General Chemistry 1 CHM201 Unit 4 Practice Test 1. An orbital s orientation in space is determined by a. the quantum number. d. the n quantum number. b. the m l quantum number. e. both the and m l quantum numbers. c. the m s quantum number. 2. Which subshell only has five orbitals? a. 2 d. 3 b. 1 e. 4 c. 5 3. A certain shell is known to have a total of 9 orbitals. Which shell is it? a. n 1 d. n 4 b. n 2 e. n 5 c. n 3 4. What is the orbital designation for an electron with the quantum numbers n 4, 3? a. 4s b. 4p c. 4d d. 4f e. cannot determine from the given information 5. Which orbital does not fill before a 3d in a multielectron atom? a. 2s d. 4s b. 2p e. 4p c. 3s 6. What is the ground state electron configuration for aluminum? a. 1s 2 2s 2 2p 6 3s 2 d. [Ne]3s 2 3p 1 b. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 4p 1 e. [Ar]3s 2 3p 1 c. [He]2s 2 2p 1
7. Which one of the following diagrams represents the ground state electron configuration of a carbon atom? a. c. b. d. 8. What is the ground-state electron configuration of an Ge 4+ ion? a. [Ar]3d 8 4s 2 d. [Ar]3d 8 4s 2 b. [Ar]3d 6 e. [Ar]3d 10 c. [Ar]3d 8 9. Which of the following atoms has no unpaired electrons in its ground state? a. Ca d. Cl b. K e. Cs c. Sc 10. Which of the following electron configurations represents an excited state? a. [Ne]3s 2 3p 5 d. [Ne]3s 2 3p 5 b. [Kr]4d 10 5s 1 5p 2 e. [Kr]4d 10 5s 1 c. [Ar]3d 10 4s 2 4p 6 11. Which of the following elements would you expect to have the greatest first ionization energy? a. C d. Be b. O e. B c. Li 12. An Si-Si bond would be classified as what type of bonding interaction? a. polar covalent d. nonpolar covalent b. metallic e. homogeneous c. ionic
13. How many valence electrons does S 2 have? a. 8 d. 4 b. 6 e. 14 c. 16 14. How many valence electrons are there in a correctly drawn Lewis structure for formamide, HCONH 2? a. 12 d. 18 b. 14 e. 20 c. 16 15. Indicate which molecule contains the largest number of nonbonding electrons. a. H 2 d. NO b. CO e. O 2 c. N 2 16. Which group of elements is listed in order of increasing electronegativity? a. F Cl Ge Sn d. Sn P Cl F b. Rb Ca Sc Cs e. Cl Br I F c. Zr V Nb Ta 17. Which of the following bonds is most polar? a. B H d. Al H b. N H e. C H c. P H 18. One of the resonance structures of the oxalate ion (C 2O 4 2 ) is shown below. How many other resonance structures, that obey the octet rule, exist for the molecule? a. 0 d. 3 b. 1 e. 4 c. 2
19. Based on consideration of formal charges, which of the following is the most stable Lewis structure for the azide ion (N 3 )? a. c. b. d. 20. Which of the following compounds is likely to have the weakest H-X bond? a. H Br (141 pm) d. H F (92 pm) b. H I (161 pm) e. H 2S (134 pm) c. H Cl (127 pm) 21. Which statement about VSEPR theory is not correct? a. VSEPR is an acronym for valence-shell electron-pair repulsion. b. In VSEPR theory, the shape or geometry of a molecule is determined by electron electron repulsion. c. The steric number of a central atom is the sum of the number of bonds and lone pairs around the atom. d. The steric number has five values from 2 to 6. e. The molecular shape or geometry always is the same as the electron-pair geometry. 22. A geometry corresponds to a steric number of. a. linear; 3 d. trigonal bipyramidal; 7 b. trigonal planar; 6 e. octahedral; 8 c. tetrahedral; 4 23. Which of the following ions is linear? a. SCN d. NH 4 + b. NO 2 e. SO 2 2 c. SO 3 2 24. Which bond is the least polar? a. H C d. H Cl b. H N e. H F c. H O
