Complete the reactions of Sn(II) and Sn(IV), and be sure that the reactions are balanced. Do not include the phases (liquid, aqueous, etc.). If no reaction occurs, leave the products side of the equation completed reaction: SnBr2+PbBr4⟶
completed reaction: SnBr4+PbBr2⟶
Select the statements that are true about the reactions.
A. PbBr4 is more stable than PbBr2.
B. The inert‑pair effect renders Sn(II) as the more stable oxidation state of tin.
C. Sn(IV) is the most stable oxidation state of tin.
D. The inert‑pair effect renders Pb(II) as the more stable oxidation state of lead.

The Inert Pair effect renders Pb(II) as the more stable oxidation state of lead

Explanation:

SnBr4 + PbBr2 ---> SnBr2 + PbBr4

SnBr2 + PbBr4 >

The Inert pair effect is mostly observed between group 15-17 in the periodic table. It leads to stability of the lower oxidation state of an element.

The reason for the Inert pair effect is that the s electrons become Inert due to poor shielding of the d and f-electrons. The Inert pair effect is a tendency of the s electrons not to participate in bonding (remain an Inert pair).

Owing to the Inert pair effect, Pb II is more stable than Pb IV

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.