chapter 6
DESCRIPTION
Chapter 6. Chapter 6 3 , 5 , 7,9,13,17,21,27,29,35,41,45, 47 ,53,55,57,59,61 , 65 , 67 , 69 Chapter 7 5,9,13,21,23,29,33,35, 37,43,45,49,53,55,. #3. True False. The frequency of radiation decreases as the wavelength increases. True - PowerPoint PPT PresentationTRANSCRIPT
Chapter 6
• Chapter 6
• 3,5,7,9,13,17,21,27,29,35,41,45,47,53,55,57,59,61,65,67,69
• Chapter 7
• 5,9,13,21,23,29,33,35,37,43,45,49,53,55,
#3
a. True
b. False. The frequency of radiation decreases as the wavelength increases.
c. True
d. False. Electromagnetic radiation and sound waves travel at different speeds.
#5
Wavelength of a. gamma rays < d. yellow (visible) light < e. red light < b. 93.1 MHz FM (radio) waves < c. 680 kHz AM (radio) waves
<
#7
mxx
sx
s
mxvcb
sx
mx
pmx
pmx
s
mxcva
411
8
120
12
8
1087.51011.5
110998.2;/.
1009.5
101
1
589.
110998.2;/.
#7 (continued)
c. No. The radiation in (a) is gamma rays and in (b) is infrared. Neither is visible to humans.
d. 6.54s x 2.998 x 108 m/s = 1.96 x 109 m
#17
photons/s 108.1J102.01
photon 1
1s
.52J
/1001.2s
m 10 2.998
10987
sJ10 626.6
19
198
9
34-
photonJ
m
hchE
#21
KE of form in theenergy excess =J10 X 7.221066.1
1066.1
101
1
120
1
s
m 10 2.998sJ10 626.6 c)
nm 275or m2.75X10Hz101.09
1
s
m 10 2.998 b)
J10 X 7.22=Hz101.09sJ10 6.626 a)
19-18
18
9
834-
715
8
19-1534-
JX
JX
nm
nm
hcE
c
hE
27 a (model for others)
mX1049.91016.3
1
s
m 10 2.998
1016.3sJ10 6.626
1009.2
1009.21
1
25
11018.2
1
n
1Rh=E
8115
8
11534-
18
1818
22i
s
c
sJX
h
E
JXJ
n f
29b
mxJx
smxxsJxEhc
JxnnREn fiH
718
834
18 fi
1056.61018.2
/10998.210626.6/
1018.2/1/1 3; n 3, 22
This is the yellow line at 656nm.
mxJx
smxxsJxnn fi
718
834
1086.41018.2
/10998.210626.6;2,4
This is the blue-green (cyan) line at 486 nm.
35
smv
mkgs
smkgv
kgm
mA
mAmhv
/1088.3
1002.1
1
106749.1
1
1
10626.6
106749.1
;1002.11
10102.1;/
3
10272
234
27
1010
45
(a) permissible, 2p
(b) forbidden, for l=0, ml can only equal 0
(c) permissible, 4d
(d) forbidden, for n=3, the largest l value is 2
53
• The 2p electron in boron is shielded from the full charge of the nucleus by the 2s electrons. (Both the 2s and 2p electrons are shielded by the 1s electrons.) Thus, the 2p electron experiences a smaller nuclear charge than the 2s electrons.
55
• The 2p electron in Ne experiences a greater effective nuclear charge. The shielding experienced by a 2p electron in the two atoms is similar, so the electron in the atom with the larger Z (ZNe=10, ZO=8) experiences the larger effective nuclear charge.
57
(a) +1/2, -1/2(b) Electrons with opposite spins are affected
differently by a strong inhomogeneous magnetic field. An apparatus similar to that in Figure 6.26 can be used to distinguish electrons with opposite spins.
(c) The Pauli exclusion principle states that no two electrons can have the same four quantum numbers. Two electrons in a 1s orbital have the same, n, l, and ml values. The must have different ms values.
