Principles of Chemistry, Chapt. 2: Atomic Structure and The Elements

I.The Structure of Atomsprotons, neutrons, and electrons

II.Atomic Structure and Properties—the Elements

atomic mass, atomic number, isotopes

III.The Mole Concept: 6.02 x 1023

IV.The Periodic Table

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Homework: Chapt. 2 Problems 26, 29, 37, 43, 75

~ 10-10 meters = 1 angstrom (Å)

_

+10-14 m

Smeared out electron charge cloud

+++

++

++

Protons and neutrons

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Most of the mass is here

Most of the Chemistry is here

Atomic Theory in a single Slide

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STM Image: Oxygen atoms at the surface of Al2O3/Ni3Al(111)

S. Addepalli, et al. Surf. Sci. 442 (1999) 3464

Electronic charge cloud surrounding the nucleus

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What’s inside the nucleus:

Particle Mass (amu) Charge

Proton (p+) 1.007 amu +1Neutron (n0) 1.009 amu 0

What’s outside the nucleus:

Electron (e-) .00055 amu -1

Note: mass ratio of electron/proton (Mp+/Me-) = 1836

For any atom: # of electrons = # of protons: Why?

Atomic Theory: Late 19th Century

Atomic theory—everything is made of atoms—generally accepted (thanks to Ludwig Boltzman, and others).

Mendeleev/periodic table—accepted, but the basis for periodic behavior not understood

What are atoms made of?

How are they held together?

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Electrical behavior: “+” attracts “-” but like charges repel

Atoms must contain smaller sub-units.

Radioactive material

Beam of , , and Electrically

charged plates

β-particles (“–”)

Gamma ray (γ) No charge, no deflection

α-particle (“+” ) Heavier, deflected less than β

–+

Alpha particle 2 n0 + 2p+

Beta particel electron (e-)Gamma photon

Electric and magnetic fields deflect the beam.• Gives mass/charge of e- = −5.60 x 10-9 g/C• Coulomb (C) = SI unit of charge

•Thomson (1897) discovered the e-:

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“Cathode rays”• Travel from cathode (-) to anode (+).• Negative charge (e−).• Emitted by cathode metal atoms.

fluorescentscreen

– high voltage + cathode ray

+

_

+

_

Essence of the Thompson Experiement (and old fashioned TV’s)

Electric field exerts Force+ plate repels +charged particles- Plate repels – charged particles

F = Eq = ma

d = displacement = ½ at2 = Eq/m (t = L/Vx)

Therefore, the greater the displacement, the lower the mass of the particle

d

x

y

Phosphor screen

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• Millikan (1911) studied electrically-charged oil drops.

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Charge on each droplet was: n (−1.60 x 10-19 C) with n = 1, 2, 3,… n (e- charge)

Modern value = −1.60217653 x 10-19 C. = −1 “atomic units”.

•These experiments give:

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Modern value = 9.1093826 x 10-28 g

= (-1.60 x 10-19 C)(-5.60 x 10-9 g/C) = 8.96 x 10-28 g

me = charge xmass

charge

• Atoms gain a positive charge when e- are lost.

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Implies a positive fundamental particle.

Hydrogen ions had the lowest mass.• Hydrogen nuclei assumed to have “unit

mass”• Called protonsprotons.Modern science: mp = 1.67262129 x 10-24 g

mp ≈ 1800 x me.

Charge = -1 x (e- charge). = +1.602176462 x 10-19 C = +1 atomic units

How were p+ and e- arranged?

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Thompson:• Ball of uniform positive charge, with small negative

dots (e-) stuck in it.• The “plum-pudding” model.

19101910 Rutherford (former Thompson graduate student) fired α-particles at thin metal foils.Expected them to pass through with minor deflections.

Rutherford Scattering Experiment

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Different Models of the Atom: different scattering results

α particles

α particles

“Plum pudding model”•+ and – charges evenly distrubted•low, uniform density of matter•No back scattering

Rutherford’s explanation of results:

Small regions of very high density+ charge in the dense regions- Charges in region around it

From wikipedia14

Some Large Deflections were osbservedSome Large Deflections were osbserved

α particles

Rutherford“It was about as credible as if you had fired a 15-inch shell at a piece of paper and it came back and hit you.”

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≈10,000 times smaller diameter than the entire atom. e- occupy the remaining space.

