The bond length of RbF is 284.00 pm. What would be the dipole moment, in D, of this compound by assuming a completely ionic bond? Remember e = 1.60x10-19 C; 1 D = 3.34x10-30 C*m
9.3
To calculate the dipole moment of a compound assuming a completely ionic bond, you need to determine the charge separation and multiply it by the bond length.
For RbF, Rb is an alkali metal with a valence of +1, and F is a halogen with a valence of -1. Therefore, the charge separation is 1- (-1) = 2.
To convert the bond length from picometers (pm) to meters (m), you need to divide it by 10^12:
284.00 pm = 284.00 × 10^-12 m
Now, multiply the charge separation by the bond length:
2 × 284.00 × 10^-12 m = 568.00 × 10^12 m
To convert the result from Coulomb-meters (C*m) to Debye (D), divide by the conversion factor:
568.00 × 10^12 m ÷ (3.34 × 10^-30 C*m) = 1.70 × 10^30 D
Therefore, the dipole moment of RbF assuming a completely ionic bond is 1.70 × 10^30 D.
To determine the dipole moment of RbF assuming a completely ionic bond, we need to first calculate the magnitude of the charge separation in the compound.
The dipole moment (μ) of an ionic bond can be calculated using the formula:
μ = q * d
where:
μ is the dipole moment
q is the magnitude of the charge
d is the bond length
In this case, RbF is an ionic compound, so the charge separation can be determined based on the ionic charges of Rb+ and F-, which are +1 and -1, respectively.
The magnitude of charge (q) can be calculated as:
q = | charge of Rb+ | + | charge of F- |
So,
q = | +1 | + | -1 |
= 1 + 1
= 2
Next, we need to convert the bond length from picometers (pm) to meters (m). We can use the conversion factor:
1 pm = 1 x 10^-12 meters
Therefore, the bond length (d) in meters can be calculated as:
d = 284.00 pm * (1 x 10^-12 m/pm)
Now, we can calculate the dipole moment using the formula:
μ = q * d
However, before calculating the dipole moment, we need to convert the charge from coulombs (C) to debye (D).
1 C = (1 / 3.34 x 10^-30) D
The dipole moment μ, in debye (D), can be calculated as:
μ = q * d * (1 / 3.34 x 10^-30) D
Substituting the values into the formula, we get:
μ = 2 * 284.00 pm * (1 x 10^-12 m/pm) * (1 / 3.34 x 10^-30) D
Simplifying the expression:
μ = (2 * 284.00 * 10^-12) / (3.34 * 10^-30) D
Calculating the numerator:
μ = (2 * 284.00 * 10^-12) D / (3.34 * 10^-30)
μ = 1.7059 * 10^-20 D
Therefore, the dipole moment of RbF, assuming a completely ionic bond, would be approximately 1.71 x 10^-20 D.