Chem Module 3

Each element has properties which are dependent on the element's electron configuration. In this section, we will study three atomic properties:

(1) ionization energy, (2) electronegativity and (3) atomic size and how these properties vary according to the element's position in the periodic table.

If 0.15 mol of I2 vapor can effuse through an opening in a heated vessel in 36 sec, how long will it take 0.15 mol of Cl2 to effuse under the same conditions?

(36 sec / unknown)2 = 253.8 / 70.9 1/tCl2 / 1/36 = √ 253.8 / 70.9 1/tCl2 /1/36 = 1.89 1/tCl2 = 1.89 x 1/36 1/tCl2 = 1/ 0.0525 = 19 sec

Mole fraction

(Mole fraction of gas 1) = X1 = n1 / nT Since the individual partial pressures are equal by the ideal gas law to: P1 = (n1) (R x T) / V and since the factor RT / V is constant for all gases in the container, then the partial pressure is proportional to moles and, therefore, the mole fraction equals the partial pressure divided by the total pressure. n1 / nT = P1 / PT also P1 = n1 / nT (PT) also P1 = X1 (PT)

angular momentum quantum number

(l) describes the shape of the orbital within a subshell and can have whole number values from 0 to n-1, where n is the principal quantum number. This quantum number is equal to the number of nodal planes which describe the shape of the orbital. The shapes of the orbital types (s, p, d and f) are shown below.

Any s orbital will have a Msubl value of

0

Orbital planes = l

0 s sphere - no lobes 1 p dumbell - 2 lobes 2 d butterfly - 4 lobes 3 f double butterfly - 8 lobes

The diagram of string standing waves shows 4 examples which differ by where the string is touched

1 (untouched) is vibrating with 2 nodes (one at each end) and a wavelength twice the length of the box. Example 2 (touched in the center) is vibrating with 3 nodes (one at each end and an additional node in the center) and a wavelength equal to the length of the box. It can be seen that vibrations are confined to inside the box and only certain wavelengths are possible.

The rate of effusion of nitrogen gas (N2) is 1.253 times faster than that of an unknown gas. What is the molecular weight of the unknown gas?

1.253)^2 = MW(unknown) / 28.014 MW(unknown)= 28.014 x (1.253)2 28.014 x 1.570009 = 43.99

Any D can hold

10 electrons in 5 orbitals

Any F can hold

14 electrons in 7 orbitals

All examples of electron config

15P 1s2 2s2 2p6 3s2 3p3 19K 1s2 2s2 2p6 3s2 3p6 4s1 28Ni 1s2 2s2 2p6 3s2 3p6 4s2 3d8 35Br 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 58Ce 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f2

Filling of subshell order

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p etc

Electron configuration of Fe (26 electrons)

1s2 2s2 2p6 3s2 3p6 4s2 3d6

SHELL SUBSHELL CONTENT MAX # OF ELECTRONS

1st 1s 2 2nd 2s + 2p 2 + 6 = 8 3rd 3s + 3p + 3d 2 + 6 + 10 = 18 4th 4s + 4p + 4d + 4f 2 + 6 + 10 + 14 = 32 5th 5s + 5p + 5d + 5f 2 + 6 + 10 + 14 = 32 6th 6s + 6p + 6d + 6f 2 + 6 + 10 + 14 = 32 7th 7s + 7p + 7d + 7f 2 + 6 + 10 + 14 = 32 etc

The volumes of gases produced in a chemical equation are directly proportional to their coefficients in the balanced equation

2 H2 (g) + O2 (g) → 2 H2O (g) If the above reaction is carried out on: 1.5 liters H2 (25°C, 2 atm) by ideal gas law n = PV / RT = (2) (1.5) / (0.0821) (298K) = 0.1226 mol H2 mol O2 = 1/2 x 0.1226 = 0.0613 mol by ideal gas law V = nRT / P = (0.0613)(0.0821)(298)/2 = 0.75 liter mol H2O = 2/2 x 0.1226 = 0.1226 mol by ideal gas law V = nRT / P = (0.1226)(0.0821)(298)/2 = 1.5 liter Therefore, the volumes of the gases H2, O2 and H2O are 1.5 : 0.75 : 1.5 which is directly proportional to their coefficients 2 : 1 : 2.

