Question

Dipole moment of certain diatomic molecule
X – Y is 0.38 D. If the X – Y bond distance is
158 pm. The percentage of electronic charge
developed on X-atom is

Answer 1

Answer:

I don't know what the answer is this questions I can't help you

Answer 2

Answer:

The percentage of electronic charge developed on X-atom is 1.5%.

Explanation:

The dipole moment is given as,

$\mu=qd$       (1)

Where,

μ=dipole moment

q=observed charge on the dipole

d=distance between the dipole

From the question we have,

μ=0.38D=0.38×10⁻³⁰Coulomb-meter

d=158pm=158×10⁻¹²m

By substituting the value of "μ" and "d" in equation (1) we get;

$0.38\times 10^-^3^0=q\times 158\times 10^-^1^2$

$q=2.4\times 10^-^2^1C$     (2)

The percentage of electronic charge on the dipole is given as,

$\%=\frac{q}{q'}\times 100$     (3)

q'=calculated charge=1.6×10⁻¹⁹C

By substituting all the required values in equation (3) we get;

$\%=\frac{2.4\times 10^-^2^1}{1.6\times 106-^1^9}\times 100=1.5\%$

Hence, the percentage of electronic charge developed on X-atom is 1.5%.

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