Chemistry Sp18 Solutions for Practice Midterm 2

Transcription

1 Copyright RJZ 2/15/18 1 no unauthorized use allowed Chemistry Sp18 Solutions for Practice Midterm 2 This material is copyrighted. Any use or reproduction is not allowed except with the expressed written permission of Dr. Zellmer. If you are taking Chem 1250 you are allowed to print one copy for your own use during the semester you are taking Chem 1250 with Dr. Zellmer. You are not allowed to disseminate this material to anyone else during the semester or in the future.

2 Copyright RJZ 2/15/18 2 no unauthorized use allowed 1) ZnO (s) + H 2 SO 4 (aq) v ZnSO 4 (aq) + H 2 O (R) ΔH rxn > 0 endothermic, energy is absorbed from the surroundings (required) Think of heat as a reactant (required). It might feel cold to the touch. ΔH rxn < 0 exothermic, energy is released to the surroundings (given off) Think of heat as a product (produced or given off). It might feel hot to the touch. A thermometer in the solution registered an increase in temperature. That means the surroundings is gaining heat. That means the system lost heat. Thus the process of ZnO(s) dissolving in H 2 SO 4 is exothermic. For the above reaction, ΔH soln < 0 so the reaction is exothermic and heat is released. ΔH soln = H final! H initial Since ΔH soln < 0, H products < H reactants Thus, products have lower enthalpy (heat content) than the reactants The reaction is carried out at constant pressure so the heat is equal to enthalpy. ΔH = q p ΔE = q v enthalpy is heat at constant pressure internal energy is heat at constant volume A (exothermic, lower, ΔH)

3 Copyright RJZ 2/15/18 3 no unauthorized use allowed 2) For calorimetry problems you should know what heat capacity, C, is and how molar heat capacity, C m, and specific heat, C s, are related to it and how they all related to heat transfer and temperature change. Heat Capacity: The amount of heat required to raise the temperature of a substance by 1 EC. Often used when you don t have pure or uniform substances (such as a styrofoam cup in exp 6). This is an extensive property (depends on how much is present, how big the cup is). C = q/δt Molar Heat Capacity: Heat Capacity of one mole of a substance (heat cap. per mol). The amount of heat required to raise the temp. of 1 g of a substance by 1 EC. This is most often used for a pure substance and is an intensive property (while the heat added is extensive). C m = C/mol Specific Heat: Heat Capacity of one gram of a substance (heat cap. per gram). The amount of heat required to raise the temp. of 1 g of a substance by 1 EC. This is most often used for a pure substance and is an intensive property (while the heat added is extensive). C s = C/gram This problem is very similar to what you did in experiment 6 in lab except the problem assumes no heat is lost to the calorimeter (it has a heat capacity of zero). Heat was gained by the H 2 O since it s temp. increased (from 22.5 EC to 25.0 EC) and heat was lost by the metal since it s temp. decreases (from 99.8 EC to 25.0 EC). q = m C s ΔT Heat = (specific (temperature change) q gained by H2O =! q lost by nickel C s = J/g@EC (the specific heat of H 2 O to 4 s.f. - from the first sentence) q H2O = C ΔT = g (4.184 J/g@EC) (25.0 EC! 22.5 EC) q H2O = 1046 J (2 s.f. since 25.0! 22.5 = 2.54, 2 s.f.) q Ni =! 1046 J & q Ni = C ΔT! 1046 J = (36.6 g) C C s,ni C (25.0 EC! 99.8 EC) = (! gcec) C C s,ni C s,ni =! 1046 J / (! gcec) = J/g@EC (technically only 2 s.f.) D

4 Copyright RJZ 2/15/18 4 no unauthorized use allowed 3) Standard Enthalpy (heat) of Formation: Change in enthalpy for a reaction which forms ONE mole of a compound from its elements with all substances in their standard states. ΔH f E Standard state: pure form at atmospheric pressure (and some temperature of interest). A specific temp. is NOT part of the definition. Thus for H 2 O at 1 atm and 25 EC it s standard state would be a liquid. For H 2 O at 1 atm and 110 EC it s standard state would be a gas. The only reaction given which produces 1 mole of a substance in its standard state from it s elements is e) Na(s) + 1/2 Cl 2 (g) + 3/2 O 2 (g) 6 NaClO 3 (s) Note: You may see fractions in these balanced eqns. for ΔH f E E

