Chapter 5 - Structure of Atoms

Chapter 5 - Structure of Atoms

Fundamental Particles (Constituents of atoms)

Nucleus Proton - mass » 1 amu ; relative charge = +1

Neutron - mass » 1 amu; relative charge = 0

Electron - mass » 0 ; relative charge = ^-1

Historical Background

Discovery of Electrons - Cathode ray tubes (see figure 5-1)

J.J. Thomson (England)

from deflection by electric & magnetic field calculated the charge to mass ratio of electron - e/m

Needed a separate measurement of e or m. Provided my Robert Millikan (US) in the famous oil drop experiment - Figure 5-2

Existence of proton theorized.

Cathode ray tubes also produce streams of particle move toward the cathode.

In fact, electrons are accelerated toward anode, strike gas atoms, "knock off" all electron, forming a cation atom ® cation + e^-

Protons - Continue

Regular pattern of e/m ratios for "canal rays" gave idea of a fundamental unit carrying positive charge - proton.

Models of the Atom

J. J. Thomsons plum pudding model

Contradicted by Ernest Rutherfords Scattering experiment - Figure 5-4 & 5-5

µ ^- particle - very dense >> gold

most µ ^- particles - no deflection

few µ ^- particles - large deflection

Consistent with an atom model where atom is mostly empty space with a very small volume containing most of the mass - nucleus.

Rutherfords Model:

Small, dense, positively charged nucleus surrounded by diffuse cloud negatively charged electrons.

Neutrons

Finally discovered in 1932 by Chadwick using alpha particle bombardment of Berylium.

Mass Number & Isotopes

H. G. J. Mosely showed that x-ray spectra (intensity of x-ray plotted against the wavelength) generated by electron bombardmat of a solid target composed of one element could be explained by correlating with Atomic Number (the number of protons in the nucleus) rather than Atomic Weight.

Properties of elements depend on Atomic No.

º # protons

All the atoms of a given element have the same number of protons (& electrons); that is, the same atomic number.

The number of neutrons can vary.

Mass Number º # protons + neutrons (nearly equal Atomic Mass)

Atoms of the same element with different Mass Numbers are called Isotopes.

Determination of Isotopic Abundance

Mass Spectrometer - Fig. 5-8

Angle of Deflection depends on:

Accelerating Voltage
Magnetic Field Strength
Mass of Particle
Charge of Particle

Can determine relative amounts of different isotopes - fig. 5-9 (Neon)

Result: 90.9% neon - 20 ¬ mass number (actually 10 protons & 10 neutrons)

0.3% neon - 21 (11 neutrons)

8.8% neon - 22 (12 neutrons)

Actual masses slightly different than mass no. neon - 20 = 19.99244 amu

Atomic Weights

Weighted average of masses of all naturally occurring isotopes.

e.g.	Isotope	% Abundance	Mass/amu
Mass no. Atomic no.	^{24 Mg 12}	78.99	23.98504
Mass no. Atomic no.	^{25 Mg 12}	10.00	24.98584
Mass no. Atomic no.	^{26 Mg 12}	11.01	25.98259

\ # protons = 12

# neutrons = 12

Atomic weight Mg = .7899 (23.98504)

+ .1000 (24.98584)

+ .1101 (25.98259) = 24.30amu

Average mass

Calculating Isotopic Abundance from Atomic Weight

Cu AW = 63.546 amu

Isotopes ⁶³Cu AW = 62.9298 amu

⁶⁵Cu AW = 64.9278 amu

Calculate the relative abundance

Fraction of ⁶³Cu º x With 2 isotopes must add to 1

Fraction of ⁶⁵Cu º 1 - x

63.546 amu = [x 62.9298 + (1-x)64.9278] amu

x = 64.9278 - 63.546 = 0.6916

64.9278 - 62.9298

^{63 Cu makes up 69.16% of natural copper}

^{65 Cu makes up 30.84 % of natural copper}Electromagnetic Radiation

Waves - repeating (periodic) pattern

Traveling Wave

length for pattern to repeat - wavelength - l

time for pattern to repeat - period - t

# repetitions per time - frequency - n ( = 1 )

velocity of wave = length for pattern = l = l n = c

time for pattern t

For electromagnetic waves c » 3.0 x 10⁸m/s

Wave - Particle Duality for Electromagnetic Radiation

Discovered that when EM radiation delivers energy to matter it does not do so continuously (as expected for a wave) but in discrete packets or "quanta" (like a particle) that depend on the frequency of the radiation.

D E = hn = hc ( h is Planck's const.)

