Chapter 10 Practice Problems

Transcription

1 Chapter 10 Practice Problems Q S S S C in S C in S = C S Q 10.2 Correct Answer: B Two oxygen atoms will have a formal charge of 1 and there will also be 4 different resonant structures. C C Bond order C- = 1/2 (# bonding e - - # antibonding e - ) = 3/2 Therefore statements 1 and 3 are true. Q 10.3 Correct Answer: A There should be one pair of electrons on the central sulphur atom in A. Solutions will be posted at 1

2 Q 10.4 Total # of valence electrons = = 18 ormal charge on = 7 [6 + ½ (2)] = 0 ormal charge on N = 5 [2 + ½ (6)] = 0 ormal charge on = 6 [4 + ½ (4)] = 0 Since the formal charge on N, and are 0, this is the most likely Lewis structure. Q 10.5 X is Nitrogen (N). Q 10.6 There are 4 lone pairs of electrons H 2 N C NH 2 Q 10.7 Correct answer: A A and C are the only neutral structures. A abides by octet rule. Solutions will be posted at 2

3 Q 10.8 Correct Answer: B Resonance structures occur when more than one Lewis structure, each with different electron distributions, can be drawn for a molecule. This usually results in molecules containing both double bonds and single bonds. (a) 3 (b) No resonance structure. CH 4 cannot have resonance structures because no double bonds are present. H H H H (c) N N N (d) HC 3 - H -1-1 H Solutions will be posted at 3

4 Q 10.9 (a) SCN - Total # of valence electrons: = 16-1 (-) (b) (-) -1 ormal charge = (# of valence electron on a free atom) (# of valence electrons assigned to the atom in the molecule) Valence electrons assigned = (# of lone pair electrons) + ½ (# of shared electrons) or S, formal charge = 6 [4 + ½ (4)] = 0 or C, formal charge = 4 [0 + ½(8)] = 0 or N, formal charge = 5 [4 + ½(4)] = -1 or S, formal charge = 6 [6 + ½ (2)] = -1 or C, formal charge = 4 [0 + ½(8)] = 0 or N, formal charge = 5 [2 + ½(6)] = 0 The first structure is preferred since N is more electronegative than S, and thus is more likely to have the negative charge. Q Correct Answer: C The formal charge of P in structures (i) and (iii) is = 5 (5 + 0) = 0. Based on formal charge, structure (i) is the most important resonance structure. The formal charge of in structure (ii) = 6 (1 + 6) = -1. The most stable structure is that with the lowest formal charge. Solutions will be posted at 4

5 Q Correct answer: C There are two lone pairs and four bonding pairs on the central S atom, giving AX 4 E 2. The lone pairs will align to maximize the distance between them. Therefore, this atom would be square planar. Q Correct answer: B C 4 is a tetrahedral molecule with all the same types of bonds at each point, so it will have a net dipole of zero. Q Species Lewis electron dot structure Shape (as given by nuclei) Hybridizatio n on central atom CS 2 S C S Linear sp Polar (P) or Nonpolar (N) species N Xe 4 Xe Square Planar sp 3 d 2 N S 3 S Trigonal planar sp 2 N S 3 2 S Trigonal Pyramidal sp 3 P S 6 S ctahedral sp 3 d 2 N Solutions will be posted at 5

6 Q H H C N H sp 2 sp 2 Answer: 4 Answer: 1 Yes Yes Solutions will be posted at 6

7 Q Molecule Lewis Structure Molecular Shape (VSEPR) Hybridization of Central Atom Polar (P) or Nonpolar (N) H 2 C C Trigonal planar sp 2 Polar H H 2 Bent sp 3 Polar HCN SiH 4 H C N H linear sp Polar tetrahedral sp 3 Non-polar H H Si H I 5 I Square - pyramidal sp 3 d 2 Polar Se 4 See-saw sp 3 d Polar Se Solutions will be posted at 7