α particles

Most of the mass and all “+” charge is concentrated in a small core, the nucleusnucleus.

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Nucleus diameter~ 10-4 Å = 10-14 metersMass ~ 10-27 Kg

Charge cloud Diameter ~ 1 ÅMass ~ 10-30 kg

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Most Chemistry involves rearrangement of outermost electrons, not nuclei

Example: H 1p+ , 1 e-

H + H H2

+

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7 Å Epitaxial Al2O3(111) film on Ni3Al(111) (Kelber group):•Grown in UHV•Uniform•No Charging

S. Addepalli, et al. Surf. Sci. 442 (1999) 3464

STM

Start with ordered films growth studiesProceed to amorphous films on Si(100)

Surface terminated by hexagonal array of O anions

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•Atomic mass > mass of all p+ and e- in an atom.•Rutherford proposed a neutral particle.

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mn ≈ mp (0.1% larger).

mn = 1.67492728 x 10-24 g.

Present in all atoms (except ‘normal’ H).

Chadwick (1932) fired -particles at Be atoms. Neutral particles, neutronsneutrons, were ejected:

~ 10-10 meters = 1 angstrom (Å)

_

+10-14 m

Smeared out electron charge cloud

+++

++

++

Protons and neutrons

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NucleusNucleus• Contains p+ and n0

• Most of the atomic mass.• Small (~10,000x smaller diameter than the atom).• Positive (each p+ has +1 charge).

• Small light particles surrounding the nucleus.• Occupy most of the volume.• Charge = -1.

Atoms are neutral. Number of e− = Number of p+

ElectronsElectrons

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A neutron can decay into a proton and electron:

n0 p+ + e-

This can cause decay of a radioactive element, e.g.,

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6C# of p+ + n0

Atomic No.(# e- = # p+

Elemental symbol (carbon)

Carbon with 6 protons and 8 neutrons is unstable (radioactive)

Carbon with 6 protons and 6 neutrons is stable (non-radioactive

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6C12

6Cradioctive stable

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An atom of 14C can undergo decay to N as a neutron turns into a proton + an emitted electron

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6C14

7 N + e-

1 p+ 1 n0 + an electron (emitted)

Same element - same number of p+

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Atomic numberAtomic number (Z) = number of p+

1 amu = 1.66054 x 10-24 g

Particle Mass Mass Charge (g) (amu) (atomic units)

e− 9.1093826 x 10-28 0.000548579 −1

p+ 1.67262129 x 10-24 1.00728 +1

n0 1.67492728 x 10-24 1.00866 0

Atomic mass unit (amu) = (mass of C atom) that contains 6 p+ and 6 n0.

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IsotopesIsotopes Atoms of the same elementsame element with different A.• equal numbers of p+

• different numbers of n0

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deuterium (D)

tritium (T)

Hydrogen isotopes: H 1 p+, 0 n011

21H 1 p+, 1 n0

31H 1 p+, 2 n0

ISOTOPES: SAME Element, Different numbers of neutrons

C12 C14

Carbon: atomic no. = 6 6 protons in the nucleus+ 6 electrons

Atomic mass = 12 amu (12 gr/mole)

Therefore , 6 protons + 6 neutrons

Atomic mass = 14 amu

Therefore, 6 protons+ 8 neutrons

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Isotopes Display the same chemical reactivities (which depend mainly on the outer arrangement of the electrons)

12C + O2 CO2

14C +O2 CO2

Isotopes display different nuclear properties

C12stable

C14 Radioactive: spontaneously emits electrons. Half-life ~ 5730 years

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Isotopes and Moles (more on this later) and isotope abundance:

1 mole = 6.02 x 1023 of anything!

1 mole of atoms = 6.02 x 1023 atoms

Molar Mass (in grams) = average atomic mass (in amu)

1 mole of H atoms = 1.008 gr.

Why not 1.000 gr?? most atoms are 1H, but some are 2H (deuterium)

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Average atomic mass of H = 1.008 amu

100 atoms have a mass of 100.8 amu

# of 2H atoms = n# of 1H atoms = 100 –n (assume these are the only two isotopes that matter)

Mass of 100 atoms = n x 2.000 +(100-n) x 1.000 = 100.8 amu

2n + 100-n = 100.8

n = 0.8 So, out of every 100 atoms , have 0.8 2H atoms Out of every 1000 atoms, have 8 2H atoms

Natural abundance of “heavy hydrogen (deuterium) is then 0.8%

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Most elements occur as a mixture of isotopes.

