chem 141 practice problems (23-30)

What is meant by a discrete line spectrum? What kind of spectra do not have discrete lines? Have you ever seen a spectrum that does not have discrete lines? Have you ever seen one with discretelines?

lesson 23: A discrete line spectrum is once that consists of discrete, or separate, lines of color. Spectra of white light do not have color. A rainbow is an example of a spectrum that does not have discrete lines. Discrete line spectra are not normally encountered outside of chemistry and physics laboratories.

Although the quantum mechanical model of the atom makes predictions for atoms of all elements, most of the quantitative confirmation of these predictions is limited to substances whose formulas are H, He+, and H2+. From your knowledge of chemical formulas and subatomic particles, can you identify a single common feature of all three substances that makes them unique and probably the easiest to investigate?

lesson 23: All species have a single electron. Species with two or more electrons are far more complex.

For hydrogen-like atoms, the energy-principal energy level relationship becomes E = -RH * Z^2/n^2, where Z is the nuclear charge. Calculate the energy of an n= 2 electron in a hydrogen-like atom with a nuclear charge of 2.

lesson 23: E = -RH * Z^2/n2 -2.18 * 10^-18J *(2)^2/(2)^2 = -2.18 *10^-18J

The maximum wavelength of electromagnetic radiation that can be used to break the chemical bond in a hydrogen molecule is 277 nm. What is the bond energy in a hydrogen molecule?

lesson 23: E =hc/λ 6.626 * 10^-34J · s * 3.00 ×10^8m/s * 1277 nm * 1 ×10^9nm/1 m = 7.18 * 10^-19J 7.18 *10^-19J * 1 kJ/1000 J * 6.022 ×10^23mol = 432 kJ/mol

Calculate the wavelength of the electromagnetic energy released when an electron falls from n= 3 to n= 2 in a hydrogen-like atom with a nuclear charge of 2 (see Question D1).

lesson 23: ∆E = -RH * Z^2 * (1/(nf)^2 -1/(ni)^2) = -2.18*10^-18J *(2)^2 * [1/(2)^2-1/(3)^2] = -1.21*10^-18J λ = hc/E = 6.626*10^-34J · s * 3.00 ×10^8m/s * 1/1.21 × 10^-18J * 1 ×10^9nm/1 m = 164 nm

One of the lines in the hydrogen emission spectrum occurs at 434.2 nm. What are the initial and final values of n that generates this line?

lesson 24: E = hc/λ 6.626*10^-34J*s * 2.998 × 10^8ms * 1/434.2 nm * 1 ×10^9nm/1 m = 4.575*10^-19J ∆E = -RH * (1/(nf)^2-1/(ni)^2) -∆E/RH = (1/(nf)^2 -1/(ni)^2) Try nf = 1: - (-4.575 ×10^-19J/2.180 ×10^-18J = 0.2099 = 1 -1/(ni)^2 ni = 1.125, not possible (not an integer) Try nf = 2: 0.2099 = 0.25 -1/(ni)^2 ni = 5; the transition is from n = 5 to n = 2

If there were three possible for values for ms, how would the periodic table be different in the first three periods? Assume that the allowed values for the other quantum numbers are unchanged.

lesson 24: In the first period, n= 1, l= 0, ml = 0, and ms would have three values, so it would have three elements. In the second period, n= 2, l= 1, ml= -1, 0, and +1, and ms would have three values each, for nine elements, plus n= 2, l= 0, ml= 0, and ms with three values, for a total of 12 elements.

Why doesn't the Bohr model work for atoms with more than one electron?

lesson 24: The Bohr model does not consider the interaction between electrons.

For each of the following cases, state the number of ground state electrons that are present and give a brief explanation of your reasoning. (a) How many electrons in a neon atom have n = 2 and ms = -1/2? (b) How many electrons in a sodium atom have l = 0? (c) How many electrons in a phosphorus atom have l = 1 and ml = +1? (d) What is the maximum number of electrons in an atom that can haven = 3? (e) What is the maximum number of electrons in an atom that can have n = 4 and l = 3?

lesson 25: (a) 4. All 8 n = 2 electrons in Ne are spin paired, so half have ms = -1/2. (b) 5. Na is 1s²2s²2p⁶3s¹, so there are 5 electrons in s orbitals. (c) 3. P is 1s²2s²2p⁶3s²3p³, so 2 electrons in 2p have ml = +1 and one of the 3p electrons has ml = +1. (d) 18. 2 in the 3s orbital, 6 in the 3p orbitals, and 10 in the 3d orbitals. (e) 14. There are seven 4f orbitals.