Any S can hold

2 electrons in one orbital

A sample of carbon dioxide (CO2) gas is collected over water at 23°C and 745 mm. The volume of the gas collected is 113.0 ml. How many moles of CO2 gas have been collected? How many grams of CO2 gas have been collected?

23 C = 21.1 mm P( CO2) = 745 - 21.1 = 723.9 x 1 atm /760 mm =0.9525 atm N (CO2) = 0.9525 (113.0 ml x 1 L /1000 ml) / (0.0821) (296 K) N (CO2) = 0.9525 (0.113) / (0.0821) (296 K) 0.1076325 / 24.3016 = 0.00443 mol of CO2 G= 0.00443 mol x 44.01 grams = 0.195 grams of CO2

Remembering that each subshell can hold a certain max # of electrons, the electrons of any atom would be arranged in a certain manner (called the electron configuration) such as follows for several elements:

26Fe = 1s2 2s2 2p6 3s2 3p6 4s2 3d6 13Al = 1s2 2s2 2p6 3s2 3p1

Describing an electron by telling certian things about it

4 quantum numbers Principal (n)- shell electron is in ex 1,2,3,4 Angular (l)-shape of orbital ex, 0-s (circle, 0 planes) 1-p (two lobes, 1 plane) 2-d (four lobes, 2 planes) 3-f (8 lobes, 3 planes) *we identify the planes not the lobes Mangnetic (M(sub)l) orientation in space of orbital -2 -->+2 dependent on orbital it ends full in Spin QN (M(sub)s) Direction spinning (-1/2 (first fill) or +1/2 (second fill))

A gas sample has an original volume of 650 ml when collected at 730 mm and 20°C. If a change is made in the gas temperature which causes the volume of the gas sample to become 840 ml at 1.05 atm, what is the new temperature? Two volumes, two pressures, one temperature given, one temperature asked for, so we use: COmbinded gas law

650 ml/1000 = 0.650 liters = Vi 730 mm/760 = 0.961 atm = Pi 840 ml/1000 = 0.840 liters = Vf 20°C + 273 = 293K = Ti 1.05 atm = Pf Put in the data: (0.961) x (0.650) = (1.05) x (0.840) (293) Tf Solve for Tf: 0.62465 x Tf = 258.426 (By cross-multiplication) Tf: = 258.426 / 0.62465 = 414 K

Any P can hold

6electrons in 3 orbitals

Example of combined gas law

A gas sample has an original volume of 650 ml when collected at 730 mm and 20°C. If a change is made in the gas temperature which causes the volume of the gas sample to become 840 ml at 1.05 atm, what is the new temperature? Two volumes, two pressures, one temperature given, one temperature asked for, so we use: 650 ml/1000 = 0.650 liters = Vi 730 mm/760 = 0.961 atm = Pi 840 ml/1000 = 0.840 liters = Vf 20°C + 273 = 293K = Ti 1.05 atm = Pf Put in the data: (0.961) x (0.650) = (1.05) x (0.840) (293) Tf Solve for Tf: 0.62465 x Tf = 258.426 (By cross-multiplication) Tf: = 258.426 / 0.62465 = 414 K

Ex of mole percent, mole fraction and partial pressures

A mixture of 4 gases consists of 5.00 moles of He, 4.00 moles of H2, 3.00 moles of CO2 and 8.00 moles of Ar. The total pressure of the mixture is 1800 mm. Determine the mole fraction of each gas in the mixture. Determine the mole percent of each gas in the mixture. Determine the partial pressure of each gas in the mixture.

