Problem 120 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following: List I 1\. thermal stability 2\. acidic nature 3\. boiling points 4\. \(\angle \mathrm{MMH}\) bond angle List II (i) \(\mathrm{H}_{2} \mathrm{~S}<\mathrm{H}_{2} \mathrm{Se}<\mathrm{H}_{2} \mathrm{Te}<\mathrm{H}_{2} \mathrm{PO}\) (ii) \(\mathrm{H}_{2} \mathrm{O}>\mathrm{H}_{2} \mathrm{~S}>\mathrm{H}_{2} \mathrm{Se}>\mathrm{H}_{2} \mathrm{Te}\) (iii) \(\mathrm{H}_{2} \mathrm{~S}<\mathrm{H}_{2} \mathrm{Se}<\mathrm{H}_{2} \mathrm{Te}<\mathrm{H}_{2} \mathrm{O}\) (iv) \(\mathrm{H}_{2} \mathrm{~S}>\mathrm{H}_{2} \mathrm{Se}>\mathrm{H}_{2} \mathrm{Te}=\mathrm{H}_{2} \mathrm{PO}\) The correct matching is 1 \(\begin{array}{lll}2 & 3 & 4\end{array}\) (a) (iii) (ii) (i) (iv) (b) (iv) (i) (ii) (iii) (c) (ii) (i) (iii) (iv) (d) (i) (ii) (iii) (iv)

Step-by-Step Solution

Answer

(d)

Step 1: Understand the Properties

We need to match the statements from List I about the properties of hydrides (thermal stability, acidic nature, boiling points, and \(\angle \text{MMH}\) bond angle) with the correct ordering from List II. The properties depend on molecular size, electronegativity, and intermolecular forces.

Step 2: Determine Thermal Stability Order

Thermal stability decreases down the group as the bond strength decreases. Thus, for List I point 1, thermal stability, the correct order is \(\mathrm{H}_2\mathrm{O} > \mathrm{H}_2\mathrm{S} > \mathrm{H}_2\mathrm{Se} > \mathrm{H}_2\mathrm{Te}\), which matches with List II point (ii).

Step 3: Determine Acidic Nature Order

Acidic strength generally increases as the hydride bond becomes weaker. Thus, for List I point 2, acidic nature, the correct order is \(\mathrm{H}_2\mathrm{S}<\mathrm{H}_2\mathrm{Se}<\mathrm{H}_2\mathrm{Te}<\mathrm{H}_2\mathrm{PO}\), which matches with List II point (i).

Step 4: Determine Boiling Points Order

Boiling points are influenced by molecular mass and hydrogen bonding. Since \(\mathrm{H}_2\mathrm{O}\) has strong hydrogen bonds, while the others increase with molecular mass, the order \(\mathrm{H}_2\mathrm{Se}<\mathrm{H}_2\mathrm{S}<\mathrm{H}_2\mathrm{O}<\mathrm{H}_2\mathrm{Te}\) is incorrect. Instead, \(\mathrm{H}_2\mathrm{O} > \mathrm{H}_2\mathrm{S} > \mathrm{H}_2\mathrm{Se} > \mathrm{H}_2\mathrm{Te}\) matches List II point (iii).

Step 5: Determine \(\angle \mathrm{MMH}\) Bond Angle Order

Bond angle decreases with increasing atomic size, from \(\mathrm{H}_2\mathrm{O}\) to \(\mathrm{H}_2\mathrm{Te}\). Thus, \(\mathrm{H}_2\mathrm{S} > \mathrm{H}_2\mathrm{Se} > \mathrm{H}_2\mathrm{Te} = \mathrm{H}_2\mathrm{PO}\) matches with List II point (iv).

Step 6: Final Matching

By analyzing each property, the correct matches are: 1-ii (thermal stability), 2-i (acidic nature), 3-iii (boiling points), 4-iv (bond angle). According to the offered options, this corresponds to option (d).

Key Concepts

Thermal Stability

In chemical compounds, thermal stability refers to the ability of a substance to remain unchanged when exposed to heat. For hydrides, thermal stability decreases as you move down the group in the periodic table. This happens because the bond between the hydrogen and the central atom becomes weaker.
Hence, the molecule breaks apart more easily with heat. In hydrides like \(\text{H}_2\text{O}\), \(\text{H}_2\text{S}\), \(\text{H}_2\text{Se}\), and \(\text{H}_2\text{Te}\), the order of thermal stability from greatest to weakest is: \[\text{H}_2\text{O} > \text{H}_2\text{S} > \text{H}_2\text{Se} > \text{H}_2\text{Te}\].

• \(\text{H}_2\text{O}\): Most stable due to strong hydrogen-oxygen bonding.
• \(\text{H}_2\text{S}\), \(\text{H}_2\text{Se}\): Less stable with weaker bonds.
• \(\text{H}_2\text{Te}\): Least stable, decomposes easily on heating.

Acidic Nature

Acidic nature in hydrides pertains to their ability to donate a proton (H⁺) during a reaction. This property increases as you move down the group because the bond between hydrogen and the central atom weakens. Therefore, hydrides release H⁺ ions more readily.
The order for acidic nature in the Group 16 hydrides is: \[\text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te} < \text{H}_2\text{PO}\].

• \(\text{H}_2\text{S}\): Weak acid, bond is relatively strong.
• \(\text{H}_2\text{Se}\), \(\text{H}_2\text{Te}\): Moderate acids, bonds weaken.
• \(\text{H}_2\text{PO}\): Strong acid, bonds are weakest making proton donation easier.

Boiling Points

Boiling points are influenced by molecular mass and interatomic interactions like hydrogen bonding. In hydrides, an increase in molecular mass often results in higher boiling points. However, hydrogen bonds, which are strong, can notably raise boiling points beyond that expected just from mass.
The order of boiling points for the hydrides of Group 16 is: \[\text{H}_2\text{O} > \text{H}_2\text{S} > \text{H}_2\text{Se} > \text{H}_2\text{Te}\].

• \(\text{H}_2\text{O}\): Exceptionally high boiling point due to strong hydrogen bonding.
• \(\text{H}_2\text{S}\): Smaller molecular weight than \(\text{H}_2\text{Se}\), but lacks strong hydrogen bonds.
• \(\text{H}_2\text{Se}\), \(\text{H}_2\text{Te}\): Heavier mass, higher boiling points than \(\text{H}_2\text{S}\).

Bond Angles

In molecular chemistry, bond angles are shaped by atomic size and electron repulsion forces. For the hydrides of Group 16, as the atom size increases from oxygen to tellurium, the bond angle decreases. This is because larger atoms have more electron cloud, pushing apart the bonds and reducing the internal angle.
The sequence for bond angles from widest to narrowest is: \[\text{H}_2\text{O} > \text{H}_2\text{S} > \text{H}_2\text{Se} > \text{H}_2\text{Te}\].

• \(\text{H}_2\text{O}\): Largest bond angle, due to small oxygen atom and strong electron pair repulsion.
• \(\text{H}_2\text{S}\), \(\text{H}_2\text{Se}\): Medium-sized bond angles, with less repulsion as atom size increases.
• \(\text{H}_2\text{Te}\): Smallest bond angle, as the large tellurium atom reduces impact of electron pair forces.