Calculate the `I-I` distance in (Å) for given compound `H_(2)C_(2) I_(2)` if `C-I` bond length is 2.35Å. `(sin60^(@)=0.866)`

Explanation

To calculate the I-I distance in the compound H₂C₂I₂, we will follow these steps:

### Step 1: Understand the Structure of H₂C₂I₂
The compound H₂C₂I₂ has a structure where two carbon atoms are double-bonded to each other (C=C), and each carbon is bonded to one iodine atom (I). The structure can be represented as:
```
H H
\ /
C=C
/ \
I I
```

### Step 2: Determine the Hybridization of Carbon
Each carbon atom in H₂C₂I₂ is bonded to two other atoms (one carbon and one iodine). The steric number for each carbon is calculated as follows:
- Number of sigma bonds = 2 (one C-C bond and one C-I bond)
- Number of lone pairs = 0

Thus, the steric number is 2, indicating that each carbon is sp² hybridized.

### Step 3: Identify the Bond Angles
In an sp² hybridized molecule, the bond angles are approximately 120 degrees. This means that the angle between the two C-I bonds is 120 degrees.

### Step 4: Analyze the Geometry
To find the I-I distance, we can visualize the geometry. The angle between the two C-I bonds is 120 degrees. If we bisect this angle, we can find the angle between the line connecting the two iodine atoms and the line connecting the carbon atoms.

### Step 5: Use the Sine Rule
We can denote:
- O as the midpoint between the two iodine atoms.
- I as the position of one iodine atom.
- C as the position of one carbon atom.

From the geometry, we can see that:
- The angle ICO = 60 degrees (half of 120 degrees).
- The angle IOC = 90 degrees.

Using the sine function:
\[ \sin(60^\circ) = \frac{IO}{CI} \]
Where:
- \( IO \) is the distance from the iodine atom to the midpoint O.
- \( CI \) is the bond length of C-I, which is given as 2.35 Å.

### Step 6: Solve for IO
Rearranging the sine equation gives:
\[ IO = CI \cdot \sin(60^\circ) \]
Substituting the values:
\[ IO = 2.35 \, \text{Å} \times 0.866 \]
Calculating this gives:
\[ IO \approx 2.0351 \, \text{Å} \]

### Step 7: Calculate the I-I Distance
The I-I distance (II) is twice the distance from one iodine atom to the midpoint O:
\[ II = 2 \times IO \]
Substituting the value of IO:
\[ II = 2 \times 2.0351 \, \text{Å} \approx 4.0702 \, \text{Å} \]
Rounding this gives:
\[ II \approx 4 \, \text{Å} \]

### Final Answer
The I-I distance in the compound H₂C₂I₂ is approximately **4 Å**.

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