The dipole moments of diatomic molecules AB and CD are 10.41D and 10.27 D, respectively while their bond distances are 2.82 and `2.67Å` respectively. This indicates that

Options

A. bonding is 100% ionic in both the molecules
B. AB has more ionic bond character than CD
C. AB has lesser ionic bond character than CD
D. bonding is nearly covalent in both the molecules

Explanation

To analyze the dipole moments and bond distances of diatomic molecules AB and CD, we will calculate the percentage ionic character of each molecule using the formula:

\[ \text{Percentage Ionic Character} = \left( \frac{\mu_{\text{observed}}}{\mu_{\text{calculated}}} \right) \times 100 \]

Where:
- \(\mu_{\text{observed}}\) is the dipole moment measured in Debye (D).
- \(\mu_{\text{calculated}}\) is the dipole moment calculated based on the assumption that the bond is 100% ionic.

### Step 1: Calculate \(\mu_{\text{calculated}}\) for AB
1. **Given Data for AB:**
- \(\mu_{\text{observed}} = 10.41 \, \text{D}\)
- Bond distance \(d = 2.82 \, \text{Å} = 2.82 \times 10^{-8} \, \text{cm}\)
- Charge \(Q = 1.6 \times 10^{-19} \, \text{C}\)

2. **Calculate \(\mu_{\text{calculated}}\):**
\[ \mu_{\text{calculated}} = Q \times d = (1.6 \times 10^{-19} \, \text{C}) \times (2.82 \times 10^{-8} \, \text{cm}) = 4.53 \times 10^{-27} \, \text{C cm} \]
Convert to Debye:
\[ 1 \, \text{D} = 3.336 \times 10^{-29} \, \text{C cm} \implies \mu_{\text{calculated}} = \frac{4.53 \times 10^{-27}}{3.336 \times 10^{-29}} \approx 13.53 \, \text{D} \]

### Step 2: Calculate Percentage Ionic Character for AB
\[ \text{Percentage Ionic Character for AB} = \left( \frac{10.41}{13.53} \right) \times 100 \approx 76.94\% \]

### Step 3: Calculate \(\mu_{\text{calculated}}\) for CD
1. **Given Data for CD:**
- \(\mu_{\text{observed}} = 10.27 \, \text{D}\)
- Bond distance \(d = 2.67 \, \text{Å} = 2.67 \times 10^{-8} \, \text{cm}\)

2. **Calculate \(\mu_{\text{calculated}}\):**
\[ \mu_{\text{calculated}} = Q \times d = (1.6 \times 10^{-19} \, \text{C}) \times (2.67 \times 10^{-8} \, \text{cm}) = 4.28 \times 10^{-27} \, \text{C cm} \]
Convert to Debye:
\[ \mu_{\text{calculated}} = \frac{4.28 \times 10^{-27}}{3.336 \times 10^{-29}} \approx 12.81 \, \text{D} \]

### Step 4: Calculate Percentage Ionic Character for CD
\[ \text{Percentage Ionic Character for CD} = \left( \frac{10.27}{12.81} \right) \times 100 \approx 80.14\% \]

### Step 5: Compare Ionic Character
- **AB:** 76.94% ionic character
- **CD:** 80.14% ionic character

### Conclusion
Since CD has a higher percentage of ionic character than AB, it indicates that CD has a stronger ionic bond compared to AB. Therefore, the bonding in AB is less ionic and more covalent compared to CD.