Question

GUYS SOLVE THIS.....
The percentage ionic character
in a compound AB if its internal nuclear distance is 127 pm
(picometers) and dipole moment is 2.06D.​

Answer 1

Answer:

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 metres

μ = 1.6 x 10^-19 x 1.2 x 10^-10 = 1.92 x 10^-29 C-m = 19.2 x 10^-30 C-m

= 5.75 D

(1 debye = 3.34 x 10^-30 C-m)

% ionic character = actual dipole x 100/ theortical dipole

= 1.92 x 100/ 5.75 = 33.4 %

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Answer 2

Percentage ionic character=experimental dipole×100/caculated diopole moment

Dipole moment=q×d

q us the charge on electron and d is the radius molecule

Dipole moment=1.6×10^-19 ×127×10^-12 (127pm=127×10^-12m)

Dipole moment=16×1.3 ×10^(-12-18)=20.8×10^-30C-m=(20.8×10^-30)/3.34×10^-30=6.23D

this is the caculated or theoratical dipole

percentage ionic character= (2.06×100)/6.23

=33.06percent