Question

Zn and Cd do not show variable valency because:
A) They are soft
B) Their d-subshells are complete
C) They have only two electrons in outermost subshells
D) Their d-subshells are incomplete

Answer 1

Hint: Zn is the last element in the first transition series of d-block elements and Cd is the last element in the second transition series of d-block elements.
- Valency shown by the element depends on the electronic configuration of the element.

Complete step by step answer:
- So in the question Zn and Cd are the elements given and both of them show a valency of +2. And the reason for why they do not show variable valency is to be stated.
- We know that Zn and Cd are d-block elements, i.e. the electrons are filled in the valence d-subshell.
All the d-blocks elements exhibit various valencies or have variable oxidation number except for the first and last elements in a transition series.
- Generally d-block elements show various oxidation states and the reason for this statement is that the ns orbital and (n-1) d orbitals are almost similar in energy, there is only a small difference in energy between the s and d orbitals. Hence the electrons from the ns orbitals and (n-1) d orbitals may be used for the bond formation in compounds.
- The valency is the measure of how much electrons it can share or donate while forming bonds.
The valency of an element is related to the electronic configuration of an element,

Here let’s write the valency of the given elements.
Zinc (Zn), has an atomic number 30 and it is placed in the extreme right side of the first transition series of the d-block.
The electronic configuration of Zn is written as:
$Zn = \left[ Ar \right]3{{d}^{10}}4{{s}^{2}}$
The Zn possess a complete d-orbital, which is highly stable and hence only the electrons from the 4s orbitals are used for the compound formation. Half-filled and completely filled d-orbitals have special stability.
So only the 2 electrons in 4s is donated by Zn and shows a valency of +2.
$Zn\to Z{{n}^{2+}}+2{{e}^{-}}$
Cadmium is the last element in the second series of d-block elements with an atomic number 48.
The electronic configuration of Cd is written as:
$Cd=\left[ Kr \right]4{{d}^{10}}5{{s}^{2}}$
- Here also the Cd is having the complete d – orbital, which possesses extra stability. And only the electrons from 5s can be given for bonding and exist with the +2 valency.
$Cd\to C{{d}^{2+}}+2{{e}^{-}}$
- Therefore, the reason for Zn and Cd not showing various oxidation states is due to the presence of completely filled d-orbital.
So, the correct answer is “Option B”.

Note: Even though the Zn and Cd are called d-block elements but technically they can’t be called as d-block elements as the removal of electrons occurs from the 4s orbital with less energy than the 3d orbitals. While considering the cases of oxidation states of d-block elements the complete filled and half-filled special stability of d-orbitals should be taken into consideration and the electrons should be distributed with care while writing the electronic configuration.