5 Copyright RJZ 2/15/18 5 no unauthorized use allowed 4) heat stoichiometry problem A 5) Use ΔH f E to determine the ΔH rxn. Based on Hess s Law. 0 o ΔH = nh f, products mh 2 NO(g) + 4 H 2 O(R) + 3 S (s) v 3 H 2 S(g) + 2 HNO 3 (R) H 2 S (g) ΔH f E = kj/mol HNO 3 (R) ΔH f E = kj/mol NO(g) ΔH f E = kj/mol H 2 O (R) ΔH f E = kj/mol ΔHE = [(3 mol)(-20.6 kj/mol) + (2 mol)( kj/mol)] & [(2 mol)(90.25 kj/mol) + (4 mol)( kj/mol) + (3 mol)(0 kj/mol)] = [-410 kj] & [ kj] = kj products reactants 0 f, reactants D

6 Copyright RJZ 2/15/18 6 no unauthorized use allowed 6) Hess s Law problem Want ΔH rxn for a reaction given ΔH rxn for other reactions. 2 BCl 3 (g) + 6 HCl(g) v B 2 H 6 (g) + 6 Cl 2 (g) ΔH rxn =? 1a) BCl 3 (g) + 3 H 2 O (R) 6 H 3 BO 3 (g) + 3 HCl(g) ΔH = kj 2a) B 2 H 6 (g) + 6 H 2 O (R) 6 2 H 3 BO 3 (g) + 6 H 2 (g) ΔH = kj 3a) 1/2 Cl 2 (g) + 1/2 H 2 (g) 6 HCl(g) ΔH = kj Remember: 1) If you multiply an eqn. by some factor the ΔH gets multiplied by that factor. 2) If you reverse an eqn. it simply changes the sign of ΔH. ΔH reverse =! ΔH forward To get the reverse of a reaction you multiply everything by!1, the coefficients and ΔH. Your coefficients in the balanced eqn. would all be negative. To get rid of the negative signs you simple turn the whole eqn. around (reverse it), just like in a math eqn. So in reality, this is #1 using a!1 as the multiplying factor. ***** continued on next page *****

7 Copyright RJZ 2/15/18 7 no unauthorized use allowed 6) (cont.) 1) Need BCl 3 as a reactant and need a coeff. of 2 (2 BCl 3 ). Can take eqn. and multiply it by 2. 1b) 2 BCl 3 (g) + 6 H 2 O (R) 6 2 H 3 BO 3 (g) + 6 HCl(g) ΔH 1b = 2 (!112.5 kj) =!225 kj 2) Need 1 B 2 H 6 molecule as a product. Can take eqn. 2a, reverse it (multiply it by!1). 2b) 2 H 3 BO 3 (g) + 6 H 2 (g) 6 B 2 H 6 (g) + 6 H 2 O (R) ΔH 2b =!(!493.4 kj) = kj 3) Need 6 Cl 2 molecule as a product. Take eqn 3a and multiply it by!12. Also, need 12 HCl as reactant to cancel the 6 HCl as a product from 1(b) and leave 6 HCl as reactants& 6 H 2 as a product to cancel 6 H 2 as a reactant from 2(b). 3b) 12 HCl(g) 6 6 Cl 2 (g) + 6 H 2 (g) ΔH 3b =!12 ( kj) = kj Add the eqns. and delh values: 1b) 2 BCl 3 (g) + 6 H 2 O (R) 6 2 H 3 BO 3 (g) + 6 HCl(g) ΔH 1b = 2 ( kj) =!225 kj 2b) 2 H 3 BO 3 (g) + 6 H 2 (g) 6 B 2 H 6 (g) + 6 H 2 O (R) ΔH 2b = kj 3b) 12 HCl(g) 6 6 Cl 2 (g) + 6 H 2 (g) ΔH 3b =!12 ( kj) = kj 2 BCl 3 (g) + 6 H 2 O (R) + 2 H 3 BO 3 (g) + 6 H 2 (g) + 12 HCl(g) 6 2 H 3 BO 3 (g) + 6 HCl(g) + B 2 H 6 (g) + 6 H 2 O (R) + 6 Cl 2 (g) + 6 H 2 (g) ΔH rxn = ΔH 1b + ΔH 2b + ΔH 3b Cancel things in eqn. which are the same on both sides, combine same things which appear on the same side and add ΔH values: 2 BCl 3 (g) + 6 HCl(g) v B 2 H 6 (g) + 6 Cl 2 (g) (desired eqn.) A ΔH rxn = (! 225 kj) + ( kj) + ( kj) = kj

