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# Answers to Chapter 5 Book problems (electron configurations and

By Hazel Lawrence,2014-11-17 00:21
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Answers to Chapter 5 Book problems (electron configurations andto,To,and,book,AND

Answers to Chapter 5 Book problems (electron configurations and calculations)

The complete pattern of electron configurations is 22626210621062141062141061s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p in auf bau order

Or 22626102 61014 2 610 14 18 261014 18 22 261s 2s 2p 3s 3p 3d 4s4p 4d 4f5s5p 5d5f5g6s 6p 6d 6f6g6h7s 7p

keeping all the principle quantum numbers together (we never really use the “g” or “h” orbitals)

- Book problems starting on page 132

o Chapter 5: 7, 9, 14, 15, 16,17, 18, 19, 20, (see diagram on page 143),

21, 29, 35, 36, 49 (see page 143 again), 50, 52, 54, 55, 56, 71,

7) a) 3p sublevel has 3 orbitals (each designated 3p or more specifically 3p,3p,3p,) xyz

b) 2s sublevel has 1 orbital

c) 4p has 3 orbitals (each called 4p)

d) 3d has 5 orbitals

e) 4f has 7 orbitals

2219) a) boron: 1s 2s 2p 1 unpaired electron 2262222 b) silicon: 1s 2s 2p 3s 3p or [Ne] 3s 3p 2 unpaired electron

1314) if frequency = 1.50 x 10 Hz

-5-9 ;= c/f = 3.0E8 / 1.5E13 = 2 x 10 m MUCH longer than red light (700 x 10 m)

1515) wavelength = 50 nm is ultra violet with a frequency of 6 x 10 Hz

16) wavelength and frequency are INVERSLY related; as one increases the other decreases

17) The electrons are attracted back to the ground state after they have been excited by heat or electricity to a higher state; these bursts of energy are quantized so they have a distinctive color pattern

18) The light emitted in an electronic transition from a higher to a lower energy level has a frequency tat is directly proportional to the energy change of the electron.

19) quantum mechanics describes the motions of atoms and subatomic particles; classical mechanics describes the motion of larger bodies.

20) electron transitions from higher levels to n=1 level

21) in order of decreasing wavelength:

C (radio), A (infrared), B (x-rays)

29) a) n=1 has 1 sublevel (the “s”)

b) n=2 has 2 sublevels (the “s” and “p”)

c) n=3 has 3 sublevels

d) n=4 has 4 sublevels

th35) “b”; 3f is an invalid state (you don’t get an “f” orbital until the 4 level

“C”; 2d is an invalid state (at level 2 you only have s and p; no d)

36) a) 2s can hold 2

b) 3p can hold 6

c) 4s can hold 2

d) 3d can hold 10

e) 4p can hold 6

f) 5s can hold 2

g) 4f can hold 14

h) 5p can hold 6

49) it is visible light called the “Balmer series”

2262650) a) 1s 2s 2p 3s 3p is Ar (Z=18) 22626102 67 1 b) 1s 2s 2p 3s 3p 3d 4s4p 4d5s is Ru (Z=44) 2262610 2610 7 261 252) c) 1s 2s 2p 3s 3p 3d4s 4p 4d4f5s 5p 5d6s is Gd (Z=64)

1054) frequency = 1150 kHz; not calculate length in centimeters so c=3.0 x 10 cm/s 4 ; = 3.0E10 / 1.15E6 = 2.61 x 10 cm

-7-555) a) 4.36 x 10 m = 4.36 x 10 cm

b) that’s 436 nm so it is in the visible range 14c) f = c / ; = 3.0E8 / 4.36E-7 = 6.88 x 10 1/s or 688 tera hertz

-5-556) a) ; = 5.890 x 10 cm and ; = 5.896 x 10 cm 1414 b) f= 5.090 x 10 hz and f = 5. 088 x 10 hz