For each of the following cases, state the number of ground state electrons that are present and give a brief explanation of your reasoning. (a) How many electrons in a magnesium atom have ms = +1/2? (b) How many electrons in a manganese atom have n = 3 and l = 1? (c) How many electrons in a sulfur atom have ml = 0? (d) What is the maximum number of electrons in an atom that can have n = 2? (e) What is the maximum number of electrons in an atom that can have n = 3 and l = 2?

lesson 25: (a) 6. All 12 electrons are spin paired, so half have +1/2 and half have -1/2. (b) 6. All of the 3p electrons have n = 3 and l = 1. (c) 9 or 10. S is 1s²2s²2p⁶3s²3p⁴, and there are 2 electrons with ml = 0 in 1s, 2s, 2p, and 3s. There must be at least one electron with ml = 0 in 3p, depending on whether the paired electrons are in the ml = 0 orbital or the ml = +1 or -1 orbitals. (d) 8. 2 in the 2s orbital and 6 in the 2p orbitals. (e) 10. There are five 3d orbitals.

Give the possible sets of (n, l, ml, ms) for the electrons in: (a) 4p2 (b) 6s2 (c) 4d5 (d) 3p4

lesson 25: a) 4,1,+1,+1/2 4,1,0,+1/2 The ml values must be different. +1 and -1 or 0 and -1 are also correct. The ms values must be the same. -1/2 and -1/2 is also correct. b) 6,0,0,+1/2 6,0,0,-1/2 c) 4,2,+,1,+1/2 4,2,0,+1/2 4,2,-1,+1/2 4,2,-2,+1/2 All ms values may be -1/2. d) 3,1,+1,+1/2 3,1,+1,-1/2 3,1,0,+1/2 3,1,-1,+1/2 The two paired electrons can both be in an orbital with ml= 0 or -1. The two unpaired electrons must be in different orbitals (must have different ml values than the paired electrons) and must have the same spin, which can be -1/2 instead of +1/2.

Give the possible sets of (n, l, ml, ms)for the electrons in the highest energy subshell of these elements: (a) Be (b) Ar (c) Mn (d) Cl.

lesson 25: a) Be: [He]2s² 2,0,0,+1/2 2,0,0,-1/2 b) Ar: [Ne]3s²3p⁶ 3,0,0,+1/2 3,0,0,-1/2 3,1,+1,+1/2 3,1,+1,-1/2 3,1,0,+1/2 3,1,0,-1/2 3,1,-1,+1/2 3,1,-1,-1/2 c) Mn: [Ar]4s²3d⁵ 4,0,0,+1/2 4,0,0,-1/2 3,2,+2,+1/2 3,2,+1,+1/2 3,2,0,+1/2, 3,2,-1,+1/2 3,2,-2,+1/2 All ms values may be -1/2. d) Cl: [Ne]3s²3p⁵ 3,0,0,+1/2 3,0,0,-1/2 3,1,+1,+1/2 3,1,+1,-1/2 3,1,0,+1/2 3,1,0,-1/2 3,1,-1,+1/2 The one unpaired electron can be in an orbital with ml = +1 or 0, but the paired electrons then need to have the other two ml values. It can have ms = -1/2 instead of +1/2.

Give the letters that are in the positions of (a) halogens and (b) alkali metals. Give the letters that are in the positions of (a) alkaline earth metals and (b) noble gases. Give the letters that correspond to transition elements. List the letters that correspond to nonmetals. List the elements D, E, and G in order of decreasing atomic size (largest atom first) List the elements Q, X, and Z in order of increasing atomic size (smallest atom first).

lesson 26: (a) Q and M (b) D and E. (a) G and A. (b) R and T. J, W, L X, Q, R, Z, M, T E > D > G Q < X < Z

Compare the ionization energies of aluminum and chlorine. Why are these values as they are?

lesson 26: Aluminum atoms have a lower nuclear charge—fewer protons in the nucleus—than chlorine atoms. It is, therefore, easier to remove a 3p electron from an aluminum atom than from a chlorine atom.

Iron forms two monatomic ions, Fe3+and Fe2+. From which sublevels do you expect electrons are lost in forming these ions? (Hint: It is possible for electrons other than those in the sand p sublevels to be involved in the formation of ions.)

lesson 26: Iron loses two electrons from the 4s orbital to form Fe2+and a third from a 3d orbital to form Fe3+. This is an example of d electrons contributing to the chemical properties of an element.

Why is the definition of atomic number based on the number of protons in an atom rather than the number of electrons?

lesson 26: The number of protons identifies an element. A neutral atom can gain or lose electrons, forming ions of that element.

