The diatomic molecule has a dipole moment of 1.98D and the bond length is 0.92Angstroms . then the value of %ionic character/112 is
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Answer:STSMS
Explanation:a
Given:
Dipole moment of the diatomic molecule = 1.98D
Bond length in Angstroms = 0.92
To find:
% of ionic character in the molecule.
Solution:
As we know that
Dipole moment (theoretical) μ = Q x d (coulomb-metre)
Q is the charge of electron = 1.6 x 10^-19 C
d is the bond length in meters
μ = 1.6 x 10^-19 x 0.92 x 10^-10 = 1.472 x 10^-29 C-m = 14.72 x 10^-30 C-m
(1 debye = 3.34 x 10^-30 C-m)
Therefore,
μ = 4.4 D
% ionic character = actual dipole x 100/ theortical dipole
= 1.98 x 100/ 4.4 = 45%
Therefore the % ionic character is 45%.