Example of Ideal gas law

A sample of CO2 gas which weighs 1.62 grams has a volume of 1.02 liters when collected at 20°C. What would be the pressure of the gas sample? CO2 C 12.011 O 15.999 x 2 44.009 1.62 / 44.009 = 0.0368 mol = n V= 1.02 L R= 0.0821 T= 20 + 273 = 293 K ? x 1.02 = 0.0368 x 0.0821 x 293 P x 1.02 = 0.885 0.885/ 1.02 = P = 0.868 atm

Gas collected over water

A sample of methane (CH4) gas is collected over water at 26°C and 745 mm. The volume of the gas collected is 55.5 ml. How many moles of CH4 gas has been collected? How many grams of CH4 gas has been collected? Find pressure at temperature on chart and then convert to atm by / 760mm: VP=25.2 mm P(CH4) = 745 - P(H2O from table) 25.2 = 719.8 x 1 atm / 760 mm = 0.947 atm Now ideal gas law to find mol; n(CH4) = PV / RT = (0.947 atm) (55.5 ml x 1 liter / 1000 ml) / (0.0821) (299K) n(CH4) = 0.00214 moles Now switch from moles to grams mol x MW grams(CH4) = moles x MW = 0.00214 moles x 16.042 grams / 1 mole = 0.0343 grams

The proton and the Neutron

After discovery of the negatively-charged electron, it was deduced that there must be a positively-charged particle present in equal numbers to the electrons so as to produce a neutral atom. In 1919-1920, Rutherford and other physicists bombarded nuclei with alpha particles (helium nuclei) transforming one atom into another. In these experiments, hydrogen nuclei were emitted, leading to the discovery that the positive charge of any nucleus was equal to a certain number of hydrogen nuclei. The charge of this particle was determined to be positive by the fact that it is attracted to a negatively-charged pole (cathode) of an electric field. Rutherford coined the name proton and used it in print in 1920. In 1932, James Chadwick bombarded Beryllium atoms with helium nuclei and observed that uncharged particles with about the same mass as a proton were emitted. This particle accounted for the difference in the mass of atoms which could not be accounted for by the mass of the atom's protons and electrons. This uncharged (neutral) particle was given the name neutron.

Vapors of Ammonia (NH3) and Hydrochloric acid (HCl) react to form a white cloud of NH4Cl vapor. If these two gases are released into a tube from opposite ends will the white cloud of NH4Cl form closer to the end of the tube from which NH3 was released or closer to the end of the tube from which HCl was released? Explain your answer.

Ammonia (NH3) with a MW of 17 will diffuse more quickly than Hydrochloric acid (HCl) with a MW of 36.5 and so the NH3 will have moved further through the tube when it meets the HCl to form the cloud of NH4Cl. So the cloud of NH4Cl will form closer to the HCl end of the tube.

Orbital shapes

An "s" type orbital is said to be spherically shaped with no nodal planes (l = 0) cutting it into one lobe. A "p" type orbital is said to be dumbell shaped with one nodal plane (l = 1) cutting it into two lobes. A "d" type orbital is said to be butterfly shaped with two nodal planes (l = 2) cutting it into four lobes. An "f" type orbital is said to be double butterfly shaped with three nodal planes (l = 3) cutting it into eight lobes. S

Orbital Content

Any "s" subshell __ Any "p" subshell __ __ __ Any "d" subshell __ __ __ __ __ Any "f" subshell __ __ __ __ __ __ __

Step 2

Do mole percent for each gas: Mole%(He) = 100[X(He)] = (100) 0.250 = 25.0% Mole%(H2) = 100[X(H2)] = (100) 0.200 = 20.00% Mole%(CO2) = 100[X(CO2)] = (100) 0.150 = 15.00% Mole%(Ar) = 100[X(Ar)] = (100) 0.400 = 40.00%

Electrons

During the 1800's a number of scientists including Sir Humphrey Davy, Michael Faraday, R. A. Milligan, G. J. Stoney, and J. J. Thompson performed experiments which led to the discovery of the electron and measurement of its properties. Davy (1807) first suggested the idea of the electron. Faraday (1832) determined the relationship between charge and quantity of element produced by that charge in an electrochemical reaction. Stoney (1891) was the first to use the term "electron". Thompson (1897) determined the mass to charge ratio of the electron and Millikan (1911) determined the value of the charge of the electron.

