Question

The dipole of HX is 1.2 D
the percent of ionic character of the bond is 25% then its Bond length is

Answer 1

bond length= (1.2)/25*100

Answer 2

Answer: $0.63\times 10^{-11}m$

Explanation:

$\mu=Q\times r$

$\mu$ = Dipole moment (theoretical)

Q = Charge of electron = $1.6\times 10^{-19} C$

r = distance of separation

$1Debye=3.34\times 10^{-30}Cm$

$\% ionic character=\frac{\text {actual dipole}}{\text { theortical dipole}}\times 100$

$25\%=\frac{\text {actual dipole}}{\text { theoretical dipole}}\times 100$

$25\%=\frac{\text {actual dipole}}{1.2D}\times 100$

${\text {actual dipole}}=0.3D$

$1Debye=3.34\times 10^{-30}Cm$

$0.3D=\frac{3.34\times 10^{-30}}{1}\times 0.3=1.002\times 10^{-30}Cm$

$1.002\times 10^{-30}Cm=1.6\times 10^{-19} C\times r$

$r=0.63\times 10^{-11}m$

Thus bond length is $0.63\times 10^{-11}m$