Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But...

88
Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2

Transcript of Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But...

Page 1: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7.1 – Electromagnetic Radiation

We may not finish today’s notes: XC HW pt if we do! But here’s ur HW:

Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2

Page 2: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

All E/M radiation travels at the same speed, c = 3 x 108 m/s

c = λ.ν

speed of wavelength frequency light (lambda) (nu)

Units:

Ex1) Strontium salts emit red light, with wavelengths around 650nm.Calculate the frequency of this light.

Page 3: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ephoton = h.v

Energy Planck’s frequency transferred constant

h = 6.626x10-34 J.s

- energy is quantized multiples of hv.

- each little packet of energy is called a quantum of energy

Ex2) Copper(I) chloride emits blue light, with wavelengths around 450nm.What is the quantum of energy associated with this wavelength?

Page 4: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

E/M radiation is energy. It travels as a wave. (wave nature)But it is quantized, as photons. (particle nature)This is called the dual nature of light

So then can particles have a wave nature?

Page 5: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

E/M radiation is energy. It travels as a wave. (wave nature)But it is quantized, as photons. (particle nature)This is called the dual nature of light

So then can particles have a wave nature?

De Broglie’s equation: Planck’s constant

wiggle wavelength mass . velocity (m.v is momentum)

Ex3) Calc the wavelength of an electron travelling at 1.0x107m/s(m = 9.11x10-31kg)

radius ofhydrogen = 0.53nm

mv

h

Page 6: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

The Bohr Model of the Atom54321

Z = nuclear chargen = energy level

Ex3) Calculate the energy required to excite the hydrogen atom from n = 1 to n = 2, and the wavelength of light required to reach this excited state from the ground state.

hc

E

2

21810178.2

n

ZJxE

Change in energy levels:

Energy levels available:

Page 7: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

The Bohr Model of the Atom54321

Z = nuclear chargen = energy level

Ex3) Calculate the energy required to excite the hydrogen atom from n = 1 to n = 2, and the wavelength of light required to reach this excited state from the ground state.

Why is it (+)?

hc

E

2

21810178.2

n

ZJxE

Change in energy levels:

Energy levels available:

Jx

xxEEE

18

2

218

2

218

12

106.1

1

11017.2

2

11017.2

Page 8: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

So where exactly is the electron?“There is a fundamental limitation to just how precisely we canknow both the position and momentum of a particle at a given time.”

Heisenberg uncertainty principle

Wave functions establish a radial probability distribution. There is a 90% probability the 1s electron is found within the radial distance that matches the innermost orbit of the Bohr model.

- called the electron cloud model

4)(

hmvx

Page 9: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Coulomb’s LawCharge is a fundamental concept. Its variable is q

Its units are coulombs (C)

Charge of electron: – 1.6 x 10-19 C Charge of proton: + 1.6 x 10-19 C

electric force constant,

Electric Force: k = 9 x 109

distance apart

Ex4) F Li

(Work = Force . distance moved)

Opposites attract.

221

d

qqkFE

Page 10: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ex5) What is the force of repulsion between 2 protons, at a distance of 2 femptometers (x10-15)?

Ex6) What is the force of attraction between a proton and an electron in the hydrogen atom, with a radius of 37pm?

Page 11: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ex5) What is the force of repulsion between 2 protons, at a distance of 2 femptometers (x10-15)?

Ex6) What is the force of attraction between a proton and an electron in the hydrogen atom, with a radius of 37pm?

Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2

Nx

CxCxxF 58

102

106.1106.1109215

19199

N

Nxx

CxCxxF

000000168.0

1068.11037

106.1106.1109 7212

19199

Page 12: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2 39. Microwave radiation has a wavelength on the order of 1.0cm. a. Calculate the frequency of a single photon of this radiation.

c = λ.ν

b. Calculate the energy of a single photon of this radiation.

c. Calculate the energy of an Avogadro’s number of photons (called an einstein) of this radiation.

Page 13: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2 39. Microwave radiation has a wavelength on the order of 1.0cm. a. Calculate the frequency of a single photon of this radiation.

c = λ.ν 3x108 m/s = (0.01m)(ν)

ν = 3x1010 Hz b. Calculate the energy of a single photon of this radiation.

