Chem 6 Practice Exam 2

Transcription

1 These problems are from past Chem 6 exams. Each exam contained a page of universal constant values and common equations; yours will, too, along with a Periodic Table! The answers follow after all the questions. 1. Here are three questions based on H atom orbital pictures. (a) One H atom orbital has three planar or conical nodes, no radial nodes, and is cylindrically symmetric about the z axis. The quantum numbers for this orbital are: n = l = m = (b) How many maxima are in the radial probability function for the 5p z wavefunction of H? (c) The three H atom orbital pictures below represent different states, but they all three have one quantum number in common. Which is it, and what is its value? What is the principal quantum number for each? n = n = n = common quantum number (circle one): n l m m s value of that quantum number =

2 2. Which element or elements in the sequence from H through Ne (a) has the largest first ionization energy (b) has the smallest atomic radius (c) has the largest electron affinity (d) has a 2+ cation with the 1s 2 configuration (e) has a stable anion with the configuration 1s 2 2s 2 2p 2 (f) is isoelectronic to He + when it is a neutral atom (g) has one unpaired e in an orbital with one nodal plane (h) can absorb light to form the excited configuration 1s 2 2s 2 3s (i) has a half-filled sub-shell (j) forms the largest anion 3. Answer the following questions about atomic electron configurations. (a) If I assign m s values to all the electrons in Ge, I find that of them have one value and of them have the other value. (b) Among the elements from H through Kr, those that are spherical in shape in their ground electron configuration is/are: (c) The electrons gained by Se when it forms the Se 2 anion enter the sub-shell(s): (d) Explain briefly why Fe readily forms both Fe 2+ and Fe 3+ ions. (e) The yellow-orange light emitted by Na atoms in, for example, sodium vapor street lights can also be absorbed by Na atoms in their ground state. This absorption excites the atom to the first excited energy electron configuration. What is this configuration? 4. I prepare a bunch of H atoms in the 4p state. If I leave them alone, they spontaneously emit light and eventually, all of them end up in the 1s ground state. (a) How many different wavelengths of light can I expect to measure from the whole bunch (with a photon detector that is sensitive to all wavelengths)? (b) Calculate the longest wavelength of them all. (Note that blind calculation of all the possibilities will take you a very long time. Think first, then let logic indicate the single wavelength calculation you need to make.) (c) Is this photon of longest wavelength sufficiently energetic to ionize a hydrogen atom in the 4p state? (Give me more than a yes/no answer. Use a simple calculation to back up your answer.)

3 5. (a) Draw a graph of the particle in a box wavefunction for the state with quantum number n = 4. Ψ 0 x L (b) If an electron is in this state and L = 2.0 Å, how much energy do I need to give the electron to excite it to the state with quantum number n = 5? 6. Some quick, short, succinct answers, please: (a) Which has the larger second ionization energy, B or C, and why? (b) Which is the largest and which is the smallest of these three isoelectronic species: K, Sc +, or Ca, and why? (c) If the 4s orbital energy was significantly higher than the 3d orbital energy (which it isn t), one of the first row transition elements would probably be an inert gas. Which one would it be and why?

