Name: Class: Date: Unit 9 Practice Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The basis of the VSEPR model of molecular bonding is. A) regions of electron density on an atom will organize themselves so as to maximize s-character B) regions of electron density in the valence shell of an atom will arrange themselves so as to maximize overlap C) atomic orbitals of the bonding atoms must overlap for a bond to form D) electron domains in the valence shell of an atom will arrange themselves so as to minimize repulsions E) hybrid orbitals will form as necessary to, as closely as possible, achieve spherical symmetry 2) The electron-domain geometry of is tetrahedral. A) CBr 4 B) PH 3 C) CCl 2 Br 2 D) XeF 4 E) all of the above except XeF 4 3) The bond angles marked a, b, and c in the molecule below are about,, and, respectively. A) 109.5, 109.5, 109.5 B) 120, 109.5, 120 C) 109.5, 109.5, 120 D) 90, 180, 90 E) 109.5, 109.5, 90 4) A molecule has the formula AB3 and the central atom is in a different plane from the surrounding three atoms. Its molecular shape is. A) tetrahedral B) trigonal pyramidal C) linear D) T-shaped E) bent 1
Name: 5) Three monosulfur fluorides are observed: SF 2, SF 4, and SF 6. Of these, is/are polar. A) SF 2 only B) SF 2 and SF 4 only C) SF 4 only D) SF 6 only E) SF 2, SF 4, and SF 6 6) The molecular geometry of the BCl 3 molecule is, and this molecule is. A) trigonal pyramidal, polar B) trigonal pyramidal, nonpolar C) trigonal planar, polar D) trigonal planar, nonpolar E) trigonal bipyramidal, polar 7) The hybridizations of iodine in IF 3 and IF 5 are and, respectively. A) sp 3, sp 3 d B) sp 3 d, sp 3 d 2 C) sp 3 d, sp 3 D) sp 3 d 2, sp 3 d E) sp 3 d 2, sp 3 d 2 8) The electron-domain geometry of the AsF 5 molecule is trigonal bipyramidal. The hybrid orbitals used by the As atom for bonding are orbitals. A) sp 2 d 2 B) sp 3 C) sp 3 d 2 D) sp 3 d E) sp 2 9) A typical double bond. A) is stronger and shorter than a single bond B) consists of one bond and one bond C) imparts rigidity to a molecule D) consists of two shared electron pairs E) All of the above answers are correct. 10) Which of the following molecules or ions will exhibit delocalized bonding? SO 2 SO 3 SO 2-3 A) SO 2, SO 3, and SO 2-3 B) SO 2-3 only C) SO 2 and SO 3 D) SO 3 and SO 2-3 E) None of the above will exhibit delocalized bonding. 11) The hybridization of the terminal carbons in the H 2 C C CH 2 molecule is. A) sp B) sp 2 C) sp 3 D) sp 3 d E) sp 3 d 2 2
Name: 12) The bond order of any molecule containing equal numbers of bonding and antibonding electrons is. A) 0 B) 1 C) 2 D) 3 E) 1/2 13) Based on molecular orbital theory, the bond order of the H H bond in the H 2 + ion is. A) 0 B) 1/2 C) 1 D) 3/2 E) 2 14) Molecular Orbital theory correctly predicts paramagnetism of oxygen gas, O 2. This is because. A) the bond order in O 2 can be shown to be equal to 2. B) there are more electrons in the bonding orbitals than in the antibonding orbitals. C) the energy of the 2p MOs is higher than that of the 2p MO D) there are two unpaired electrons in the MO electron configuration of O 2 E) the O O bond distance is relatively short 15) Of the following, only appears to gain mass in a magnetic field. A) C 2 B) N 2 C) F 2 D) O 2 E) Li 2 16) The bond order of a homonuclear diatomic molecule can be decreased by. A) removing electrons from a bonding MO or adding electrons to an antibonding MO B) adding electrons to a bonding MO or removing electrons from an antibonding MO C) adding electrons to any MO D) removing electrons from any MO E) The bond order of a homonuclear diatomic molecule cannot be decreased by any means. 17) For a molecule with the formula AB 2, the molecular shape is. A) linear or bent B) linear or trigonal planar C) linear or T-shaped D) T-shaped E) trigonal planar 18) The molecular geometry of the SiH 2 Cl 2 molecule is. A) trigonal planar B) tetrahedral C) trigonal pyramidal D) octahedral E) T-shaped 3