E = h.v

c. Calculate the energy of an Avogadro’s number of photons (called an einstein) of this radiation.

Page 14: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2 39. Microwave radiation has a wavelength on the order of 1.0cm. a. Calculate the frequency of a single photon of this radiation.

c = λ.ν 3x108 m/s = (0.01m)(ν)

ν = 3x1010 Hz b. Calculate the energy of a single photon of this radiation.

E = h.v E = (6.626x10-34J.s)(3x1010 1/sec)

E = 1.99x10-23J c. Calculate the energy of an Avogadro’s number of photons

(called an einstein) of this radiation.

Page 15: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

43. It takes 7.21 x 10-19 J of energy to remove an electron from an iron atom. What is the maximum wavelength of light that can do this?

IE

e-1

nucleus

E

hchcE

Page 16: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

43. It takes 7.21 x 10-19 J of energy to remove an electron from an iron atom. What is the maximum wavelength of light that can do this?

IE

e-1

nucleus

7.21x10

)s)(3x10J(6.626x1019

sm834

JE

hchcE

Page 17: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2 45. Calculate the de Broglie wavelength for each of the following.

a. a proton with a velocity 90.% of the speed of light

b. a 150g ball with a velocity of 10 m/s

mv

h

mv

h

Page 18: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2 49. Calculate the wavelength of light emitted when each of the following transition occur in the hydrogen atom.

a. n=3 n=2 n=4

n=3

n=2

n=1JxJxE

JxJxE

192

218

2

192

218

3

1043.52

110178.2

1041.23

110178.2

E

hchcE

Page 19: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

53. Does a photon of visible light (λ ≈ 400nm to 700nm) have sufficient energy to excite an electron in hydrogen atom from the n=1 to the n=5 energy state?

JxJxE

JxJxE

202

218

5

182

218

1

107.85

110178.2

1018.21

110178.2

Page 20: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2 1. What is the force of attraction between a sodium ion and a chloride ion in a salt crystal, where the distance apart is 0.94 nm?

2. What is the force of attraction between a barium ion and a chloride ion in a salt crystal, where the distance apart is 1.21 nm?

An atom of sodium has one 3s electron outside a closed shell, and it takes only 5.14 electron volts of energy to remove that electron. The chlorine lacks one electron to fill a shell, and releases 3.62 eV when it acquires that electron (it's electron affinity is 3.62 eV). This means that it takes only 1.52 eV of energy to donate one of the sodium electrons to chlorine when they are far apart. When the resultant ions are brought closer together, their electric potential energy becomes more and more negative, reaching -1.52 eV at about 0.94 nm separation. This means that if neutral sodium and chlorine atoms found themselves closer than 0.94 nm, it would be energetically favorable to transfer an electron from Na to Cl and form the ionic bond.

2

21

d

qqkFE

Page 21: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7.2 – The Periodic Table and Electron Configurations

Page 22: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Battleship Positions: Ex1) C Ca Au

Page 23: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Aufbau Principle: As protons are added to the nucleus, electrons are added to the energy levels in order of increasing energy of the orbitals.Hund’s rule: max out ur unshared pairs

Electron Configurations and Orbital Filling Diagrams:Ex2) Oxygen:

Aluminum:

Ag:

Page 24: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ex3) EC and OFD for sodium:

for potassium:

Page 25: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ex4) Na: [Ne]3s1

K: [Ar]4s1

Valence electrons – the ones in the outermost principle quantum #.(outermost shell)These are important involved in bonding.

Elements of the same group have the same # valence electrons.

Ex5) OFDs – Ca:Sc:Ti:V:Cr:

Cu:Ch7 HW#2 p342 50b, 69(1st 4), 71

Page 26: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 HW#2 p342 50b, 69(1st 4),7150. Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom.b. n=5 → n=4

n=5 n=4 n=3 n=2 n=1

JxJxE

JxJxE

192

218

4

192

218

5

1036.14

110178.2

10871.05

110178.2

Page 27: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

50. Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom.b. n=5 → n=4

n=5 n=4 n=3 n=2 n=1

JxJxE

JxJxE

192

218

4

192

218

5

1036.14

110178.2

10871.05

110178.2

54 EE

Page 28: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

69. The elements Si, Ga, As, and Ge are all used in the manufacture of various semiconductor devices. Write the expected electron configuration for these atoms.

