The study of diple
moment of a molecule is useful to explain the
shape of a molecule and also to predict a
The net dipole moment of a polyatomic
molecule is the resltant of the different bond
moments present in that molecule . The values
are generalluy expressed in Debye or in the S.I.
units in terms of Coulomb meter (C m)
A covalent molecule , X-Y is found to have a
dipole moment of ` 1.5 xx10^(-29)` C m and a bond
length of 150 pm . The per cent ionic character of the bond will be

Options

A. `50%`
B. `62.5%`
C. `75%`
D. `90%`

Explanation

To determine the percent ionic character of the covalent bond in the molecule X-Y, we can follow these steps:

### Step 1: Understand the relationship between dipole moment, charge, and bond length.
The dipole moment (μ) of a bond can be calculated using the formula:
\[ \mu = q \times r \]
where:
- \( \mu \) is the dipole moment,
- \( q \) is the charge (in coulombs),
- \( r \) is the bond length (in meters).

### Step 2: Convert bond length from picometers to meters.
Given the bond length is 150 pm (picometers), we convert it to meters:
\[ 150 \text{ pm} = 150 \times 10^{-12} \text{ m} = 1.5 \times 10^{-10} \text{ m} \]

### Step 3: Use the charge of an electron.
The charge of an electron is approximately:
\[ q = 1.602 \times 10^{-19} \text{ C} \]

### Step 4: Calculate the theoretical dipole moment.
Using the values from Steps 2 and 3, we can calculate the theoretical dipole moment:
\[ \mu_{\text{theoretical}} = q \times r \]
\[ \mu_{\text{theoretical}} = (1.602 \times 10^{-19} \text{ C}) \times (1.5 \times 10^{-10} \text{ m}) \]
\[ \mu_{\text{theoretical}} = 2.403 \times 10^{-29} \text{ C m} \]

### Step 5: Compare the observed dipole moment with the theoretical dipole moment.
The observed dipole moment of the molecule X-Y is given as:
\[ \mu_{\text{observed}} = 1.5 \times 10^{-29} \text{ C m} \]

### Step 6: Calculate the percent ionic character.
The percent ionic character can be calculated using the formula:
\[ \text{Percent Ionic Character} = \left( \frac{\mu_{\text{observed}}}{\mu_{\text{theoretical}}} \right) \times 100 \]
Substituting the values:
\[ \text{Percent Ionic Character} = \left( \frac{1.5 \times 10^{-29}}{2.403 \times 10^{-29}} \right) \times 100 \]
\[ \text{Percent Ionic Character} \approx 62.5\% \]

### Final Answer:
The percent ionic character of the bond X-Y is approximately **62.5%**.

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