Consider the block of elements in Periods 2 to 6 and Groups 5A to 7A (15 to 17). In Group 7A/17 are the halogens, a distinct chemical family with many properties shared by the different elements. The elements in Group 6A/16 have enough similarity to be considered a family; they are called the chalogen family. In Group 5A/15, however, family similarities are weak, and those that exist belong largely to nitrogen and phosphorus, and to some extent, arsenic (Z = 33). Why do you suppose family similarities break down in Group 5A/15?

lesson 26: The smaller atoms in Group 5A/15 tend to complete their octets by gaining or sharing electrons, which is a characteristic of nonmetals. Larger atoms in the group tend to lose their highest-energy s electrons and form positively charged ions, a characteristic of metals.

Carbon does not form a stable monatomic ion. Suggest a reason for this.

lesson 26: To form a monatomic ion, carbon would have to lose four electrons. The fourth ionization energy of any atom is very high.

Suggest reasons why the ionization energy of magnesium is greater than the ionization energy of sodium, aluminum, and calcium.

lesson 26: We expect magnesium to have a greater ionization energy than sodium and calcium because of the trends in ionization energy explained in the text. Aluminum has a greater nuclear charge than magnesium, but the additional electron is "by itself" in a 3p orbital, which makes it easier to remove.

Xenon (Z = 54) was the first noble gas to be chemically combined with another element. Xenon and krypton are present in nearly all of the small number of noble gas compounds known today. Note the ionization energy trend begun with the other noble gases in Figure 26.2.What do these facts suggest about the relative reactivities of the noble gases and the character of noble gas compounds?

lesson 26: Xenon has the lowest ionization energy of the noble gases and apparently the greatest reactivity. This is the same ionization energy trend seen in all groups in the periodic table.

Imagine a three-dimensional set of x-y-z axes where the x axis comes out of and goes behind the plane of the page, the y axis is oriented toward the top and bottom of the page, and the z axis crosses the page to the left and right. Consider the z axis to be the bond axis, an imaginary straight line that connects the bonded nuclei. For each of the following pairs of orbitals, describe how the orbital overlap occurs and state whether the bond is sigma or pi. (a) s and pz (b) px and px (c) py and py (d) pz and pz (e) a sp hybrid from sand pz orbitals and pz (f) a sp2 hybrid formed from s, px, and pz orbitals and s

lesson 27: (a) A sigma bond with the p orbital oriented left and right, with one lobe overlapping the spherical s orbital. (b) A pi bond with the p orbitals coming in and out of the page, with both lobes overlapping side to side. (c) A pi bond with the p orbitals oriented up and down, with both lobes overlapping side to side. (d) A sigma bond with the p orbitals oriented left and right, with one lobe of one p orbital overlapping one lobe of the other p orbital. (e) A sigma bond with the sp hybrid orbital oriented left and right, with the fat lobe pointing toward and overlapping one lobe of the p orbital, which is also oriented left and right. (f) A sigma bond with the sp2 hybrid orbital oriented left and right, with the fat lobe pointing toward and overlapping the spherical s orbital.

Which orbitals of each atom overlap in forming a bond between bromine and oxygen?Explain.

lesson 27: 4p from bromine ([Ar]4s23d104p5) and 2p ([He]2s22p4) from oxygen. The half-filled 4p orbital in the bromine atom will overlap one of the half-filled 2p orbitals in the oxygen atom.

Is there any such thing as a completely nonpolar bond?If yes, give an example.

lesson 27: A bond between identical atoms is completely nonpolar. Their attractions for the bonding electrons are equal.Examples include H2, F2, Cl2, ....

Describe the orbitals and orbital overlap used for bonding in the nitrogen molecule. :N ≡ N:

lesson 27: A nitrogen atom has five valence electrons, [He]2s22p3. The 2s orbital and one of the 2p orbitals will hybridize to form two sp hybrid orbitals. One of the sp hybrid orbitals will be half-filled with one electron and overlap with a sp hybrid orbital of the other nitrogen atom to form the sigma bond. The other sp hybrid orbital will be filled with two electrons that are the unshared pair. The two remaining 2p orbitals are each half-filled and overlap with 2p orbitals on the other nitrogen atoms to form the two pi bonds.

Which bond, F—Si or O—P, is more polar? Explain. You may look at a full periodic table in answering this question, but do not look at any source of electronegativity values.

lesson 27: An F—Si bond is more polar than an O—P bond. F has a higher electronegativity than O, and Si has a lower electronegativity than P, based on their relative positions in the periodic table (high at the upper right, low at the lower left. Therefore, the difference in electronegativities is largest for F—Si, which makes it the more polar bond.

If you did not have an electronegativity table, could you predict the relative electronegativities of elements whose positions are B/A in the periodic table? What about elements whose positions are X\Y ? In both cases, explain why or why not.

lesson 27: Electronegativities are highest at the upper right corner of the periodic table and lowest at the lower left corner. Therefore, the electronegativity of A is higher than the electronegativity of B. Because X is higher in the table than Y, the electronegativity of X should be larger than that of Y, but because Y is farther to the right, the electronegativity of X should be smaller than Y. Therefore, no prediction can be made for X and Y.