Si:

Ga:

As:

Ge:

Page 29: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

69. The elements Si, Ga, As, and Ge are all used in the manufacture of various semiconductor devices. Write the expected electron configuration for these atoms.

Si: 1s22s22p63s23p2 [Ne]3s23p2

Ga: 1s22s22p63s23p64s23d104p1 [Ar]4s23d104p1

As:

Ge:

Page 30: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

71. Write the expected electron configurations for each of the following atoms: Sc, Fe, P, Cs, Eu, Pt, Xe, Br.

Sc:

Fe:

P:

Cs:

Eu:

Pt:

Xe

Br:

Page 31: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

71. Write the expected electron configurations for each of the following atoms: Sc, Fe, P, Cs, Eu, Pt, Xe, Br.

Sc: [Ar]4s23d1

Fe: [Ar]4s23d6

P: [Ne]3s23p3

Cs: [Xe]6s1

Eu:

Pt:

Xe

Br:

Page 32: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Lab7.1 Spectra

- due tomorrow

- Ch7 HW#4 due @ beginning of period

Page 33: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Don’t copy, just listen…and readEx5) An atom of sodium has one 3s electron, and it takes only 5.14 eV of energy to remove that electron. The chlorine lacks one electron to fill a shell, and releases 3.62 eV when it acquires that electron (it's electron affinity is 3.62 eV). This means that it takes only 1.52 eV of energy to donate one sodium electron to chlorine when they are far apart. When the resultant ions are brought closer together, their electric potential energy becomes more and more negative, reaching -1.52 eV at about 0.94 nm separation. This means that if neutral sodium and chlorine atoms found themselves closer than 0.94 nm, it would be energetically favorable to transfer an electron from Na to Cl and form the ionic bond.

There is a minimum at 0.236 nm separation and then a steep rise in potential which represents a repulsive force.

(Electron clouds and quantum mechanics)

An atom of sodium has one 3s electron outside a closed shell, and it takes only 5.14 electron volts of energy to remove that electron. The chlorine lacks one electron to fill a shell, and releases 3.62 eV when it acquires that electron (it's electron affinity is 3.62 eV). This means that it takes only 1.52 eV of energy to donate one of the sodium electrons to chlorine when they are far apart. When the resultant ions are brought closer together, their electric potential energy becomes more and more negative, reaching -1.52 eV at about 0.94 nm separation. This means that if neutral sodium and chlorine atoms found themselves closer than 0.94 nm, it would be energetically favorable to transfer an electron from Na to Cl and form the ionic bond.

v

Page 34: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.
Page 35: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7.3 – Periodic TrendsIonization Energy – energy required to

remove an electron

The first ionization energy is the energy required to remove one electron,

leaving a +1 charge behind.

The second ionization energy is the energy to remove a second electron (once the first has already been removed), leaving a +2 charge behind.

• And so on, and so on…

See Table 7.5 on p328 for a list of first, second and third ionization energies for … GOOGLE IT!

Page 36: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

In general, the ionization energy increases from left to right across a period and from bottom to top in a group on the PT.

Page 37: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ex1) Order the following elements from lowest to highest

1st ionization energy: Na, P, Cl, F

Page 38: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

When graphed, IE shows interesting trends:

1. Groups:

2. Periods:

3: Bumps:

Be able to predict and explain the trends in ionization energies.

A similar graph can be found in Fig 7.31 p328.

Page 39: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ex2) Consider the electron configurations: 1s22s22p63s23p1

How do the 1st, 2nd,3rd, and 4th IE’s compare?

Page 40: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ex3) Consider the following 3 electron configurations: 1s22s22p6 1s22s22p63s1 1s22s22p63s2

Which atom has the largest 1st ionization energy, and which has the smallest 2nd ionization energy?

Page 41: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Electron Affinity - the energy change associated with the addition of an electron(how much the atom wants an electron.)