The Atom

Early ideas of the structure of the atom ranged from Thompson's (1904) "plum-pudding" model with negative charges distributed throughout an area made up of positive charge to Ernest Rutherford's (1909) planetary atom model in which electrons occupy most of the volume of an atom around a small, dense, positively-charged nucleus to Niels Bohr's (1913) model of electrons orbiting the positively-charged nucleus in distinct energy levels.

Diffraction

Electromagnetic radiation wave properties show diffraction. he slight bending of waves around the edge of any object in their path. Light is bent around tiny water droplets found in clouds producing light and dark bands that constitute the coronas surrounding the sun or moon.

Interference

Electromagnetic radiation wave properties show this; the combination of waves to form alternating areas of increased and decreased amplitude. This is the principle that creates beats in sound wave production.

Determine the mass of Ar in 1.00 liter of a gas mixture at 25°C which contains 0.300 atm of Ne and has a total pressure of 4.00 atm.

FInd pressure of Ar: 4.00 (which is the total pressure, PT) = PNe + PAr (and since PNe = 0.300) So 4.00 = 0.300 + PAr So PAr = 4.00 - 0.300 = 3.700 X(Ne) = 0.300 / (4.00) = 0.075 X(Ar) = 3.700 / (4.00) = 0.925 nT = P V / R T = (4.00) (1.00) / (0.0821) (298K) = 0.1635 mole X(Ar) = n(Ar) / nT n(Ar) = X(Ar) nT n(Ar) = (0.925) (0.1635) = 0.15124 g(Ar) = n(Ar) [MW(Ar)] = (0.15124) (39.95) = 6.042 g

Gas Volume Stoichiometry

Find MOlecular weights Use gram given to calculate mol by dividing the number of grams by the molar mass to get mol of known. Then use coefficients to Multiply/divide by coefficent to get other mols EXThe combustion of ethanol (C2H5OH) takes place by the following reaction equation. C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g) What is the volume of CO2 gas produced by the combustion of excess ethanol by 23.3 grams of O2 gas at 25ºC and 1.25 atm? (46) (32) (44) (33) C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g) 0.233 mol 3/1 x 0.233 2/1 x 0.233 3/1 x 0.233 0.233 mol 0.699 mol 0.466 mol 0.699 mol 10.7 g 22.4 g 20.5 g 23.07 g PV = nRT 1.25 (V) = 0.4854 (0.0821) (298) 1.25 (V) = 11.876 / 1.25 V = 9.5 L EXThe formation of hydrazine (N2H4) from its elements takes place by the following reaction equation. N2 (g) + 2 H2 (g) → N2H4 (g) What are the volumes of N2 gas and H2 gas required to form 28.5 grams of N2H4 at 30ºC and 1.50 atm? 28.014) (2.016) (32.0452) N2 (g) + 2 H2 (g) → N2H4 (g) 0.8906 mol 2/1 x 0.89 0.8906 mol 1.781 mol V N2 = (1.50) (V) = (0.8906) (0.0821) (303) V N2 = 14.77 L V H2= (1.50) (V) = (1.781) (0.0821) (303) V H2 = 29.54 L

Step 3:

Find partial pressures: P(He) = X(He) (1800 mm) = 0.250 (1800 mm) = 450 mm P(H2) = X(H2)(1800 mm) = 0.200 (1800 mm) = 360 mm P(CO2) = X(CO2)(1800 mm) = 0.150 (1800 mm) = 270 mm P(Ar) = X(Ar) (1800 mm) = 0.400 (1800 mm) = 720 mm

How to do: Step 1

First do mole fraction for each gas: X(He) = 5.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.250 X(H2) = 4.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.200 X(CO2) = 3.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.150 X(Ar) = 8.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.400

Shell = m

First number is the shell it is in

About gases

Gases (think of air as an example) are very light, are easily expandable, are easily compressible, and completely fill a container into which they are released. These properties are due to the fact that the individual molecules of the gas are very far apart from one another, are not attracted to one another and are in constant, very rapid motion. This description of gases is known as the kinetic-molecular theory of gases.