How many cis-trans isomers are possible for 10, 12-hexadecadien-1-ol, CH3(CH2)2CH=CHCH=CH(CH2)8CH2OH? Explain.

lesson 27: Four. There are two carbon-carbon double bonds in the molecule, and each can have cis-transisomerism. Thus, the molecule can be trans/trans, trans/cis, cis/trans, or cis/cis at each of the two double bonds.

Arrange the following bonds in order of decreasing polarity: Na—O; Al—O; S—O; K—O; Ca—O. If the polarity of any two bonds cannot be positively placed relative to each other based on periodic trends, explain why.

lesson 27: K—O(3.4 -0.8 = 2.6) > Na—O (3.4 -0.9 = 2.5) > Ca—O(3.4 -1.0 = 2.4) > Al—O(3.4 -1.6 = 1.8) > S—O(3.4 -2.6 = 0.8). Without electronegativity values, the group-to-group spread predicts Na—O is more polar than Ca—O, but the period-to-period spread predicts Ca—O is more polar than Na—O.

In 2003, a team of German scientists reported that they measured the C-N-C bond angle of triisopropylamine, a molecule with the general structure NR3, as 119.2°. What bond angle is ideally predicted by VSEPR theory? What hybrid orbitals are utilized by the central nitrogen atom for bonding? What do you suppose is responsible for the unusual bond angle?

lesson 27: The ideal bond angle for four electron pairs (3 single bonds and 1 unshared electron pair) around a central atom is 109.5°. This implies a sp3 hybrid orbital set. The larger-than-expected bond angle occurs because the R groups are unusually large and they repel one another, forcing the C-N-C bond angle to be larger than it is when the R groups are small.

Use a vertical energy scale to illustrate the relative energies of atomic orbitals in which the valence electrons of an isolated, unbonded beryllium atom reside. Also show unfilled orbitals in the same principal energy level. Use arrows to illustrate the electrons within these orbitals. Then show the relative energies of the hybrid orbitals that form as a beryllium chloride molecule begins to form. Also show any unhybridized atomic orbitals (if any). Use arrows to illustrate the electrons within these orbitals. Label each box that represents an orbital with the appropriate designations.

lesson 27: see vertical energy scale visual

(a) Use a vertical energy scale to illustrate the relative energies of atomic orbitals in which the valence electrons of an isolated, unbonded nitrogen atom reside. Also show unfilled orbitals in the same principal energy level. Use arrows to illustrate the electrons within these orbitals. Then show the relative energies of the hybrid orbitals that form as an ammonia molecule begins to form. Also show any unhybridized atomic orbitals (if any). Use arrows to illustrate the electrons within these orbitals. Label each box that represents an orbital with the appropriate designations. (b) Repeat the exercise in part (a) for an isolated oxygen atom that becomes part of a water molecule.

lesson 27: see vertical energy scale visual two

The way that molecules interact with one another depends on the highest-energy occupied molecular orbital (HOMO) of one species and the lowest-energy unoccupied molecular orbital (LUMO) of another species (this is similar to Lewis acid-base theory). Identify the HOMO and LUMO in: (a) dilithium (b) dicarbon

lesson 28: see C LUMO/HOMO diagram

The way that molecules interact with one another depends on the highest-energy occupied molecular orbital (HOMO) of one species and the lowest-energy unoccupied molecular orbital (LUMO) of another species (this is similar to Lewis acid-base theory). Identify the HOMO and LUMO in: (a) nitrogen (b) fluorine

lesson 28: see C LUMO/HOMO diagram

Draw just the portion of the molecular orbital diagram that illustrates the relative energies of the atomic and hybrid orbitals and the resulting molecular orbitals for each bond described. Sketch the atomic and hybrid orbitals and the resulting bonding and antibonding orbitals. (a) The C-H bond in CH3Cl (b) The C-Cl bond in CH3Cl

lesson 28: see D LUMO/HOMO diagram

Draw molecular orbital diagrams and state the bond order of NO-and NO+

lesson 28: see D LUMO/HOMO diagram

Draw the molecular orbital diagram and state the bond order of OF-. Assume that the molecular orbital diagram is similar to that of NO.

lesson 28: see D LUMO/HOMO diagram

Sulfur dioxide and nitrous acid Nitric acid and nitrogen trifluoride Sulfur trioxide and carbonic acid OSCl₂ and O2SCl₂ (lewis diagram)

lesson 29: see 29C diagram

Benzene, C₆H Toluene, C₆H₅-CH Para-xylene, CH₃-C₆H₄-CH₃(the -CH₃ groups are opposite one another) Naphthalene, C₁₀H₈ (two fused rings) (lewis diagram)

lesson 29: see 29D diagram