Review Assignment #6

Chapter 6: Electronic Structure of Atoms

4   a.False. Electromagnetic radiation is capable of passing through water. (That's why you can see through it.)
     b. True.
     c. True.
     d. False. The glow from a fireplace and the energy within a microwave oven are forms of electromagnetic radiation; the blast from a foghorn is not.

6. Beginning with the shortest wavelengths, the order will be: X-Rays < orange light < infra-red (from the toaster oven) < radio wave < 50 Hz radiation
 

9. 436 nm light is in the blue portion of the visible spectrum.

16. The relationship between energy and frequency of a given photon is given by E = hn.

For the AM radio wave:

For the FM radio wave:

So, the FM radio waves are more energetic by about two orders of magnitude.

19. First, find the frequency that corresponds to the energy required:

The corresponding wavelength is:
 

This light is in the ultraviolet portion of the electromagnetic spectrum.

27a. The energy of an emitted photon is given by:

 

This photon is emitted by the n = 5 ®n=1 transition; it is deep in the UV.

b.

This photon is emitted by the n = 6 ®n= 5 transition; it is in the infra-red.

c. The energy of an absorbed photon is equal in magnitude but opposite in sign to that of an emitted photon. So,

 

This photon is absorbed and results in the n = 4 ® n= 5 transition; it is in the infra-red.

a.

 

A note on units: the definition of a joule is:

So in the above problem, a conversion factor can be used to cancel joules:

b.

c.

 

^{42        a. When n = 3, possible values of l and m_l are:}

^{l: 0, 1, 2
m_l = -2, -1, 0, 1, 2}

^{b. When n = 5, possible values of l and m_lare:}

^{l: 0, 1, 2, 3, 4
m_l = -4, -3, -2, -1, 0, 1, 2, 3, 4}

50. (a) The number of nodes for any orbital is always equal to n - 1, where n is the principal quantum level. (b) So, there will be one node in a 2p_x orbital and there will be two nodes in the 3s orbital. (c) The angular nodes in p orbitals are defined by two dimensional planes that include the nucleus; therefore there will be a region of electron density above the plane, another below the plane, but no electron density in the plane itself. (d) In the hydrogen atom, the energies of the orbitals depends solely on the principal quantum number. So the order of energies for the listed orbitals would be: 2s = 2p < 3s < 4d < 5s.
 

60a) when n = 2, there are four possible orbitals: the 2s (where l = 0, m_l = 0) and three 2p orbitals (l = 1, m_l = -1, 0 and 1). There are four electrons that can have n = 2 and m_s = -½; that is, we are only counting electrons with "down" spins.

b) When n = 5 and l = 3, there is a total of seven orbitals possible (m_l = -3, -2, -1, 0, 1, 2, 3; these are the 5f orbitals); since each orbital can have two electrons, there is a total; of fourteen electrons with n = 5 and l = 3.

c) The three quantum numbers n = 4, l = 3 and m_l = -3 define a single 4f orbital. Only two electrons can have the same n, l and m_l values.

d) The three quantum numbers n = 4, l = 1 and m_l = 1 define a single 4p orbital. Only two electrons can have the same n, l and m_l values.

64. Only a is correct. The correct orbital diagrams are:
 

a)

Sr

[Ar]

­¯

­

__

__

__

__

4s

3d

 

b)

V

[Ar]

­¯

­_

­_

­_

___

__

4s

3d

 

c)	Si	[Ne]		­¯		­_		­_		__
				3s				3p

 

d)

Sn

[Kr]

­¯

­¯

­¯

­¯

­¯

­¯

­_

­_

__

5s

4d

5p

 

66.

74a.

b. For one photon:

For one mole of photons,

c. UV-B radiation is more energetic. It has shorter wavelengths and therefore greater frequency; energy is proportional to frequency.

d. Yes, since UV-B radiation is more energetic, it is more capable of causing chemical reactions that can damage living tissue.

