The dipole moments of a diatomic molecule AB is10.4D and bond distance is 2.82`A^@`, calculate the % ionic character ofAB

Explanation

To calculate the percentage ionic character of the diatomic molecule AB, we can follow these steps:

### Step 1: Understand the given values
- The dipole moment (μ) of the molecule AB is given as 10.4 D (Debye).
- The bond distance (d) is given as 2.82 Å (angstroms).

### Step 2: Convert the bond distance from angstroms to meters
1 Å = \(1 \times 10^{-10}\) m, so:
\[ d = 2.82 \, \text{Å} = 2.82 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the theoretical dipole moment (μ) for a completely ionic bond
The theoretical dipole moment can be calculated using the formula:
\[ \mu = q \times d \]
where:
- \(q\) is the charge of an electron, approximately \(1.6 \times 10^{-19}\) C.
- \(d\) is the bond distance in meters.

Substituting the values:
\[ \mu = (1.6 \times 10^{-19} \, \text{C}) \times (2.82 \times 10^{-10} \, \text{m}) \]
Calculating this gives:
\[ \mu = 4.51 \times 10^{-29} \, \text{C m} \]

### Step 4: Convert the theoretical dipole moment from C m to Debye
1 D = \(3.33 \times 10^{-29}\) C m, so:
\[ \mu_{\text{theoretical}} = \frac{4.51 \times 10^{-29} \, \text{C m}}{3.33 \times 10^{-29} \, \text{C m/D}} \approx 13.52 \, \text{D} \]

### Step 5: Calculate the percentage ionic character
The percentage ionic character can be calculated using the formula:
\[ \text{Percentage Ionic Character} = \left( \frac{\mu_{\text{observed}}}{\mu_{\text{theoretical}}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Ionic Character} = \left( \frac{10.4 \, \text{D}}{13.52 \, \text{D}} \right) \times 100 \]
Calculating this gives:
\[ \text{Percentage Ionic Character} \approx 76.92\% \]

### Final Answer:
The percentage ionic character of the diatomic molecule AB is approximately **76.92%**.

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