The dipole moment of compound AB is 10.92 D and that of compound CD is 12.45 D. The bond length AB is 2.72 Å and that of CD is 2.56 Å then for these compound true statement is

Options

A. More ionic nature in AB
B. More ionic nature in CD
C. Equal in both
D. Not predicted

Answer

To determine the ionic nature of compounds AB and CD based on their dipole moments and bond lengths, we can calculate the percentage ionic character for each compound using the formula:

\[ \text{Percentage Ionic Character} = \left( \frac{\mu}{e \cdot d} \right) \times 100 \]

Where:
- \( \mu \) = dipole moment in Coulomb-meters
- \( e \) = charge of an electron = \( 1.6 \times 10^{-19} \) Coulombs
- \( d \) = bond length in meters

### Step 1: Convert dipole moment from Debye to Coulomb-meters
1 Debye = \( 3.336 \times 10^{-29} \) Coulomb-meters.

**For compound AB:**
\[ \mu_{AB} = 10.92 \, \text{D} = 10.92 \times 3.336 \times 10^{-29} \, \text{C m} = 3.644 \times 10^{-28} \, \text{C m} \]

**For compound CD:**
\[ \mu_{CD} = 12.45 \, \text{D} = 12.45 \times 3.336 \times 10^{-29} \, \text{C m} = 4.151 \times 10^{-28} \, \text{C m} \]

### Step 2: Convert bond lengths from Ångstroms to meters
1 Ångstrom = \( 1 \times 10^{-10} \) meters.

**For compound AB:**
\[ d_{AB} = 2.72 \, \text{Å} = 2.72 \times 10^{-10} \, \text{m} \]

**For compound CD:**
\[ d_{CD} = 2.56 \, \text{Å} = 2.56 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the percentage ionic character for compound AB
Using the formula:
\[ \text{Percentage Ionic Character}_{AB} = \left( \frac{3.644 \times 10^{-28}}{(1.6 \times 10^{-19}) \cdot (2.72 \times 10^{-10})} \right) \times 100 \]

Calculating the denominator:
\[ (1.6 \times 10^{-19}) \cdot (2.72 \times 10^{-10}) = 4.352 \times 10^{-29} \]

Now substituting back:
\[ \text{Percentage Ionic Character}_{AB} = \left( \frac{3.644 \times 10^{-28}}{4.352 \times 10^{-29}} \right) \times 100 \approx 83.8\% \]

### Step 4: Calculate the percentage ionic character for compound CD
Using the formula:
\[ \text{Percentage Ionic Character}_{CD} = \left( \frac{4.151 \times 10^{-28}}{(1.6 \times 10^{-19}) \cdot (2.56 \times 10^{-10})} \right) \times 100 \]

Calculating the denominator:
\[ (1.6 \times 10^{-19}) \cdot (2.56 \times 10^{-10}) = 4.096 \times 10^{-29} \]

Now substituting back:
\[ \text{Percentage Ionic Character}_{CD} = \left( \frac{4.151 \times 10^{-28}}{4.096 \times 10^{-29}} \right) \times 100 \approx 101.3\% \]

### Conclusion
Since the percentage ionic character for compound CD is higher than that for compound AB, we conclude that compound CD has more ionic character.

### Final Answer
The true statement is that compound CD has more ionic nature than compound AB.