5.112 Principles Of Chemical Science Worksheet With Answers - Problem Set #4 Solutions - 2011 Page 2
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16Problem set #4 solutions for 5.112, Fall 2011
Page 2 of 16
⎛
⎞
0
0
( )
For the 3.25 A
bond length:
e
3.25 A
⎝ ⎜
⎠ ⎟
QR
µ =
=
= 15.6 D
0
0
−1
−1
0.2082 C A
D
0.2082 e A
D
(b) Calculate the Coulombic potential energy for each molecule. (3 points)
q
q
We need to remember the equation for Coulombic potential: ΔE
=
1
2
, where q is the
4 πε
coulomb
r
0
charge on the ion multiplied by the fundamental charge (e).
Now we can calculate the Coulombic potential energy for each molecule:
(
)
(
)
0
-19
-19
1.60218 x 10
C
1.60218 x 10
C
q
q
For the 1.75 A
bond length:
ΔE
=
=
1
2
(
)
(
)
4 πε
4 π 8.85419 x 10
coulomb
-12
-1
2
-1
-10
r
J
C
m
1.75 x 10
m
0
= 1.31829 x 10
-18
J
-18
=1.32 x 10
J
(
)
(
)
0
-19
-19
1.60218 x 10
C
1.60218 x 10
C
q
q
For the 3.25 A
bond length:
ΔE
=
=
1
2
(
)
(
)
4 πε
4 π 8.85419 x 10
coulomb
-12
-1
2
-1
-10
r
J
C
m
3.25 x 10
m
0
= 7.09875 x 10
-19
J
-19
=7.10 x 10
J
Question 4 (out of 4 points)
Estimate the percent ionic character of the bond in each of the following species. All the
species are unstable or reactive under ordinary laboratory conditions, but are observed in
interstellar space.
Average Bond Length (Å)
Dipole Moment (D)
(a) OH
0.980
1.66
(b) CH
1.131
1.46
(c) CN
1.175
1.45
(d) C
1.246
0
2
The percentage ionic character is calculated by determining the δ (fraction of unit charge on each
atom) and multiplying δ by 100.
(a) OH
⎡
⎤
0
-1
⎣ ⎢
⎦ ⎥
0.2082A
D
[
]
δ =
= 0.353
For OH, the percentage ionic character is 35.3 %.
1.66 D
0
0.980 A
(b) CH
⎡
⎤
0
-1
⎣ ⎢
⎦ ⎥
0.2082A
D
[
]
δ =
= 0.269 For CH, the percentage ionic character is 26.9 %.
1.46 D
0
1.131 A
(c) CN
⎡
⎤
0
-1
⎣ ⎢
⎦ ⎥
0.2082A
D
[
]
δ =
= 0.257 For CN, percentage ionic character is 25.7 %
1.45 D
0
1.175 A
(d) C
For C
the percentage ionic character is 0 %.
2
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