25. Identify the hybridization of the three carbon atoms labeled in the following molecule. I II III I II III a. sp 2, sp 3, sp 2 d. sp 2, sp, sp 3 b. sp 3, sp 2, sp e. sp 2, sp 3, sp c. sp, sp 3, sp 2
CHM201 Unit 4 Practice Test Answer Section MULTIPLE CHOICE 1. ANS: B From the descriptions of the quantum numbers: n - Principle quantum number indicates shell and relative size of orbital (distance from the nucleus); - Angular momentum quantum number indicates the shape of the orbital; m l - Magnetic quantum number indicates the orbital s orientation in space; m s - Spin magnetic quantum spin describes the electron s spin. PTS: 1 DIF: Easy REF: 7.6 OBJ: Identify the shell, subshells, and orbitals associated with an atomic energy level. MSC: Remembering 2. ANS: A There are five orbitals allowed by the subshell, determined by the angular momentum quantum number. Each orbital s orientation is determined by the magnetic quantum number, m l. The possible orbitals is constrained by the angular momentum quantum number. The possible orbitals in a subshell (m l values) may range from -l to +l. If there are only 5 orbitals, the ml values must be -2, -1, 0, +1, +2, and no other value is possible. This means that l must equal 2. These are d-subshells (l =2), and there can only be 5 orientations. PTS: 1 DIF: Easy REF: 7.6 OBJ: Identify the number of orbitals and electrons associated with particular values of quantum numbers. 3. ANS: C
There are 9 orbitals in the shell, and orbitals are designated by the magnetic quantum number, m l, which is, in turn determined by the angular momentum quantum number, l, or subshell. The number of subshells is determined by the principal quantum number, n. We are trying to determine the total number of orbitals, so all possible subshells must be counted. For n = 1, l = 0 and m l = 0 For n = 2, l = 0, and m l = 0 For n = 2, l = 1, and m l = -1, 0, +1 (1 orbital, 1s) only one orbital. (1 orbital, 2s). (3 orbitals, 2p) for a total of 4 orbitals. For n = 3, l = 0, and m l = 0 For n = 3, l = 1, and m l = -1, 0, +1 For n = 3, l = 2, and m l = -2, -1, 0, +1, +2 (1 orbital, 3s). (3 orbitals, 2p). (5 orbitals, 2p) for a total of 9 orbitals. An easier way is to take the square root of the total number of orbitals in the shell. This conforms with the rules for quantum numbers. PTS: 1 DIF: Medium REF: 7.6 OBJ: Identify the number of orbitals and electrons associated with particular values of quantum numbers. 4. ANS: D We are given the principal and angular momentum quantum numbers, n 4, l = 3 The orbital designation includes the shell (n) and the subshell (l ), or orbital type. The shell is represented by the numerical value of n, and the subshell is represented by the letter designation corresponding to the numerical value of l as follows: l = 0 is an s orbital. l = 1 is an p orbital. l = 2 is an d orbital. l = 3 is an f orbital.
