Chem 6A 2011 (Sailor) QUIZ #7 Name: Student ID Number: Section Number: VERSION A KEY Some useful constants and relationships: Specific heat capacities (in J/g. K): H 2 O (l) = 4.184; Al (s) = 0.900; Cu (s) = 0.387; Steel (s) = 0.45 101.325 J = 1 L. atm 1 atm = 760 Torr 1J = 1kg. m 2 /s 2 1 ev = 1.6022 x 10-19 J R = Ideal gas constant: 0.08206 L. atm. mol -1. K -1 = 8.31451 J. mol -1. K -1 Avogadro constant: 6.022 x 10 23 mole -1 Planck's constant = h = 6.6261 x 10-34 J. s c = speed of light: 3.00 x 10 8 m/s R H = 1.097 x 10-2 nm -1 C 2 = second radiation constant = 1.44 x 10-2 K. m q = C p mδt Tλ max = 1 5 C Emitted power (W) 2 = (constant)t 4 Surface area (m 2 ) e = mc 2 c = λν 1 λ = R 1 H 2 n 1 2 E = hν E = hc E(in Joules) = 2.18 10 18 Z 2 1 n 2 λ n 2
Diamagnetic Cd 2+ Chem 6A 2011 (Sailor) QUIZ #7 1. Rank each set of elements in order of decreasing size. The first one is done for you as an example. (10 pts) (a) Li, Na, K K > Na > Li (b) Sr, Ba, Ca Ba > Sr > Ca (5 pts if correct, all or nothing) (c) O, O -, O 2- O 2- > O - > O (5 pts if correct, all or nothing) 2. Fill in the electrons in the valence level orbital diagrams below to provide a ground state electron configuration for each gas phase atom or ion that is consistent with the magnetic property given and label the valence s, d, and p shells with the correct principle quantum number. The first one is done as an example. (20 pts) Paramagnetic Br 4 s 3 d 4 p 5s 4d 5p Paramagnetic Cu 4s 3d 4p 2 pts for correct quantum shell numbers, all or nothing. 8 points for correct electron configuration, all or nothing This was homework problem 8.61 in the book. 3. Identify each element from the given electron configuration for its neutral atom. Indicate the number of unpaired electrons in each. The first one is done as an example. (20 pts) This was homework problem 8.30 in the book. element Electron configuration # unpaired electrons P [Ne] 3s 2 3p 3 3 Sr Si Ne - Fe [Kr] 5s 2 [Ne] 3s 2 3p 2 [He] 2s 2 2p 6 [Ar] 4s 2 3d 6 0 (2 pts) 2 (2 pts) 0 (2 pts) 4 (2 pts)
Chem 6A 2011 (Sailor) QUIZ #7 Name: Student ID Number: Section Number: VERSION B KEY Some useful constants and relationships: Specific heat capacities (in J/g. K): H 2 O (l) = 4.184; Al (s) = 0.900; Cu (s) = 0.387; Steel (s) = 0.45 101.325 J = 1 L. atm 1 atm = 760 Torr 1J = 1kg. m 2 /s 2 1 ev = 1.6022 x 10-19 J R = Ideal gas constant: 0.08206 L. atm. mol -1. K -1 = 8.31451 J. mol -1. K -1 Avogadro constant: 6.022 x 10 23 mole -1 Planck's constant = h = 6.6261 x 10-34 J. s c = speed of light: 3.00 x 10 8 m/s R H = 1.097 x 10-2 nm -1 C 2 = second radiation constant = 1.44 x 10-2 K. m q = C p mδt Tλ max = 1 5 C Emitted power (W) 2 = (constant)t 4 Surface area (m 2 ) e = mc 2 c = λν 1 λ = R 1 H 2 n 1 2 E = hν E = hc E(in Joules) = 2.18 10 18 Z 2 1 n 2 λ n 2
