The dipole moment of `LiH` is `1.964 xx10^(-29)Cm` and the interatomic distance between `Li` and `H` in this molecule is `1.596Å` .What is the per cent ionic character in `LiH`.

Options

A. `75.0`
B. `76.8`
C. `79.8`
D. 100

Explanation

To find the percentage ionic character in lithium hydride (LiH), we can follow these steps:

### Step 1: Identify the given values
- **Dipole moment (μ)**: \(1.964 \times 10^{-29} \, \text{Cm}\)
- **Interatomic distance (d)**: \(1.596 \, \text{Å} = 1.596 \times 10^{-10} \, \text{m} = 1.596 \times 10^{-8} \, \text{cm}\)

### Step 2: Calculate the theoretical dipole moment (μ theoretical)
The theoretical dipole moment can be calculated using the formula:
\[ \mu_{\text{theoretical}} = q \times d \]
where:
- \(q\) is the charge of an electron, approximately \(1.602 \times 10^{-19} \, \text{C}\)
- \(d\) is the interatomic distance in centimeters.

Substituting the values:
\[ \mu_{\text{theoretical}} = (1.602 \times 10^{-19} \, \text{C}) \times (1.596 \times 10^{-8} \, \text{cm}) \]
Calculating this gives:
\[ \mu_{\text{theoretical}} = 2.557 \times 10^{-29} \, \text{Cm} \]

### Step 3: Calculate the percentage ionic character
The percentage ionic character can be calculated using the formula:
\[ \text{Percentage ionic character} = \left( \frac{\mu_{\text{experimental}}}{\mu_{\text{theoretical}}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage ionic character} = \left( \frac{1.964 \times 10^{-29}}{2.557 \times 10^{-29}} \right) \times 100 \]
Calculating this gives:
\[ \text{Percentage ionic character} \approx 76.8\% \]

### Final Answer
The percentage ionic character in LiH is approximately **76.8%**.

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