A balloon deflates as the gas molecules effuse through the pore of a balloon. Which balloon would deflate more quickly, one filled with H2 or He? Explain your answer.

H2 with a MW of 2 will effuse more quickly than He which has a MW of 4 and therefore the balloon filled with H2 would deflate more quickly.

CaCO3 (s) + 2 HCl (l) → CaCl2 (s) + CO2 (g) + H2O (l)

If the above reaction is carried out on 28.5 g of CaCO3 at 25°C and 1 atm, what amount of HCl will be required and what amounts of CaCl2, CO2, and H2O will be produced? The answers will be determined in terms of moles, grams and volumes (for any gases). (MW = 100) (MW = 36.5) (MW = 111) (MW = 44) (MW = 18) CaCO3 (s) + 2 HCl (l) → CaCl2 (s) + CO2 (g) H2O (l) 28.5 grams 20.81 grams 31.64 grams 12.54 grams 5.13 grams ↓ ↑ ↑ ↑ ↑ 0.285 mol → 2/1 x 0.285 mol → 1/1 x 0.285 mol → 1/1 x 0.285 mol → 1/1 x 0.285 mol V = nRT / P = (0.285)(0.0821)(298)/1 = 6.97 liter

The Bohr Atom

In 1913, Niels Bohr, a young Danish physicist came to work in the laboratory of Ernest Rutherford who, a few years earlier, had proposed the planetary model of the atom. He was asked to solve some of the problems with the model mainly that it predicted an unstable atom since the orbiting electrons would lose energy and quickly fall into the nucleus. Bohr used facts known about light emission by atoms to develop his new theory which is summarized as follows: Electrons can occupy only certain stable orbits around the nucleus each of which represents a definite energy level, these energy levels are said to be quantized or restricted to certain values rather than a continuous set of values. Light of a definite energy or frequency is emitted or absorbed by an atom when an electron jumps from one orbit to another orbit. When Bohr's theory was applied to the hydrogen atom, it produced the exact emission results given by the actual atom. In 1922, Bohr received the Nobel Prize in Physics for this work. However, Bohr's model only worked for hydrogen, which is the simplest atom.

Quantum Mechanical Model of the Atom

In 1926, Erwin Schrodinger published the Schrodinger equation in which he described the behavior and energies of electrons using an equation similar to those used to describe electromagnetic waves. The equation was used to calculate properties of the hydrogen atom electron and these values agreed with those experimentally observed. The Schrodinger equation is very complex and exact solutions are impossible for multi-electron atoms but approximations again give results which agree with observed values. The Schrodinger equation solutions produce three quantum numbers which describe an electron in three-dimensional space.

Ni28 1s2 2s2 2p6 3s2 3p6 4s2 3d8

Last to fill= 3d8 _↑↓_ _↑↓_ _↑↓_ _↑_ _↑_ -2 -1 0 +1 +2 m= 3 l= 2 planes M sub l= 0 Msubs= +1/2 Outermost= 4s2 m= 4 l = 0 Msubl= 0 Msubs= +1/2

So, depending on the orbital in which the electron exists, the l quantum number will be as follows:

Orbital Type Angular momentum quantum number (l) value (also the # of nodal planes) s 0 p 1 d 2 f 3

ml

Orientation in space of orbital -l -->0 --> +l So an l=1 (d), Msubl will be -1 --> +1

Determine the total pressure of a mixture of 0.400 mole of He and 0.600 mole of Ne in a 2.00 liter container at 25°C.

P = n R T / V = (0.400 + 0.600) (0.0821) (298K) / 2.00 = 12.23 atm

combined gas law

P1 x V1 / T1 = P2 x V2 / T2

Effusion

Process by which a gas escapes through a small opening into a vacuum

Electron shells

The electrons are first organized into layers (called shells) which are described starting with the innermost layer closest to the nucleus; 1st shell, 2nd shell, 3rd shell, etc. Each of these shells can accommodate a maximum number of electrons depending on its subshell (see next paragraph) content. Each of these shells is subdivided into subsections (called subshells) which are described by the letters "s", "p", "d", and "f". Any "s" subshell can hold a maximum of two (2) electrons, any "p" subshell can hold a maximum of six (6) electrons, any "d" subshell can hold a maximum of ten (10) electrons, and any "f" subshell can hold a maximum of fourteen (14) electrons.

