1 D | = | 10–18 statC.cm |
= | 10–10 esu.A° | |
= | (1/299,792,458) × 10–21 C.m | |
∼ | 3.33564 × 10– 30 C.m | |
∼ | 1.10048498 × 1023 qPlP | |
∼ | 0.393430307 ea0 | |
∼ | 0.20819434 eA° |
The dipole moment of a diatomic molecule is equal to 0.82 D. If its bond length is 1.4 A°, calculate the fraction of the electronic charge existing on each atom in the molecule. [1 Debye = 10–18 esu cm]?
μ | = | 0.82 D |
= | 0.82 × 10–18 esu cm | |
r | = | 1.4 A° |
= | 1.4 × 10– 18 cm [1 A° = 10– 18 cm] | |
Let the fraction of the electronic charge(q) be x. Now we know that | ||
μ | = | x × r |
x | = | 0.82 × 10– 18 esu cm / 1.48 × 10–8 cm |
= | 0.585 × 10–10 esu | |
we know that the electronic charge on isolated ion = 4.8 × 10– 10 esu. | ||
Thus, since out of 4.8 × 10– 10 esu electronic charge existing on each atom = 0.58 × 10–10 esu, | ||
out of 100 esu electronic charge existing on each atom | = | (0.585 × 10–10 esu / 4.8 × 10–10) × 100 |
= | 12.2%(Answer) | |
The answer can also be found by using the relation: % of electronic charge existing on each atom | ||
= | (μ(esu cm) / r(cm) × (4.8 × 10– 10 esu)) × 100 % | |
= | (0.82 × 10– 18 esu cm / ((1.4 × 10– 8 cm) × (4.8 × 10– 10 esu ))) × 100 | |
= | 12.2 % |