Name: 19) The O-S-O bond angle in SO 2 is slightly less than. A) 90 B) 109.5 C) 120 D) 180 E) 60 20) The hybridization of orbitals on the central atom in a molecule is sp. The electron-domain geometry around this central atom is. A) octahedral B) linear C) trigonal planar D) trigonal bipyramidal E) tetrahedral 21) The electron-domain geometry of the AsF 6 - ion is octahedral. The hybrid orbitals used by the As atom for bonding are orbitals. A) sp 2 d 2 B) sp 3 C) sp 3 d D) sp 3 d 2 E) sp 2 22) There are and bonds in the H C C H molecule. A) 3 and 2 B) 3 and 4 C) 4 and 3 D) 2 and 3 E) 5 and 0 23) There is/are bond(s) in the molecule below. A) 7 B) 6 C) 2 D) 1 E) 0 4
Name: 24) The central atom in the ICl 4 - ion has nonbonded electron pair(s) and bonded electron pair(s) in its valence shell. A) 4, 2 B) 0, 4 C) 2, 4 D) 2, 2 E) 6, 0 25) The electron-domain geometry and molecular geometry of the nitrite ion are and, respectively. A) trigonal bipyramidal, trigonal planar B) tetrahedral, trigonal pyramidal C) trigonal bipyramidal,t-shaped D) trigonal planar, bent E) T-shaped, trigonal planar 26) Using the VSEPR model, the molecular geometry of the central atom in SO 3 is. A) linear B) trigonal planar C) tetrahedral D) bent E) trigonal pyramidal 27) Of the molecules below, only is polar. A) CO 2 B) CH 4 C) PF 5 D) SnF 3 E) H 2 28) According to valence bond theory, which orbitals overlap in the formation of the bond in Br 2? A) 3s B) 3p C) 4s D) 4p E) 3d 29) There is/are hybridized s atomic orbital(s) in an sp-hybridized carbon atom. A) 4 B) 0 C) 3 D) 1 E) 2 30) Based on molecular orbital theory, the bond order of the N N bond in the N 2 molecule is. A) 0 B) 1/2 C) 3 D) 2 E) 1 5
Name: 31) In molecular orbital theory, the bond order of the He He bond in He 2 is. A) 0 B) 2 C) He He D) 1 E) 1/2 Short Answer 1) What is the molecular geometry of a molecule that has three bonding and two non-bonding domains? 2) In the valence shell of an atom there are six electron domains. They will be arranged in a(n) geometry. 3) What are the three bond angles in the trigonal bipyramidal structure? 4) Three molecules have similar electron domains, but different molecular shapes. Why? 5) The 1s hydrogen orbital overlaps with the iodine orbital in HI. 6) A covalent bond in which overlap regions lie above and below an internuclear axis is called a(n). 7) The sensation of vision results from a nerve impulse that is triggered by the separation of retinal from. 8) In molecular orbital theory, the stability of a covalent body is related to its. 9) Each molecular orbital can accommodate, at most, two electrons with their spins paired. This is called the. 10) The more unpaired electrons in a species, the stronger is the force of magnetic attraction. This is called. True/False Indicate whether the statement is true or false. 1) Possible shapes of AB 3 molecules are linear, trigonal planar, and T-shaped. 2) Boron trifluoride has three bonding domains, and its electron domain geometry is trigonal planar. 3) Electron domains for single bonds exert greater force on adjacent domains than the electron domains for multiple bonds. 4) The quantitative amount of charge separation in a diatomic molecule contributes to the dipole moment of that molecule. 6
