Chapter 9 Molecular Geometry and Bonding Theories

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1 Chapter 9 Molecular Geometry and Bonding Theories 1. or a molecule with the formula AB 2 the molecular shape is. (a). linear or trigonal planar (b). linear or bent (c). linear or T-shaped (d). T-shaped Explanation: When answering questions like these always try to imagine a molecule that would fit the generic (AB 2 ) formula. In this case you could use the examples of C 2 or H 2. Thus in the case of C 2, the molecule would be linear since the central atom would not have any lone pairs while with water the shape would be bent due to the lone pairs on the central atom. 2. According to VSEPR theory, if there are five electron domains in the valence shell of an atom, they will be arranged in a(n) geometry. (a). octahedral (b). trigonal bipyramidal (c). tetrahedral (d). trigonal planar Explanation: The trigonal bipyramidal arrangement of electron domains would be the ideal one to minimize the repulsions involved. 3. The electron-domain geometry and molecular geometry of iodine trichloride are and respectively. (a). trigonal bipyramidal, trigonal planar (b). tetrahedral, trigonal pyramidal (c). trigonal bipyramidal, T-shaped (d). octahedral, trigonal planar Explanation: The Lewis structure of I 3 shows that the two lone pairs occupy the equatorial positions in this trigonal bipyramidal electron domain geometry. The molecule itself is T-shaped. The lone pairs occupy equatorial positions and are 120 apart, experiencing the least amount of repulsions, compared to axial positions. I Copyright 2006 Dr. Harshavardhan D. Bapat 1

2 4. According to valence bond theory, which orbitals on bromine atoms overlap in the formation of the bond in Br 2? (a). 3s (b). 3p (c). 4s (d). 4p Explanation: According to the valence bond theory, orbital overlap between two atoms leads to a bond formation. The orbitals that overlap are usually the outermost valence orbitals. In Br atoms, the valence electrons are held in the 4s and 4p orbitals with the 4p being the outermost orbital. 5. The electron-domain geometry of a sulfur-centered compound is trigonal bipyramidal. The hybridization of the central sulfur atom is. (a). sp (b). sp 2 (c). sp 3 d (d). sp 3 Explanation: A trigonal bipyramidal electron domain geometry indicates that there are 5 electron domains available. These 5 electron domains can be made possible only if a sp 3 d set of hybrid orbitals is available on the central atom. 6. The hybridization of orbitals on the central atom in a molecule is sp. The electrondomain geometry around this central atom is. (a). octahedral (b). linear (c). trigonal planar (d). tetrahedral Explanation: The presence of sp hybrid orbitals indicates that the central atom can form only 2 bonds, making the electron domain geometry linear. 7. The hybridization of orbitals on the central atom in a molecule is sp 2. The electrondomain geometry about this central atom is. (a). tetrahedral (b). linear (c). trigonal planar (d). trigonal bipyramidal Copyright 2006 Dr. Harshavardhan D. Bapat 2

3 Explanation: The presence of the sp 2 hybrid orbitals on the central atom indicate that the electron domains can be arranged only in a trigonal planar geometry. 8. The hybridization of the carbon atom in carbon dioxide is. (a). sp (b). sp 2 (c). sp 3 (d). sp 3 d Explanation: The Lewis structure of the molecule shows that it is a linear molecule with no lone pairs on the central atom. As a result of this the central atom can have only sp hybrid orbitals on it. C 9. In order to produce sp 3 hybrid orbitals, s atomic orbital(s) and p atomic orbitals (s) must be mixed. (a). one, two (b). one, three (c). one, one (d). two, two Explanation: The sp 3 hybrid orbitals are formed by the combination of one s orbital and three p orbitals. This information can be deduced by adding the number of orbitals (one s + three p ) present in the name of the hybrid orbitals. 10. There are σ and π bonds in the H C C H molecule. (a). 3 and 2 (b). 3 and 4 (c). 4 and 3 (d). 2 and 3 Explanation: The single bonds between the C and H atoms are counted as sigma (σ) bonds while the triple bond between the 2 C atoms is made of one sigma and two pi (π) bonds. 11. There are σ and π bonds in the H 2C C CH 2 molecule. (a). 4, 2 (b). 6, 2 (c). 2, 2 (d). 2, 6 Copyright 2006 Dr. Harshavardhan D. Bapat 3

