the dipole moment, in debyes, and -1, respectively 10. a) The hond length for HF
ID: 694310 • Letter: T
Question
the dipole moment, in debyes, and -1, respectively 10. a) The hond length for HF is 0.92 A. Calculate that would result if the charges on H andFwe (ID= 3.34 x 10"'cm: Q = 1.60 x 10'19 C) b) The dipole moment for HF is 1.82 D and the bond length of HF is 0.92 A What is the magnitude of the charge on the H and F atoms in terms of the charge of a single electron? The charge on a single electron is 1.60 x 10" C 11. Calculate the formal charge for each of the atoms in NO: 12. Draw a Lewis structure for: a) NF b) CClExplanation / Answer
Answer:
10) (a)
The charge on each atom is the electronic charge, e: 1.60 x 10-19 C. The separation in HF (bond length) is 0.92 Å. and the forml charge on H and F are +1 and -1.
By analogy to the calculation in the text above, the dipole moment is
Dipole momentum u=Qr=(1.6x10-19 C)(0.92 Å)(10-10 m/1 Å)(1 D/3.34x10-30 C-m) (D=unit of dipole moment debyes)
u=4.407 D.
(b) Here we have gien with u=1.82 D, and the value of r=0.92 Å, and we want to calculate the value of Q:
u=Qr => Q=u/r=[(1.82 D)x(3.34x10-30 C-m/1D)/(0.92 Å)(10-10 m/1 Å)]=6.607x10-20 C
We can readily convert this charge to units of e:
Charge in e=(6.607x10-20 C)(1e/1.6x10-19 C)=0.412 e
Thus, the experimental dipole moment indicates the following charge separation in the HF molecule:
H0.412+-F0.412-
Because the experimental dipole moment is less than that calculated in part (a), the charges on the atoms are less than a full electronic charge.
As per guide lines I have answered first question.
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