8 Q Solutions: i) N N N ii) Trigonal planar iii) 3 iv) sp 2 v) X = C and Z = B Solutions will be posted at 8

9 Q Species Lewis Structure Shape Hybridization on Central atom Polar/Nonpolar P 5 AX 5 Trigonalbipyramidal sp 3 d N B 3 AX 3 Trigonalplaner sp 2 N P 3 AX 3 E Tetrahedral sp 3 P Br 3 AX 3 E 2 T-shaped sp 3 d P Br 2 + AX 2 E 2 Angular sp 3 P Se 4 AX 4 E Seesaw sp 3 d P Br 5 AX 5 E Squarepyramidal sp 3 d 2 P Solutions will be posted at 9

10 Q C 1 is sp 3 with (tetrahedral) C 2 is sp 2 with 120 (trigonal planar) A is sp 2 hybridized B is sp 3 hybridized Nitrogen is sp 3 hybridized with a bond angle of There are 9 bonds (single bonds) There is 1 bond (double bond) Q Correct answer C I 5 = 42 electrons. AB 5 E 1 n the picture below, there is also a lone pair of electrons around the central iodine. This molecule is polar, because it has a net dipole towards the fluorine (in the plane). I Q Correct answer - E There are 14 sigma bonds (single bonds) and 3 pi bonds Recall that a double bond equals 1 sigma bond and 1 pi bond. Recall that a triple bond equals 1 sigma bond and 2 pi bonds. Solutions will be posted at 10

11 Q Correct Answer: D D and E are the only answers that sum to give a -1 anion (eliminate A, B, C). or nitrogen: = -1, sulfur: = 0, carbon: 4 4 =0 Q Answer: C Sodium iodide, NaI An ionic compound is generally the result of a compound composed of a non-metal and a metal. Q Answer: D Iron is a metal, and has primarily metallic bonding delocalization of electrons Q Answer: B luorine is the most electronegative atom in the periodic table. Electronegative increases from bottom to top, and increases from left to right. Q Correct answer D There are 9 sigma bonds (single bonds) and 1 pi bond (double bond) Recall that a double bond equals 1 sigma bond and 1 pi bond. Q Correct answer C The oxygen is AB 2 E 2 which is tetrahedral family and sp 3 Solutions will be posted at 11

12 Q Correct answer B This carbon behaves as trigonal planar, and therefore sp 2 hybridization Q Correct Answer A Eliminate answers B,D (B has negative 2 charge, D has +2) C is incorrect because it puts negative formal charge on the less electronegative atom. In E, carbon octet is exceeded. Q Correct answer B (is false) N 3 - has 24 electrons Lewis Structures (resonance): N N H N As a result, N- bond length is identical in all N- bonds (making B the false answer). Nitrogen atom has no lone pairs of electrons, and is an AB 3 type molecule (trigonal planar), and hence sp 2 hybridized. Solutions will be posted at 12

13 Q*10.30 Correct answer: C PCl 2 3 can have several orientations; however one set cannot exist because we are told the molecule is polar! PCl 2 3 = = 40 electrons Cl Cl P Cl non-polar Cl P polar In the polar molecule, there is at least one -P- angle of 90 C Q Correct Answer: A Carbon = = 0 Sulfur = = 0 Nitrogen = 5 4 = +1 xygen = = -1 Solutions will be posted at 13

14 Chapter 11 Molecular rbital Theory Chapter 11 Practice Problems Q 11.1 or 2 (total 12 electrons) Bond rder ( 2 ) = (8 4)/2 = 2 Structure: = or 2 (total 14 electrons) Bond rder ( 2 ) = (8 6)/2 = 1 Structure: Q 11.2 or N 2 (total 10 electrons) Bond rder (N 2 ) = (8 2)/2 = 3 Structure: N N or B 2 (total 14 electrons) Bond rder ( 2 ) = (4 2)/2 = 1 Structure: B B Solutions will be posted at 14