4 Answers 1. (a) One H atom orbital has The quantum numbers for this orbital are: n = 4 (total number of nodes plus 1) l = 3 (total number of planar nodes) m l = 0 (since it has cylindrical symmetry) (b) How many maxima are in the radial probability function for the 5p z wavefunction of H? 4 (5p has n 1 = 5 1 = 4 nodes in all, one of these is a plane since it s a p wavefunction with l = 1, so it has three radial nodes and thus four maxima in the radial distribution function since each node has electron density on each side of it) (c) From left to right, the pictures are: n = 3 (spherically symmetric, two radial nodes, n = = 3 or 3s wavefunction) n = 2 (one planar node only, l = 1, so n = = 2, aligned along z axis, so its the 2p z ) n = 3 (two conical planar nodes, l = 2 so n = = 3, aligned along z axis, so its the 3d z 2) common quantum number m value of that quantum number = 0 (since all three are cylindrically symmetric about z) 2. (a) has the largest first ionization energy He (b) has the smallest atomic radius H (or He either was accepted) (c) has the largest electron affinity F (d) has a 2+ cation with the 1s 2 configuration Be (e) has a stable anion with the configuration 1s 2 2s 2 2p 2 B (f) is isoelectronic to He + when it is a neutral atom H (g) has one unpaired e in an orbital with one nodal plane B and F (h) can absorb light to form the excited configuration 1s 2 2s 2 3s B (i) has a half-filled sub-shell H, Li, and N (j) forms the largest anion Li 3. (a) If I assign I find that 17 of them have one value and 15 of them have the other value. (b) Among the elements those that are spherical in shape H, He, Li, Be, N, Ne, Na, Mg, P, Ar, K, Ca, Cr, Mn, Zn, As, Kr (filled or 1/2 filled subshells). (c) The electrons gained by Se when it forms the Se 2 anion enter the sub-shell(s) 4p. (d) Explain briefly why Fe readily forms both Fe 2+ and Fe 3+ ions. Fe has the [Ar] 4s 2 3d 6 configuration. The easily removed 4s 2 electrons lead to Fe 2+, and if one more is lost, we have Fe 3+ with the [Ar] 3d 5 configuration, which has a stable half-filled subshell. 4. I prepare a bunch of H atoms in the 4p state all of them end up in the 1s ground state. (a) How many different wavelengths Six because n is the only quantum number that affects the H atom s energy. Thus, the possible transitions are n = 4 n = 3, 4 2, 4 1, 3 2, 3 1, and 2 1.

5 (b) Calculate the longest wavelength The longest wavelength means the smallest energy change for the atom from among these six possibilities. That s the n = 4 to n = 3 transition. The atom s energy change in this transition is J (1/4 2 1/3 2 ) = J, and we equate this to the energy of the photon, hν = hc/λ. Solving J = hc/λ for λ, the wavelength, we find λ = 1880 nm. (c) Is this photon sufficiently energetic to ionize a hydrogen atom in the 4p state? The ionization energy of a H atom in a state with n = 4 is ( J)/4 2 = J, which is greater than the energy of the longest wavelength (which we found in part (b) to be only J). Thus, this photon cannot ionize an n = 4 H atom. 5. (a) Draw a graph of the particle in a box wavefunction for n = 4. The number of nodes in a particle in a box wavefunction is n 1; so, we have 3 nodes. That means a graph of sin(4π x/l), which is a two-cycle sine wave extending from x = 0 to x = L. (b) If an electron is in this state We need to give the electron an energy increase Δ E = E 5 E 4 where E 5 is the energy of the n = 5 level (which is 5 2 h 2 /8mL 2 ) and E 4 is the energy of the n = 4 level (which is 4 2 h 2 /8mL 2 ). With m = m e = electron mass and L = 2.0 Å = m, we find Δ E = J. 6. (a) Which has the larger second ionization energy, B or C, and why? The first ionization of each removes a 2p electron, but the second removes a 2s electron from B, but only another 2p electron from C. The change in subshell with B means it has the higher second ionization energy. (b) Which is the largest and which is the smallest of K, Sc +, or Ca, and why? All three have the same number of electrons, but K has the fewest number of protons in its nucleus to hold these electrons in place (making it the largest) while Sc + has the largest number of protons (making it the smallest). Ca is in between. (c) If the 4s orbital energy was significantly higher than the 3d orbital energy (which it isn t), one of the first row transition elements would probably be an inert gas. Which one would it be and why? After Ar, we would add the next electron (in K) to the 3d subshell rather than the 4s. This would continue for a total of ten elements (which takes us to Ni) until we had filled the n = 3 shell. This filled shell configuration for Ni would tend to make it an inert gas.