Question

Which of the following factors favors the formation of complexes by d-block elements?
(A) Small size
(B) High nuclear charge
(C) Presence of low energy vacant orbitals to accept lone pair of electrons donated by ligands
(D) All of the above

Answer 1

Hint: The d block elements are found in the group \[3,4,5,6,7,8,9,10,11\] and $12$ of the periodic table. These are also known as transition metals. The d orbital is filled with the electronic shell $n - 1$. There are a total 40 d block elements.

Complete step by step solution:
Now, let us look at the periodic table first.

The elements which lie in the middle of the group II A and group IIB elements in the periodic table are d block elements. They are known as transition elements as they are the elements that lie between the metals and non-metals of the periodic table.
These transition metals usually have high melting and boiling points. This is mainly because they have completely filled d orbitals because of which no unpaired electron is available. Because of the unavailability of unpaired electrons, these metals do not undergo covalent bonding.
The d block elements are small in size and generally have high electron positive density.
They consist of ($n - 1$) d free orbitals to accept the free electrons from the ligand and hence, form complexes easily.

Hence option D is correct.

Note: The d block elements show color because of d-d transition. When visible light falls on the transition metal compound or ion, the unpaired electrons present in the lower energy d orbitals get promoted to higher energy d –orbitals, and this is known as d-d transition which occurs due to absorption of visible light. Since, the energy involved in d-d transition is quantized, only a definite wavelength gets absorbed and the remaining wavelength gets transmitted. Therefore, the transmitted light shows color complementary to the absorbed colour.