the frequency (ν) of the wave

The number of crests that pass a given point within one second

Rate of effusion

The rate of effusion of ammonia gas (NH3) is 2.45 times faster than that of an unknown gas. What is the molecular weight of the unknown gas? [r(NH3) / r(unknown)]2 = MW(unknown) / MW(NH3) (2.45)2 = MW(unknown) / 17 MW(unknown) = (2.45)2 x 17 = 102

Orbital Diagrams

The subshells are further divided into subdivisions (called orbitals). Each orbital can hold two electrons, so an "s" subshell will contain one orbital, a "p" subshell will contain three orbitals, a "d" subshell will contain 5 orbitals and an "f" subshell will contain 7 orbitals. These orbital subdivisions have certain shapes

Quantum Numbers and Atomic Orbitals

The three quantum numbers derived mathematical solution of the Schrodinger equation are the principal quantum number (n), the angular momentum quantum number (l) and the magnetic quantum number (ml). A fourth quantum, the spin quantum number, (ms), is determined from equations used to describe the magnetism of the electron.

About combined gas law

This equation states that if the initial pressure (atm) of a gas is multiplied by its initial volume (liters) and is divided by its initial temperature (K) and a change is made in the pressure or the temperature causing a corresponding change in the volume, the final pressure (atm) multiplied by the final volume (liters) and divided by the final temperature (K) will still equal the same numerical value. This equation makes it possible to calculate one of these values if the other five are known. This equation should be used to do any gas problem in which two temperatures, volumes or pressures are given or asked for. If one of the quantities (pressure, volume or temperature) remains constant or is not given, that quantity is dropped from both sides of the equation.

Law of partial pressures

This results in a law known as the law of partial pressures which states that the total pressure of gases in a container is equal to the sum of the partial pressures of the individual gases present in the container expressed by the following equation: PT = P1 + P2 + P3 ......... where PT = the total pressure of gases in the container and P1, etc equal the partial pressures of each individual gas.

Electromagnetic Radiation

Waves are energy transport disturbances that move through matter (although as we will learn later some waves do not require matter to travel through) temporarily displacing particles from an original position (to which position the particles then return) transporting energy but not the matter. The particles of the medium (i.e. water or air molecules) oscillate about a fixed position as the wave disturbance moves from one point to another point.

Similarly, each electron orbit must be equal to a whole number times the wavelength of the electron as shown in the diagram (A)

Waves in orbits which are not equal to a whole number times the electron wavelength (B) will destroy themselves by interference. Because the energy of an electron depends on its orbit, only certain energy values are possible, energy is quantized.

The equation (ν x λ = c) which relates wavelength, frequency and velocity of electromagnetic radiation makes it possible to calculate frequency if wavelength is known (or vice versa) as follows. Examples:

What is the wavelength of a radio wave with v = 6 x 107 Hz? λ = c/v = 2.998 x 108 m/s / 6 x 107 cycles/s = 4.997 m What is the frequency of infrared light with λ = 2.5 x 10-5 m? v = c/λ = 2.998 x 108 m/s / 2.5 x 10-5 m = 1.2 x 1013 Hz

Name quantum numbers Draw it out first then find the values: EX: 2p4 Ex: 3s2

_↑↓ _ _↑_ _↑_ m=2 p=1 Msubl= -1 Msubs= +1/2 _↑↓ _ m= 3 l = 0 Msubl= 0 Msubs= +1/2

One wave (or cycle) per second is called

a Hertz (Hz), after Heinrich Hertz who established the existence of radio waves. A radio wave with 1,000,000 cycles that pass a point in one second has a frequency of 106 Hz.