Name: 5) XeF 4 is a polar molecule. 6) Hybridization is the process of mixing atomic orbitals as atoms approach each other to form a bond. 7) Electrons in core orbitals contribute to atom bonding. 8) Nitrogen is colorless because the minimum energy to excite an electron is in the ultraviolet section of the spectrum. 7
Unit 9 Practice Answer Section MULTIPLE CHOICE 1) ANS: D PTS: 1 DIF: 1 REF: Page Ref: 9.2 2) ANS: E PTS: 1 DIF: 2 REF: Page Ref: 9.2 3) ANS: C PTS: 1 DIF: 2 REF: Page Ref: 9.2 4) ANS: B PTS: 1 DIF: 2 REF: Page Ref: 9.2 5) ANS: B PTS: 1 DIF: 2 REF: Page Ref: 9.3 OBJ: 9.3; G2 6) ANS: D PTS: 1 DIF: 1 REF: Page Ref: 9.3 OBJ: 9.3; G2 7) ANS: B PTS: 1 DIF: 1 REF: Page Ref: 9.5 8) ANS: D PTS: 1 DIF: 1 REF: Page Ref: 9.5 9) ANS: E PTS: 1 DIF: 1 REF: Page Ref: 9.6 10) ANS: C PTS: 1 DIF: 1 REF: Page Ref: 9.6 11) ANS: B PTS: 1 DIF: 2 REF: Page Ref: 9.6 12) ANS: A PTS: 1 DIF: 3 REF: Page Ref: 9.7 OBJ: 9.7; G2 13) ANS: B PTS: 1 DIF: 4 REF: Page Ref: 9.7 OBJ: 9.7; G2 14) ANS: D PTS: 1 DIF: 4 REF: Page Ref: 9.7-9.8 OBJ: 9.7, 9.8; G2 15) ANS: D PTS: 1 DIF: 4 REF: Page Ref: 9.8 OBJ: 9.8; G2 16) ANS: A PTS: 1 DIF: 4 REF: Page Ref: 9.8 OBJ: 9.8; G2 17) ANS: A PTS: 1 DIF: 1 REF: Page Ref: 9.1 18) ANS: B PTS: 1 DIF: 1 REF: Page Ref: 9.2 19) ANS: C PTS: 1 DIF: 1 REF: Page Ref: 9.2 20) ANS: B PTS: 1 DIF: 1 REF: Page Ref: 9.5 21) ANS: D PTS: 1 DIF: 2 REF: Page Ref: 9.5 1
22) ANS: A PTS: 1 DIF: 1 REF: Page Ref: 9.6 23) ANS: D PTS: 1 DIF: 1 REF: Page Ref: 9.6 24) ANS: C PTS: 1 DIF: 1 REF: Page Ref: 9.2 25) ANS: D PTS: 1 DIF: 1 REF: Page Ref: 9.2 26) ANS: B PTS: 1 DIF: 1 REF: Page Ref: 9.2 27) ANS: D PTS: 1 DIF: 2 REF: Page Ref: 9.3 OBJ: 9.3; G2 28) ANS: D PTS: 1 DIF: 1 REF: Page Ref: 9.4 OBJ: 9.4; G2 29) ANS: D PTS: 1 DIF: 1 REF: Page Ref: 9.5 30) ANS: C PTS: 1 DIF: 4 REF: Page Ref: 9.8 OBJ: 9.7, 9.8; G2 31) ANS: C PTS: 1 DIF: 1 REF: Page Ref: 9.7-9.8 OBJ: 9.7, 9.8; G2 SHORT ANSWER 1) ANS: T-shaped PTS: 1 DIF: 4 REF: Page Ref: 9.2 2) ANS: octahedral PTS: 1 DIF: 5 REF: Page Ref: 9.2 3) ANS: 90, 120, 180 PTS: 1 DIF: 4 REF: Page Ref: 9.2 4) ANS: different numbers of non-bonding domains PTS: 1 DIF: 4 REF: Page Ref: 9.2 5) ANS: 5p PTS: 1 DIF: 2 REF: Page Ref: 9.5 2
6) ANS: bond PTS: 1 DIF: 2 REF: Page Ref: 9.6 7) ANS: opsin PTS: 1 DIF: 2 REF: Page Ref: 9.6 8) ANS: bond order PTS: 1 DIF: 2 REF: Page Ref: 9.7 OBJ: 9.7; G2 9) ANS: Pauli principle PTS: 1 DIF: 4 REF: Page Ref: 9.8 OBJ: 9.8; G2 10) ANS: paramagnetism PTS: 1 DIF: 1 REF: Page Ref: 9.8 OBJ: 9.8; G2 TRUE/FALSE 1) ANS: F PTS: 1 DIF: 2 REF: Page Ref: 9.1 2) ANS: T PTS: 1 DIF: 2 REF: Page Ref: 9.2 3) ANS: F PTS: 1 DIF: 1 REF: Page Ref: 9.2 4) ANS: T PTS: 1 DIF: 1 REF: Page Ref: 9.3 OBJ: 9.3; G2 5) ANS: F PTS: 1 DIF: 3 REF: Page Ref: 9.3 OBJ: 9.3; G2 6) ANS: T PTS: 1 DIF: 1 REF: Page Ref: 9.5 7) ANS: F PTS: 1 DIF: 1 REF: Page Ref: 9.6 8) ANS: T PTS: 1 DIF: 3 REF: Page Ref: 9.8 OBJ: 9.8; G2 3
Unit 9 Practice [Answer Strip] ID : A _ B 5) _ A 12) _ C 19) _ C 24) _ D 1) _ D 6) _ B 13) _ B 20) _ D 25) _ B 7) _ D 14) _ D 21) _ B 26) _ E 2) _ C 3) _ D 8) _ D 15) _ A 22) _ D 27) _ E 9) _ A 16) _ D 23) _ D 28) _ B 4) _ C 10) _ A 17) _ D 29) _ B 18) _ C 30) _ B 11)
Unit 9 Practice [Answer Strip] ID : A _ C 31) _ F 5) _ T 6) _ F 7) _ T 8) _ F 1) _ T 2) _ F 3) _ T 4)