n = 4, and l = 3, so the orbital designation is 4f. This conforms with the naming conventions. PTS: 1 DIF: Easy REF: 7.6 OBJ: Identify valid sets of quantum numbers for an electron in an atom. MSC: Remembering 5. ANS: E We are asked about Aufbau ordering according to the energy of electrons. We must use the Aufbau ordering chart to determine which of the selections has higher energy than the 3d subshell. Of the selections, only 4p comes after 3d in the filling order. 4p electrons are more energetic than 3d electrons. PTS: 1 DIF: Easy REF: 7.7 OBJ: Characterize s, p, and d orbitals in terms of their nodes. 6. ANS: D Aluminum is atomic number 13, so it has 13 electrons. The electron configuration of the ground state of an atom is determined by applying the Aufbau chart. According to the filling order: 1s 2 2s 2 2p 6 3s 2 3p 1. Since the electron configuration of the nearest noble gas, neon, is 1s 2 2s 2 2p 6, we can write the condensed notation:
[Ne] 3s 2 3p 1 The Aufbau requirements have been met. PTS: 1 DIF: Medium REF: 7.8 OBJ: Apply the Pauli exclusion principle, the aufbau principle, and Hund s rule to determine electron configurations of atoms. 7. ANS: B In which orbital diagram are all the rules for Aufbau modeling followed? We use the Aufbau ordering chart to determine the electron configuration for carbon. Then we must insert the electrons into the orbital diagram in compliance with the Pauli Exclusion Principle and Hund s Rule. Electron configuration: 1s 2 2s 2 1 2p 3. a. Violates Hund s Rule (parallel spins). b. Violates none of the rules or principles. c. Does not conform to the electron configuration derived from Aufbau. d. Does not conform to the electron configuration and violates Hund s Rule. Only one of them works. PTS: 1 DIF: Easy REF: 7.8 OBJ: Apply the Pauli exclusion principle, the aufbau principle, and Hund s rule to determine electron configurations of atoms. 8. ANS: E Germanium is atomic number 32, and we are considering the ion with a 2+ charge. The electron configuration of the ground state of an atom is determined by applying the Aufbau chart.
According to the filling order, the configuration of the germanium atom is: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 2 After losing the valence electrons (4s 2 4p 2 ), we are left with: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 Since the electron configuration of the nearest noble gas, argon, is 1s 2 2s 2 2p 6 3s 2 3p 6, we can write the condensed notation: [Ar] 3d 10 The Aufbau requirements have been met after losing four electrons from the valence shell just like its cousin, carbon. PTS: 1 DIF: Easy REF: 7.9 OBJ: Determine the electron configurations of cations and anions of main group elements. 9. ANS: A Calcium is atomic number 20, potassium 19, scandium is 21, chlorine is 17, and cesium is 55. We can eliminate any atoms with an odd number of electrons, since they can t possibly all be paired. That leaves only calcium (20). We should verify the configuration of calcium. According to the filling order, the configuration of the calcium atom is: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 or [Ar] 4s 2 The s-orbitals are lower energy than p-orbitals, so they fill completely before any p-orbitals. Two electrons in an s- orbital must be paired according to the Pauli Exclusion Principle. Our initial suspicion is verified. PTS: 1 DIF: Easy REF: 7.9 OBJ: Identify the number of unpaired electrons in an atom or ion. 10. ANS: B
We are given 5 electron configurations Compare each to the Aufbau ground-state configuration for the number of electrons present. [Ne]3s 2 3p 5 as predicted for the ground state [Kr]4d 10 5s 1 5p 2 not as predicted for the ground state. It should be [Kr]4d 10 5s 2 5p 1 [Ar]3d 10 4s 2 4p 6 as predicted for the ground state [Ne]3s 2 3p 5 as predicted for the ground state [Kr]4d 10 5s 1 as predicted for the ground state An s-shell must fill completely before adding p-shells. PTS: 1 DIF: Easy REF: 7.9 OBJ: Identify excited state electron configurations. 11. ANS: B Five elements from the same period are given. Periodic trends in the table for first ionization energies are known to increase from left to right in the periodic table. Ordering the elements as they appear in the table, Li, Be, B, C, O. The first ionization energies should be as follows: Li<Be<B<C<O, so oxygen should be the greatest. This conforms to the periodic trend. PTS: 1 DIF: Easy REF: 7.11 OBJ: Order atoms according to their first ionization energy based on their positions in the periodic table. MSC: Understanding 12. ANS: D