The maximum displacement of a particle at a peak or trough is called

amplitude

About ionizing radiation and non ionizing radiation

can be harmful to human tissues. Ionizing radiation includes gamma rays and X-rays and some wavelengths of ultraviolet light. Radio transmission waves, microwaves, infrared light, and visible light are normally considered non-ionizing radiation, though high intensity beams of these can have similar properties as ionizing radiation. Ionizing radiation is found naturally in the environment in naturally occurring radioactive materials and cosmic rays. Man-made sources include artificially-produced radioisotopes and X-ray tubes which are used, under controlled conditions, in medical treatment and diagnosis. Uncontrolled exposure to ionizing radiation can cause fetal mutation, radiation sickness, cancer, and death.

Graham's Law

described by the following equation where r1 / r2 equals the relative rates of effusion and MW is the molecular weight of each gas. Note that the rate of effusion is inversely proportional to the molecular weights of the gases. This means that the rate of effusion is faster for a gas with a lower molecular weight.

spin quantum number (ms)

describes the direction of spin of the electron in its orbital and has only two values, -1/2 and +1/2. Just as a wheel can spin only clockwise or counter-clockwise, the electron has only two spin directions (corresponding to the two ms values) which are described as up and down and diagramed with up or down arrows. In 1924, Wolfgang Pauli proposed this spin characteristic to account for the presence of previously unexplained lines in the emission spectra of atoms. He also is responsible for the Pauli exclusion principle which states that no two electrons in an atom can have the same four quantum numbers. So, as electrons occupy the orbitals of a subshell, they each occupy a separate orbital (starting with lowest ml value, each having the same spin direction; the same ms value = -1/2) until all orbitals are half-filled (with one electron each) and then the electrons begin to fill each orbital (starting with lowest ml value) until all orbitals are filled (these second electrons to fill each orbital have a ms value = +1/2).

Magnetic quantum number (ml)

describes the orientation in space of an orbital and can have values of -l to 0 to +l where l is the angular momentum quantum number. Each ml value corresponds to one of the orbitals within each type of subshell. We know that an "s" subshell has one orbital whose ml value would be 0. The three "p" orbitals would have individual ml values of -1, 0 and +1. The five "d" orbitals would have individual ml values of -2,-1, 0, +1 and +2. The seven "f" orbitals would have individual ml values of -3, -2, -1, 0, +1, +2 and +3. The magnetic quantum numbers for each orbital in each type of subshell are shown in the chart below. Values are listed from most negative to most positive to indicate lowest to highest energy.

principal quantum number

describes the size and energy of an orbital and corresponds with the shell number, having positive whole number values (1, 2, 3, etc.). All electrons in the first shell (those would be 1s electrons) have n = 1, all electrons in the second shell (2s and 2p electrons) have n = 2, all electrons in the third shell (3s, 3p and 3d electrons) have n = 3, all electrons in the fourth shell (4s, 4p, 4d and 4f electrons) have n = 4, etc.

The size of the elements gets greater as you proceed

down a vertical column

Electromagnetic radiation also shows particle properties. Photoelectric effect

e particles of light are called photons, just as the atom is the smallest unit of matter and the electron is the smallest unit of electricity. Unlike the electron, which has one unit of negative electric charge, the photon is uncharged. The particle property of light is called the photoelectric effect and occurs when visible light interacts with alkali metals causing electrons (called photoelectrons) to be emitted. This property is taken advantage of in automatic door openers.

As electrons fill the orbitals of a subshell

each electron fills into a separate orbital (shown by an up arrow) until all orbitals contain one electron, then the remaining electrons as they start to pair up with the electrons already in each orbital, must reverse their spin (shown by a down arrow) to be able to enter the orbital and pair up.

Short wavelength =

high frequency

Nuclear charge gets greater as you proceed from

left to right across a horizontal row

Long wavelength =

low frequency

mechanical waves

must have a medium through which they travel Examples of mechanical waves are sound waves and water waves.

Electron Configuration

n an earlier section, we learned how to determine how many electrons were present in an atom (# of electrons = # of protons = atomic #). Now we will learn how chemists describe the manner in which those electrons are arranged within the area they occupy around the nucleus. Electrons are not arranged in a haphazard or random manner but in a very organized way which is described by a set of letters and numbers.

