Assignment 09 A. 2- The image below depicts a seesaw structure. Which of the following has such a structure?

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1 Assignment 09 A 1- Give the total number of electron domains, the number of bonding and nonbonding domains, and the molecular geometry, respectively, for the central atom of P 3. a) four electron domains, two bonding domains, and two nonbonding domains, tetrahedral b) four electron domains, one bonding domain, and three nonbonding domains, linear c) four electron domains, three bonding domains, and one nonbonding domain, trigonal planar d) three electron domains, three bonding domains, and zero nonbonding domains, trigonal planar e) four electron domains, three bonding domains, and one nonbonding domains, trigonal pyramidal (There are three bonded domains and one nonbonded domain for a total of four electron domains about the P atom. Since only three of the four domains are bonding, the molecular shape is trigonal pyramidal.) 2- The image below depicts a seesaw structure. Which of the following has such a structure? a) XeF 4 b) BF 4 c) SiF 4 d) SeF 4 (SeF 4 has a seesaw structure due to four bonding domains and one lone nonbonded pair of electrons on the selenium atom.) 3- Which of the following molecules contains polar bonds but is nonpolar? a) F 2 b) H 2 O c) CF 4 d) SO 2 e) NF 3 (This is a tetrahedral molecule with the polar C F bonds symmetrically arranged about the C atom.) 4- Which hybrid orbital-type and molecular geometry are incorrect? a) sp 3 d: trigonal bipyramidal b) sp 3 : tetrahedral c) sp 3 d 2 : octahedral d) sp 2 : trigonal pyramidal e) sp: linear (The sp 2 hybrid orbital set is trigonal planar; there are three sets of orbital lobes.) 5- Which molecule has a nitrogen atom with sp 2 hybridization? a) nitrosyl cation, NO b) trimethylamine, :N(CH 3 ) 3 c) acetonitrile, CH 3 CN: d) nitromethane, CH 3 NO 2 e) ammonium cation, NH 4 (The Lewis structure for this molecule shows four bonds on the nitrogen (two single and one double), so it has three electron domains and sp 2 hybridization.) 6- How many and bonds are in C 2 H 2 in which the two carbon atoms are adjacent and each carbon atom has

2 one hydrogen atom attached to it? a) three bonds and two bonds b) two bonds and one bond c) four bonds and one bond d) three bonds and one bond e) one bond and one bond (There is one sigma bond for the C C bond and one for the C H bonds, as well as two C C pi bonds.) 7- According to molecular orbital theory, predict the bond order for H 2. a) 1.5 b) 2 c) 2.5 d) 1 e) 0.5 (This ion has two bonding electrons and one antibonding electron, giving it a bond order of 0.5.) 8- Which of the following ions are diamagnetic? N a) N b) N 2 c) N d) O 2 2 e) N 2 f) O 2 (Both the O 2 2 and the Be 2 2 ions contain no unpaired electrons; thus, they are diamagnetic.) 9- Are the following molecules polar? (i) (ii) H a) More information is needed on bond angles. b) Only i is polar. c) Neither is polar. d) Only ii is polar. e) Both i and ii are polar. (In this molecule, there are only three atoms, and they are not symmetrically arranged; hence, the dipoles do not cancel.) 10- Describe the characteristic electron-domain geometry of each of the following numbers of electron domains about a central atom, respectively: 3, 4 a) 3: trigonal planar, 4: T-shaped b) 3: linear, 4: trigonal planar c) 3: bent, 4: trigonal planar d) 3: trigonal pyramidal, 4: tetrahedral e) 3: trigonal planar, 4: bent f) 3: trigonal planar, 4: tetrahedral (Three electron domains will be arranged in trigonal planar fashion, and four electron domains will be arranged in tetrahedral fashion.) 11- The image below depicts a tetrahedral structure. Which of the following does not have such a structure?

3 a) SiBr 4 b) SBr 4 c) GeF 4 d) Al 4 e) PH 4 (SBr 4 has four bonding domains and one nonbonding domain about the central S atom, giving the molecule a see-saw molecular shape.) 12- What are the bond angles about carbon in the following structure? O a) b) 90 c) 60 d) 180 e) 120 (There are three electron domains about the C atom, which results in bond angles of 120º.) 13- S 2 has a nonzero dipole moment, but Be 2 has a dipole moment of zero. How can you explain the difference? a) S is more electronegative than Be. b) S has two nonbonding pairs of electrons. c) S is more massive than Be. d) S can expand its octet. e) Be forms a double bond with one of the atoms. (The two nonbonding domains on the S cause this to be a bent molecule, while Be 2 is a linear molecule.) 14- Which of the following molecules is/are polar? C 4, SF 4, CS 2, NF 3, SO 3, PF 5 a) CS 2 and PF 5 b) NF 3 only c) CS 2 and SF 4 only d) PF 5 only e) SF 4 and NF 3 only f) SF 4 only (Both of these are polar molecules because of an asymmetric distribution of bonding and nonbonding electron domains.) 15- Indicate the designation for the hybrid orbitals formed from each of the following combinations of atomic orbitals: (i) one s and two p (ii) one s, three p, and one d a) sp 2 and spd b) sp and sp 2 d c) sp and spd d) sp 2 and sp 2 d e) sp 2 and sp 3 d (The first combination involves one s and two p orbitals, while the second combination involves one s, three p, and one d orbitals.)

