The transition elements have an unparalleled tendency to form coordination compounds with the Lewis bases, which are called as ligands.  

CO³⁺ + 6NH₃ ————→ [CO(NH₃)₆]³⁺

Fe²⁺ + 6CN^– ————→ [Fe(CN)₆]^4–

s and p block elements form very few complexes. The reason transition elements are so good at forming complex is that they have small, highly charged ions and have vacant low energy orbitals to accept lone pairs of electrons donated by ligands.

Size of Atoms and Ions  

The covalent radii of the elements decrease from left to right across a row in the transition series. This is because of the poor screening by the d electrons due to which, the nuclear charge attracts all of the electrons more strongly, hence a contraction in size occurs.

The elements in the first group in the d-block show the excepted increase (due to the addition of extra shell) in size Sc → Y → La. However in the subsequent groups there is an increase between first and second members, but hardly any increase between second and third elements. This is due to lanthanide contraction (discussed in f-block elements).

Exercise 4:Why have the transition elements high enthalpy of hydration?
Colour
Many compounds of transition elements are coloured in contrasts to those of s and p block elements.

In compound state due to the surrounding groups (ligands), the d-orbitals of transition elements are not degenerate but split into two groups of different energy. Thus it is possible to promote electrons from one group to another group. This corresponds to fairly small amount of energy difference and so light is absorbed in visible region. Some compounds of transition metals are white, for example ZnSO₄ and TiO₂. In these compounds it is not possible to promote the electrons within the d-level.

Illustration 7:Why Zn²⁺ salts are white while Ni²⁺ salts are blue?
Solution:Zn²⁺ has completely filled d-orbitals (3d¹⁰) while Ni²⁺ has incompletely filled d-orbitals (3d⁸).
 

Illustration 8:Why Zn²⁺ salts are white while Cu²⁺ salts are blue?

Illustration 9:Giving reasons indicate which one of the following would be coloured?

Cu⁺, VO²⁺, Sc³⁺, Ni²⁺ (At. Nos Cu = 29, V = 23, Sc = 21, Ni = 28)

Solution:Ni²⁺ due to incompletely filled d-orbitals.

Exercise 5:[Ti(H₂O)₆]³⁺ is coloured while [Sc(H₂O)₆]³⁺ is colourless. Explain.