Bond dissociation energy of `E-H^(*)` bond of the order. A. ` O` B. `S` C. `Se` D. `Te` Choose the correct from the options given below:

Options

A. `D gt C gt B gt A`
B. `B gt A gt C gt D`
C. `A gt B gt D gt C`
D. `A gt B gt C gt D`

Answer

To determine the bond dissociation energy of the `E-H` bond in the hydrides of group 16 elements (O, S, Se, Te), we can analyze the trend in bond dissociation energies based on the size of the atoms involved.

### Step-by-Step Solution:

1. **Identify the Hydrides**: The hydrides of group 16 elements we are considering are:
- H₂O (water)
- H₂S (hydrogen sulfide)
- H₂Se (hydrogen selenide)
- H₂Te (hydrogen telluride)

2. **Understand the Trend in Atomic Size**: As we move down the group from oxygen to tellurium, the atomic size increases due to the addition of electron shells. This means that:
- Oxygen (O) is the smallest,
- Sulfur (S) is larger than oxygen,
- Selenium (Se) is larger than sulfur,
- Tellurium (Te) is the largest.

3. **Relate Atomic Size to Bond Length and Strength**: The bond length increases as the size of the atom increases. A longer bond length generally leads to a weaker bond because the atoms are held together less tightly. Therefore, we can expect:
- H-O bond (in H₂O) to be the strongest,
- H-S bond (in H₂S) to be weaker than H-O bond,
- H-Se bond (in H₂Se) to be weaker than H-S bond,
- H-Te bond (in H₂Te) to be the weakest.

4. **Order of Bond Dissociation Energies**: Based on the above reasoning, we can conclude the order of bond dissociation energies for the `E-H` bonds is:
- H₂O > H₂S > H₂Se > H₂Te

5. **Final Answer**: Therefore, the correct order of bond dissociation energy for the `E-H` bond is:
- A (O) > B (S) > C (Se) > D (Te)

### Conclusion:
The correct option is **A > B > C > D**, which corresponds to the order of bond dissociation energies of the `E-H` bonds in the hydrides of oxygen, sulfur, selenium, and tellurium.