8 Copyright RJZ 2/15/18 8 no unauthorized use allowed 7) The rules for the values for the quantum numbers are given in section 6.5 (and 6.7 concerning the spin q.n.). n = 1, 2, 3,..., 4 (integer values starting at 1) principal quantum number (shell number) R = 0, 1, 2,..., (n-1) angular momentum (azimuthal) q.n. (subshell q.n. and defines shape of orbitals w/in a subshell) m R = -R,..., 0,..., +R (values increase by 1) magnetic q.n. (describes orientation of orbital in space - orbitals are degenerate unless in an applied magnetic field) m s = + ½ and - ½ (only one with non-integer values) spin magnetic q.n. (electrons spin around an axis, only has two directions of spin, up or down ) Correct q.n.: For (3) when n = 4 and R = 3 the possible values of m R are m R = -3, -2, -1, 0, 1, 2, 3 and m s =! ½ ok For (4) when n = 8 and R = 6 the possible values of m R are m R = -6,..., 0,..., 6 and m s = + ½ ok Incorrect q.n.: For (1) when n = 2 and R = 0, 1 (R can t be 2 when n = 2) For (2) when n = 3 and R = 1 can t have m R = + 2 or m s = 1 (m s can be only + ½ and!½) A (1 and 2)

9 Copyright RJZ 2/15/18 9 no unauthorized use allowed 8) The frequency of radiation is related to wavelength, c ν = λ Within this distance of 1.6 x 10!7 m: The wave has 4.5 wavelengths & a single wavelength is 3.6 x 10!8 m and the frequency is 8.4 x Hz c 3.00 x 10 8 m/s ν = = = x s -1 λ 3.6 x 10!8 m = 8.4 x Hz h c E = hν = λ h = Planck s constant = x 10!34 JCs E = (6.626 x 10!34 JCs) (8.437 x s -1 ) = x 10!18 J = 5.6 x 10!18 J C

10 Copyright RJZ 2/15/18 10 no unauthorized use allowed 9) The de Broglie eqn. relates the frequency, wavelength and mass of a particle: h λ = m v λ = wavelength of the particle m = mass of particle v = velocity of particle h = Planck s constant = x 10!34 JCs (1 J = 1 kgcm 2 /s 2 ) This eqn. shows that wavelength and velocity are inversely proportional (when one is large the other is small) and wavelength is also inversely prop. to mass. Want the speed of a muon with a rest mass of amu and a a wavelength of 3.98 x 10!3 nm x 10!24 g 1 kg? kg = amu x x = x 10!28 kg 1 amu 10 3 g B h x 10!34 JCs v = = = 8.87 x 10 5 m/s m λ (1.877 x 10!28 kg) (3.98 x 10!12 m) 10) Look at figures 6.17 (energy levels in H atom) and 6.24 (energy levels for many-electron atoms). In the H atom the subshells in a shell are all degenerate (have same energy). Also, the energy levels increase with the principal quantum number, n, and do not cross. In many-electron atoms the interaction between the electrons causes the subshell energies to change and the various subshells within a shell have different energies. The energies of the subshells increase with increasing azimuthal (angular momentum) number R. The orbital energies within a subshell are still degenerate. This also causes a reordering of subshells with different principal quantum numbers, n. For example, the 4s subshell is lower in energy than the 3d. There are many other examples of this type of reordering. A