Convincing evidence form "Photoelectric Effect"

no e^- for frequency too low
If n high enough #e^- depends

Time

One intensity but not n

Figure 5-13 (discussion)

Atomic Spectra

Light can be emitted from a gas in a cathode ray tube - only a few wavelengths are observed - emission spectra

If white light passes through as only a few wavelengths absorbed - absorption spectrum

Fig. 5-14, 15

Particular wavelengths ® Particular Energies Conclusion: electrons are absorbing or emitting energy by moving between distinct levels.

Rydberg Eq. 1 = R (1 - 1 )

l n²₁ n²₂ for H₂ gas

n₁= 1,2,3,… n₂ = 1,2,3,… R is Rydberg Constant = 1.097 X 10⁷m^-1

Bohr Theory

& energy µ 1 (actually = hcR from Rydberg)

n² n²

(please feel free to skip "enrichment" section)

Wave - Particle Duality for Electrons

Electrons also behave like waves in some experiments l = h/mn (Louis de Broglie)

Davisson & Germer (Bell Labs) demonstrated diffraction (a property of waves) of electrons by a crystal of nickel.

Bohr theory also explained by the idea that electron orbits were standing waves.

Bohr theory only partly successful. Electrons in atoms are described by wave functions but more complicated - solutions to the Schrödinger Equation. Not just one number necessary to specify energy state of electron but for r, n, e, m_e,m_s.

Quantum Numbers

Principal quantum no. n= 1, 2, 3, 4,….

Azimuthal quantum no. l = 0, 1, 2,…, n - 1

e.g. for n= 3, l can be 0, 1, 2,

magnetic quantum no. m_l= ^-l, ^-l + 1, …0, …l - 1, l

e.g. for l = 2, m_lcan be ^-2, ^-1, 0, 1, 2

spin quantum no. m_s= + 1/2 , ^- 1/2

no 2 electrons can have the same four quantum numbers. - Pauli Exclusion Principal

n, l, m_l specify a particular sub orbit then have m_s= + 1.2 & ^-1/2 so 2 electrons in each sub orbit.

Atomic Orbitals - Notation

Shells: n = 1, 2, 3, 4, …

Orbitals: l = 0, 1, 2, 3, …

Orbital s, p, d, f ….

So, orbital 2p Table 5-4

n = 2 l = 1

# sub orbitals = 2l + 1 (number of possible m_l)

e.g. p orbital l = 1 Þ 3 orbitals (m_l= ^- 1, 0, 1)

Pictures of atomic orbitals for hydrogen atom - figure 5-20 to 5-27

Spin Quantum No. - in an electric field electrons behave as if they have a magnetic field with two orientations, m_s= + 1/2 or ^- 1.2

Favorable energetic interaction allows 2 e^- with opposite m_sto occupy the same sub orbital - so 2 e^- in each orbital.

Multi Electron Atom (i.e. not hydrogen)

Approximation: electrons occupy the hydrogen atom orbitals - (total e^- density is superposition - sum)

"Build up" electronic structure by filling H- atom orbitals from lowest to highest.

Fot multi electron stoms there is a specific "filling order" - Aufbau Order

1s then 2s then 2p then 3s the 3p, 4s, 5d, 4p, 5s, 4d, …..LEARN THESE fig 5-29

example: Na = ll electrons

Na ¯ ¯ ¯ ¯ ¯ or 1s²2s²2p⁶3s¹

1s 2s 2p 2p 2p 3s

m_s= ^+-1/2 m_l = ^-1

⁺1 Note: this is called an "unpaired" e^-

Para Magnetism & Diamagnetism

Atoms/Ions with unpaired electrons are attracted into magnetic fields - Paramagnetic

Atoms with no unpaired electrons are repelled by a magnetic field - Diamagnetic

Hund's Rule: Fill all suborbitals of a given orbital (e.g. zp) singly before pairing.

Example: Carbon

_{6 C
¯

¯

_ 1s²2s²2p²
1s 2s 2p}

Note: the elements with completely filled s & p orbitals are the Noble Gases. This is a very stable configuration and the noble gases are very unreactive.

