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Solutions to various chemistry problems related to bond lengths, polarity, and Lewis structures. Topics covered include determining the bond order and type, classifying bonds as nonpolar covalent, polar covalent, or ionic, calculating partial charges, and arranging atoms in order of electronegativity. Additionally, it includes drawing Lewis structures for various molecules and ions.
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SOLUTIONS - CHAPTER 6 Problems (Problems 1-11 cover material for Exam 2)
a) C=O For bonds between the same two atoms, the higher the bond order the shorter the bond. b) NN For bonds between the same two atoms, the higher the bond order the shorter the bond. c) C-O The bond orders are the same and the bonds are both single bonds with carbon. Since an O atom is smaller than an S atom, the C-O bond will be shorter than the C-S bond.
c) The NB bond in H 3 NBCl 3 Two different nonmetals, and so polar covalent. EN = 1.
d) The CF bond in CF 4 Two different nonmetals, and so polar covalent. EN = 1.
The dipole moment for a diatomic molecule is given by the expression
= Qr where = dipole moment Q = charge R = distance of separation
So Q = /r
It is easiest to convert everything in the above equation into MKS units.
= 1.24 D 3.336 x 10-30^ Cm = 4.137 x 10-30^ Cm 1 D
r = 1.57 Å 10 -10^ m = 1.57 x 10-10^ m 1 Å
So Q = (4.137 x 10-30^ Cm)/( 1.57 x 10-10^ m) = 2.64 x 10-20^ C
The percent ionic character in a bond in a diatomic molecule is given by the expression
% ionic character = (observed) x 100% (calculated using discrete charge)
= Qr r = 1.76 Å 10 -10^ m = 1.76 x 10-10^ m 1 Å
Q = 1.6022 x 10-19^ C (for discrete +1 and -1 charges)
So (calculated using discrete charge) = (1.6022 x 10-19^ C)(1.76 x 10-10^ m)
= 2.820 x 10-29^ Cm 1 D = 8.45 D 3.336 x 10-30^ Cm
So % ionic character = 1.42 D x 100% = 16.8 % 8.45 D
The electronegativities of the atoms in the above bonds are as follows
C 2.5 Cl 3.0 H 2.
So C-C EN = 0.0 C-Cl EN = 0.9 C-H EN = 0.
And so C-C < C-H < C-Cl.
(Burdge, 6.22) Draw Lewis structures for the following molecules or ions: a) NCl 3 c) H 2 O 2 e) CN- b) OCS d) CH 3 COO-^ f) CH 3 CH 2 NH 3 +
(Burdge, 6.74 a,b,c) Draw Lewis structures for the following organic molecules a) tetrafluoroethylene, C 2 F 4 b) propane, C 3 H 8 c) butadiene, CH 2 CHCHCH 2
(Burdge, 6.29) Draw Lewis structures for the following ions. Show formal charge. a) NO 2 +^ c) S 2 2- b) SCN-^ d) ClF 2 +
(Burdge, 6.34) Draw three resonance structures for the chlorate ion (ClO 3 - ). Show formal charges.
In each of the above resonance structures Cl has a formal charge of +1, and the two single bonded oxygens have a formal charge of -1.
The structure on the left is the most important resonance structure, as it keeps the formal charges close to zero and puts the negative formal charge on oxygen, the more electronegative atom. The structure in the middle also minimizes formal charges, but is not as good as the first structure as it puts a negative formal charge on a N atom. The structure on the right is the worse resonance structure, as all of the formal charges differ from zero, and one formal charge is -2.
(Burdge, 6.50) Until the early 1960s the noble gases were not known to form compounds. Since then, a few compounds of Kr, Xe, and Rn - and one compound of Ar - have been made in the laboratory. Draw Lewis structures for the following molecules. In each case Xe is the central atom. a) XeF 2 c) XeF 6 e) XeO 2 F 2 b) XeF 4 d) XeOF 4
Draw the Lewis structure for a) BCl 3 , b) NO
Note that in b the less electronegative atom (N) has the “missing” electron.