11 Copyright RJZ 2/15/18 11 no unauthorized use allowed 11) For energy changes in a hydrogen atom we have the following eqn., 1 1 Δ E Hydrogen =! (hcr H ) ( -----! ) n f 2 n i Δ E Hydrogen =! (2.18 x 10!18 J) ( -----! ) n f 2 n i 2 Δ E Hydrogen =! (hcr H ) (1/n f 2-1/n i2 ) For a transition from n i = 3 to n f = 1, Δ E Hydrogen =! (hcr H ) (1/n f 2-1/n i2 ) =! (hcr H ) (1/1 2! 1/3 2 ) =! 8/9 (hcr H ) This is negative because this is the energy change for an electron falling to a lower energy level which releases energy in the form of a photon. While this is negative the energy of the photon would be positive. C

12 Copyright RJZ 2/15/18 12 no unauthorized use allowed 12) The correct statements are 2, 3 & 4: The number used to designate a shell: 2) is placed before a letter indicating the subshell (e.g. 1s, 2s, 2p, etc.). 3) provides information about the maximum number of orbitals in a shell (# orbitals in shell n = n 2 ). 4) provides information about the average distance of an electron in the shell from the nucleus. The corrected answers for 1 & 5 would be: 1) provides information about the number of orbitals in a shell; # of orbitals in a shell = n 2 (not subshell). 5) can be any integer š 1. The following table shows values of q.n. R, corresponding letters, # orb. and max. # e- in subshell R = subshell letter s p d f g h... # orbitals in subshell max. # e - in subshell B (2, 3 & 4 are correct) 13) They all have a electrons in a d subshell. D 32Ge [Ar] 4s 2 3d 10 4p 2 (has e - in a 3d subshell) 37Rb [Kr] 5s 1 (has e - in a 3d subshell) 49In [Kr]5s 2 4d 10 5p 1 (has e - in 3d and 4d subshells) 56Ba [Xe]6s 2 (has e - in 3d and 4d subshells) 14) Need to write out the e- configuration for each atom and then look at which electrons will come out when it forms an ion. Remember, the valence e! come out first when forming ions. Cr [Ar] 4s 1 3d 5 (exception to filling) C Cr 3+ [Ar] 3d 3

13 Copyright RJZ 2/15/18 13 no unauthorized use allowed 15) Paramagnetic substances have unpaired electrons are attracted to a magnetic field Diamagnetic substances have no unpaired electrons and are actually weakly repelled by a magnetic field. 1) 30Zn [Ar] 4s 2 3d 10 0 unpaired electrons Diamagnetic 2) 29Cu [Ar] 4s 1 3d 10 1 unpaired electrons Paramagnetic 3) 28Ni [Ar] 4s 2 3d 8 2 unpaired electrons Paramagnetic E (Cu and Ni) 16) IE (ionization energy) is the energy required to remove an electron from a neutral atom or an ion. EA (electron affinity) is the energy change which occurs when an electron is added to a gaseous atom. c) F: The first EA (energy associated with the addition of 1 e! to a neutral gaseous atom) of most atoms is negative but not all. The first EA for ALL the noble gases is positive since adding an e! to an atom with filled shells and subshells requires energy since these e! configuration are the most stable. For group 2Athe atoms have a filled s-subshell. To add an electron it has to go into the higher energy p subshell. For the smaller group 2 atoms this requires energy. Nitrogen also has a positive EA since it s p subshell is ½-filled. That is the next most stable e! configuration. In order to add another e! it has to go into an orbital which already has an electron and that leads to electron-electron repulsion. This requires energy for N since it is small and the electrons can t spread out to reduce the extra repulsion. C