4 Explanation: Here the C atoms on the outside are each bonded to 2 H atoms by single bonds and then boded to the middle carbon with a single bond, making the total sigma bonds = 6. The number of pi bonds on the other hand is 2, bonding the middle C atom to the outside C atoms. 12. In counting the electron domains around the central atom in VSEPR theory, a is not included. (a). nonbonding pair of electrons (b). single covalent bond (c). core level electron pair (d). double covalent bond Explanation: According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, it is the electrons that are in the valence (outermost) shell that contribute towards bonding. The core electrons do not participate in bonding at all. 13. The -C- bond angle in the C 3 2- ion is approximately. (a). 90 (b) (c). 180 (d). 120 o Explanation: The Lewis structure of the C 3 2- ion shows that the three atoms are arranged around the central C atom in a trigonal planar manner. Since the central C atom does not have any lone pairs the bond angles would not be affected and hence be exactly C 14. The -C- bond angle in the C 2 molecule (C is the central atom) is slightly. (a). greater than (b). less than (c). less than 120 (d). greater than 120 Explanation: The Lewis structure of the C 2 molecule shows that there is a double bond between the and the central C atom. A double bond represents a Copyright 2006 Dr. Harshavardhan D. Bapat 4

5 slightly larger electron domain which disrupts the ideal 120 -C- bond angle, making it smaller. C 15. The molecular geometry of the left-most carbon atom in the molecule below is. H C H H C (a). trigonal planar (b). tetrahedral (c). trigonal bipyramidal (d). octahedral H Explanation: When working with a molecule that has more than one central atom always concentrate on counting the number of electron domains around the particular central atom in question. Here the left most C atom has 4 electron domains around it and no lone pairs. As a result of this the geometry around it would be tetrahedral. 16. The molecular geometry of the right-most carbon in the molecule below is. H H C C H (a). T-shaped (b). trigonal planar (c). tetrahedral (d). octahedral H Explanation: When working with a molecule that has more than one central atom always concentrate on counting the number of electron domains around the particular central atom in question. Here the right most C atom has 3 electron domains around it and no lone pairs. As a result of this the geometry around it would be trigonal planar. Copyright 2006 Dr. Harshavardhan D. Bapat 5

6 17. An electron domain consists of (a). a nonbonding pair of electrons (b). a single bond (c). a multiple bond (a). a and b only (b). b and c only (c). a and c only (d). a, b, and c Explanation: Electron domains are regions where valence electrons are most likely to be present. These electrons may be present as a non-bonding pair or a bonding pair of electrons. 18. According to VSEPR theory, if there are three electron domains on a central atom, they will be arranged such that the angles between the domains are. (a). 90 (b). 180 (c) (d). 120 Explanation: According to the VSEPR theory the best arrangement of a given number of electron domains is the one that minimizes the repulsions between the pairs. Thus for 3 electron domains the best arrangement will be the one with angles of According to VSEPR theory, if there are four electron domains on a central atom, they will be arranged such that the angles between the domains are. (a). 120 (b) (c). 180 (d). 90 Explanation: According to the VSEPR theory the best arrangement of a given number of electron domains is the one that minimizes the repulsions between the pairs. Thus for 4 electron domains the best arrangement will be the one with angles of Copyright 2006 Dr. Harshavardhan D. Bapat 6

7 20. The electron-domain geometry and the molecular geometry of a molecule of the general formula AB n are. (a). never the same (b). always the same (c). sometimes the same (d). not related to each other Explanation: The electron domain geometry of a molecule describes the arrangement of the electron domains in the molecule while the molecular geometry describes the arrangement of the actual atoms in the molecule. The presence of non-bonding electron pairs can affect the molecular geometry making it different than the electron domain geometry. 21. The electron-domain geometry and the molecular geometry of a molecule of the general formula AB n will always be the same if. (a). there are no lone pairs on the central atom (b). there is more than one central atom (c). n is greater than four (d). the octet rule is obeyed Explanation: The presence of lone pairs on the central atom can alter the electron domain geometry of the molecule. The two geometries can be the same only if there are no lone pairs present on the central atom. 22. or molecules of the general formula AB n, n can be greater than four. (a). for any element A (b). only when A is boron or beryllium (c). only when A is an element from the third period or below the third period (d). only when A is Kr Explanation: The value of n can be greater than 4 only the element A is from the third or higher period of the periodic table. Elements from these periods can have more than 8 electrons around their atoms (have expanded octets). 23. f the molecules below, only is polar. (a). C 4 (b). CH 4 (c). Si 4 (d). Se 4 Explanation: The polarity of a molecule is closely related to its electron domain and molecular geometries. Both the electron domain and molecular geometries of C 4, Copyright 2006 Dr. Harshavardhan D. Bapat 7

