Which of the following elements shows the inert pair effect?

The inert pair effect refers to the tendency of the two outermost s-electrons (usually from the ns² subshell) to remain non-bonding or inert during chemical reactions. This effect becomes more pronounced as we move down a group in the p-block elements of the periodic table. It is primarily observed in the heavier elements of groups 13, 14, and 15.

The inert pair effect arises due to the poor shielding of the nuclear charge by inner d- and f-electrons, which causes the ns² electrons to be held more tightly by the nucleus. As a result, these electrons do not participate readily in bonding, leading to lower oxidation states becoming more stable for heavier elements.

In Group 14, elements such as tin (Sn) and lead (Pb) show this effect clearly. Tin can exhibit oxidation states of +2 and +4, while lead shows +2 as its more stable state due to a stronger inert pair effect. In contrast, lighter elements of the same group like carbon and silicon do not show this effect because their ns² electrons are easily involved in bonding due to weaker nuclear attraction.

For example, tin forms both SnCl₂ (tin(II) chloride) and SnCl₄ (tin(IV) chloride), but the +2 oxidation state becomes relatively more stable in heavier congeners like lead. This stability trend down the group reflects the increasing inertness of the s-electron pair.

Therefore, among the given elements — boron, carbon, silicon, and tin — the element that exhibits the inert pair effect is tin (Sn). This concept is significant in understanding the variation of oxidation states, metallic character, and chemical reactivity across the p-block elements in inorganic chemistry.