Two silicon atoms are bonded. What is the classification of the bond? Compare the electronegativities of the atoms to determine the polarity (or lack thereof) of the bond. Since the two atoms are identical, their electronegativities are the same and the ΔEN = 0. The bond is nonpolar covalent. Same element would favor neither in the attraction of electrons. PTS: 1 DIF: Easy REF: 8.1 OBJ: Characterize different types of bonds: ionic, covalent, and metallic. MSC: Understanding 13. ANS: A We are asked to determine the number of valence electrons in an ion (S 2- ). Find the number of valence electrons in the atom and add or subtract electrons to compensate for the charge on the ion. Sulfur is a representative element of group VIA, so it has 6 valence electrons. The charge of 2- adds two electrons to the valence shell for a total of 8. An anion generally has 8 valence electrons. PTS: 1 DIF: Easy REF: 8.2 OBJ: Identify the number of valence electrons in main group elements. MSC: Understanding 14. ANS: D
We have the following formula, HCONH 2. Determine the total number of valence electrons. Find the number of valence electrons in each atom, add them up and add or subtract electrons to compensate for any ionic charge (there is no charge here). 3 H atoms = 3 x 1e - = 3 e - 1 C atom = 4 e - 1 O atom = 6 e - 1 N atom = 5 e - 3 + 4 + 6 + 5 = 18 e - This is a simple count. PTS: 1 DIF: Easy REF: 8.2 OBJ: Determine the number of valence electrons in a molecule or ion. 15. ANS: E Of five molecules, which has the most electrons that are not involved in bonds? We must draw the Lewis structure for each and count the nonbonded electrons. H 2 (no unbonded e - ) CO (4 unbonded e - ) N 2 (4 unbonded e - ) NO (7 unbonded e - ) (violates the octet rule odd number of e - ) O 2 (8 unbonded e - ) Molecular oxygen has the most unbonded e -. PTS: 1 DIF: Medium REF: 8.2 OBJ: Construct Lewis structures for molecules.
16. ANS: D Determine the correct ascending order of electronegativity in a list of elements. Apply the periodic tendencies with regard to electronegativity. F < Cl < Ge < Sn Fluorine is the most electronegative element, so this can t be right. Rb < Ca < Sc < Cs Cesium is below rubidium in the PT, so its electronegativity is lower. Zr < V < Nb < Ta Tantalum and niobium are below vanadium in the PT. Sn < P < Cl < F These four follow the periodic tendencies. Cl < Br < I < F Bromine and iodine are below chlorine in the PT.. Only one ordering follows the periodic tendencies PTS: 1 DIF: Medium REF: 8.3 OBJ: Arrange elements according to their electronegativity. 17. ANS: B MSC: Understanding Determine the relative polarities of the bonds in question. Since all the bonds are with hydrogen, the element most different in electronegativity from hydrogen (2.1) will form the most polar bond. Apply the periodic tendencies regarding electronegativity to determine this. The five elements, ordered according to electronegativity are: Al < B < P < C < N. Although boron and phosphorous are close, neither is close to nitrogen, so N-H is the most polar bond. Electronegativity differences determine polarity. PTS: 1 DIF: Easy REF: 8.3 OBJ: Rank bonds in order of their polarity. MSC: Understanding
18. ANS: D Determine the number of resonance structures in oxalate ion. There are several arrangements of the two double bonds. Count them exhaustively. Label the four oxygen atoms 1 through 4 and list the possible combinations for the double bond locations: 1 & 2, 1 & 3, 1 & 4, 2 & 3, 2 & 4 and 3 & 4. One is already listed, and two of the combinations violate the octet rule, so there are three more that meet our criteria. There are four legal double bond combinations, that leaves three that aren t listed. PTS: 1 DIF: Easy REF: 8.4 OBJ: Draw resonance Lewis structures when needed to describe a molecule. 19. ANS: C Choose the most stable of four proposed Lewis structures for azide ion, N 3-. We must check the legitimacy of the structures and then determine the formal charges on all the atoms. Legal structure. Formal charges are, left to right, -2, +1, 0. Illegal structure too many electrons. Legal structure. Formal charges are, left to right, -1, +1, -1.
