Ch 8-9 Practice Test Answer Key

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Ch 8-9 Practice Test Answer Key 18. Which of the atoms below is least likely to violate the octet rule? a) Be- deficient likely b) P- can expland c) S- can expand d) B- deficient is likely e) F- usually wants to follow octet rule to bond 19. How many electrons are shown in the Lewis structure of perchlorate ion, ClO4? a) 30 b) 31 c) 32 d) 50 e) 51 Cl would bring 7 e-, each of the 4 O s would bring 6 (24 total) and -1 overall adds an electron for a total of 32

20. Which of the following is the correct Lewis structure for SOCl2? (Consider formal charge) a) you can t expand on atoms that are not central, and that is don eon the Cl with the = here. It would have formal charges, too b) This would orient the atoms correctly, would give -1 on the O and S would be +1. Overall charges work, and has the correct number of electrons. It is the only viable option here c) no- you d NEVER put Cl in the center; you d put the LEAST electronegative atom in the center, also, formal charges much? d) no- you d NEVER put Cl in the center; you d put the LEAST electronegative atom in the center. also, formal charges much? e) no formal charges and the correct central atom but incorrect # electrons. 21. As the bond order of a carbon-carbon bond increases, which one of the following decreases? a) # of electrons between the carbon atoms (increases by two when BO increases by 1) b) vibrational frequency of bond vibrations (more vibration and rotation with fewer bonds) c) bond energy (bond strength) (reverse relationship- energy increases with BO and BS (yes, I see that I put BS) d) bond length 22. Which of the following bonds is most polar? (remember that differences are absolute value and that the partially negative end would be at the more electronegative material) a) N Cl (3-3) so Δ is 0: Non-polar

b) C N (2.5-3) so Δ is 0.5, polar c) S S (2.5-2.5 )so Δ is 0: Non-polar d) Br Br (2.8-2.8) so Δ is 0: Non-polar e) S O so Δ is (2.5-3.5) so Δ is 1: polar, and more polar than b) above 23. Which one of the following molecules is a polar molecule? a) linear with no dipole b) the N-F bond has an electronegativity difference of 1, and are not symmetrical across the whole molecule (it pulls down from the N between all of the Fs) c) linear with no net dipole d) symmetrical so all dipoles cancel 24. Which of the following molecular shapes has six atoms joined to a central atom? a) linear d) trigonal bipyramid b) tetrahedral e) planar triangular c) octahedral 25. Which molecular shape has bond angles which are not all the same? a) linear- 180 b) tetrahedral- 109.5 c) octahedral- 90 d) planar triangular- 120 (trigonal planar is a better name) e) trigonal bipyramid 26. What is the shape of BrI3? (28 electrons) you end up with 2 lone pairs on Br (AX3E2) a) square planar d) pyramidal b) T-shaped e) bent c) distorted tetrahedral

27. What is the shape of the IF4 ion? (36e- with 4 bonds and 2 lone pairs, AX4E2 a) square planar d) octahedral b) tetrahedral e) T-shaped c) square pyramidal 28. Which of the following is a polar species? a) CO2 linear all polar so equal b) PCl5 TBP, polar but all equal c) ICl2 linear with TBP parent (AX2E3) so it is all equal d) TeCl4 AX4E, so TBP and not equal b/c lone pair prevents it; dipole moment between all Cls e) CCl4 tetrahedral all polar so equal 29. Which hybridization is associated with a steric number of 3? (3 domains) a) sp- linear d) sp 3 d- TBP b) sp 2- trigonal pyramidal e) sp 3 d 2 - octahedral c) sp 3 - tetrahedral 30. The molecule BrF3 has a steric number of on the central atom? a) 3 b) 4 c) 5 b/c two lone pairs on the Br, making it TBP d) 6 31. What is the hybridization of Br in BrF3? a) sp d) sp 3 d- see #29 above b) sp 2 e) sp 3 d 2 c) sp 3 32. How many equivalent sp 3 d orbitals are there? a) 3 b) 1 c) 5 d) 6 33. What type of hybridization is associated with a square planar molecular shape? (see #29 for parent geometries related to the hybrid orbitals- sq planar has a TBP parent) a) sp 3 d) sp 3 d b) sp 2 e) sp 3 d 2 c) sp

(EC- not on the test) 34. A double bond contains sigma bond(s) and pi bond(s). a) 0, 2 b) 1, 2 c) 2, 0 d) 1, 1 35. What angle exists between orbitals in sp 3 d 2 hybrid orbitals? (octahedral parent so all are the same) a) 90.0 d) 120.0 b) 180.0 e) 78.5 c) 109.5 36. Which of the following elements is most likely to display sp 3 d hybridization? It must be one that expands, so it can only be P a) oxygen d) carbon b) nitrogen e) boron c) phosphorus Previous AP Test Questions to try 1982 D (a) Draw the Lewis electron-dot structures for CO 3 2-, CO 2, and CO, including resonance structures where appropriate. (b) Which of the three species has the shortest C-O bond length? Explain the reason for your answer. It would be where there is the greatest bond order, CO. (c) Predict the molecular shapes for the three species. Explain how you arrived at your predictions. CO 3 2- would be trigonal planar to spread out the electrons between 3 domains

CO 2 would be linear to spread the electrons between 2 domains, and CO would be linear as well (can t be anything else with only 2 atoms) 1990 D Use simple structure and bonding models to account for each of the following. (a) The bond length between the two carbon atoms is shorter in C 2 H 4 double bond between the Cs than in C 2 H 6. Single bond between Cs (b) The H-N-H bond angle is 107.5º, in NH 3. The lone pairs of electrons take up space, and more space than the N-H bond, pushing the 3 bonding domains down from the parent tetrahedral geometry (3 bonding and one non-bonding domain) (c) The bond lengths in SO 3 are all identical and are shorter than a sulfur-oxygen single bond. Resonance structures account for the fact that there is more than a single bond but less than a double bond. The electrons are delocalized (d) The I 3 - ion is linear. The lone pairs of electrons on the central I cause the side I s to be as far apart as possible- 180 degrees 1996 D Explain each of the following observations in terms of the electronic structure and/or bonding of the compounds involved. (a) At ordinary conditions, HF (normal boiling point = 20ºC) is a liquid, whereas HCl (normal boiling point = -114ºC) is a gas. Stronger attractions because a greater dipole exists in Hf than in HCl (b) Molecules of AsF 3 are polar, whereas molecules of AsF 5 are nonpolar. The 6Fs would be completely symmetraical in all planes as an octahedral shape, whereas the AsF5 can not be (TBP shape) (c) The N-O bonds in the NO 2 - ion are equal in length, whereas they are unequal in HNO 2. NO2 has resonance structures where HNO2 does not. (I had this backwards in the last version)- (d) For sulfur, the fluorides SF 2, SF 4, and SF 6 are known to exist, whereas for oxygen only OF 2 is known to exist. Because S can expand whereas O can not

1997 D Consider the molecules PF 3 and PF 5. (a) Draw the Lewis electron-dot structures for PF 3 and PF 5 and predict the molecular geometry of each. (b) where there is no dipole overall because the bonds are equally spread on the molecule (b) Is the PF 3 molecule polar, or is it nonpolar? Explain. Polar because 1.9 Eneg difference is not cancelled out by geometry (pull down between all 3 Fs_- see image above (c) On the basis of bonding principles, predict whether each of the following compounds exists. In each case, explain your prediction. (i) NF5 No- N can not exand (ii) AsF5 yes because As can expand (TBP geometry)