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Magnesium is a mixture of:

24Mg 25Mg 26Mg

number of p+ 12 12 12

number of n0 12 13 14

mass / amu 23.985 24.986 25.982

For most elements, the percent abundancepercent abundance of its isotopes are constant (everywhere on earth).

The periodic table lists an average atomic weight.

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ExampleExample

Boron occurs as a mixture of 2 isotopes, 10B and 11B. The abundance of 10B is 19.91%. Calculate the atomic weight of boron.

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Atomic weight for B = 1.994 + 8.817 amu = 10.811 amu

Atomic mass = Σ(fractional abundance)(isotope mass)

(11.0093 amu) = 8.817 amu11B 80.09100

(10.0129 amu)10B 19.91100

= 1.994 amu

% abundance of 11B = 100% - 19.91% = 80.09%

Boron occurs as a mixture of 2 isotopes, 10B and 11B. The abundance of 10B is 19.91%. Calculate the atomic weight of boron.

B10.811

Boron

5 Atomic number (Z)

Symbol

Name

Atomic weightAtomic weight

Periodic table:

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A counting unit – a familiar counting unit is a “dozen”:

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1 dozen eggs = 12 eggs

1 dozen donuts = 12 donuts

1 dozen apples = 12 apples

1 mole (mol) = Number of atoms in 12 g of 12C• Latin for “heap” or “pile”• 1 mol = 6.02214199 x 1023 “units”• Avogadro’s numberAvogadro’s number

A green pea has a ¼-inch diameter. 48 peas/foot.

(48)3 / ft3 ≈ 1 x 105 peas/ft3. V of 1 mol ≈ (6.0 x 1023 peas)/(1x 105 peas/ft3) ≈ 6.0 x 1018 ft3

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height = V / area, 1 mol would cover the U.S. to:

U.S. surface area = 3.0 x 106 mi2

= 8.4 x 1013 ft2

6.0 x 1018 ft3

8.4 x 1013 ft2=7.1 x 104 ft = 14 miles !

1 mole of an atom = atomic weight in grams.

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1 Xe atom has mass = 131.29 amu

1 mol of Xe atoms has mass = 131.29 g

There are 6.022 x 1023 atoms in 1 mol of He andand 1 mol of Xe – but they have different masses.

1 He atom has mass = 4.0026 amu

1 mol of He has mass = 4.0026 g

… 1 dozen eggs is much heavier than 1 dozen peas!

ExampleExampleHow many moles of copper are in a 320.0 g sample?

Cu-atom mass = 63.546 g/mol (periodic table)

Conversion factor: 1 mol Cu63.546 g

= 1

nCu = 320.0 g x 1 mol Cu63.546 g = 5.036 mol Cu

n = number of moles

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Calculate the number of atoms in a 1.000 g sample of boron.

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nB = (1.000 g) 1 mol B10.81 g

= 0.092507 mol B

B atoms = (0.092507 mol B)(6.022 1023 atoms/mol)

= 5.571 1022 B atoms

Dimensional Analysis and Problem SolvingSpecial Homework Problem: Due Tues. Recitation

Density = mass/volume

Problem:

Assume that a hydrogen atom has a spherical diameter of 1 angstrom

Assume that the nucleus (1 proton) has a diameter of 10-4 angstrom

1.Calculate the densities of the nucleus, and of the electron charge cloud in kg/m3

2.Calculate the ratio of the two densities: R = dnucleus/delectron cloud

Mass of proton = 1.67 x 10-27 kg

Mass of electron = 9 x 10-31 kg

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Summarizes• Atomic numbers.• Atomic weights.• Physical state (solid/liquid/gas).• Type (metal/non-metal/metalloid).

Periodicity• Elements with similar properties are

arranged in vertical groups.

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In the USA, “A” denotes a main group element…

…”B” indicates a transition element.

International system uses 1 … 18.

The Periodic Table

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Main group metal

Transition metal

Metalloid

Nonmetal

The Periodic Table

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A period is a horizontal row

Period number

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A group is a vertical column Group 7AHalogens

Group 8ANoble gases

Group 2AAlkaline

earth metals

Group 1AAlkali metals (not H)

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Alkali metals (group 1A; 1)Alkaline earth metals (group 2A; 2)

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• Grey … silvery white colored.

