Answer:
The value is [tex]x = 0.151 \ e [/tex]
Explanation:
From the question we are told that
The bond length is [tex]l = 1.93\ \r a = 1.93 *1 *10^{-10} =1.93 *10^{-10}\ m [/tex]
The bond dipole moment is [tex] \mu = 1.40 d = 1.40 * 3.33564 *10^{-30} = 4.6699 *10^{-30} \ C \cdot m[/tex]
Generally the dipole moment is mathematically represented as
[tex]\mu = Q * l[/tex]
Here Q is the partial negative charge on the bromine atom
So
[tex]Q = \frac{\mu}{ l}[/tex]
=> [tex]Q = \frac{4.6699 *10^{-30}}{ 1.93 *10^{-10} }[/tex]
=> [tex]Q = 2.42 *10^{-20} C [/tex]
Generally
1 electronic charge(e) is equivalent to [tex]1.60*10^{-19} C[/tex]
So x electronic charge(e) is equivalent to [tex]Q = 2.42 *10^{-20} C [/tex]
=> [tex]x = \frac{2.42 *10^{-20}}{1.60*10^{-19} }[/tex]
=> [tex]x = 0.151 \ e [/tex]
