Chem 222 Intro to Inorganic Chemistry Summer 2011

Transcription

1 Chem 222 Intro to Inorganic Chemistr Summer 2011 Problem Set 1 Answer Ke Wednesda, Ma 11, Draw and label the,, 3p, and 3d orbitals, placing them correctl on cartesian aes (,,). and : 3p: lobes can be signs/or shading p p p 3d: lobes can be signs/or shading d d d d 2 d (RC&O 2.20) Which atom should have the larger covalent radius, fluorine or chlorine? Give our reasoning. Fluorine has electron configuration 2 2 5, while chlorine has electron configuration p 5. Chlorine should have a larger covalent radius because its outermost electrons are in the 3p orbitals, for which the radial distribution function etends much further out from the nucleus than do the orbitals. The larger the value of n, the larger the orbital. Note that this trumps the fact that the Z eff felt b the outermost electrons of Cl is higher than that eperienced in F, as Z eff does not take into account the distance between the charges. 3. (a) (RC&O 1.5) Identif the orbital that has n = 5 and l = 1. For the principle quantum number n = 5, the angular momentum quantum number, l, can have values 4,3,2,1,0. The number l=0 alwas corresponds to the s orbital; the l=1 alwas corresponds to the p orbitals. Thus, the above quantum numbers correspond to a 5p orbital. (b) (RC&O 1.6) Idenitf the orbital that has n = 6 and l = 0. See eplanation for 3(a): this is the 6s orbital.

2 Chem 222 Intro to Inorganic Chemistr Summer (H&S 1.23) Write down the si sets of quantum numbers that describe the electrons in a degenerate set of 5p atomic orbitals. Which pairs of sets of quantum numbers refer to spinpaired electrons? For n=5, the p orbitals correspond to l=1, and m l = 1,0,-1. The si sets of quantum numbers that describe the degenerate set of 5p orbitals are: n=5, l=1, m l = 1, m s =1/2; n=5, l=1, m l = 1, m s =1/2: these electrons are spin-paired n=5, l=1, m l = 0, m s =1/2; n=5, l=1, m l = 0, m s =1/2: these electrons are spin-paired n=5, l=1, m l = 1, m s =1/2; n=5, l=1, m l = 1, m s =1/2: these electrons are spin-paired 5. (H&S 1.24) For a neutral atom, X, arrange the following atomic orbitals in an approimate order of their relative energies (not all orbitals are listed):,, 6s, 4p, 3p, 3d, 6p,. = < < < 3p< 3d< 4p< 6s <6p 6. (H&S 1.25) Using the concepts of shielding and penetration, eplain wh a ground state configuration of 2 1 for a Li atom is energeticall preferred over 2 1. Ground state configurations are alwas the lowest energ arrangement possible (Aufbau principle). Because the orbital penetrates the orbital better than the orbital does (i.e. has a higher probabilit close to the nucleus than ), a 1 electron feels a greater Z eff than does a 1 electron. A 1 electron eperiences more shielding from the nucleus b the filled orbital than does a 1 electron. Because the 1 electron feels a greater Z eff, this is a lower energ configuration. See the radial distribution functions in Fig from the Introduction notes to see wh this is the case. 7. (a) (H&S 1.26) For each of the following atoms, write down a ground state electronic configuration and indicate which electrons are core and which are valence: (i) Na, (ii) F, (iii) N, (iv) Sc. (i) Na: [ ] (core) 1 (valence) (ii) F: [ 2 ] (core) 2 5 (valence) (iii) N: [ 2 ] (core) 2 3 (valence) (iv) Sc: [ p 6 ] (core) 4s 2 3d 1 (valence) (i) (b) (H&S 1.27) Draw energ level diagrams (see class notes on electron configuration) to represent the ground state electronic configurations of the atoms in (a). (ii) (iii) (iv)

3 Chem 222 Intro to Inorganic Chemistr Summer s 3d 3p 8. (RC&O 1.12) Write noble-gas core ground state electron configurations for (i) calcium, (ii) chromium, (iii) lead. (i) Ca: [Ar] 4s 2 (ii) Cr: [Ar] 4s 1 3d 5 (recall etra stabilit derived from half-filled shells) (iii) Pb: [Xe] 6s 2 4f 14 5d 10 6p 2 9. (H&S 1.30) Draw energ level diagrams to show the ground state electronic configurations of onl the valence electrons in an atom of (i) fluorine, (ii) aluminum, (iii) magnesium. (i) F ([He] 2 5 ) (ii) Al ([Ne] 2 3p 1 ) (iii) Mg ([Ne] 2 ) 3p 10. (H&S 1.31) The ground state electronic configuration of a group 16 element is of the tpe [X]ns 2 np 4 where X is a group 18 element. How are the outer four electrons arranged, and what rules are ou using to work out this arrangement (i.e. eplain our reasoning)? The four electrons in the degenerate np orbitals are arranged in this wa: Hund s rule states that for a set of degenerate orbitals electrons will not spin pair until each orbital in the set contains one electron, and these single electrons will have parallel spins. Thus the first three electrons have parallel spins and singl occup the three p-orbitals. The fourth electron takes the opposite spin and pairs with another electron in one of the three orbitals (impossible to sa which one). (Recall this is dictated b the Aubau principle, which sas that orbitals are filled in the order of lowest

