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Question: Why is \[B{i^{3 + }}\] more stable than \[B{i^{5 + }}\] ? A. Due to inert pair effect B. Due to ...

Why is Bi3+B{i^{3 + }} more stable than Bi5+B{i^{5 + }} ?
A. Due to inert pair effect
B. Due to inert gas electron configuration
C. Due to half-filled configuration
D. Due to lanthanide contraction

Explanation

Solution

As we go down the group in a periodic table, stability of higher oxidation states decreases due to more penetration of s-orbital in the nucleus. Therefore, the lower oxidation states become more stable than the higher ones as we go down the group.

Complete step-by-step solution:
The non-participation of the two s-electrons in bonding due to the high energy needed for unpairing them or exciting them from ground state, is called inert pair effect. It is most common among groups 13, 14 and 15 elements. When the s-electrons remain paired, the lower oxidation state is more favoured than the characteristic oxidation state of the group.
The electrons present in the outermost s-orbital do not take part in the formation of a chemical bond. The tendency of these two s-electrons to stay inert and not react is called inert pair effect. The main valency is the number of electrons present in the p-subshell of outermost electrons.
These types of elements have inner d or f electrons. Due to the poor shielding of d and f orbitals, the effective nuclear charge on outer s-orbital electrons acts strongly and are strongly pulled by the nucleus. This prevents them from taking part in the bonding and so they become inert.
In case of bismuth, its electronic configuration is [Xe]4f145d106s26p3[Xe]4{f^{14}}5{d^{10}}6{s^2}6{p^3}. It loses three electrons from p-orbital and becomes Bi3+B{i^{3 + }} and if it loses two more electrons of s-orbital to become Bi5+B{i^{5 + }}. But due to inert pair effect, loss of s-electrons is not feasible and so Bi3+B{i^{3 + }} more stable than Bi5+B{i^{5 + }}.

Hence, the correct option is (A).

Note: The elements present in the bottom of the group have f-subshell electrons present that are diffused due to their shape of the f-orbitals. They effectively shield the s-electrons from the pull of the nucleus and make it inert for bonding.