Question

Which hydride is not a reducing agent among the following?

• A. ${\mathrm{H}}_2\mathrm{Se}$
• B. ${\mathrm{H}}_2\mathrm{O}$
• C. ${\mathrm{H}}_2\mathrm{Te}$
• D. ${\mathrm{H}}_2\mathrm{S}$

Answer 1

The correct option is B

${\mathrm{H}}_2\mathrm{O}$

The explanation for the correct option:

B) ${\mathrm{H}}_2\mathrm{O}$

• Higher the tendency to lose the electrons, the higher the reducing character.
• As we go down from top to bottom in a group, the tendency to lose electrons increases due to the increase in atomic size.
• The electrons are added to different shells and hence are no longer under the influence of the nucleus. This increases the reducing power.
• Applying the above concept, ${\mathrm{H}}_2\mathrm{O}$ has the least reducing power as oxygen is the first element in group-$16$, and has the smallest size.
• It has the least tendency to lose electrons and hence cannot act as a reducing agent.

The explanation for the incorrect options:

A) ${\mathrm{H}}_2\mathrm{Se}$

• Selenium lies below oxygen and sulfur and has a considerable atomic size.
• The tendency to lose electrons is more as the outer electrons are not under the influence of the nucleus and can be lost easily.
• It can act as a reducing agent as it has considerable reducing power.

C) ${\mathrm{H}}_2\mathrm{Te}$

• Tellurium lies below selenium and has more atomic size than the elements remaining above it.
• The tendency to lose electrons is more as the outer electrons are not under the influence of the nucleus and can be lost easily.
• It can act as a reducing agent as its reducing power is more.

D) ${\mathrm{H}}_2\mathrm{S}$

• Sulfur lies below oxygen and has more atomic size than oxygen.
• The tendency to lose electrons is more as the outer electrons are not under the influence of the nucleus and can be lost easily.
• It can act as a reducing agent as its reducing power is more.

Hence, option (B) is the correct answer. ${\mathrm{H}}_2\mathrm{O}$ cannot act as a reducing agent.