• Highly reactive.• Never found as native

metals.• Form alkaline solutions.

Transition Elements (groups 1B – 8B)

• Also called transition metals.• Middle of table, periods 4 – 7.• Includes the lanthanides & actinides.

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Lanthanides and Actinides• Listed separately at the bottom.• Chemically very similar.• Relatively rare on earth.

(old name: rare earth elements)

Groups 3A to 6AGroups 3A to 6A• Most abundant elements in the Earth’s crust

and atmosphere.• Most important elements for living organisms.

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Halogens (group 7A; 17)Halogens (group 7A; 17)• Very reactive non metals.• Form salts with metals.• Colored elements.

Noble gases (8A; 18)Noble gases (8A; 18)• Very low reactivity.• Colorless, odorless gases.

Atoms are very small.• 1 tsp of water contains 3x as many atoms as

there are tsp of water in the Atlantic Ocean!

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Impractical to use pounds and inches...

Need a universal unit system• The metric system.• The SI system (Systeme International) - derived

from the metric system.

• A decimal system.• Prefixes multiply or divide a unit by multiples of ten.

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Prefix Factor Example

mega M 106 1 megaton = 1 x 106 tons

kilo k 103 1 kilometer (km) = 1 x 103 meter (m)

deci d 10-1 1 deciliter (dL) = 1 x 10-1 liter (L)

centi c 10-2 1 centimeter (cm) = 1 x 10-2 m

milli m 10-3 1 milligram (mg) = 1 x 10-3 gram (g)

micro μ 10-6 1 micrometer (μm) = 1 x 10-6 m

nano n 10-9 1 nanogram (ng) = 1 x 10-9 g

pico p 10-12 1 picometer (pm) = 1 x 10-12 m

femto f 10-15 1 femtogram (fg) = 1 x 10-15 g

1 pm = 1 x 10-12 m ; 1 cm = 1 x 10-2 m

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How many copper atoms lie across the diameter of a penny? A penny has a diameter of 1.90 cm, and a copper atom has a diameter of 256 pm.

x 1 x 10-2 m 1 cm

= 7.42 x 107 Cu atoms

1 pm1 x 10-12 m

x1.90 cm = 1.90 x 1010 pm

Number of atoms across the diameter:

1.90 x 1010 pm x 1 Cu atom 256 pm

LengthLength 1 kilometer = 0.62137 mile1 inch = 2.54 cm (exactly)1 angstrom (Å) = 1 x 10-10 m

VolumeVolume 1 liter (L) = 1000 cm3 = 1000 mL= 1.056710 quarts

1 gallon = 4 quarts = 8 pints

MassMass 1 amu = 1.66054 x 10-24 g1 pound = 453.59237 g = 16 ounces1 ton (metric) = 1000 kg1 ton (US) = 2000 pounds

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Example: What is the volume of a 2 gallon container in Liters?

1 gallon x 4 quarts/gallon = 4 quarts

4 quarts x 1Liter/1.057 quarts = 3.784 Liters (L)

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165 mg dL

A patient’s blood cholesterol level measured 165 mg/dL. Express this value in g/L

1 mg = 1 x 10-3 g ; 1 dL = 1 x 10-1 L

x 1 x10-3 g 1 mg

= 1.65 g/Lx 1 dL1 x10-1 L

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All measurements involve some uncertainty.

Reported numbers include oneone uncertain digit.

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Consider a reported mass of 6.3492 g• Last digit (“2”) is uncertain• Close to 2, but may be 4, 1, 0 …• Five significant figuressignificant figures in this number.

Read numbers from left to right.Count all digits, startingstarting with the 1st non-zero digit.

AllAll digits are significant exceptexcept zeros used to position a decimal point (“placeholders”).

0.00024030 5 sig. figs. (2.4030 x 10-4)

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placeholders significant

significant

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Round 37.663147 to 3 significant figures.

Examine the 11stst non-significant digit non-significant digit. If it:

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• > 5, round up.• < 5, round down.• = 5, check the 2nd non-significant digit.

round up if absent or odd; round down if even.

last retained digit

1st non-significant digit

Rounds up to 37.72nd non-

significant digit