79. For the reaction, O₃ (g) + hn ® O₂ (g) + O (g)

The change in enthalpy is given by:

DH°_reaction = S (DH°_f)_products - S (DH°_f)_reactants

DH°_reaction = [DH°_f(O₂) + DH°_f(O)] - [DH°_f(O₃)]

DH°_reaction = [0 + 247.5] - [142.3] = 105.2 kJ/mol

This is the energy required to convert one mole of O₃ to the corresponding products. The energy per O₃ molecule is given by:

The frequency of light that corresponds to this energy is:

The wavelength is

This wavelength lies in the infrared. This is much longer than expected since O₃ absorbs UV radiation, not infrared. The mechanism involved is more complicated than a simple photon absorption and bond cleavage. See the discussion on page 689 for a more complete discussion of the role of ozone in filtering ultraviolet radiation in the atmosphere.

84a. The uncertainty in the velocity measurement is given as 0.2%, or (0.002) (2.2 ´ 10⁵ m/s) = 440 m/s. The resulting uncertainty in momentum, Dp, is the product of the electron mass and the uncertainity in the momentum:

 

According to the Heisenberg Uncertainty Principle,

So, the minimum uncertainty in the position, Dx, can be found by rearranging the above relationship as:

b. By the same process we have:

The relatively low uncertainty of the argon atom's momentum comes at the cost of a relatively large uncertainty in its position; note that 2 ´ 10⁴ m is many orders of magnitude greater than the atomic radius of the atom itself.

Chapter 7: Periodic Properties of the Elements

13. (a) Atomic size decreases as you move from left to right across the periodic table due to increasing effective nuclear charge; that is, as the charge of the nucleus increases, it has a stronger attraction to the valence electrons. (b) Atomic size increases as you move from top to bottom in a given column as the valence electrons populate quantum levels of increasingly high energy. (c) The order of size, from smallest to largest is: F< S < P < As.

15. By the same reasoning as above, the helium atom will be smaller than the hydrogen atom since its nucleus has a greater charge and attracts the electrons more effectively. The helium atom is smaller than the neon atom because the neon atom has electrons in the second principal quantum level, which is more diffuse than the first.

18a. Ga ® Ga⁺ + e^-                                      DH = IE₁

        Ga⁺® Ga²⁺ + e^-                       DH = IE₂

 
b. Rh³⁺ ® Rh⁴⁺ + e-               DH = IE₄

Scandium:

3^rd Ionization energy: Sc²⁺ ® Sc³⁺ + e^- DH = IE₃ = 2,390 kJ/mol

4^th Ionization energy: Sc³⁺ ® Sc⁴⁺ + e^- DH = IE₄ = 7,090 kJ/mol

The electron configuration of Sc2+ is: [Ar] 3d¹. So the electron removed in the third ionization process comes from the relatively high energy 3d orbital. On the other hand, the electron configuration of Sc³⁺ is [Ar], or [Ne] 3s²3p⁶. The electron removed in the fourth ionization process originates from a much lower energy 3p orbital, that it, the fourth ionization process is the first in which low-level core electrons are being removed. The very strong attraction to the nucleus of the core electrons accounts for the very much enthalpy change of the fourth ionization process compared to the third

Titanium

3^rd Ionization energy: Ti²⁺ ® Ti³⁺ + e^- DH = IE₃ = 2,650 kJ/mol

4^th Ionization energy: Ti³⁺ ® Ti⁴⁺ + e^- DH = IE₄ = 4,170 kJ/mol

In the case of titanium (named for the Titans, the giants who were born of the Greek earth goddess, Gaea, and who ruled the earth until overthrown by the Olympian gods), the electron configuration of Ti²⁺ is [Ar] 3d². For Ti³⁺: [Ar] 3d¹. In neither case does the removal of an electron involve the removal of a core electron; as a result, there is not the dramatic difference in IE₃ and IE₄ that was seen in the case of scandium. In case you're interested, the fifth ionization does involve the removal of a core and shows a large increase in ionization energy.

5^th Ionization energy: Ti⁴⁺ ® Ti⁵⁺ + e^- DH = IE₅ = 9,570 kJ/mol

21. As you go down the Group 8A elements, the ionization energies steadily decrease. This is because the valence electrons occupy increasingly high energy quantum levels and are not as strongly attracted to the nucleus. For the same reason, the atomic radii of the elements show a gradual increase as you go down the column.
 

