Question

Which is not correctly matched? (1) Basic strength of oxides \(\mathrm{Cs}_{2} \mathrm{O}<\mathrm{Rb}_{2} \mathrm{O}<\mathrm{K}_{2} \mathrm{O}<\mathrm{Na}_{2} \mathrm{O}<\mathrm{Li}_{2} \mathrm{O}\) (2) Stability of peroxides \(\mathrm{Na}_{2} \mathrm{O}_{2}<\mathrm{K}_{2} \mathrm{O}_{2}<\mathrm{Rb}_{2} \mathrm{O}_{2}<\mathrm{Cs}_{2} \mathrm{O}_{2}\) (3) Stability of bicarbonates \(\mathrm{LiHCO}_{3}<\mathrm{NaHCO}_{3}<\mathrm{KHCO}_{3}<\mathrm{RbHCO}_{3}<\) \(\mathrm{CsHCO}_{3}\) (4) Melting point \(\mathrm{NaF}<\mathrm{NaCl}<\mathrm{NaBr}<\mathrm{Nal}\) (1) 1 and 4 (2) 1 and 3 (3) 1 and 2 (4) 2 and 3

Short answer

The correct answers are options 1 and 4.

Step-by-step solution

Understanding Chemical Trends

Identify the chemical trends for each pair and understand their expected behavior.

Check Basic Strength of Oxides (Option 1)

Basic strength of oxides increases up the group in the periodic table. Thus, the correct order should decrease from \(\text{Li}_2 \text{O}\) to \(\text{Cs}_2 \text{O}\). The given order is incorrect.

Check Stability of Peroxides (Option 2)

Stability of peroxides increases down the group. The given order is correct since \(\text{Na}_2 \text{O}_2 \lt \text{K}_2 \text{O}_2 \lt \text{Rb}_2 \text{O}_2 \lt \text{Cs}_2 \text{O}_2\).

Check Stability of Bicarbonates (Option 3)

Stability of bicarbonates increases down the group. The given order is correct since \(\text{LiHCO}_3 \lt \text{NaHCO}_3 \lt \text{KHCO}_3 \lt \text{RbHCO}_3 \lt \text{CsHCO}_3\).

Check Melting Points (Option 4)

Melting point decreases down the group of halides due to the increase in ionic size. The correct order should not be \(\text{NaF} \lt \text{NaCl} \lt \text{NaBr} \lt \text{NaI}\); \(\text{NaF} \gt \text{NaCl} \gt \text{NaBr} \gt \text{NaI}\) is correct. So the given order is incorrect.

Key concepts

Basic strength of oxides

The basic strength of oxides is an important property to understand, especially for students preparing for the IIT JEE Chemistry exam. Basic oxides are those that react with acids to form salt and water, commonly found in alkali metals. In the periodic table, the basicity of oxides generally increases up the group. This behavior is primarily due to the decreasing size of the metal cations as we go up the group, which increases the oxide ion's attraction to the cation, making the metal oxide more basic. With these points in mind, the order \(\text{Cs}_2 \text{O} < \text{Rb}_2 \text{O} < \text{K}_2 \text{O} < \text{Na}_2 \text{O} < \text{Li}_2 \text{O}\) given in the exercise is incorrect. The correct trend should start from lithium oxide as the most basic to cesium oxide as the least basic.

Stability of peroxides

Peroxides contain the \(\text{O}_2^{2-}\) ion and are found in compounds with alkali metals. The stability of peroxides increases as you move down the group in the periodic table. This trend emerges because larger cations better stabilize the large peroxide ion due to their low charge density. Hence, in alkali metals, you will observe the order \(\text{Na}_2 \text{O}_2 < \text{K}_2 \text{O}_2 < \text{Rb}_2 \text{O}_2 < \text{Cs}_2 \text{O}_2\). This means cesium peroxide is the most stable while sodium peroxide is the least stable. The given exercise correctly represents this order.

Stability of bicarbonates

Bicarbonates or hydrogen carbonates \(\text{HCO}_3^{-}\) in their stability also follow similar periodic trends. As you move down the group in the alkali metal series, the stability of bicarbonates increases. This trend occurs because larger metal cations can better stabilize the bicarbonate anion. In simpler terms, the larger the ion, the less polarizing it is, making the crystal structure more stable. The given order \(\text{LiHCO}_3 < \text{NaHCO}_3 < \text{KHCO}_3 < \text{RbHCO}_3 < \text{CsHCO}_3\) accurately reflects this increasing stability from lithium bicarbonate to cesium bicarbonate.

Melting point trends

Melting points of alkali metal halides follow a specific trend due to variations in ionic size and lattice energy. Melting points generally decrease down a group as the ionic size increases, which results in weaker ionic bonds in the lattice structure. A correct order for melting points should reflect the following: \(\text{NaF} > \text{NaCl} > \text{NaBr} > \text{NaI}\). This order shows that \(\text{NaF}\) has the highest melting point due to the smaller and more tightly held fluoride ion, while \(\text{NaI}\) has the lowest due to the larger iodide ion. The given exercise incorrectly states \(\text{NaF} < \text{NaCl} < \text{NaBr} < \text{NaI}\), which does not align with the expected trend.