14 Copyright RJZ 2/15/18 14 no unauthorized use allowed 17) (see #16 above for some of the statements in this problem). For d) The EA of O is less negative than the EA for S Electron affinity, EA, is the energy change which occurs when an electron is added to a gaseous atom. It measures the attraction, or affinity, of the atom for the added electron. In general EA becomes more negative up a group. However, when going from row 3 to row 2 elements the EA is more positive (or less negative). That s because row 2 elements are small and can t spread out the resulting extra negative charge as well so not as much energy is released when the electron is added (more repulsion). C 18) Simply look at the electronegativities of the atoms. Remember, EN increases bottom to top and left to right in the periodic table. Inc. ionization energy & electronegativity Also remember the ordering of a number of the most used atoms and the fact there is a big drop in EN going from the 2 nd row to the 3 rd row and then from the 3 rd row down the EN decreases but not by much. E Since all the molecules are binary compounds involving H and one other atom one would expect the most polar bond to be the one involving the most EN atom attached to the H atom. F is the most electronegative of the elements in the answers (S < N = Cl, < O < F). Thus, one would expect HF to have the most polar bond. The S is the least EN of the elements in the answers so one would expect H 2 S to have the least polar bonds.

15 Copyright RJZ 2/15/18 15 no unauthorized use allowed 19) SeO 3 2& Se (6A) O (6A)!2 chg 1) A = 1(6e!) + 3(6e!) + (2e!) = 26 val e! 2) Draw skeleton structure. The more EN O atoms attached to Se. This accounts for 6 e -. 3) Put 6 e - on each O to fulfill octet. This uses 18 e -. 4) # e - left = 26 e - - (6 e e - ) = 2 e - The 2e- go on the Se to fulfill octet to begin with. This would give the structure in (d), which conforms to the octet rule. However, we want the structure which conforms the best to the Formal Charge rules. Formal Charge on the atoms in SeO 3 2! lpe - assigned to the atom they re on. Divide e - in bonds equally between atoms (each atom gets ½ the e - involved in the bond). Subtract this total from the valence e - on the atom. Ex: For Se: 6 - (2 + 4) = 0 For O: 6 - (6 + 1) =!1 (for O on top with single bond) For O: 6 - (4 + 2) = 0 (for O with double bond) For O: 6 - (6 + 1) =!1 (for O at the bottom left with single bond) FC should add up to give overall chg., in this case -2 (as they do) and the more negative FC s are on the more electronegative atoms. Also, the absolute values add up to 2. The structure in (d) is the correct structure for it conforming to the octet rule. In this structure the FC on the Se is +1 and the FC s on the O atoms are all -1. The absolute values add up to 4 so based on this the structure in (d) isn t as good as the one above since the absolute values of the FC s for the one above add up to only 2. ***** continued on next page *****

16 Copyright RJZ 2/15/18 16 no unauthorized use allowed 19) (Cont.) E Resonance structures: 3 resonance structures conforming to the octet rule and 3 resonance structures conforming to the species shown above. The pair of e & representing the π-bond can be placed in 3 different positions (between the Se and any of the three O atoms). This is true for either (d) or (e). All 3 Se-O bonds are equivalent, the same bond length and bond strength. The bond is between a single and double bond. Each Se-O bond is essentially 1 1/3 of a bond (the 1/3 of a bond coming from having 1 pair of e & spread over all three different S-O bonding regions. For either (d) or (e) the structure is the same: Trigonal pyramidal, angles of approx E (regardless if you use the structure based on the octet rule or formal charge rule) 20) To fulfill the octet rule on sulfur there has to be a double bond between the S and an O atom. The double bond could be drawn in 2 different positions leading to 2 resonance structures. The double bond isn t moving or flipping back and forth between the two regions. There aren t two separate molecules. The two bonds are identical (have the same bond length and electron distribution). There s simply no easy way to draw one single Lewis Structure for this molecule. What it represents is that the extra pair of electrons represented in the double bond is actually being shared equally in both regions (actually shared by all three atoms in a molecular orbital which spans all three atoms). It s like the extra bond is shared equally to give a bond order of 1.5. B 2 resonance structures ( double bond can be in 1 of the 2 different positions) Bent, angles of approx. 120E