8 CH 4 and Si 4 are tetrahedral. The molecular geometry of Se 4 is see-saw while the electron-domain geometry is trigonal bipyramidal due to a lone pair on the central Se atom. As a result of the lone pair on the central atom, this molecule becomes polar. 24. f the molecules below, which one is nonpolar? Se (a). B 3 (b). N 3 (c). I 3 (d). PBr 3 Explanation: With the exception of the B 3 molecule all the other molecules will have one or more lone pairs on the central atom. As a result of the lone pair(s) the molecule will then be polar. In B 3 the polar B- bonds will cancel each other, making the molecule non-polar. 25. Three sulfur fluorides are observed: S 2, S 4, and S 6. f these, is/are polar. (a). S 2 only (b). S 2 and S 4 only (c). S 4 only (d). S 6 only Explanation: The polarity of a molecule is closely related to its electron domain and molecular geometries. Here, S 6 has an octahedral geometry and does not have any electrons on the central S atom. As a result of this the polar S- bonds cancel each other s pulls. In both the other fluorides, the central S atom has lone pairs on it making them polar molecules. 26. The molecular geometry of the Be 2 molecule is, and this molecule is. (a). linear, nonpolar (b). linear, polar (c). bent, nonpolar (d). bent, polar Explanation: The polarity of a molecule is closely related to its electron domain and molecular geometries. The Lewis structure of this molecule clearly shows that the Copyright 2006 Dr. Harshavardhan D. Bapat 8

9 molecule is linear and since the central Be atom does not have any lone pairs on it, the polar Be- bonds will negate each other making it a non-polar molecule. Be 27. or molecules with only one central atom, how many lone pairs on the central atom guarantees molecular polarity? (a). 1 (b). 2 (c). 1 or 2 (d). Any number greater than 1. Explanation: The polarity of a molecule is closely related to its electron domain and molecular geometries. The presence of one lone pair of electrons on the central atom makes the molecule polar, even if other bonds in the molecule may cancel each other. 28. According to valence bond theory, which orbitals overlap in the formation of the bond in HBr? (a). 1s on H and 4s on Br (b). 1s on H and 4p on Br (c). 1s on H and 3p on Br (d). 2s on H and 4p on Br Explanation: According to the valence bond theory bonds are formed due to overlaps between valence orbitals. In the case of the HBr molecule the valence electrons are contained in the 1s orbital for H and in the 4p orbital for the Br. The 4s orbital is full and does not have any room. 29. The combination of two atomic orbitals results in the formation of molecular orbitals. (a). 1 (b). 2 (c). 3 (d). 0 Explanation: The combination of two atomic orbitals always leads to the formation of a bonding and an antibonding molecular orbital. Copyright 2006 Dr. Harshavardhan D. Bapat 9

10 30. The sp 3 d 2 atomic hybrid orbital set accommodates electron domains. (a). 2 (b). 3 (c). 6 (d). 5 Explanation: The number of electron domains indicated by a set of hybrid orbitals can be easily found by adding the number of atomic orbitals present in the set. Here we have one s + three p + two d (= 6) orbitals present making 6 electron domains available. 31. The sp 2 atomic hybrid orbital set accommodates electron domains. (a). 2 (b). 3 (c). 4 (d). 5 Explanation: The number of electron domains indicated by a set of hybrid orbitals can be easily found by adding the number of atomic orbitals present in the set. Here we have one s + two p (= 3) orbitals present making 3 electron domains available. 32. The hybridization of iodine in I 3 and I 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 Explanation: The Lewis structures of these molecules indicate that I 3 will have a trigonal bipyramidal electron-domain geometry while I 5 will have an octahedral geometry. To be able to have these geometries the molecules will need the hybrid orbitals to accommodate the electron domains. I I Copyright 2006 Dr. Harshavardhan D. Bapat 10

11 33. The hybridizations of bromine in Br 5 and of arsenic in As 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 Explanation: The Lewis structures of these molecules indicate that the Br 5 will have an octahedral electron-domain geometry where as the As 5 will have trigonal bipyramidal geometry. The central Br atom will thus need 6 electron domains to accommodate the 5 Br- bonds and the lone pair. The central As atom on the other hand will need only 5 electron domains to bond with the 5 atoms. Br As Consider the following species when answering questions 34 and 35: (i) P 3 (ii) C 4 (iii) Te 4 (iv) Xe 4 (v) S In which of the molecules does the central atom utilize d orbitals to form hybrid orbitals? (a). (i) and (ii) (b). (iii) only (c). (i) and (v) (d). (iii), (iv), and (v) Explanation: Molecules with central atoms from the third period and beyond can have an expanded octet; meaning more than 4 pairs of electrons around them. To do this the atoms use their empty d orbitals. f all the species here, only Te, Xe and S fit this requirement and hence will utilize the d orbitals to form hybrid molecular orbitals. Copyright 2006 Dr. Harshavardhan D. Bapat 11

12 35. Which of the molecules has a see-saw shape? (a). (i) only (b). (ii) only (c). (iii) only (d). (iv) only Explanation: The Lewis structures of the species involved clearly show that only the Te 4 will have a see-saw shape. P C Xe S Copyright 2006 Dr. Harshavardhan D. Bapat 12

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