Question

Why do the transition elements have higher enthalpies of atomization? In 3d series (Sc to Zn), which element has the lowest enthalpy of atomization and why?

Answer 1

Hint: In case of transition elements due to presence of electrons at d orbitals , which is closer to the outermost shell of the metal. They show variable oxidation state. With increasing the number of electrons of the d orbitals (up to 5 electrons), the metallic bonds of their solids get stronger.

Complete step by step answer:
From Sc ($$\text{Z = 21}$$) to Zn($$\text{Z = 30}$$) the number of d orbital electrons increases. But the metallic bond strength of the transition metal depends upon the unpaired electrons. Higher the unpaired electrons in d orbital, higher will be the bond strength of the transition metal.
Now, after Mn the unpaired electron number decreases, as a result bond strength decreases.
Now with increasing the bond strength , higher will be the energy required for atomization.
In case of Mn due to the highest number of unpaired electrons the atomization energy for Mn is also highest. On the other hand, in case of Zn due to all paired d electrons the atomization energy is least for Zn.

Note:
When a transition metal forms complex, due to different extent of interaction with ligands the d orbitals divided into several energy states.d-d transition of the transition metals is basically excitation of electrons by absorbing energy from lower energy level to higher energy level. After these transitions, electrons immediately come back with emission of some amount of radiation, for which the complex shows colour. This colour is the complementary colour of the light absorbed in excitation.