17 Copyright RJZ 2/15/18 17 no unauthorized use allowed 21) Formal Charge on the atoms in SCN! Divide e - in bonds equally between atoms (each atom gets ½ the e - involved in the bond). lpe - assigned to the atom they re on. Subtract this total from the valence e - on the atom. Ex: For O: 6 - (6 + 1) =!1 For C: 4 - (4) = 0 For N: 5 - (2 + 3) = 0 FC should add up to give overall chg., in this case!1 (as they do) B

18 Copyright RJZ 2/15/18 18 no unauthorized use allowed 22) IF 4! I: 7A F: 7A I F 1) A = 1(7e-) + 4(7e-) + (1e-) = 36 val e! 2) Draw skeleton structure. The more EN F atoms attached to I. This accounts for 8 e! leaving 28 e!. 3) Put 6 e! on each F to fulfill octet. This uses 24 e!. 4) # e - left = 36 e!- (8 e!+ 24 e!) = 4 e! These go on the I atom, in the equatorial positions. This gives 12 e! on the I atom, more than an octet. Iodine is in the middle I has 4 bonds & 2 lpe! in this case since it is in the middle (when halogens are on the outside they have only 1 bond, F is always on the outside, never in the middle) The ED geometry around the I is octahedral and the molecular geometry is square planar with bond angles of exactly 90E. B (2 lone pairs of electrons on the I)

19 Copyright RJZ 2/15/18 19 no unauthorized use allowed 23) SF 6 and BCl 3 There are several instances when the octet rule is not obeyed. There are cases when there is less than an octet and there s an even number of electrons. This occurs for Be (group 2A) and B and Al (group 3A). Be gets only 4 e! around it and forms 2 bonds (group 2A) B and Al get only 6 e! around them and form 3 bonds (group 3A) C CBe C CB C (Al like B since they are in the same group, 3A) 2 bonds 3 bonds 4 e! 6 e! There s also the cases when an atom can get more than an octet of electrons. This occurs for atoms from the 3 rd row down. It does NOT occur for atoms in row 2. Examples are PCl 5, SF 6, XeF 2 (3 lone pairs and 2 atoms), XeF 4 (2 lone pairs and 4 atoms) and others. There are some systems with an odd # of electrons. These will not get an octet on one of the atoms and there will be an unpaired electron. This occurs most often with Nitrogen containing compounds (group 5A) and halogens (other than F and when they re in the middle surrounded by other atoms, particularly O atoms). NO 2 has 17 valence electrons and it winds up with an unpaired e! in the molecule and it s usually shown on the N atom. D While H has less than an octet it conforms to the octet rule, what I also called the noble-gas rule. It winds up with 2 e! around it to look like He, a noble gas. Some books will state H conforms to the octet rule even though it doesn t get eight electrons around it and some state it doesn t.

20 Copyright RJZ 2/15/18 20 no unauthorized use allowed 24) Use bond enthalpies to determine the ΔH rxn. You will see bond enthalpies as BE(bond) or D(bond). ΔH = BE broken BE formed H H H H H C C O H + H Br H C C Br + H O H H H H H C - H = 414 kj H - Br = 366 kj C - C = 348 kj C - Br = 276 kj C - O = 358 kj O - H = 463 kj ΔHE = [(5) D(C - H) + (1) D(C - C) + (1) D(C - O) + (1) D(O - H) + (1) D(H - Br)] & [(5) D(C - H) + (1) D(C - C) + (1) D(C - Br) + (2) D(O - H) ] ΔHE = [(5) (414 kj) + (1) (348 kj) + (1) (358 kj) + (1) (463 kj) + (1) (366 kj)] & [(5) (414 kj) + (1) (348 kj) + (1) (276 kj) + (2) (463 kj)] = [3605 kj] & [3620 kj] =! 15 kj There is an easier way to do this. Since the C-C and C-H bonds being broken are also being reformed you can ignore them (they simply cancel out in the end). The same for one of the O-H bonds. ΔHE = [(1) D(C - O) + (1) D(H - Br)] & [(1) D(C - Br) + (1) D(O - H) ] ΔHE = [(1) (358 kj) + (1) (366 kj)] & [(1) (276 kj) + (1) (463 kj)] =! 15 kj A