24    a) Ca will have a larger ionization energy than K since it has a larger effective nuclear charge.

b) C will have a larger ionization energy than Si since its valence electrons are in a lower energy (i.e., more stable) quantum level.

c) Br will have a larger ionization energy than Mn due to its greater nuclear charge.

d) Xe will have a larger ionization energy than Sn due to its greater nuclear charge.

28. Recall that a negative electron affinity indicates an exothermic (that is, favorable) process. To see why the addition of an electron to bromine is favorable, consider the electron configuration.

 
Br: [Ar] 4s²3d¹⁰4p⁵             Br^- : [Ar] 4s²3d¹⁰4p⁶
 

In this case the added electron enters a 4p orbital, which is not filled in a neutral atom of bromine. A neutral atom of krypton, on the other hand, has a filled valence shell, so an added electron will enter the higher energy 5s orbital and will not be strongly attracted to the nucleus.

Kr: [Ar] 4s²3d¹⁰4p⁶             Kr^- : [Ar] 4s²3d¹⁰4p⁶5s¹
 

1st Ionization energy: C ® C⁺ + e^-           DH = IE₁
2nd Ionization energy: C⁺ ® C²⁺ + e^-       DH = IE₂
3^rd Ionization energy: C²⁺ ® C³⁺ + e^-        DH = IE₃
4^th Ionization energy: C³⁺ ® C⁴⁺ + e^-       DH = IE₄
5th Ionization energy: C⁴⁺ ® C⁵⁺ + e^-        DH = IE₅

6^th Ionization energy: C⁵⁺® C⁶⁺ + e^-       DH = IE₆

The electron configuration of neutral carbon is: 1s²2s²2p². Thus, the first four electrons removed are from the second principal quantum level. These show a gradual increase since the electrostatic attraction between the nucleus and the electrons increases as each sequential electron is removed. This is because every time an electron is removed, the electron-electron repulsion of the remaining electrons decreases and they are increasingly pulled toward the nucleus.

The large increase in ionization energy is seen at the fifth ionization step because this process involves removal of a core electron from the first quantum level. These are much closer to the nucleus and are held very strongly.

67

(a) The first ionization energy of phosphorus is greater than that of sulfur. This is one of the "exceptions" to the general trend of increasing ionization energy from left to right across a given row. It is explained by the fact that phosphorus has no unpaired electrons in the 3p orbitals (its configuration is [Ne] 3s²3p³; so each of the 3p orbitals is half-filled). Sulfur, on the other hand, has two electrons in one of the 3p orbitals (its configuration is [Ne] 3s²3p⁴). As a result, sulfur has greater electron-electron repulsion in the doubly-occupied 3p orbital which more than offsets its greater nuclear charge.
 

(b) The electron affinity for nitrogen is lower for nitrogen than either carbon or oxygen. The electron configurations for these elements are:

C: [He] 2s²2p²
N: [He] 2s²2p³
O: [He] 2s²2p⁴

The additional electron added to carbon goes into an unoccupied 2p orbital and does not greatly increase electron-electron repulsion. In contrast however, an additional electron in nitrogen must enter a partially filled 2p orbital; since there is already an electron in the region of space, electron-electron repulsion is increased and this serves to decrease the electron affinity. In the case of oxygen, the added electron also must enter a partially filled orbital and also must increase the electron electron repulsion. However, the magnitude of this effect is expected to be the same as that seen for nitrogen, but this is more than offset by the increase in nuclear charge which serves to attract the electron to the atom.
 

(c) The electron configurations for O⁺ and F⁺ are shown below:

The removal of an electron from F⁺ involves taking an electron from a filled orbital, that is, one that has significant electron-electron repulsion. The removal of an electron from O⁺ does would involve the removal of an electron from a half-filled orbital; there is no electron-electron repulsion making acting to make this process more favorable. This is entirely analogous to the fact that the first ionization of oxygen is lower than the first ionization energy of nitrogen.

(d) The electron configurations of Mn²⁺, Cr²⁺ and Fe²⁺ are shown below.

Mn²⁺ has a higher effective nuclear charge than Cr²⁺, explaining its higher third ionization energy. Fe²⁺ has paired electrons in a 3d orbital; this gives rise to electron-electron repulsion which makes ionization more favorable.

