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.

The correct decreasing order of bond dissociation enthalpies for the following compound is:

$H_2O> H_2S > H_2Se > H_2Te$

The correct answer is Option 1. $H_2O> H_2S > H_2Se > H_2Te$.

Let us delve into the bond dissociation enthalpies of the hydrides of Group 16 elements (\(H_2O\), \(H_2S\), \(H_2Se\), and \(H_2Te\)) and the factors that influence their strengths.

Bond dissociation enthalpy (BDE) is the energy required to break a specific bond in a molecule, leading to the formation of two radicals or atoms. A higher BDE indicates a stronger bond.

Importance: BDE can give insights into the stability and reactivity of compounds. Stronger bonds are generally associated with higher stability.

Analyzing the Hydrides

\(H_2O) (Water)

Bond: The \(O–H\) bond in water is exceptionally strong.

Factors:

Electronegativity: Oxygen has a high electronegativity (3.44 on the Pauling scale), leading to a strong attraction between the hydrogen and oxygen atoms.

Small Size: The small atomic radius of oxygen allows for effective overlap of orbitals, resulting in a strong bond.

BDE: \(H_2O\) has the highest bond dissociation enthalpy among the group due to these factors.

\(H_2S\) (Hydrogen Sulfide)

Bond: The \(S–H\) bond is weaker than the \(O–H\) bond.

Factors:

Lower Electronegativity: Sulfur's electronegativity (2.58) is significantly lower than oxygen's, resulting in a less polar bond.

Larger Size: Sulfur is larger than oxygen, leading to longer bond lengths and decreased overlap of the orbitals, which weakens the bond.

BDE: The bond dissociation enthalpy of \(H_2S\) is lower than that of \(H_2O\).

\(H_2Se\) (Hydrogen Selenide)

Bond: The \(Se–H\) bond is weaker than the \(S–H\) bond.

Factors:

Even Lower Electronegativity: Selenium has an electronegativity of 2.55, which is slightly lower than sulfur's.

Increased Size: Selenium is larger than sulfur, which further weakens the bond due to even less effective orbital overlap.

BDE: \(H_2Se\) has a lower bond dissociation enthalpy than \(H_2S\).

\(H_2Te\) (Hydrogen Telluride)

Bond: The \(Te–H\) bond is the weakest among the hydrides discussed.

Factors:

Lowest Electronegativity: Tellurium has an electronegativity of 2.1, the lowest in the group, resulting in an even less polar bond.

Largest Size: Tellurium is the largest of the group 16 elements, leading to the longest bond lengths and the weakest overlap of orbitals.

BDE: \(H_2Te\) has the lowest bond dissociation enthalpy, making it the least stable hydride in this series.

Summary of Bond Dissociation Enthalpy Order

The general trend in bond dissociation enthalpy for these compounds can be summarized as follows:

Decreasing Order: \(\text{H₂O} > \text{H₂S} > \text{H₂Se} > \text{H₂Te}\)

Conclusion

In conclusion, the bond dissociation enthalpy is highest for water \((H_2O)\) due to strong \(O–H\) bonds, followed by hydrogen sulfide \((H_2S)\), hydrogen selenide \((H_2Se)\), and finally hydrogen telluride \((H_2Te)\) which has the weakest bond. This trend reflects the combined effects of electronegativity, atomic size, and bond length, highlighting the importance of these factors in determining bond strengths across a group in the periodic table.