ti0 and there is no unpaired electron in the complex. So, it is colourless.