15 Q 11.3 H 2 has a total of 2 electrons. In the M diagram, both electrons reside in the bonding orbital giving H 2 a bond order of 1. He 2 has a total of 4 electrons. In the M diagram, the first two electrons will occupy the bonding orbital however the next two will occupy antibonding orbitals. The bond order for He 2 will be zero and therefore non existent. Q* 11.4 A) CN- has 10 valence electrons ill as appropriate: 2s (2) < * 2s (2) < 2p (4) < 2p (2) < * 2p < * 2p CN + = 8 electrons; B = 2 CN = 9 electrons; B = 2½ CN - = 10 electrons; B = 3 Shortest bond order = longest bond Solutions will be posted at 15

16 Q* 11.5 B = 8-8/2 = 0 Therefore, Ne 2 does not exist Solutions will be posted at 16

17 Q* 11.6 Bond order for C is larger than bond order for N, therefore C- bond is shorter. Q* 11.7 Q 11.8 Correct answer: C (false) rbitals are conserved: the number of Ms will always be the same as the number of A used to construct them. Solutions will be posted at 17

18 Q 11.9 Correct answer C Bond order = (8-2)/2 = 6/2 = 3 No unpaired electrons; therefore diamagnetic. Q Correct answer E Bond rder = (8-6)/2 = 2/2 = 1 No unpaired electrons, therefore diamagnetic Solutions will be posted at 18

19 Q Correct answer B 2 + has the same molecular orbital diagram as 2, except one electron is removed! The * 2p therefore has 1 valence electron. Solutions will be posted at 19

20 Q Correct answer D B 2 has 6 valence electrons, so B 2 - has 7 2s (2) < * 2s (2) < 2p (3) Q Correct answer D N 2 has 10 valence electrons, so N 2 + will have 9 valence electrons N 2 2s (2) < * 2s (2) < 2p (4) < 2p (2) B = (8-2)/2 = 3 N 2 + 2s (2) < * 2s (2) < 2p (4) < 2p (1) B = (7-2)/2 = 2.5 As a result, bond order will decrease, and the unpaired electron makes it paramagnetic. Solutions will be posted at 20

21 Q Correct answer C B 2 2s (2) < * 2s (2) < 2p (2) B = (4-2)/2 = 1 C 2 2s (2) < * 2s (2) < 2p (4) B = (6-2)/2 = 2 N 2 2s (2) < * 2s (2) < 2p (4) < 2p (2) B = (8-2)/2 = 3 2 2s (2) < * 2s (2) < 2p (2)< 2p (4) < * 2p (2) B = (8-4)/2 = 2 2 2s (2) < * 2s (2) < 2p (2)< 2p (4) < * 2p (4) B = (8-6)/2 = 1 Q Correct answer B CN + = 8 electrons; B = 2 CN = 9 electrons; B = 2½ CN - = 10 electrons; B = 3 N + = 10 electrons B = 3 ill as appropriate: 2s < * 2s < 2p < 2p Shortest bond order = longest bond Solutions will be posted at 21

22 enster Extra Problems Q I Answer D Corundum is a crystalline form of aluminum oxide (Al 2 3 ) with traces amount of iron, titanium and chromium. Q II Answer E Q III E There are over 500 chemicals found in natural apples Q IV Answer D N 2 is dinitrogen oxide, and is referred to as laughing gas. Q V Answer: A Q VI Answer B Solutions will be posted at 22

23 Chapter 24 Coordination Chemistry Chapter 24 Problems Q 24.1 Correct Answer: A H 2 is neutral ligand, Cl is each -1, Cr = +3, so overall is -1 Q 24.2 Correct Answer: D H 2 is neutral ligand, Co = +2, so overall complex is +2. Q 24.3 Correct answer: A NH 3 is neutral, Br = -1, Pt = +2, therefore overall = zero Q 24.4 Correct answer: A Coordination number = # ligands = 4 S 4 = -2, ammonia = 0, therefore Cu = +2 Q 24.5 Correct Answer: E Coordination number = # ligands = 2 CN = -1, overall complex = -1, therefore, Ag = +1 Q 24.6 Correct Answer: C Coordination number = # ligands = 4+2 = 6 H 2 = neutral, Cl = -1, counterion = -1, therefore Cr = +3 Q 24.7 Correct Answer: B Coordination number = # ligands = 4 CN = -1(4) = -4, counterion = +1(4) = +4, therefore Ni = 0 Solutions will be posted at 23

