141 Valence Bond Molecular Orbital Theory Worksheet key PDF

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Valence Bond Molecular Orbital Theory Worksheet Answers...

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Chapter 9 Worksheet Key 1) Determine the hybridization of the central atom for each of the following molecules: a. PCl5 Cl Cl

Cl P Cl

Cl

EG = 5 (5 BP 0 LP)

Trigonal bipyramidal so sp3d hybridization on P

EG = 6 (6 BP 0 LP)

Octahedral so sp3d2 hybridization on S

EG = 4 (4 BP 0 LP)

Tetrahedral so sp3 hybridization on S

EG = 4 (4 BP 0 LP)

Tetrahedral so sp3 hybridization on P

EG = 3 (2 BP 1 LP)

Trigonal planar so sp2 hybridization on N

b. SF6 F F F

S

F

F F

c. SO42-

2O O

S

O

O

d. PO43-

3O O

P

O

O

e. NO2O

N

O

2) Determine the hybridization of EACH atom, and draw the bonding using valance bond theory. Remember to label the sigma and pi bonds. a. HCN

H=s

C = sp N = sp

Diagram will have a triple bond between C and N consisting of σ bond via end-to-end overlap of C sp orbital and N sp orbital, plus two π bonds formed by side-to-side overlap of unused p orbitals on C and N. The H is bonded to C via σ bond formed by end-to-end overlap of H s orbital and C sp orbital. b. NO2-

Central N is sp2

O double bonded to N is sp2

O single bonded to N is sp3

Diagram will have a double bond between N and one of the O atoms (sp2 hybridized) consisting of σ bond via end-to-end overlap of N sp2 orbital and O sp2 orbitals, plus one π bond formed by side-to-side overlap of unused p orbitals on O and N. The other O is bonded to N via single σ bond formed by end-to-end overlap of N sp2 orbital and sp3 orbital of O. Note the lone pair on N is in the third N sp2 hybrid orbital. c. CHCl3

All are sp3

Diagram will have three σ bonds formed by end-to-end overlap of C sp3 and Cl sp3 hybrid orbitals. H is bonded to C via σ bond formed by end-to-end overlap of remaining C sp3 orbital and H s orbital. d. BF3

B = sp2 F = sp3

Three σ bonds formed by end-to-end overlap of B sp2 and F sp3 hybrid orbitals. 3) Use molecular orbital theory to determine the bond order of the following bonds. Draw a MO diagram, determine the valence electron MO configuration. a. He-He (See text or powerpoint slides for Ch 9 for MO diagram) Valance MO electron configuration is (σ1s)2 (σ*1s)2 Bond order = (2-2)/2 = 0 b. N-O

(See text or powerpoint slides for Ch 9 for MO diagram)

Valence MO electron configuration is (σ2s)2 (σ*2s)2 (σ2p)2 (π2p)4 (π*2p)1 Bond order = (8 – 3)/2 = 2.5 c. B-N Valence MO electron configuration is (σ2s)2 (σ*2s)2 (σ2p)2 (π2p)2 Bond order = (6-2)/2 = 2 d. C-O Valence MO electron configuration is (σ2s)2 (σ*2s)2 (σ2p)2 (π2p)4 Bond order = (8-2)/2 = 3

4) Order the bonds in Q3 from strongest to weakest.

C-O > N-O > B-N > He-He (order of decreasing bond order)

5) Draw the Lewis structures of the following molecules from Q3, and compare the bond order to your answers in Q3. Do they match? a. BN B

N

Bond order = 3/1 = 3 so no this does not match. Note Lewis also does not predict that BN is paramagnetic (MO diagram has 2 unpaired electrons in π2p MO)

b. CO C

O

Bond order is 3/1 = 3 so yes this matches. Note Lewis structure puts +1 formal charge on O and -1 formal charge on C

6) Use molecular orbital theory to predict the bond orders and whether the following molecules will be diamagnetic or paramagnetic. a. B2 Valence MO electron configuration is (σ2s)2 (σ*2s)2 (π2p)2 Bond order = (4-2)/2 = 1 Paramagnetic (unpaired electrons in π2p) b. C2 Valence MO electron configuration is (σ2s)2 (σ*2s)2 (π2p)4 Bond order = (6-2)/2 = 2 Diamagnetic (all paired electrons) c. NF Valence MO electron configuration is (σ2s)2 (σ*2s)2 (σ2p)2 (π2p)4 (π*2p)2 Bond order = (8-4)/2 = 2 Paramagnetic (unpaired electrons in π*2p)...