Group 16 Elements

Group 16 elements, also known as the chalcogens, generally exhibit oxidation states of -2, +2, +4, and +6. Oxygen ( ) is the most electronegative element in this group and readily forms the -2 oxidation state. As you move down the group, the ability of the elements to form the -2 oxidation state decreases due to increasing atomic size and decreasing electronegativity.

Among the options given:

• Oxygen ( ): Commonly shows a -2 oxidation state.
• Selenium ( ): Can show a -2 oxidation state.
• Tellurium ( ): Also capable of a -2 oxidation state.
• Polonium ( ): Generally does not form a -2 oxidation state due to its metallic nature and instability.

The correct answer is that Polonium ( ) does not typically show a -2 oxidation state due to its position at the bottom of Group 16 and its tendency to exhibit metallic properties rather than non-metallic ones.

Given below are two statements:

Statement I: The boiling point of hydrides of Group 16 elements follow the order H2O > H2Te > H2Se > H2S.

Statement II: On the basis of molecular mass, H2O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H2O, it has higher boiling point.

In the light of the above statements, choose the correct answer from the options given below:

Explanation:

Statement I:

The boiling points of the hydrides of Group 16 elements generally increase with increasing molecular mass. However, water (H2O) is an exception due to strong hydrogen bonding. The correct order is H2O > H2Te > H2Se > H2S. Therefore, Statement I is correct.

Statement II:

As stated, based on molecular mass alone, H2O should have the lowest boiling point. However, the extensive intermolecular hydrogen bonding in water significantly increases its boiling point, making it higher than the other hydrides in the group. Thus, Statement II is also correct.

Conclusion:

Both Statement I and Statement II are True.