X(g) + e- X-(g)

Page 42: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

In general, the electron affinity increases from left to right

across a period (very important, very apparent)

and from bottom to top in a group on the PT (not much change).

Ex) Who has the greatest E.F.: O, F, Cl?

Ch7 HW#3 p343 85,Bonus1,Bonus2, 97

Page 43: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 HW#3 p343 85,Bonus1,Bonus2, 9785. Arrange the atoms in order of increasing first ionization energy. a. Be, Mg, Ca b. Te, I, Xe c. Ga, Ge, In

Page 44: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Bonus 1) The first seven ionization energies for an unknown element are listed. Which group might this element come from ?

1st 2nd 3rd 4th 5th 6th 7th

IE 1005 2260 3375 4565 6850 8490 27,000

Page 45: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Bonus 2) A table of 1st thru 4th ionization energies is listed for unknown elements W, X, Y, and Z. Match these letters with elements Na, Mg, Al, and K, using only a periodic table and your knowledge of periodic trends. W X Y Z

1st 495 419 735 580

2nd 4560 4210 1445 1815

3rd 5620 5109 7730 2740

4th 6780 6370 9100 11,600

Page 46: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

97. Order the atoms in each of the following sets from the least exothermic electron affinity to the most.

a. S Se

b. F, Cl, Br, I

Page 47: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7.4 - Atomic and Ionic Radius

Page 48: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

It may seem strange, but despite the fact that atomic mass increases from left to right across the periodic table, the atomic size, or atomic radius, actually decreases.

This is because the increased number of protons (positive charges) in the nucleus creates a greater electromagnetic pull on the electrons (negative charges) orbiting around the outside.

Fig 7.35 p333 for actual values.

Page 49: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ionic Radii Cation: Lost electron(s), Positive charge (Na+, Mg2+, etc.) - Tends to occur in metals. - Decreases the atomic radius of the atom.

Atom: _____

+12

Page 50: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ionic Radii Cation: Lost electron(s), Positive charge (Na+, Mg2+, etc.) - Tends to occur in metals. - Decreases the atomic radius of the atom.

Atom: _Mg_ +1 Ion: ______

+12 +12

Page 51: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ionic Radii Cation: Lost electron(s), Positive charge (Na+, Mg2+, etc.) - Tends to occur in metals. - Decreases the atomic radius of the atom.

Atom: : _Mg_ +1 Ion: _Mg+_ +2 Ion: ____

+12 +12 +12

Page 52: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Cation: Lost electron(s), Positive charge (Na+, Mg2+, etc.) - Tends to occur in metals. - Decreases the atomic radius of the atom.Atom: : _Mg_ +1 Ion: _Mg+_ +2 Ion: _Mg+_

+12 +12 +12

Anion: Gained electron(s), Negative charge (F-, S2-, etc.)- Tends to occur in nonmetals.- Increases the atomic radius of the atom.

Atom: ____ –1 ion: ____ –2 ion: ____

+16

Page 53: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Cation: Lost electron(s), Positive charge (Na+, Mg2+, etc.) - Tends to occur in metals. - Decreases the atomic radius of the atom.Atom: : _Mg_ +1 Ion: _Mg+_ +2 Ion: _Mg+_

+12 +12 +12

Anion: Gained electron(s), Negative charge (F-, S2-, etc.)- Tends to occur in nonmetals.- Increases the atomic radius of the atom.

Atom: _S_ –1 ion: _S-1_ –2 ion: ____

+16 +16

Page 54: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Cation: Lost electron(s), Positive charge (Na+, Mg2+, etc.) - Tends to occur in metals. - Decreases the atomic radius of the atom.Atom: _Mg_ +1 Ion: _Mg+_ +2 Ion: _Mg+_

+12 +12 +12

Anion: Gained electron(s), Negative charge (F-, S2-, etc.)- Tends to occur in nonmetals.- Increases the atomic radius of the atom.

