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SOLUTIONS - CHAPTER 6 Problems (Problems 1-11 cover material for Exam 2)
- Which of the following bonds will have the smaller value for bond length? a) C-O or C=O b) N=N or NN c) C-O or C-S
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.
- (Burdge, 6.20) Classify the following bonds as nonpolar covalent, polar covalent, or ionic. Explain. a) The CC bond in H 3 CCH 3 Covalent (both bonded atoms are the same, so EN = 0 b) The KI bond in KI Ionic. Metal + nonmetal. EN = 1.7.
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.
- (Burdge, 6.12) The radical species ClO has a dipole moment of 1.24 D and a Cl-O bond distance of 1.57 Å. Determine the magnitude of the partial charges in ClO.
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
- (Burdge, 6.14) The measured dipole moment of bromine monofluoride, BrF, is 1. D, and the Br-F bond distance is 1.76 Å. Determine the percent ionic character in the bond in BrF.
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
- Arrange the following bonds in order from least polar to most polar: C-C, C-Cl, C-H.
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.
- Arrange the following atoms in order from lowest electronegativity to highest electronegativity. a) As, N, P As < P < N Electronegativity increases from bottom to top within a group b) Ca, Cs, K Cs < K < Ca Cs < K as electronegativity increases from bottom to top within a group; K < Ca as electronegativity increases from left to right within a period c) Ge O, Se Ge < Se < O Ge < Se as electronegativity increases from left to right within a period; Se < O as electronegativity increases from bottom to top within a group
(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.
- (Burdge, 6.38) Draw three resonance structures for the molecule NNO, where N is the central atom. Indicate formal charges, and rank the resonance structures in order of most important to least important.
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.
