Question

Bond dissociation energy of "E-H" bond of the " $H _2 E$ " hydrides of group $16$ elements (given below), follows order.
A. $O$ B. $S$ C. $Se$ D. $Te$
Choose the correct from the options given below:

• A. $D > C > B > A$
• B. $B > A > C > D$
• C. $A>B>D>C$
• D. $A>B>C>D$

Answer 1

The Correct Option is D

The question concerns the bond dissociation energy (BDE) of the "E-H" bond in hydrides of Group 16 elements, arranged in the order of oxygen (O), sulfur (S), selenium (Se), and tellurium (Te). We need to determine the correct order of BDE for these hydrides.

The bond dissociation energy is the energy required to break the bond between hydrogen and another element. For hydrides of Group 16 elements, the BDE order typically decreases down the group. This is because as we move down the group, the size of the element (E) increases, resulting in a longer and weaker E-H bond that requires less energy to dissociate.

Here's the explanation sorted by groups:

• Oxygen (O): Forms the smallest and strongest bond among these elements, leading to the highest BDE for the O-H bond.
• Sulfur (S): Larger than oxygen; therefore, the S-H bond is weaker compared to the O-H bond.
• Selenium (Se): Larger than sulfur; the Se-H bond is even weaker.
• Tellurium (Te): Largest among the four elements, making the Te-H bond the weakest.

Based on the size and strength differences, the BDE should decrease in the order of: \(A > B > C > D\), which corresponds to \(O > S > Se > Te\). This is why the correct answer is:

$A > B > C > D$

Therefore, option $A > B > C > D$ is the correct answer.