Answer the question on basis of passage given below:

All the members of Group 15 elements form covalent hydrides with hydrogen as $NH_3, PH_3, AsH_3, SbH_3, BiH_3$. Group 16 elements form covalent hydrides with hydrogen as $H_2O, H_2S, H_2Se, H_2Te, H_2PO$. Group 17 elements form covalent hydrides with hydrogen as $HF, HCI, HBr, HI$. Based on these facts answer the following question.

Which of the following is the correct order of reducing strength of hydrides of Group 16 elements?

$H_2O<H_2S <H_2Se< H_2Te$

The correct answer is Option (3) → $H_2O<H_2S <H_2Se< H_2Te$.

The reducing strength of a compound refers to its ability to donate electrons (or hydrogen atoms in the case of hydrides) to another substance, reducing that substance. In the case of Group 16 hydrides (\(H_2O\), \(H_2S\), \(H_2Se\), \(H_2Te\)), their reducing strength increases as we move down the group from oxygen to tellurium. This is because of two key factors: bond dissociation energy and electronegativity.

Key Concepts:

Bond Dissociation Energy:

This is the energy required to break the bond between hydrogen and the Group 16 element \((O, S, Se, Te)\) in the hydride. As we move down the group, the size of the atoms increases, and the bond length between hydrogen and the element increases, making the bond weaker. As the bond weakens, it becomes easier to break the bond and release hydrogen, which is a key part of reducing behavior. The lower the bond dissociation energy, the stronger the reducing agent.

The bond strength decreases in the following order:

\(H–O > H–S > H–Se > H–Te\)

This means that the bond is weakest in \(H_2Te\) and strongest in \(H_2O\).

Electronegativity:

Electronegativity is the tendency of an atom to attract electrons towards itself. Oxygen is the most electronegative element in Group 16, followed by sulfur, selenium, and tellurium. A more electronegative element holds onto its electrons more tightly, making it less likely to donate electrons or act as a reducing agent.

As electronegativity decreases down the group, the elements become more likely to lose electrons, increasing the reducing strength.

The electronegativity of the Group 16 elements decreases in the order:

\(O > S > Se > Te\)

This trend shows that oxygen is less willing to donate electrons (or hydrogen), and tellurium is the most willing.

Hydrides of Group 16 Elements:

\(H_2O\) (Water):

Oxygen is very electronegative, and the \(O-H\) bond is strong, so water is a weak reducing agent. It is poor at donating hydrogen or electrons.

\(H_2S\) (Hydrogen sulfide):

Sulfur is less electronegative than oxygen, and the \(S-H\) bond is weaker than the \(O-H\) bond. This makes hydrogen sulfide a stronger reducing agent than water.

\(H_2Se\) (Hydrogen selenide):

Selenium is even less electronegative than sulfur, and the \(Se-H\) bond is weaker than the \(S-H\) bond. Hydrogen selenide is an even stronger reducing agent than \(H_2S\).

\(H_2Te\) (Hydrogen telluride):

Tellurium is the least electronegative of these elements, and the \(Te-H\) bond is the weakest. Hydrogen telluride is the strongest reducing agent among the Group 16 hydrides because it can easily donate hydrogen.

Trend in Reducing Strength:

As we move down Group 16, the reducing strength increases because:

I. The bond dissociation energy decreases, making it easier to release hydrogen.

II. The electronegativity decreases, making the element more willing to donate electrons.

The Correct Order: Based on the above analysis, the reducing strength of the hydrides increases in the following order:

\(\mathbf{H₂O < H₂S < H₂Se < H₂Te}\)

Water \((H_2O)\) is the weakest reducing agent.

Hydrogen sulfide \((H_2S)\) is stronger than water but weaker than the heavier hydrides.

Hydrogen selenide \((H_2Se)\) is a stronger reducing agent than H₂S.

Hydrogen telluride \((H_2Te)\) is the strongest reducing agent among them due to the weak Te-H bond and low electronegativity of tellurium.

Conclusion:

The correct order of reducing strength for the hydrides of Group 16 elements is: \(\mathbf{H₂O < H₂S < H₂Se < H₂Te}\)