Solutions to Problem Set #1

Chemistry 1311 Problem Set 11. Calculate the ratio of the energies of the n=3 to n=1 transitions of the hydrogen atomand the Li 2+ ion.The energy of a transition is given by E = -13.6Z 2 eV (1/n f 2 -1/n i 2 ). Since both n valuesfor the transitions in question are the same the ratio of the energies is determined onlyby the ratio of the Z 2 values or (1) 2 /(3) 2 = 1/9. Thus the energy of the n=3 to n=1transition in Li 2+ ion is 9 times greater than that for H atom.2. How many orbitals can have the designations: 5p; 3d xz ; 4d; n=5, m o =3?The point here is how many 5p orbitals are there? (3) How many 3d xz orbitals? (1) Howmany 4d orbitals? (5) How many orbitals with n=5, m o =3? (Ordinarily one would notask this particular question, but as written the correct answer is 2, since one of the 5fand one of the 5g orbitals will have m o =33. How many electrons in a given can have the set of quantum numbers, n=3, o=2?An o value defines a subshell, in this case the 3d subshell. Since there are five dorbitals, each of which can hold two electrons, so that a total of ten electrons can havethis set of quantum numbers.4. Calculate the ratio of the ionization energies for each of the following species to thatof the hydrogen atom using the Bohr model: He + , Fe 25+ . Note how the value dependsupon Z 2 .This problem basically involves the same concepts as problem 1. The ionizationenergy of any one-electron species is the same as a transition of an electron from n=1to n=infinity or E = 13.6Z 2 . Thus the ratio of the ionization energies of any pair of oneelectronspecies is simply given by the ratio of the squares of their nuclear charges. Hto He 1+ = (1) 2 /(2) 2 = 1/4; H to Fe 25+ = (1) 2 /(26) 2 = 1/676. Although there is no such thingas a Fe 25+ ion the calculated ionization energy approximates what is required to ionizean electron from the 1s orbital of an iron atom or more normal +2 or +3 ion. Certaintypes of spectroscopy involve irradiation of a sample with high energy radiation thationizes a core electron. Photons emitted as a 2s or 2 p electron drops to fill the 1svacancy are detected.5. What is meant by the statement that "a 4s electron is more penetrating than a 3delectron."This simply refers to the difference in radial distribution for the 3d and 4s orbitals. The4s orbital has a significant portion of its electron density near the nucleus, which thus is"penetrating" the core electron density (the electrons in levels of lower principalquantum number). An alternative way to view this is that the core electrons screen(shield) electrons in the 3d level more than those in the 4s level.

6. Write the electron configurations of S 1+ , O 1- , Fe. Indicate the number of unpairedelectrons for each species.S 1+ [Ne] 3s 2 3p x 1 3p y 1 3p z 1 (3 unpaired electrons)O 1- [He] 2s 2 2p 5 (one unpaired electron)Fe [Ar] 4s 2 3d 2 3d 1 3d 1 3d 1 3d 1 (4 unpaired electrons)7. Give an example of an excited state for C and Na.Any arrangement of electrons other than that of the ground state constitutes an excitedstate. For carbon this may take the form 2s 1 2p 4 , 2s 2 2p 1 3s 1 , or even 2s 2 2p x 2 or 2p 2configuration with nonparallel spins. Similar considerations apply to the Na atom.8. Use Excel or another spreadsheet or plotting program to generate plots of the radialdensity (surface density) function 4πr 2 R 2 for the 2s orbital for the hydrogen atom andthe Li 2+ ion.Beyond the obvious that the twocurves have the same shape andeach displays the single radial nodethat is expected from the 2-Zr/a oterm in the wave function, the effectof the higher nuclear charge of theLi 2+ ion is responsible for theconstriction of the electron densityin this ion compared to thehydrogen atom. The reduction insize and the increase in electrondensity of the Li 2+ orbital areapparent in the smaller r and highervalue of the surface density. Notethat there will be n-o-1 radial nodesfor any orbital.161412HydrogenLithium 2+10864200 1 2 3 4 5 6r9. Calculate the energy required to remove an electron from the n=4 level of the He +ion.The energy of any transition in a one-electron species is given by E = -13.6Z 2 eV (1/n f 2 -1/n i 2 ). Since n f is infinity for ionization, then the energy is E = 13.6]2 2 /4 2 = 3.4 eV. Thevalue is positive since energy must be added to the system to achieve ionization.10. Describe the trends of the following properties on going from left to right, and fromtop to bottom, in the periodic table: a. electronegativity; b. atomic radius; c. covalentbondenergy; d. common oxidation numbers. In what blocks of the periodic table (ifany) do you find “anomalous” counter trends and account for the anomaly.