24 Q 24.8 Correct Answer: E Correct name is tetrachlorocobaltate(ii) ion, (underline for emphasis only) Q 24.9 Correct Answer: A Chloride because counterion, 4 = tetra, di = 2 Q Correct Answer: E Nickelate because anionic, oxidation number = 0, tetracyano Q Correct answer: A B is not by alphabetical order, oxidation number is incorrect in C and D Q Correct answer: D Bromide = counterion, (en) = neutral, so cobalt = +3. must be balanced by 3 bromide ions Q Correct Answer: A CN = -1, Ni = 0, therefore 4 K are required. Ligands on inside, counterion (K) on outside K (counterion) is positive (cation), so goes on left hand side. Q Correct Answer: A Coordination number = # Ligands = 2+2 = 4 Cl = -1, CN = -1, Therefore Cd = +4, and counterion is -1(2). Q [ecl 6 ] must equal -3. Balance charge 3(+2) = 6; 2(x) = -6 to give neutral charge. x = -3. Solutions will be posted at 24

25 Q Correct Answer: -1 NH 4 is +1 and counterion to give neutral charge, therefore complex must equal -1. Q Correct answer: B V = +4, = 2-, CN = -1, therefore overall charge must be equal to -2 Q Correct answer: A There are two cyanide ions complexed with copper(i). This gives the complex a charge of 1-, which means there must be one potassium ion. Q Correct Answer: E Rh = +1, C = 0, Br = -1(2); therefore overall ion complex = -1, so 1 potassium ions are required to make it neutral. Q Correct answer: B A is incorrect (Coordination number is 3 of Ni(en) 3 ) en is bidentate (making C incorrect) hexaamminenickel(3) ion is incorrect, it needs roman numerals 3 sulphate groups are required; cross charges, to give [Ni(en) 3 ] 2 (S 4 ) 3 Q Correct Answer: C Ionization isomer is one where the counter-ion is switched. So switch the bromo for the chloro in (c). Solutions will be posted at 25

26 Q Correct answer: D Cu 2+ is the oxidation state. Cu 2+ is d 9. Irrevalent whether it is high spin or low spin system. 1 unpaired electron Q Correct answer: B Name tells you oxidation state of Mn(II) = 2 Shape = octahedral Mn 2+ = d 5 CN = strong field; t 2g 5 e g 0 1 unpaired electron Solutions will be posted at 26

27 Q Correct answer: E [CoCl(NH 3 ) 5 ] Cl 2 pentaamminechlorocobalt(iii) chloride Co 3+ = d 6 Coordination number = 6 = octahedral t 2g 6 e g 0 0 unpaired electrons (strong field) Q Correct answer: D Cs[eCl 4 ] [ecl 4 ] = -1 e = +3 e 3+ d 5 Cl is a weak-field ligand; so 5 unpaired electrons Solutions will be posted at 27

28 Q Correct answer: A Cl = -1, en = neutral, NH 3 = neutral; therefore Ni = +2 Coordination number = 6 (en is bidentate); Ni 2+ d 8 It does not matter whether it is high field or low field, d 8 gives 2 unpaired electrons in either situation. Solutions will be posted at 28

29 Q Correct Answer: C Each are Ni 2+ d 8 Cl is a weak-field ligand, while CN is a strong field ligand Q Correct answer: E Revisit the spectrochemical series. A D are all strong field. Br is weak field, and would result in a high spin system with unpaired electrons (paramagnetic). Co 3+ = d 6 Solutions will be posted at 29