21 Copyright RJZ 2/15/18 21 no unauthorized use allowed 25) PO 3 3& P: 6A O: 6A P O 3e& 1) A = 1(5e-) + 3(6e-) + (3e-) = 26 val e! 2) Draw skeleton structure. The more EN O atoms attached to P This accounts for 6 e -. P is in the middle 3) Put 6 e - on each O to fulfill octet. This uses 18 e -. 4) # e - left = 26 e - - (6 e e - ) = 2 e - These go on the P atom to fulfill octet. P has 3 bonds & 1 lpe - in this case since it is in the middle O has 1 bond & 3 lpe - (Oxygen and other group 6A elements usually have 2 bonds and 2 lpe - or 1 bond 3 lpe -, particularly when surrounding another atom, like Cl, Br, I, N and S) Shape and bond angle 3 atoms (bonds) & 1 lpe - on phosphorus trigonal pyramidal, angles of. or < 109.5E There are 4 things around P (3 O s and 1 lpe - ). These 4 things arrange themselves in a tetrahedral structure (electron-pair geometry) around P and bond angles start out at 109.5E. However, since all 4 things on P are not identical the O & P & O angles are E but not exactly 109.5E. Three atoms and 1 lpe - on P gives a trigonal pyramidal shape around the P with bond angles of E. Ignore the lpe - when giving the molecular shape. E

22 Copyright RJZ 2/15/18 22 no unauthorized use allowed 26) SO 3 2! and AsCl 3 In order for a molecule to be trigonal pyramidal there has to be 3 atoms and 1 lpe & around the central atom. trigonal pyramidal: 3 atoms and 1 lpe & surrounding A angle < 109.5E or.109.5e *1) SO 3 2! 3 atoms and 1 lpe & on S (4 things total) trigonal pyramidal angle E (neither polar nor nonpolar since it s an ion) 2) SeO 2 2 atoms and 1 lpe & on Se (3 things total) Bent angle. 120 E polar 3) CS 3 2! 3 atoms and no lpe & on C (3 things total) trigonal planar angle = 120E (exactly) (neither polar nor nonpolar since it s an ion) 4) BCl 3 3 atoms and no lpe & on B (3 things total) trigonal planar angle = 120E (exactly) nonpolar *5) AsCl 3 3 atoms and 1 lpe & on As (3 things total) trigonal pyramidal, like NH 3, PCl 3 angle E polar D (1 & 5, SO 3 2& and AsCl 3, are trigonal pyramidal)

23 Copyright RJZ 2/15/18 23 no unauthorized use allowed 27) Shape around: C atoms: N atom: tetrahedral - 4 atoms attached to the C atoms and no lone pairs, bond angles around C are E and it is polar around the C atoms. trigonal pyramidal - 3 atoms and 1 lone pair of electrons on the N atom. Since the electron-pair arrangement around the N atom is tetrahedral but all 4 things on the N atom are not identical the molecule will have bond angles of E and it will be polar around the N atom. Around the N atom it looks like NH 3 but with one of the H atoms replaced with the CH 3 CH 2 - group. B