65
(a) Rb: [Kr]5s1
(b) Se: [Ar]4s23d104p4
(c) Zn: [Ar]4s23d10
(d) V: [Ar]4s23d3
(e) Pb: [Xe]6s24f145d106p2
(f) Yb: [Xe]6s24f14
67
• A) 3 unpaired electrons in 4p orbitals• B) 2 unpaired electrons in 5p orbitals (one
paired set electrons)• C) 3 unpaired electrons in the 5p• D) 1 unpaired electron in the 5s (remember
this is one element that disobeys the filling order
• E) 2 unpaired electrons in the 5d
5
• Krypton has a larger nuclear charge (Z=36) than argon (Z=18). The shielding of the n=3 shells by the 1s and 2s electrons in the two atoms is approximately equal, so the n=3 electrons in Kr experience a greater effective nuclear charge and are thus situated closer to the nucleus.
13
(a) Atomic radii decrease moving left to right across a row and (b) increase from top to bottom within a group.
(c) F<S<P<As. The order is unambiguous according to the trends of increasing atomic radius moving down a column and to the left in a row of the table.
21
• Moving from He to Rn in group 8A, first ionization energies decrease and atomic radii increase. The greater the atomic radius, the smaller the electrostatic attraction of an outer electron for the nucleus and the smaller the ionization energy of the element.
29 Li + 1e- → Li-; [He]2s1 [He]2s2
Adding an electron to Li completes the 2s subshell. The added electron experiences essentially the same effective nuclear charge as the other valence electron, except for the repulsion of pairing electrons in an orbital. There is an overall stabilization; ΔE is negative.
An extra electron in Be would occupy the higher energy 2p subshell. This electron is shielded from the full nuclear charge by the 2s electrons and does not experience a stabilization in energy; ΔE is positive.
Be + 1e- → Be-[He]2s2 [He]2s22p1
33
• Metallic character increases moving down a family and to the left in a period.
(a) Li
(b) Na
(c) Sn
(d) Al
35
• Ionic: MgO, Li2O, Y2O3;
• Molecular: SO2, P2O5, N2O, XeO3
• Ionic compounds are formed by combining a metal and a nonmetal; molecular compounds are formed by two or more nonmetals.
37
• When dissolved in water, an “acidic oxide” also called an acidic anhydride produces an acidic solution.
7.43
• A) Ca and Mg are both metals, they tend to lose electrons and form cations. Ca is more reactive because it has a larger atomic radius. With a larger atomic radius, the valence electrons are further from the nucleus due to electron shielding with limited nuclear charge thus require less energy to pull away an electron
7.43b
• K and Ca are both metals, they tend to lose electrons and form cations when they react. K is more reactive because it has a lower ionization energy. The 4s electrons in K is less tightly held because it experiences a smaller nuclear charge with the same amt of shielding.
7.45
• A) K + 2H2O →2KOH + H2
• B) Ba + 2H2O →2Ba(OH)2 + H2
• C) 6Li + N2 →2Li3N
• D) 2Mg + O2 →2MgO
7.49
• F Cl
• [He]2s22p5 [Ne]3s23p5
• -1 -1
• 1681kj/mol 1251kJ/mol
• Reacts quickly reacts slower
• -328kJ/mol -349 kJ/mol
• .71 Angstroms .99 Angstroms
7.53
• A) 2O3 → 3O2
• B) Xe + F2 →XeF2
• Xe + 2F2 →XeF4
• Xe + 3F2 →XeF6
• C) 2NaCl + 2H2O →2NaOH + Cl2 + H2
• D) S + 2Li →Li2S
7.55
• A) Te has more metallic character• B) At room temp, oxygen molecules are diatomic
and exist in the gas phase. Sulfur molecules are 8 membered rings and exists as a solid.
• C) Chlorine is generally more reactive thn bromine because Cl atoms have a greater (more exothermic) tendency to gain electrons (EA) than Br due to its small atomic radius.