Illegal structure too many electrons. Between the two legal structures, the charges in a. have a wider spread of formal charges. The charges in c. are symmetrically distributed, and they are smaller. Nature favors smaller, more well-distributed formal charges. PTS: 1 DIF: Difficult REF: 8.5 OBJ: Use formal charge to identify the most stable or preferred Lewis structure. 20. ANS: B Given the bond lengths, we must determine which bond is weakest. a. H Br (141 pm) b. H I (161 pm) c. H Cl (127 pm) d. H F (92 pm) e. H 2S (134 pm) We know that the longer the bond length, the weaker the bond. The longest bond length is the hydrogen to iodine bond, so it is the weakest of this group. Conforms to the rule. PTS: 1 DIF: Medium REF: 8.7 OBJ: Based on Lewis structures, arrange bonds in order of relative bond energy (strength). 21. ANS: E Rational:
Molecular shape is determined by the arrangement of atoms around the central atom, not necessarily by the arrangement of electron pairs. PTS: 1 DIF: Easy REF: 9.2 OBJ: Describe how the electron-pair geometry of a molecule and the steric number of a central atom are determined. MSC: Remembering 22. ANS: C Which geometry matches the steric number of its selection. a. linear geometry corresponds to a steric number of 2. b. trigonal planar geometry corresponds to a steric number of 3. c. tetrahedral geometry corresponds to a steric number of 4. d. trigonal bipyramidal geometry corresponds to a steric number of 5. e. octahedral geometry corresponds to a steric number of 6. Only selection c. has geometry matching the steric number. Conforms to the rule. PTS: 1 DIF: Easy REF: 9.2 OBJ: Relate the five possible steric numbers to the corresponding names of the electron-pair geometries. MSC: Remembering 23. ANS: A Which geometry matches selection. This must be determined from the Lewis structure (but there are some shortcuts). a. SCN - has carbon as the central atom, and it forms a double bond with each of the other two atoms. Its steric number is therefore 2, which indicates linear geometry.
b. NO 2 - has nitrogen as its central atom. It forms a single bond with one oxygen atom, a double bond with the other, and it has one unbonded (lone) pair. Its steric number is greater than 2, which indicates geometry that isn t linear. c. SO 3 - has sulfur as its central atom, and it forms single bonds with three oxygen atoms. Its steric number is greater than 2, which indicates geometry that isn t linear. d. NH 4 + has nitrogen as its central atom, and it forms single bonds with four hydrogen atoms. Its steric number is greater than 2, which indicates geometry that isn t linear. e. SO 2 - has sulfur as its central atom, and it forms single bonds with two oxygen atoms. It has two unbonded pairs. Its steric number is greater than 2, which indicates geometry that isn t linear. It is not necessary to consider lone pairs on the molecules bonded to more than two atoms. PTS: 1 DIF: Medium REF: 9.2 OBJ: Determine electron pair and molecular geometries for a molecule. 24. ANS: A Which bond is the least polar among the five selections. This must be determined from relative electronegativities, ΔEN. a. H C ΔEN = EN(C) EN(H) = 2.5 2.1 = 0.4 b. H N ΔEN = EN(N) EN(H) = 3.0 2.1 = 0.9 c. H O ΔEN = EN(O) EN(H) = 3.5 2.1 = 1.4 d. H Cl ΔEN = EN(Cl) EN(H) = 3.0 2.1 = 0.9 e. H F ΔEN = EN(F) EN(H) = 4.0 2.1 = 1.9 Carbon is closest to hydrogen on the periodic table, so their electronegativities must also be closer. PTS: 1 DIF: Easy REF: 9.3 OBJ: Identify bonds that are polar. 25. ANS: E
What is the hybridization on each of the labeled carbon atoms in the structure? Hybridization is determined by the number of electrons present in hybridized orbitals. A simpler method is to determine the steric number (#of unbonded pairs + #of atoms bonded to the central atom). Note that double and triple bonds are counted as one bonded atom. Carbon atom I is bonded to three atoms (C-C single bond, C-C double bond and C-H single bond) with no unbonded pairs. This is a steric number of three and is therefore sp 2. Carbon atom II is bonded to four atoms (C-C single bond, C-N single bond and two C-H single bonds) with no unbonded pairs. This is a steric number of four and is therefore sp 3. Carbon atom III is bonded to two atoms (C-C triple bond and C-N single bond) with no unbonded pairs. This is a steric number of two and is therefore sp. This conforms with hybridization rules. PTS: 1 DIF: Easy REF: 9.4 OBJ: Identify the hybridization of atoms in a molecule.