24 Copyright RJZ 2/15/18 24 no unauthorized use allowed 28) *1) PCl 3 trigonal pyramidal: 4 things around P and they are not all identical (3 Cl and 1 lpe! ) lpe & on P ( E angles) Polar (like :NH 3, :NF 3, :PH 3, :PF 3, etc.) *2) OF 2 bent: 4 things around O and not all identical (2 F and 2 lpe! ) lpe & on O ( E angles).... Polar (like H 2 O:, H 2 S:, etc.) 3) AlF 3 trigonal planar (120E angles, exactly) 3 things around Al and all identical (and no lpe & ) Nonpolar (like BH 3, BF 3, AlCl 3 etc.) 4) BeCl 2 linear (180E angles) the 2 atoms on Be are identical Nonpolar *5) SiHCl 3 tetrahedral (.109.5E angles) all 4 atoms on C are not identical Polar (like CH 2 Cl 2, CH 3 Cl, CH 2 F 2, CHCl 3, etc.) Note: The symmetric shapes (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral and square planar) are the only ones that can give nonpolar. For linear, trigonal planar and tetrahedral all surrounding atoms have to be identical to each other for the molecule to be nonpolar. For CHCl 3 (#5) all surrounding atoms are NOT identical so the molecule is polar. Also, the molecules (1 & 2) with lpe! are polar. Generally, if there are 1 or more lpe - on the central atom, the molecule will be polar (at least for molecular shapes coming from ED geometries involving 2, 3 or 4 ED). The exceptions to this are for the linear molecular geometry arising from the trigonal bipyramidal ED geometry and the square planar molecular geometry arising from the octahedral ED geometry. These can be nonpolar if all the atoms surrounding the central atom are identical. E (1, 2 & 5 are Polar)

25 Copyright RJZ 2/15/18 25 no unauthorized use allowed 29) sp sp sp 3 sp sp sp 3 sp sp sp 2 sp 2 sp 3 sp 2 sp 2 sp 3 H & C / C & CH 2 & C / C & CH 2 & C / C & CH = CH & CH 2 & CH = CH & CH 3 The carbon atoms in the triple bond form 2 sp hybrid orbitals each with two p orbitals remaining. The sp orbitals form a σ bond with hydrogen and/or with carbon atoms (total of 2 σ bonds). The other two bonds between the carbon atoms results from a side-by-side overlap of the two p orbitals of each carbon atom to form two pi (π) bonds. The molecule above has 4 sp 3, 4 sp 2 and 6 sp hybridized C atoms. The molecule above has 3 triple bonds. Since each triple bond has two sp hybridized carbon atoms, there are a total of 6 sp hybridized carbon atoms. (Remember, each triple bond has 2 sp hybridized orbitals and two p orbitals). The double-bonded C atoms each have only 1 double bond to them so there are 4 sp 2 hybridized carbon atoms (Remember, each normal double bond has 2 sp 2 hybridized carbon atoms; see exception below). Normally carbon atoms with double bonds are sp 2 hybridized. However, there is an exception when a carbon atom has 2 double bonds to it (the 2 nd carbon from the right). In this case the carbon atom is sp hybridized (see below).! C / C! sp hybrid carbon atoms \ * / * / C = C = C sp hybrid carbon atom (= C = in middle, = C is sp 2 ) / \ \ \ / C = C / \ sp 2 hybridized carbon (2 single bonds and double bond to carbon) *!C! sp 3 (4 single bonds) 2 things attached to sp hybridization a central atom linear, 180E 3 things attached to sp 2 hybridization a central atom trigonal planar, 120E 4 things attached to sp 3 hybridization a central atom tetrahedral, 109.5E C (6 sp hybridized C atoms)

26 Copyright RJZ 2/15/18 26 no unauthorized use allowed 30) sp 2 hybrid orbitals give an overall trigonal planar geometry with angles of 120E. When an atom has 3 things around it the atom generally is using sp 2 hybrid orbitals for bonding and lone pairs. The bent molecular shape with bond angles close to 120E arises when an atom is bonded to 2 other atoms and has 1 lpe & on it (3 things total). An example is SO 2. The S atom is sp 2 hybridized and uses two of the hybrid orbitals to form the bonds to the O atoms and 1 sp 2 hybrid orbital is used for the lpe & on the S atom. sp sp 2 sp 3 linear, 180E trigonal planar, 120E (molecular shapes: trigonal planar, bent) tetrahedral, 109.5E (molecular shapes: tetrahedral, trigonal pyramidal, bent) B