Explain color & complex formation tendency of transition metal ions?

Hint: Color in transition metal compounds is generally due to electronic transitions of two principal types:
1.Charge transfer transitions
2.d-d transitions

Complete step by step answer:
Unpaired electron of d-d transition is the reason for the color of transition. Large number of complexes form transition elements.
d-d transition of unpaired electrons is shown by the given complex for the explanation.
 ${e_g}$ ${[Ti{({H_2}O)_6}]^{ + 3}}$ , Ti atom is reacted with six moles of water and forms a complex. Here Ti atom has +3 oxidation state. The electronic configuration of the outer state is $3{d^1}$ . Ti has one unpaired electron which is excited from ${t_{2g}}$ to ${e_g}$ by absorbing yellow light and appears violet coloured.
Large number of complexes is the reason for transition elements. Small size cations have High effective nuclear charge. They accept unpaired electrons from ligands and form coordinate bonds. The example of this complexation is $Ni{(CO)_4}$ .
Transition metals have great tendencies. This tendency is useful to the formation of complexes because of vacant d-orbitals.
The factors to form complex are vacant orbitals, small size of cations and higher nuclear charge.
Here charge transfer from the ligands to transition metal causes color and formation of complex.

Note:
Electron transition between d-orbitals comes from the colored transition metals. If the substance is colorless then the d-orbitals are full or empty and it is not a transition element. Due to the unpaired electron of d-orbital is the reason for the transition metal which forms the color. The transition metal absorbs visible light and causes colored nature. Here incomplete valence shell orbitals form coordination complexes by the transition metal ions.