Question

Determine the partial negative charge on the bromine atom in a c−br bond. the bond length is 1.93 å and the bond dipole moment is 1.40 d . express your answer using 3 significant figures. the partial negative charge on the bromine atom = previous answersrequest answer incorrect; try again; 4 attempts remaining provide feedback.

Answer 1

Answer:

The value is $x = 0.151 \ e$

Explanation:

From the question we are told that

The bond length is $l = 1.93\ \r a = 1.93 *1 *10^{-10} =1.93 *10^{-10}\ m$

The bond dipole moment is $\mu = 1.40 d = 1.40 * 3.33564 *10^{-30} = 4.6699 *10^{-30} \ C \cdot m$

Generally the dipole moment is mathematically represented as

$\mu = Q * l$

Here Q is the partial negative charge on the bromine atom

So

$Q = \frac{\mu}{ l}$

=> $Q = \frac{4.6699 *10^{-30}}{ 1.93 *10^{-10} }$

=> $Q = 2.42 *10^{-20} C$

Generally

1 electronic charge(e) is equivalent to $1.60*10^{-19} C$

So x electronic charge(e) is equivalent to $Q = 2.42 *10^{-20} C$

=> $x = \frac{2.42 *10^{-20}}{1.60*10^{-19} }$

=>     $x = 0.151 \ e$