`Cl_(2)+underset((Excess))(F_(2)) overset(573K)to(A)`
Shape of compound (A) is

Options

A. Linear
B. Tetrahedral
C. T- shape
D. Trigonal bipyramidal

Answer

To determine the shape of the compound formed from the reaction of chlorine gas (Cl₂) with excess fluorine gas (F₂) at a temperature of 573 K, we can follow these steps:

### Step 1: Identify the Reaction
The reaction given is:
\[ \text{Cl}_2 + \text{Excess} \, \text{F}_2 \overset{573K}{\rightarrow} \text{A} \]

### Step 2: Determine the Product
When chlorine reacts with fluorine, the product formed is chlorine trifluoride (ClF₃). Thus, compound A is:
\[ \text{A} = \text{ClF}_3 \]

### Step 3: Calculate Valence Electrons
Next, we need to calculate the total number of valence electrons in ClF₃:
- Chlorine (Cl) has 7 valence electrons.
- Each fluorine (F) has 7 valence electrons, and there are 3 fluorine atoms.

Total valence electrons:
\[ \text{Total} = 7 (\text{from Cl}) + 3 \times 7 (\text{from F}) = 7 + 21 = 28 \]

### Step 4: Determine Bond Pairs and Lone Pairs
To find the number of bond pairs and lone pairs:
1. **Bond Pairs**: Divide the total valence electrons by 2 to find the number of electron pairs:
\[ \text{Total pairs} = \frac{28}{2} = 14 \text{ pairs} \]

2. **Bond Pairs**: In ClF₃, chlorine forms 3 bonds with fluorine atoms. Thus, there are 3 bond pairs.

3. **Lone Pairs**: The remaining pairs are lone pairs:
\[ \text{Lone pairs} = 14 - 3 = 11 \text{ pairs} \]
However, we need to consider the total number of valence electrons used in bonding. Each bond uses 2 electrons, so:
\[ \text{Lone pairs} = \frac{(28 - 6)}{2} = \frac{22}{2} = 11 \text{ pairs} \]

### Step 5: Determine the Geometry
In ClF₃, we have:
- 3 bond pairs (from Cl-F bonds)
- 2 lone pairs (remaining pairs on the central chlorine atom)

This arrangement corresponds to a **trigonal bipyramidal** geometry, where:
- The 3 bond pairs occupy the equatorial positions.
- The 2 lone pairs occupy the axial positions.

### Step 6: Determine the Shape
The presence of 2 lone pairs in axial positions leads to a **T-shaped** molecular geometry. The shape of the compound A (ClF₃) is therefore:
\[ \text{Shape of compound A} = \text{T-shaped} \]

### Final Answer
The shape of compound A (ClF₃) is T-shaped.

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