Examples: Neon, Argon

_{10 Ne
¯

¯

¯

¯

¯

1s 2s 2p

_{18 Ar [Ne]
¯

¯

¯

¯
_ _ _ _ _ _
3s 3p 3d 4s

stands for the Neon Even though 3d orbitals are empty this
configuration above. Is still very stable

Noble Gas Configuration: [ ] n s²n p⁶}}Exceptions to the Aufbau Order

Chromium

_{24 Cr [Ar]

3d 4s

Copper

_{29 Cu [Ar]
¯

¯

¯

¯

¯

3d 4s

explanation: special stability to half-filled & filled sets of suborbitals (d)
further complications for higher Atomic no. need to remember only these two.

ra}}

Some problems from the text 5.8

Radius H atom = .0529 nm = .0529 x 10^-9m

= 5.29 x 10^-11m

Volume H atom = 4/3 p (5.29 x 10^-11m

= 6.20 x 10^-o31 m ³

radius of H nucleus (proton) = 1.5 x 10^-15m

Volume H nucleus = 4/3 p (1.5 x 10^-15m)³= 1.4 x 10^-44m³

Vol nucleus = 1.4 x 10^-44m³ = 2.3 x 10^-14or 2.3 x 10^-12%

Vol atom 6.20 x 10^-31m³

Mass Spectrum 5-27

Aw = 0.7215 (84.9117 amu) + 0.2785 (86.9085 amu)

= 85.47 amu Ü look in Periodic Table element is Rubidium - Rb

Lyman Series (a series of lines in emission spectrum)

Transitions to the n=1 energy state

Calculate wavelengths of first 3 lines

n = 4

n = 3

n = 2 Energy µ 1 = transitions from

n² higher energy to

n = 1 lower - emit a

"quantum" of radiation

use Rydberg equation to calculate the wavelengths for each transition

(1) 1 = R(1 - 1) n₁= 1 n₂ = 2

l ₁ 1² 2²

= 1.097 x 10⁷m^-1(3/4) = 8.728 x 10⁶m^-1

^{l
₁ = 1 = 1.215 x 10^-7m
8.228 x 10⁶m^-1}

= 121.5 x 10^-9m = 121.5 nm

Note: D E ₁= hc

l ₁

5-60 continued

(2) 1 = 1.097 x 10⁷m^-1(1- 1)

l ₂ 1² 3²

= 9.751 x 10⁶ m^-1

l ₂ = 1 = 1.026 x 10^-7m = 102.6 nm

9.751 x 10⁶m^-1

5-83 How many e^- could have n = 5 ?

n = 5

l = 0, 1, 2, 3, 4

m_l= 0, ^-1, 0, 1, ^-2, ^-1, 0, 1 ,2

*sub orbitals 1, 3 5

total # orbitals = 1 + 3 + 5 + 7 + 9 = 25

2e^- / orbital \ total # e^- = 50

Any of following violate Pauli Exclusion Principle?

(a) 1s² ¯ n = 1 l = 0 m_l= 0 m_s= ⁺1/2

1s n = 1 l = 0 m_l= 0 m_s= ^-1/2

(b) 1s²2p¹ ¯ _ _ _

1s 2s 2p

electron configurations above plus

n = 2 l = 1 m_l= 0 m_s= ⁺1/2

This is not a ground state but an example of an excited state - Pauli?

1s means n = 1 l = 0 m_e= 0 ¬ only allowed value

\ n = 1 l = 0 m_l= 0 m_s= ⁺1/2

n = 1 l = 0 m_l= 0 m_s= ^-1/2 these are the same

n = 1 l = 0 m_l= 0 m_s= ⁺1/2

Violates the Exclusion principle

Paramagnetic?

F atomic # 9

_{9 F
¯

¯

¯

¯

1 unpaired e^- Þ
paramag}

1s 2s 2p (1s² 2s² 2p⁵)

(b) ₁₈Ar [Ne] ¯ ¯ ¯ ¯ no unpaired e^- Þ diamagnetic

([Ne] 3s²3p⁶)

(17 e^-s!) 3s 3p ([Ne] 3s²3p⁵)

(d) ₃₀Zn [Ar] 3d¹⁰ 4s² no unpaired e^-

(e) ₁₆S^2- [Ne] 3s² 3p⁶ no unpaired e^-

(18 e^-s!) note: this is the same configuration as Argon.

Periodic Table

Table 5-5

Chemical Properties are largely determined by the configuration of the highest energy electrons - the outermost occupied shell.
Elements with the same configuration but different principal quantum numbers have similar chemical behaviour.

e.g. ₉F [He] 2s² 2p⁵

₁₇Cl [Ne] 3s² 3p⁵

HF, HCl - acids

NaF, NaCl - soluble salts

Or ₆C [He] 2s²2p²

₁₄Si [Ne] 3s²3p²

CO₂ , Si O₂

CH₃ CH₃ CH₃

ç ç ç

… ¾ C¾ O ¾ C ¾ O ¾ C ¾ … a poly ether

ç ç ç

CH₃ CH₃ CH₃

ç ç ç

… ¾ Si¾ O ¾ Si ¾ O ¾ Si ¾ … a poly siloxane

ç ç ç

CH₃ CH₃ CH₃