VII Atomic Structure and Quantum Theory B

VII Atomic Structure and Quantum Theory B

Atomic Models

Quantum Numbers and

Bonding

Atomic Models

• Thomson Model (Early 1900’s)– “plum pudding” model– Atoms is cloud of positive charge with

nuggets of negative charge embedded in it.

Atomic Models

• Rutherford Model (1909)– Devised experiment to test Thomson Model– Stream of alpha particles () (He2+) at metal foil - if

Thomson correct - expected to pass through with minimum dispersion. But saw major deflection, some even coming right back at the source.

– Rutherford concluded:1) mass of atom is concentrated in a small

region in the center of the atom (nucleus) and has a positive charge.

origin of nuclear atom2) nucleus surrounded by large void

containing electrons moving at random.

Bohr Model (1913)

• Was attempting to explain line spectra

• Used a combination of classical and quantum physics

• Treated only H atom (one electron system)

Bohr Model (1913)

Assumptions

1) Only certain set of allowable circular orbits for an electron in an atom

2) An electron can only move from one orbit to another. It can not stop in between. So discrete quanta of energy involved in the transition in accord with Planck (E = h) 3) Allowable orbits have unique properties particularly that the angular momentum is quantized.

mvr nh2

Bohr Model (1913)

• Equations derived from Bohr’s Assumption

• Radius of the orbit

r1

r2

r3

n=1 n=2

n=3

rn n 2a o

Z

a o h 2

4 2me2

n = orbit number

Z = atomic number

h = Planck’s constantm = mass of electrone = charge on electron

Bohr Model (1913)

For H:

For He+ (also 1 electron)

r1 12ao

1 ao Called Bohr radius

r2 22ao

1 4ao

r1 12ao

2

12ao Smaller value for the radius.

This makes sense because of the larger charge in the center

For H and any 1 electron system:

n = 1 called ground staten = 2 called first excited staten = 3 called second excited stateetc.

Bohr Model (1913)

Which of the following has the smallest radius?

A) First excited state of H

B) Second excited state of He+

C) First excited state of Li+2

D) Ground state of Li+2

E) Second excited state of H

Bohr Model (1913)

Which of the following has the smallest radius?

A) First excited state of H

B) Second excited state of He+

C) First excited state of Li+2

D) Ground state of Li+2

E) Second excited state of H

r2 22ao

1 4ao

r3 32ao

2

92ao

r2 22ao

3

43ao

r1 12ao

3

1

3ao

r3 32ao

1 9ao

Bohr Model (1913)

• Energy of the Orbit

E n AZ2

n 2

For H, A = 2.18 x 10-18 J

E1 A12

12

A

E 2 A12

22

1

4A

E 3 A12

32

1

9A

What’s happening to the energy of the orbit as the orbit number increases?

Energy is becoming less negative, therefore it is increasing. The value approaches 0.

E0 (have formed ion) Completely removed the electron from the atom.

Bohr Model (1913)

• Energy differences Between Orbits

• For H: electron moves from n = 1 to n = 2

Is energy absorbed or emitted?

How much energy is needed?

the amount of energy needed is the difference between the energy of

orbit 1 and orbit 2.

Bohr Model (1913)

E1 A

E 2 A

4

E = E 2 - E1 A

4 ( A)=+

3

4A

+ sign shows that energy was absorbed.

xJ) = 1.64 x 10-18 J

What is E when electron moves from n = 2 to n = 1?

E = E1 - E 2 A (A

4)=-

3

4A

xJ) = - 1.64 x 10-18 J

• What is the wavelength of the photon needed to move an electron from n = 1 to n = 2 in the H atom?

recall: h = 6.62 x 10-34J sec, = c

• 1.64 x 10-18 m

• 2.47 x 1015 m

• 6.62 x 10-34 m

• 1.21 x 10-7 m

• 2.18 x 10-18 m

• What is the wavelength of the photon needed to move an electron from n = 1 to n = 2 in the H atom?

