Question

Which one of the following is correct about stability of the given ions? (a) \(\mathrm{Sn}^{4+}>\mathrm{Sn}^{2+}\) (b) \(\mathrm{Pb}^{4+}>\mathrm{Pb}^{2+}\) (c) \(\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}\) (d) \(\mathrm{Si}^{2+}>\mathrm{Si}^{4+}\)

Short answer

Option (c) \\(\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}\\) is correct.

Step-by-step solution

Understanding the Concept of Stability

Consider the oxidation states of these elements and their positions in the periodic table. Elements tend to be more stable in oxidation states that correspond with a full or half-full outer electron shell configuration.

Examining Tin's Oxidation States

Tin (Sn) is located in Group 14. It can exist in two common oxidation states: +2 and +4. Among these, Sn prefers to lose 4 electrons to achieve a more stable electronic configuration similar to that of noble gases, hence \(\mathrm{Sn}^{4+}\) is more stable than \(\mathrm{Sn}^{2+}\).

Analyzing Lead's Oxidation States

Lead (Pb) is also in Group 14 and can exhibit both +2 and +4 oxidation states. However, due to the inert pair effect, \(\mathrm{Pb}^{2+}\) is more stable than \(\mathrm{Pb}^{4+}\). The inert pair effect leads to a reluctance of Pb to lose all its s-electrons.

Evaluating Silicon's Oxidation States

Silicon (Si), another Group 14 element, can exist in +2 and +4 states. However, \(\mathrm{Si}^{4+}\) is the preferred and more stable state as it achieves a full p-orbital set similar to that of a noble gas.

Verification

Analyze the given options based on the steps above. The correct comparison is where \(\mathrm{Pb}^{2+}\) is more stable than \(\mathrm{Pb}^{4+}\), which corresponds to option (c).

Key concepts

Stability of Ions

The stability of ions is a crucial concept in chemistry, helping to explain why certain ions are more prevalent and persistent in nature than others. Ion stability often correlates with their electron configuration, particularly when it involves achieving a full or half-full electron shell.
For elements like Sn and Pb, which belong to Group 14 of the periodic table, they can exhibit variable oxidation states, specifically +2 and +4. However, a key factor in determining the stability of these ions lies in how closely their electronic configurations approach those of noble gases.
Generally, ions where the atom’s highest energy orbitals are filled tend to be more stable. For instance, \(\text{Sn}^{4+}\) is more stable than \(\text{Sn}^{2+}\), as Sn losing 4 electrons results in a noble gas configuration. In contrast, the stabilization of \(\text{Pb}^{2+}\) over \(\text{Pb}^{4+}\) is an example of how other effects, such as the inert pair effect, come into play.

Periodic Table

The periodic table is an essential tool in understanding elemental behavior, including ion stability. It organizes elements based on increasing atomic number and recurring chemical properties. These properties are chiefly determined by an element's electron configuration.
Group 14 of the periodic table, where Sn, Pb, and Si belong, typically features oxidation states of +2 and +4. The position of these elements helps predict their chemical reactivity and stable ionic forms.
For instance, moving down the group from Sn to Pb, the stability trend for oxidation states shifts due to the increasing influence of the inert pair effect. This is why \(\text{Pb}^{2+}\) is more stable compared to \(\text{Pb}^{4+}\), contrasting the stability pattern seen in \(\text{Sn}\). Silicon \(\text{Si}\), meanwhile, conforms more to the expected +4 oxidation state, consistent with its position and electronic structure.

Inert Pair Effect

The inert pair effect is a phenomenon observed in heavier elements of the p-block, where the s-electrons tend to resist participation in bonding. This effect influences the stability of oxidation states, especially in elements like lead (Pb).
As a result, \(\text{Pb}^{2+}\) becomes more stable than \(\text{Pb}^{4+}\) due to the reluctance of the +4 oxidation state to form, even though, nominally, losing four electrons would lead to a noble gas-like configuration.
This effect becomes more pronounced downward the group as atomic size increases and the electrons become more "shielded" from the nucleus by other electrons, thus stabilizing the lower oxidation state due to lesser involvement of ns electrons. Hence, inert pair effect exemplifies another dimension of understanding ionization and stability trends across the periodic table.