Atom: _S_ –1 ion: _S-1_ –2 ion: _ S-2_

+16 +16 +16

Ch7 HW#4 p343 83,87,89+Bonus

Page 55: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 HW#4 p343 83,87,89+Bonus83. Arrange the following groups of atoms in order of increasing size.

a. Be, Mg, Ca

b. Te, I, Xe

c. Ga, Ge, In

Page 56: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

87. In each of the following sets, which atom or ion has the smallest radius?

a. Li, Na, K

b. P, As

c. O+, O, O-

d. S, Cl, Kr

e. Pd, Ni, Cu

Page 57: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

89. In 1994 at an American Chemical Society meeting, it was proposed that element 106 be named seaborgium, Sg, in honor of Glenn Seaborg, discoverer of the first transuranium element. a. Write the expected electron configuration for element 106.

b. What other element would be most like element 106 its properties?

c. Write the formula for a possible oxide and a possible oxyanion of element 106.

Page 58: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Bonus1) In each of the following sets, which atom or ion has the smallest radius?

a. Aluminum or the aluminum ion?

b. Bromine or the bromide ion?

c. O2-, F-, Ne, or Na+1?

Page 59: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Bonus1) In each of the following sets, which atom or ion has the smallest radius?

a. Aluminum or the aluminum ion? Al+3

b. Bromine or the bromide ion? Br

c. O2-, F-, Ne, or Na+1?

Page 60: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Bonus1) In each of the following sets, which atom or ion has the smallest radius?

a. Aluminum or the aluminum ion? Al+3

b. Bromine or the bromide ion? Br

c. O2-, F-, Ne, or Na+1

+8 +9 +10 +11

Page 61: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7.5 – Properties of Groups “It’s the # of valence electrons that primarily determines

an atom’s chemistry.”

Page 62: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Elements can be classified as being metal, nonmetal, or metalloid.

Metals: - Good conductors of heat/electricity. Lose electrons

- Ductile (can be made into a wire) to form (+) ions

- Most are malleable (can be molded)

- Tend to be solid at room temp.

Page 63: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Elements can be classified as being metal, nonmetal, or metalloid.

Nonmetals: - Mostly poor conductors of heat/electricity.

- Tend to be gaseous at room temp.

- Not malleable

- gain electrons to become (–) ions

Page 64: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Elements can be classified as being metal, nonmetal, or metalloid.

Metalloids: - Have properties of both metals and nonmetals.

- Behavior often controlled by changing conditions

+/ – usually determined by whom they are with.

Page 65: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Sub-categories of metals and nonmetals:

Good copy p334! Ch7 HW#5 p340+ 121,122,123,124,136

Page 66: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 HW#5 p340+ 121,122,123,124,136121) The successive ionization energies for an unknown element are:

I1 = 896kJ/mol I2 = 1752 kJ/mol

I3 = 14,807 kJ/mol I4 = 17948 kJ/molTo which family in the periodic table does the unknown element most

likely belong?

Page 67: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

122) An unknown element is a nonmetal and has a valence electron

configuration of ns2np4

a) How many valence electrons does this element have?b) What are some possible identities for this element?c) What is the formula of the compound this element would form

with potassium?d) Would this element have a larger or smaller radius than barium?e) Would this element have a greater or smaller ionization energy

than fluorine?

Page 68: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

123) Elements with very large ionization energies also tend to have highly exothermic electron affinities. Explain. Which group of elements would you expect to be an exception to this statement?

Page 69: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

124) The changes in electron affinity as one goes down a group in the periodic table are not nearly as large as the variations in the ionization energies. Why?

Page 70: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

136) While Mendeleev predicted the existence of several undiscovered elements, he did not predict the existence of noble gases, the lanthanides, or the actinides. Propose reasons why Mendeleev was not able to predict the existence of noble gases.

Page 71: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

Ch7 Rev p 341 44,47,50,52,54,76 (c,d),84, 86, 88, 90, 98,10244. The ionization energy of gold is 890.1 kJ/mol. Is light with a wavelength of 225nm capable of ionizing a gold atom (removing an electron) in the gas phase.

Page 72: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

44. The ionization energy of gold is 890.1 kJ/mol. Is light with a wavelength of 225nm capable of ionizing a gold atom (removing an electron) in the gas phase.

mxJE

hchcE 7

18s

m834

1034.11.48x10

)s)(3x10J(6.626x10

electronevery for kJx1048.1se'6.02x10 1mol

1mol kJ 890.1 21-23

Page 73: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

47. Calculate the de Broglie wavelength of an electron moving with

a velocity that is 0.001.c.

