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`, given that the `C-I` bond length is 2.35 Å, we can follow these steps:

### Step 1: Understand the Molecular Geometry
The compound `H₂C₂I₂` has a structure where each carbon is bonded to two iodine atoms. The carbon atoms are sp² hybridized, which means the bond angles around each carbon are approximately 120 degrees.

### Step 2: Draw the Structure
Draw the structure of `H₂C₂I₂`. Each carbon atom is at the center of a triangle formed with the two iodine atoms. The bond angle between the two `C-I` bonds is 120 degrees.

### Step 3: Identify the Relevant Triangle
In the triangle formed by the `C-I` bonds, we can denote:
- `C` as the carbon atom,
- `I₁` and `I₂` as the two iodine atoms.

The angle between the two `C-I` bonds (∠I₁CI₂) is 120 degrees. To find the distance between the two iodine atoms (`I-I`), we can drop a perpendicular from the carbon to the line joining the two iodine atoms, creating two right triangles.

### Step 4: Calculate the Length of `IO`
We can denote:
- `CI` as the bond length (2.35 Å),
- `IO` as the distance from the carbon to the midpoint of the `I-I` bond.

Using the sine function in the triangle:
- The angle `ICO` (where O is the midpoint of `I-I`) is half of 120 degrees, which is 60 degrees.

Using the sine definition:
\[ \sin(60^\circ) = \frac{IO}{CI} \]
Substituting the known values:
\[ 0.866 = \frac{IO}{2.35} \]
Now, solving for `IO`:
\[ IO = 0.866 \times 2.35 \]
Calculating this gives:
\[ IO \approx 2.035 \, \text{Å} \]

### Step 5: Calculate the `I-I` Distance
Since `IO` is the distance from carbon to the midpoint of the `I-I` bond, the total `I-I` distance is:
\[ I-I = 2 \times IO = 2 \times 2.035 \approx 4.07 \, \text{Å} \]

### Final Answer
The `I-I` distance in `H₂C₂I₂` is approximately **4.07 Å**.

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