a. electronegativity. Electronegativity increase going from left to right in a period (row)and from top to bottom in any family (column) of the periodic table. The former effect isa result of generally increasing electron affinities and ionization energies on going fromleft to right. The latter effect is a result of generally decreasing electron affinities andionization energies on going down a family. There are no anomalies inelectronegativities.b. atomic radius. Atomic radii decrease on going from left to right in a period andincrease on going from top to bottom in a family. The former effect is a result ofsteadily increasing a steadily increasing effective nuclear charge, which constricts theelectron density and results in a smaller atom. The latter effect is a result of theincrease in principle quantum level for each period and the greater distance throughwhich the effective nuclear charge must operate. There are no particular anomalies inatomic radii.c. covalent bond energies. Covalent bond energies are largely restricted to bondsbetween elements in the p-block and hydrogen. First one must differentiate betweenhomonuclear and heteronuclear bonds and for simplicity let us only consider singlebonds. In general both homonuclear and heteronuclear cases bond energies decreaseas one goes down a family because the atomic sizes are increasing and orbitals arebecoming larger and more diffuse (lower electron density). Therefore the distancesbetween nuclei are greater and bond energies are lower. In general, the trend inhomonuclear bond energies is an increase on going from left to right, except in thesecond period where they decrease. The general effect is smaller orbital, thereforegreater overlap, which leads to a stronger bond. The reason for the anomaly in thesecond period is the increase in electron-electron repulsion due to lone pairs. Forheteronuclear bonds, the trend is generally increasing bond energy with increasingdifference in electronegativity in a period because of the contribution of ionic bonding(recall the original method of computing electronegativity). This means that C-F and Si-F are stronger than O-F and S-F, respectively. Energies decrease generally on going toheavier members of a family, except between the second and third row fluorides wherelone pair repulsions result in weaker bonds for the former, because of the increasingsizes of the atoms even though the electronegativity differences are obviously greater.d. oxidation state. Here one needs to differentiate between positive and negativeoxidation states and to decide what is meant by “normal”. But in general the commonoxidation states of the s- and p-block elements will result in either a rare gasconfiguration (octet) or filled subshell configuration. Transition metal ions represent aspecial case. Normal positive oxidation states for the s-block and transition metalelements might be considered to be those commonly found in hydrated ions, whichusually result from ionization of the s electrons, and sometimes an additional d electronfor the transition elements. However, once manganese is reached, all of the +3 ionsexcept for iron are so highly oxidizing that their hydrated forms are not stable. The highoxidizing potential of the +3 ions results from the increase in effective nuclear charge.The reason that hydrated iron(III) is stable is a result of the special stability associatedwith a half-filled shell. Positive oxidation states in the p block elements vary

considerably but there is a rational pattern for groups 13-15 where they equal either thenumber of p electrons or the number of p+s electrons. This means that the oxidationstates of group 13 are +1 and +3, group 14 are +2 and +4, and for group 15, +3 and +5.The situation is a bit different for groups 16 and 17 where the same pattern does nothold. For group 16 there are three oxidation states +2, +4 and +6 and for group 17four, +1, +3 + 5 and +7. However, what is true for all p-block elements is that the stablepositive oxidation states are those that do not result in any unpaired spins, since thiswould normally lead to formation of a sigma bond between two such moieties. Negativeoxidation states, which exist only for the p-block elements, result from addition ofsufficient electrons to generate filled valence shells (rare gas configurations). Thus forgroup 14, -4 (few examples here), group 15, -3, group 16, -2 and group 17, -1.Obviously all electrons are paired in these forms.11. Within each of the following groups arrange the species in order of increasingradius and in order of increasing ionization energy:O 2- , F - , Na + , Mg 2+Li, C, F, Ne.Explain the reason for the order of each series.radius: O 2- > F - >Na + >Mg 2+ The order in this isoelectronic group of ions is a result of theincreasing effective nuclear charge binding the valence electrons as one progressesthrough the series.ionization energy: O 2- < F - F> Ne These are not isoelectronic, but the decrease in size is a result ofa steadily increasing effective nuclear charge which results in constriction of theelectron density and therefore size of the atom. The situation with F and Ne is a bitconfusing because some tables, including table 17.11 in Oxtoby suggest that Ne islarger than F. However, the problem here is one of method of determination. If radiiare computed then the radius of Ne is found to be less than that of F. There areextensive data and an outstanding discussion of atomic sizes in WebElements, one ofthe weblinks provided.ionization energy: Li< C< F< Ne As in the first series the trend in ionization energymust always be opposite the trend in size.12. Draw sketches of the d xz and p z orbitals on a labeled Cartesian coordinate system.Indicate the phase (sign of the wave function) in each portion of the orbital.

zzxxThe shaded portion of each orbital corresponds to the positive sign (phase). Note thatthe d orbital has a center of symmetry whereas the p orbital does not.