4 16- Arrange the following bonds in terms of decreasing strength (bond order): O 2, O, O 2 2. a) O 2 2 b) O 2 2 c) O 2 2 d) O 2 2 e) O 2 2 (The bond orders are 2.5, 2, 1.5, and 1 respectively.) 17- Give the bond order and number of unpaired electrons in C 2 and N 2. a) C 2 bond order 2, 0 unpaired electrons; N 2 bond order 2.5, 0 unpaired electrons. b) C 2 bond order 1, 2 unpaired electrons; N 2 bond order 3.5, 1 unpaired electron c) C 2 bond order 2, 2 unpaired electrons; N 2 bond order 2.5, 1 unpaired electron d) C 2 bond order 2, 1 unpaired electron; N 2 bond order 2.5, 1 unpaired electron e) C 2 bond order 2, 0 unpaired electrons; N 2 bond order 2.5, 1 unpaired electron (The 2p orbitals are lower in energy than the 2p orbitals in C 2, producing essentially two pi bonds; N 2 is also correct.) 18- Give the total number of electron domains, the number of bonding and nonbonding domains, and the molecular geometry for the central atom of O 2. a) four electron domains, one bonded domain, and three nonbonding electron domains, linear b) two electron domains, two bonded atoms, and zero nonbonding domains, bent c) four electron domains, two bonded atoms, and two nonbonding electron domains, tetrahedral d) two electron domains, two bonded domains, and zero nonbonding domains, linear e) four electron domains, two bonding, and two nonbonding, bent (There are two atoms bonded to the and two nonbonded domains, giving a total of four electron domains. The shape for four electron domains and only two bonded atoms is bent.) 19- In which one of the following are there five electron domains (bonding and nonbonding) around the central atom? a) Si 4 b) Al 6 3 c) I 2 d) PH 4 e) NF 3 (This ion contains a central I atom with two bonding domains and three nonbonding domains. The "octet" of I is expanded.) 20- Which identification of shape is incorrect? a) Al 6 3 ; octahedral b) Sb 5 ; trigonal bipyramidal c) I 3 ; trigonal pyramidal d) F 5 ; square pyramidal e) Se 4 ; seesaw (I 3 has a T-shaped molecular structure, with three bonding domains and two nonbonding domains on the central I atom.) 21- In which molecule do we find bond angles of 109.5? a) BF 3 b) SF 4 c) C 4 d) Br 5 e) XeF 4

5 (C 4 is tetrahedral with bond angles of ) 22- Which of the following is nonpolar? (i) CH 4 (ii) CHBr 3 (iii) CBr 2 2 (iv) CH 3 Br a) CH 4 b) CHBr 3 c) CBr 2 2 d) CH 3 Br e) All of the above are nonpolar. (This is a tetrahedral molecule with four identical atoms bonded to the carbon; hence, all of the dipoles cancel.) 23- Indicate the hybrid-orbital set used by the central atom in each of the following polyatomic ions, respectively. (i) BH 4 (ii) PF 6 a) sp 3 and sp 3 d 2 b) sp 3 and sp 3 d 3 c) sp 2 and sp 3 d 2 d) sp 2 and sp 2 d 2 e) sp 3 and sp 2 d 2 (BH 4 has four electron domains and is sp 3 hybridized. PF 6 has six electron domains and is sp 3 d 2 hybridized.) 24- SiF 6 2 exists, but CF 6 2 does not. Why? Consider the concept of hybrid orbitals in your answer. a) Silicon can hybridize using nearby 3d orbitals. Such orbitals are not available to carbon since it is located in the second period of the periodic table. b) Carbon sp 3 cannot accommodate six ligands. c) Carbon is never found in 2 ions. d) Silicon is less electronegative, so its hybrid orbitals are lower in energy. e) Silicon is unusual, and hybridization is irrelevant. (Silicon can form sp 3 d 2 hybrids, whereas carbon cannot.) 25- The energy-level diagram for molecular orbitals of second-row diatomic molecules predicts which of the following to be paramagnetic? O 2, N 2, B 2 a) N 2 and B 2 b) O 2 and B 2 c) N 2 only d) O 2 only e) N 2 and O 2 (Both of these molecules contain two unpaired electrons.)

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