Midterm Exam I. CHEM 181: Introduction to Chemical Principles September 24, 2015 Key

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1 Midterm Exam I CHEM 8: Introduction to Chemical Principles September 24, 205 Key. A Li 2+ ion in an unknown, highly excited electronic state, first emits a photon at a wavelength of 4.36 µm, and following that, emits a second photon at a wavelength of 0.25 nm. What is the initial state and final state of this Li 2+ ion? You can t use the Rydberg equation from your equation sheet as-is: ( λ = R ) n 2 n 2 2 The +3 nuclear charge of the lithium nucleus also has to be included, so: ( = R Z2 ) λ n 2 n 2 2 = ( m ) ( 3 2) ( ) n 2 n 2 2 ) = m ( n 2 n 2 2 n 2 n 2 2 = m λ For labeling, say we start at n a, go to n b, and end at n c. It s easier to solve the n b n c step first: n 2 c n 2 b = m m = n c (the final state) must be for this to work at all, so n 2 b = n 2 b = n 2 b = 82.4 n b 9

2 It s true that 82 8, but this is from round-off errors (the wavelength of the photon to more significant figures is nm.) Now that we ve got n b, we can solve for n a = m n 2 b n 2 a m = n 2 a = = n 2 a n a = 0 The initial state is n = 0, and we end up at n =. 2. Write down the ground-state electron configurations for the following atoms or ions. You can either include all electrons or use core/valence notation; e.g., sodium could be written as either Na: s 2 2s 2 2p 6 3s or Na: [Ne] 3s In cases where it is difficult to predict the exact electron configuration, you can write two configurations along with a brief (one sentence) explanation. (If there is only one reasonable configuration, do not include extras you will lose points for doing so.) (a) Cl: [Ne]3s 2 3p 5 (b) Ni + : [Ar]4s 3d 8 or [Ar]4s 0 3d 9 It s ok to miss the second of these but not the first (though my bet is that the second is what actually happens). (c) V : [Ar]4s 2 3d 4 or [Ar]4s 3d 5 I think the first is much better, because you know that if you ionize V, it s the 3d electron that leaves. We ve got the opposite of the behavior we 2

3 see for positive ions the effective nuclear charge is low for the number of electrons and it seems unlikely that V is isoelectronic with Cr. (d) Os 2+ : [Xe]6s 0 5d 6 or [Xe]6s 5d 5 (FYI: Ni + and V are atypical states found in relatively few compounds, though both can be produced in the laboratory.) 3. N 2 and N + 2 both have multiple stable excited states with electron configurations that are different from the ground-state configuration. Use the following information about energies and bond lengths to figure out these electron configurations. (All answers that are logical and self-consistent will be marked as correct. You can use the MO diagram for N 2 on the next page.) Starting points: σ 2pz is the highest-energy orbital with electrons in it, so it must be a σ 2pz electron that leaves to make N + 2, so its ground state must be (22400). The 0.8 aj state in N + 2 is significantly lower in energy than anything that N 2 does, so it must correspond to a configuration with no close analogue in N 2 : (223200). This also tells us that quantitatively, the π bonding orbitals are 0.8 aj higher than the σ 2pz. (This is a good approximation, though not exact, as orbital energies will be slightly different between N 2 and N + 2. The.77 aj state in N 2 keeps bond length the same, so we re looking for a configuration with the same bond order. This has to happen by moving a σ 2s anti-bonding electron into another anti-bonding orbital: (2420) or (2420). The.00 and.8 states in N 2 have reduced bond order (longer bonds), so will have one fewer bonding and one more anti-bonding electron. If we say the lower-energy of these has one fewer σ 2pz electron and one more π electron, then the higher-energy one would have the same π electron, but will have lost a π bonding electron instead this fits with the 0.8 aj spacing we d think these two levels would have. There are other self-consistent answers that will be marked as correct. actual, measured electron configurations are shown in the table. The 3

4 N 2 energy (aj) bond length (nm) σ 2s σ 2s π 2px, π 2py σ 2pz π 2p x, π 2p y σ 2p z N + 2 energy (aj) bond length (nm) σ 2s σ 2s π 2px, π 2py σ 2pz π 2p x, π 2p y σ 2p z

5 4. The following diagram shows the shapes and energies of the molecular orbitals for BeH 2, which can be described as combinations of the Be 2s and 2p orbitals with the H atoms s orbitals: H Be H a a nb Energy Be 2p Be 2s H s b b Do the following: (a) Label each orbital (b for bonding, a for antibonding, and nb for nonbonding). The two molecular orbitals with the same energy look exactly like beryllium 2p orbitals, and they must have the same energy as these orbitals (which allows us to locate the energy of Be 2p as well.) They are non-bonding the electrons do not care whether the Be is part of a BeH 2 molecular or not. (b) Place the appropriate number of electrons into the molecular orbitals (draw them on top of the horizontal lines used for energy levels.) 5

6 The molecule has six electrons total, but two of these are core s electrons on the beryllium, and not involved in molecular orbitals at all. This diagram needs to be filled with the four valence electrons (2 from Be, from each H). (c) On the same energy scale as the molecular orbitals, indicate (use horizontal lines) the energies of the 2s and 2p orbitals for atomic beryllium and the s orbital for atomic hydrogen. Explain below how you determined these energies: The Be 2p must be at the same energy as the non-bonding molecular orbitals. The 2s must be lower in energy than the 2p, but must be higher in energy than the lowest-energy bonding orbital, which involves constructive interference between (and thus a drop in energy from) Be 2s and H s. The s orbitals of the H atoms must be lower in energy than the first antibonding orbital (destructive between s and 2s) and higher than the second bonding orbital (constructive between s and 2p.) However, because H is more electronegative than Be, we expect the valence electron of H to be lower in energy than the valence electrons of Be that is, H s should be lower in energy than Be 2s. 6

Name: Block: Date: Atomic Radius: the distance from the center of the nucleus to the outer most electrons in an atom.

Name: Block: Date: Atomic Radius: the distance from the center of the nucleus to the outer most electrons in an atom. Name: Block: Date: Chemistry 11 Trends Activity Assignment Atomic Radius: the distance from the center of the nucleus to the outer most electrons in an atom. Ionic Radius: the distance from the center