The bond angle around central atom is maximum for

Options

A. `H_(2)O`
B. `H_(2)S`
C. `H_(2)Se`
D. `H_(2) Te`

Explanation

To determine which hydride has the maximum bond angle around the central atom, we can follow these steps:

### Step 1: Identify the Hydrides
We need to identify the hydrides in question. The common hydrides of the elements in the same group are:
- H2O (Water)
- H2S (Hydrogen Sulfide)
- H2Se (Hydrogen Selenide)
- H2Te (Hydrogen Telluride)

### Step 2: Understand the Concept of Bond Angles
The bond angle is influenced by the repulsion between the bonding pairs of electrons. As the central atom's electronegativity decreases, the repulsion between the hydrogen bond pairs also decreases, which can lead to a decrease in the bond angle.

### Step 3: Analyze Electronegativity and Atomic Size
Electronegativity is inversely related to atomic size. As you move down a group in the periodic table, the size of the atoms increases, leading to a decrease in electronegativity. The elements in question (O, S, Se, Te) belong to the same group (Group 16).

### Step 4: Compare the Electronegativity of the Central Atoms
- Oxygen (O) has the highest electronegativity.
- Sulfur (S) has a lower electronegativity than oxygen.
- Selenium (Se) has an even lower electronegativity than sulfur.
- Tellurium (Te) has the lowest electronegativity among these.

### Step 5: Determine the Bond Angles
Due to the decreasing electronegativity and increasing atomic size:
- H2O (bond angle ~ 104.5°)
- H2S (bond angle ~ 92°)
- H2Se (bond angle ~ 90°)
- H2Te (bond angle ~ 90°)

### Step 6: Conclusion
Since H2O has the highest electronegativity and the smallest size, it experiences the greatest repulsion between the hydrogen bond pairs, leading to the maximum bond angle of approximately 104.5 degrees.

### Final Answer
The bond angle around the central atom is maximum for **H2O**.

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