E h hc

hc

E

(6.62 x 10 34 J sec)(3.00 x 108 m/sec)

1.64 x 10-18 J 1.21 x 10-7 m

So, light with this wavelength is absorbed

When the electron goes back to n = 1, light with the same energy and wavelength will be emitted.

So: Ephoton = |E|transition = h = h(c/)

h = Planck’s constant = 6.62 x 10-34 J sec

c = speed of light = 3.00 x 108 m/sec

When E is positive, the photon is absorbed

When E is negative, the photon is emitted

E = - A1

n f2

1

n i2

Ionization Energy

energy needed to remove the outermost electron from an atom in its ground state.

For H:electron moves from n = 1 to n = ∞

E∞= 0, E1 = - 2.18 x 10-18 J

Therefore, the ionization energy for H is

2.18 x 10-18 J

DeBroglie Postulate (1924)

Said if light can behave as matter, i.e. as a particle, then matter can behave as a wave. That is, it moves in wavelike motion.

So, every moving mass has a wavelength () associated with it.

where h = Planck’s constant

v = velocity

m = mass

hmv

What is the in nm associated with a ping pong ball (m = 2.5 g) traveling at 35.0 mph.

A) 1.69 x 10-32 B) 1.7 x 10-32

C) 1.69 x 10-23 D) 1.7 x 10-23

hmv

6.62x10 34 kgm2

s2s

2.5g 1kg1000g

35.0mi.hr

1.609kmmi

1000mkm

1hr3600s

1.69 x10 32m1.7 x10 23nm

Heisenberg Uncertainty Principle

To explain the problem of trying to locate a subatomic particle (electron) that behaves as a wave

Anything that you do to locate the particle, changes the wave properties

He said: It is impossible to know simultaneously both the momentum(p) and the position(x) of a particle with certainty

Mathematically:

uncertainty

So: the better you know the position, the speed is less well known and vice versa, and anything done to find one changes the other

xp h45.27 x10 35

Schrodinger Model(1926)

developed theory that treated electrons as a matter waves

[new field: wave mechanics =

quantum mechanics]

model combined certain

allowable quantities of energy

that corresponds to certain

allowable 3D wave patterns of an electron in an atom

H = Hamiltonian operator

(set of math operations)

= Wave function

E = Energy

describes a particle with a wave pattern, when H is carried out on the function (the solution) you get an allowable energy value, so each solution is associated with a given atomic orbital

E

= wave function = atomic orbital

gives the probability of finding an electron in a certain volume from the nucleus

Look at probability density. How likely are we to “find” an electron at a particular distance from the nucleus.

So when solve the wave equation and get probability, yields 3 quantum numbers used to describe the atomic orbital: size(n), shape(l), location (ml)

Quantum Numbers and Atomic Orbitals

n = principal quantum number (shell)describes orbital sizespecifies primary energy level

greater n, higher E (like Bohr’s E)l = angular momentum quantum number (subshell)(azimuthal

quantum number)describes orbital shape l = 0 = s orbital

l = 1 = p orbital inc E l = 2 = d orbital within l = 3 = f orbital n

n1

0n -1

Section 20-26

Section 27-29,50 - 52

Section 53-59

ml = magnetic quantum numberdescribes orbital orientation

the direction is spacealso tells the number of orbitals of same energy (degenerate orbitals)

So, the combination of n, l, and ml, completely describes a specific orbital, its size , shape and orientation.

To make a complete picture, a 4th quantum number added, ms.

ms = spin quantum number - describes the electron spin in the orbital

m

ms1/2 or

• m l quantum numbers gives the number of each type of orbital

• So: for an s orbital, l = 0 and m l = 0, so there is only 1 s type orbital for each n level which can hold 2 electrons

• For a p orbital, l = 1, so m l = -1,0,or +1 thus there are 3 distinct p orbitals for each n level which can hold a total of 6 electrons

• For a d orbital, l = 2, so m l =-2, -1,0,or +1,+2 thus there are 5 distinct d orbitals for each n level which can hold a total of 10 electrons

Shapes of Atomic Orbitalss orbital p orbitals

d orbitals

Pauli Exclusion Principle• No two electrons in the same atom can have the

same four quantum numbers• Each electron has a unique address• Designated by a set of quantum numbers (n, l ,m

l,ms)

• So: (1,0,0, +1/2) indicates an electron in a 1s orbital• While (3,1, -1, +1/2) would be a 3p electron

Which of the following does not represent a 3d electrons?