72 (c) Rn should have the highest ionization energy. (d) Fr will have the lowest first ionization energy. (e) At will have the most negative electron affinity. (f) Fr will have the largest atomic radius.
 

78      a) magnesium nitride = Mg₃N₂

b) The reaction is: Mg₃N₂ + 3 H₂O ® MgO + 2 NH₃

c) The total number of mass of magnesium can be found from the mass of pure MgO that is obtained after the reaction of the mixture with water:

By the conservation of mass, this is equal to the number of moles of Mg in the mixture. We can define the mass of the two components in the mixture as follows:

Let x = mass MgO

Let y = mass Mg₃N₂

The total number of moles of magnesium can be related to the masses of the two components by:

But since x + y = 0.470, we can substitute for y in the above expression, giving us one equation and one unknown, which we can solve.

 

x = 0.388

Thus, the mass of MgO is 0.388 g and the mass of Mg₃N₂ is 0.082 g.

d) 3 Mg + 2 NH₃ ® Mg₃N₂ + 3 H₂

Since complete reaction of the ammonia yields the smaller amount of hydrogen in the above calculation, NH₃ is the limiting reactant and 0.456 g of H₂ will be produced.

e) DH° = [DH°_f (Mg₃N₂) + 3 DH°_f (H₂)] - [3 DH°_f (Mg) + 2 DH°_f (NH₃)]

DH° = [(-461.08 kJ) + 3 (0 kJ)] - [3 (0 kJ) + 2 (-46.19 kJ)] = -368.7 kJ
 

12. The lattice energies are: NaF (910 kJ/mol), MgO (3795 kJ/mol). This large difference is due to the stronger electrostatic attraction that exists between ions with charges of +2 and -2 (Mg²⁺ and O^2-), compared with those of charges +1 and -1 (Na⁺ and Cl^-). Recall that the potential energy of two ions is given by:

Here, Q⁺ is the charge of the cation, and Q^- is the charge of the anion and k is a constant. So, provided the distance between the ions is similar, the potential energy of the MgO ion pair would be greater than that of NaCl by a factor of four. This is reflected in the lattice energies.

The smaller difference in lattice energies for SrCl₂ and MgCl₂ is accounted for by the fact that Sr²⁺ is a larger ion than Mg²⁺; thus the value of d in the above equation will be greater for SrCl₂, resulting in a smaller lattice energy
 

KBr < KCl < LiCl < CaO

This is in agreement with the tabulated data.
 

The steps are:
 

Vaporization of Rb	Rb (s) ® Rb (g)	DH° = DH°f (Rb, g) = 85.8 kJ = DH1
Ionization of Rb	Rb (g) ® Rb+ (g) + e-	DH° = IERb = 403 kJ = DH2
Formation of Cl	½ Cl2 (g) ® Cl (g)	DH° = DH°f (Cl, g) = 121.7 kJ = DH3
Electron Affinity of Cl	Cl (g) + e-® Cl- (g)	DH° = EACl = -349 kJ = DH4
Crystal Formation	Rb+ (g) + Cl-(g) ® RbCl (s)	DH° = DH5 = -DHlattice
Formation of RbCl	Rb (s) + ½ Cl2 (g) ® RbCl (s)	DH° = DH°f (RbCl) = -430.5 kJ = DHtotal

The last reaction is the sum of the first five, so the DH of the last reaction is the sum of the individual DH values for the component reactions. So,

DH_total = DH₁ + DH₂ + DH₃ + DH₄ + DH₅

Since DH₅ is the unknown, we can rearrange the equation:

DH₅ = DH_total- DH₁ - DH₂ - DH₃ - DH₄

DH₅ = -430.5 - 85.8 - 403 - 121.7 - (-349) = -692 kJ

This is the enthalpy change for the crystal formation reaction listed above. The lattice energy is defined as the change in enthalpy for the reverse reaction,

RbCl (s) ® Rb⁺ (g) + Cl^- (g) DH = DH_lattice = -DH₅ = +692 kJ

This is less than that for NaCl, as would be expected given the larger size of the Rb⁺ ion compared to Na⁺.