Paramagnetic Ru 2+ Chem 6A 2011 (Sailor) QUIZ #7 1. Rank each set of elements in order of decreasing size. The first one is done for you as an example. (10 pts) (a) Li, Na, K K > Na > Li (b) N, As, P As > P > N (5 pts if correct, all or nothing) (c) S, S -, S 2- S 2- > S - > S (5 pts if correct, all or nothing) 2. Fill in the electrons in the valence level orbital diagrams below to provide a ground state electron configuration for each gas phase atom or ion that is consistent with the magnetic property given and label the valence s, d, and p shells with the correct principle quantum number. The first one is done as an example. (20 pts) Paramagnetic Br 4 s 3 d 4 p 5s 4d 5p Paramagnetic Ge 4s 3d 4p 2 pts for correct quantum shell numbers, all or nothing. 8 points for correct electron configuration, all or nothing This was homework problem 8.61 in the book. 3. Identify each element from the given electron configuration for its neutral atom. Indicate the number of unpaired electrons in each. The first one is done as an example. (20 pts) This was homework problem 8.30 in the book. element Electron configuration # unpaired electrons P [Ne] 3s 2 3p 3 3 Rb Al Cl Co [Kr] 5s 1 [Ne] 3s 2 3p 1 [Ne] 3s 2 3p 5 [Ar] 4s 2 3d 7 3 (2 pts)
Chem 6A 2011 (Sailor) QUIZ #7 Name: Student ID Number: Section Number: VERSION C KEY Some useful constants and relationships: Specific heat capacities (in J/g. K): H 2 O (l) = 4.184; Al (s) = 0.900; Cu (s) = 0.387; Steel (s) = 0.45 101.325 J = 1 L. atm 1 atm = 760 Torr 1J = 1kg. m 2 /s 2 1 ev = 1.6022 x 10-19 J R = Ideal gas constant: 0.08206 L. atm. mol -1. K -1 = 8.31451 J. mol -1. K -1 Avogadro constant: 6.022 x 10 23 mole -1 Planck's constant = h = 6.6261 x 10-34 J. s c = speed of light: 3.00 x 10 8 m/s R H = 1.097 x 10-2 nm -1 C 2 = second radiation constant = 1.44 x 10-2 K. m q = C p mδt Tλ max = 1 5 C Emitted power (W) 2 = (constant)t 4 Surface area (m 2 ) e = mc 2 c = λν 1 λ = R 1 H 2 n 1 2 E = hν E = hc E(in Joules) = 2.18 10 18 Z 2 1 n 2 λ n 2
Paramagnetic Mo 2+ Chem 6A 2011 (Sailor) QUIZ #7 1. Rank each set of elements in order of decreasing size. The first one is done for you as an example. (10 pts) (a) Li, Na, K K > Na > Li (b) Mg, Be, Ca Ca > Mg > Be (5 pts if correct, all or nothing) (c) Se, Se -, Se 2- Se 2- > Se - > Se (5 pts if correct, all or nothing) 2. Fill in the electrons in the valence level orbital diagrams below to provide a ground state electron configuration for each gas phase atom or ion that is consistent with the magnetic property given and label the valence s, d, and p shells with the correct principle quantum number. The first one is done as an example. (20 pts) Paramagnetic Br 4 s 3 d 4 p Paramagnetic Se 4s 3d 4p 5s 4d 5p 2 pts for correct quantum shell numbers, all or nothing. 8 points for correct electron configuration, all or nothing This was homework problem 8.61 in the book. 3. Identify each element from the given electron configuration for its neutral atom. Indicate the number of unpaired electrons in each. The first one is done as an example. (20 pts) This was homework problem 8.30 in the book. element Electron configuration # unpaired electrons P [Ne] 3s 2 3p 3 3 K S I Ni [Ar] 4s 1 [Ne] 3s 2 3p 4 [Kr] 5s 2 4d 10 5p 5 [Ar] 4s 2 3d 8 2 (2 pts) 2 (2 pts)