A. (3,2,1,+1/2)

B. (3,2,-1,+1/2)

C. (3,1,1,+1/2)

D. (3,2,2,-1/2)

E. (3,2,0,+1/2)

Energy LevelsFor H atom All other atoms4s_4p_ _ _ 4d_ _ _ _ _4f _ _ _ _ _ _ _ 4s _

3d _ _ _ _ _

3p _ _ _

3s _ 3p _ _ _ 3d _ _ _ _ _ 3s _

2p _ _ _ (degenerate orbitals)

2s _ 2p _ _ _(degenerate orbitals) 2s _

1s _ 1s _

Inc E Inc E

Aufbau Principle - For an atom in its ground state ( the lowest energy configuration) fill the lowest energy orbital first then go up in energy until all the electrons are used.

1s2s2p3s3p3d4s4p4d…..

Spectroscopic NotationElectron Configuration

H(1e) 1s1

He(2e) 1s2

Li(3e) 1s22s1

Be(4e) 1s22s2

B(5e) 1s22s22p1

C(6e) 1s22s22p2

N(7e) 1s22s22p3

Ne(10e) 1s22s22p6

Na(11e) 1s22s22p63s1

Si(14) 1s22s22p63s23p2 Ar(18) 1s22s22p63s23p6 K(19) 1s22s22p63s23p6 4s1

Predicting Orbital Filling

Diagonal Rule

1s2s 2p3s 3p 3d4s 4p 4d 4f5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f

Using The Periodic Table To Predict Orbital Filling

s dp

f

1234567

s1

d1 d6

p1 p5

4f

5f

Exam 2

Frequency

0

5

10

15

20

25

30

35

40

0 8 16 24 32 40 48 56 64 72 80 88 96 104

112

120

128

score

Frequency

Range = 10 - 120Average = 58.7Median = 57

Ar Sc V Fe Zn As Kr Y Sn Xe

Sc(21) [Ar]4s23d1 = [Ar]3d14s2 V(23) [Ar]4s23d3

Zn(30) [Ar]4s23d10 As(33) [Ar]4s23d104p3 Kr(36) [Ar]4s23d104p6

Y(39) [Kr]5s24d1

What would be the correct electron configuration for P?

A. [Ne]3s5

B. [Ar]3s23p3

C. [Ar]4s23p3

D. [Ne]4s23p3

E. [Ne]3s23p3

O Mg Ar Sc V Fe Zn As Br Kr Y Sn Xe

Electron Configuration of Ions

Anions Add extra electrons in lowest available orbital

Br [Ar]4s23d104p5

Br - [Ar]4s23d104p6

O

O -2

1s22s22p4

1s22s22p6

Chloride ion would have the same electron configuration as?

A. Ne

B. Ar

C. Kr

D. S

E. P

O Mg Ar Sc V Fe Zn As Br Kr Y Sn Xe

Cations Cations form by losing e from the highest n levelFirst. Not necessarily the last e added.

Mg 1s22s22p63s2

Mg+2 1s22s22p6

Fe [Ar]4s23d6

Fe+2 [Ar]4s03d6 = [Ar]3d6 Fe+3 [Ar]3d5

SnSn+2

Sn+4

[Kr]5s24d105p2 [Kr]5s24d10 [Kr] 4d10

Which of the following would have the same electron configuration as Sn4+?

A) Cd

B) Cd2+

C) Ge4+

D) Ag+

E) Sb3-

Which of the following would have the same electron configuration as Sn4+?

A) Cd

B) Cd2+

C) Ge4+

D) Ag+

E) Sb3-

Sn+4 [Kr] 4d10

[Kr]5s24d10

[Kr]5s04d10