Question

The stable oxidation state of Bismuth is:

Answer 1

Hint: The stability of higher oxidation states reduces as we move down the periodic table due to greater s-orbital penetration in the nucleus. As we progress through the group, the lower oxidation states become more stable than the higher ones.

Complete Step by Step Solution:
The inert pair effect describes the non-participation of the two s-electrons in bonding due to the large energy required to excite them from their ground state. It's most frequent among elements in groups 13, 14, and 15. The lower oxidation state is preferred over the group's characteristic oxidation state when the s-electrons stay coupled.

The electrons at the outermost s-orbital do not participate in chemical bond formation. The inert pair effect describes the tendency of these two s-electrons to remain inactive and not react. The number of electrons in the p-subshell of the outermost electrons is the major valency.

Inner d or f electrons are found in these elements. The effective nuclear charge on outer s-orbital electrons operates aggressively and is substantially pulled by the nucleus due to the inadequate shielding of d and f orbitals. They are rendered inactive as a result of their inability to participate in the bonding process.

Bismuth has an electrical configuration that is \[\left[ {Xe} \right]4{f^{14}}5{d^{10}}6{s^2}6{p^3}\]. It becomes \[B{i^{3 + }}\] after losing three electrons from the p-orbital, and \[B{i^{5 + }}\] after losing two additional electrons from the s-orbital. Loss of s-electrons is not possible due to the inert pair effect, hence \[B{i^{3 + }}\] is more stable than \[B{i^{5 + }}\].

Therefore, \[B{i^{3 + }}\] is more stable than \[B{i^{5 + }}\].

Note: Because of the geometry of the f-orbitals, the elements near the bottom of the group have f-subshell electrons that are diffused. They successfully shelter the s-electrons from the nucleus's attraction and render them bonding inactive.