4 Chem 222 Intro to Inorganic Chemistr Summer 2011 energ: the energetic cost of spin-pairing is lower than the energetic cost of placing the fourth electron in the (n1)s orbital. Another wa to eplain this is in terms of shielding effects: the Z eff felt b an np electron is sufficientl higher than the Z eff felt b an (n1)s electron to outweigh the etra pairing energ required.) 11. (RC&O 2.22) Suggest a reason wh the covalent radius of hafnium (144 pm) is almost identical to that of irconium (145 pm), the element above it in the periodic table. This corresponds to a contraction in radius that is observed for elements in the sith period (relative to that predicted b Schrödinger s wave equation). For these heavier atoms, relativistic effects on the innermost core electrons result in a decreased orbital sie, particularl for the s and p orbitals. In addition to this relativistic phenomenon, the 6 th period transition metals eperience the lanthanoid contraction : the electrons in Hf s filled 4f orbitals are etremel poor at shielding the 6s and 5d electrons, so the Z eff felt b these electrons is large enough to reduce the covalent radius of Hf to approimatel the same value as that of Zr. 12. (RC&O 2.25) Using Slater s rules, calculate the effective nuclear charge on one of the 3d electrons compared to that on one of the 4s electrons for an atom of manganese. Mn has electron configuration p 6 4s 2 3d 5 or [ 2 ] [ 2, 6 ] [ 2 3p 6 ] [3d 5 ] [4s 2 ] For one of the five 3d electrons, the screening or shielding, S, predicted b Slater s rules is: S = 2(0) 4(0.35) 8(1.00) 8(1.00) 2(1.00) = = 19.4 Therefore Z eff = Z S = = 5.6 for a 3d electron For one of the 4s electrons: S = 1(0.35) 5(0.85) 8(1.00) 8(1.00) 2(1.00) = = 22.6 Therefore Z eff = ZS = = 2.4 for a 4s electron. NOT: this eplains in an approimate wa wh the 4s electrons are the first to leave when the 1 st row transition metals are ionied. 13. (RC&O 2.26) Using Slater s rules, calculate the effective nuclear charge on a 3p electron in (a) aluminum and (b) chlorine. plain how our results relate to: (i) the relative atomic radii of the two atoms and (ii) the relative first ioniation energies of the two atoms. (a) Al has electron configuration p 1 or [ 2 ] [ 2, 6 ] [ 2 3p 1 ] For the 3p electron: S = 2(0.35) 8(0.85) 2(1.00) = = 9.50 Therefore Z eff = Z S = = 3.50 (b) Cl has electron configuration p 5 or [ 2 ] [ 2, 6 ] [ 2 3p 5 ] For a 3p electron: S = 6(0.35) 8(0.85) 2(1.00) = = Therefore Z eff = Z S = = 6.10

5 Chem 222 Intro to Inorganic Chemistr Summer 2011 NOT: the outermost (3p) electrons in chlorine eperience a higher Z eff than the 3p electron in aluminum. This is consistent with the smaller atomic radius of Cl (r cov = 99 pm, vs 130 pm for Al). It is also consistent with the much higher 1 st ioniation energ for Cl (1251 kj/mol, vs kj/mol for Al). 14. (RC&O 2.27) Which element should have the higher first ioniation energ, silicon or phosphorus? Give our reasoning. Because phosphorus is further to the right in the third period than silicon, its 3p electrons should eperience a slightl higher Z eff. That means phosphorus should have the higher 1 st ioniation energ. (Check a tetbook or other reference does it?) 15. (RC&O 2. 32) Which element, sodium or magnesium, should have an electron affinit closer to ero? Give our reasoning. lectron affinities (M e- M-) for elements are mostl negative (i.e. adding an electron to the neutral element is an eothermic process). If an element has less electron affinit (i.e. the incoming electron feels a lower Z eff ), the eothermicit is reduced, which makes the value closer to ero. For Na, the incoming electron will occup the orbital, where it will eperience a relativel high Z eff because it is not eperiencing much shielding b the other electron. For Mg the orbital is full and is relativel good at shielding the incoming 3p electron, which will eperience a slightl lower Z eff. Thus, despite the general increase in electron affinities across a period, Mg should ehibit a comparable or slightl lower electron affinit relative to Na: i.e. its A 1 will be closer to ero. 16. (H&S 2.15) Using the electronegativit data in Table 2.2 determine which of the following covalent single bonds is polar and (if appropriate) in which direction the dipole moment will act. (i) N-H; (ii) F-Br; (iii) C-H; (iv) P-Cl; (v) N-Br. Note: Pauling electronegativit values are well documented and constant, so ou could have also gotten these values from our first ear tet or the lecture notes. (i) P N = 3.0, P H = 2.2: the N-H bond is polar in the sense N H, thus the dipole vector along this bond will point toward the N. (ii) P F = 4.0, P Br = 3.0: the F-Br bond is polar in the sense F Br, thus the dipole vector along this bond will point toward the F. (iii) P C = 2.6, P H = 2.2: while the C-H bond is slightl polar in the sense C H and the (small) dipole vector along this bond will point toward the C, this is small enough (<0.5) that it is considered non polar. (iv) P P = 2.2, P Cl = 3.2: the P-Cl bond is polar in the sense P Cl, thus the dipole vector along this bond will point toward the Cl. (v) P N = 3.0, P Br = 3.0: the N-Br bond is not polar. Note that the polar vector is described in the IUPAC manner (pointing towards the negative pole), which is different from the description in H and S tetbook. Definitions/Concepts: atomic orbital, degenerate, shielding, penetration, boundar surface, core electrons, valence electrons, Aufbau principle, Pauli eclusion principle, nodal plane, Hund s rule, effective nuclear charge, shielding, 1 st ioniation energ, electron affinit, electronegativit, bond polarit, dipole moment.