Page 74: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

47. Calculate the de Broglie wavelength of an electron moving with

a velocity that is 0.001.c.

mxxkg

sJ

sm

9831-

-34

1042.2)103001.0)((9.11x10

6.626x10

Page 75: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

50. Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom.a. n=4 → n=2

c. n=5 → n=3

Page 76: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

50. Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom.a. n=4 → n=2c. n=5 → n=3 n=5 n=4 n=3 n=2 n=1

JxJxE

JxJxE

192

218

2

192

218

4

1043.52

110178.2

1036.14

110178.2

Page 77: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

50. Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom.a. n=4 → n=2c. n=5 → n=3 n=5 n=4 n=3 n=2 n=1

Infrared

mxx

xx 620

834

1084.4)101.4(

)103)(10626.6(

Jx

JxJxEE20

191924

101.4

)1036.1(1043.5

JxJxE

JxJxE

192

218

2

192

218

4

1043.52

110178.2

1036.14

110178.2

Page 78: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

50. Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom.a. n=4 → n=2c. n=5 → n=3 n=5 n=4 n=3 n=2 n=1

JxJxE

JxJxE

192

218

3

192

218

5

1041.23

110178.2

10871.05

110178.2

Page 79: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

50. Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom.a. n=4 → n=2c. n=5 → n=3 n=5 n=4 n=3 n=2 n=1

Infrared

mxx

xx 620

834

1029.1)1054.1(

)103)(10626.6(

Jx

JxJxEE20

191935

1054.1

)1041.2(1087.0

JxJxE

JxJxE

192

218

3

192

218

5

1041.23

110178.2

10871.05

110178.2

Page 80: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

52. Using vertical lines, indicate the transitions from Exercise 50 on an energy-level diagram for the hydrogen atom.

(previous slide)

Page 81: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

54. An excited hydrogen atom emits light with a frequency of

1.141x1014 Hz to reach the energy level for which n = 4. In what principal quantum level did the electron begin.

Page 82: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

54. An excited hydrogen atom emits light with a frequency of

1.141x1014 Hz to reach the energy level for which n = 4. In what principal quantum level did the electron begin.

ΔE = h.vΔE = (6.626x10-34J.s)(1.141x1014 1/sec)

ΔE = 7.56x10-20J

2 1.8 n

1

4

110178.21056.7

i

221820

initialnJxJx

Page 83: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

76. Using only the periodic table, write the expected ground-state electron configurations for:

c. an element with three unpaired 5d electrons.

d. the halogen with electrons in the 6p atomic orbitals.

Page 84: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

84. Arrange the following groups atoms in order of increasing size.a. Rb, Na, Be

b. Sr, Se, Ne

c. Fe, P, O

86. Arrange the atoms in Exercise 83 in order of increasing first ionization energy.

Page 85: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

88. In each of the following sets, which atom or ion has the smallest ionization energy?

a. Cs, Ba, La

b. Zn, Ga, Ge

c. In, P, Ar

d. Tl, Sn, As

e. O, O-, O2-

Page 86: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

90. Predict some of the properties of element 117 (the symbol is Uus, following conventions proposed by the International Union of Pure and Applied Chemistry, or IUPAC). a. What will be its electron configurations?

b. What element will it most resemble chemically?

c. What will be the formula of the neutral binary compounds it forms with sodium, magnesium, carbon and oxygen.

d. What oxyanions would you expect Uus to from?

Page 87: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

98. Order the atoms in each of the following sets from the least exothermic electron affinity to the most.

a. N, O, F

b. Al, Si, P

Page 88: Ch7.1 – Electromagnetic Radiation We may not finish today’s notes: XC HW pt if we do! But here’s ur HW: Ch7 HW#1 p341 39,43,45,49a,53,Bonus1,Bonus2.

102. Use data in this chapter to determine

a. the ionization energy of Cl-.

b. the ionization energy of Cl.

c. the electron affinity of Cl+