A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25°C. Determine the partial pressures of He and Ne in the flask. What is the total pressure?

n(He) = g(He) / [MW(He)] = (0.52 mg x 1 g / 1000 mg) / 4.002 = 0.000129935 mol n(Ne) = g(Ne) / [MW(Ne)] = (2.05 mg x 1 g / 1000 mg) / 20.18 = 0.000101586 mol PT = n R T / V = (0.000129935 + 0.000101586) (0.0821) (298K) / 1.00 = 0.005664 atm PT = 0.005664 atm x 760 mm/1 atm = 4.305 mm X(He) = 0.000129935 / (0.000129935 + 0.000101586) = 0.5612 X(Ne) = 0.000101586 / (0.000129935 + 0.000101586) = 0.4388 P(He) = X(He) (4.305 mm) = 0.5612 (4.305 mm) = 2.416 mm P(Ne) = X(Ne) (4.305 mm) = 0.4388 (4.305 mm) = 1.889 mm

Within a horizontal row (period) properties are affected by

nuclear charge difference between elements.

Surface waves

particles move in a circular motion; an example of these is the waves on the surface of an ocean unlike those within the depths of the ocean which are longitudinal waves.

The high points along the wave are called

peaks or crests

Diffusion

process by which a gas spreads out through a space which may be a vacuum or occupied by another gas to occupy the space uniformly.

# orbitals for each letter (when drawing it, how many lines there will be to fill)

s- 1 p- 3 d-5 f- 7

Within a vertical column (group) properties are affected by

size differences between elements

Wave theory of an electron

the electron energy levels in an atom are quantized, as theorized by Bohr in 1913. In 1924, Louis de Broglie, a French graduate student explained this quantization was due to the wave properties of electrons. De Broglie proposed that electrons behave as standing waves because they are held in place by the positively-charged nucleus. Standing waves, like guitar string vibrations, are confined to a region of space and do not move from place to place as travelling waves like water waves do. A diagram of standing waves created on a string which is attached at each end within a box is shown below.

Transverse waves model the line running through the wave is called

the equilibrium or rest position

Longitudinal wave

the particles move in a direction parallel to wave direction. Sound waves travelling through air are longitudinal waves whereas electromagnetic waves such as light are transverse waves

Transverse wave

the particles move in a direction perpendicular wave direction

Pressure will ____________ if temperature _________________

the pressure of a sample of gas will increase if the temperature of the gas sample is increased and the pressure of a sample of gas will decrease if the temperature of the gas sample is decreased.

Period (p) of a wave is

the time it takes for one cycle to pass and the units are always in terms of time. The faster a wave moves the smaller is its period. The speed (c) of electromagnetic radiation in a vacuum is a constant 2.998 x 108 meters per second (m/s) and is equal to the product of its frequency times its wavelength.

volume will ___________ if pressure _________________

the volume of a sample of gas will expand (increase) if the pressure on the gas sample is decreased and the volume of a sample of gas will contract (decrease) if the pressure on the gas sample is increased. [The pressure unit we will use with gases is atmosphere (atm) scale, which is mm/760].

If volume ________ Temp _________________

the volume of a sample of gas will expand (increase) if the temperature of the gas sample is increased and the volume of a sample of gas will contract (decrease) if the temperature of the gas sample is decreased. [The temperature scale used with gases is the Kelvin (K) scale, which is °C + 273]. [The volume scale used with gases is the liter scale, which is ml/1000].

The distance between a point on the wave and the same point in the next cycle of the wave EX trough to trough is called

the wavelength

Electromagnetic waves can travel

through a vacuum while other waves Examples of electromagnetic waves are light, microwaves, x-rays, and TV and radio transmission waves. The reason electromagnetic waves can travel through a vacuum is that they are composed of electric and magnetic fields vibrating at right angles to one another.

The low points along the wave are called

troughs

The symbol lamba (λ) is used to represent

wavelength

Mole percent

which expresses the percentage of a component relative to the total: Mole % = (100) X1 = 100 (n1 / nT)

Ideal gas law

which relates the pressure, volume and temperature of a gas to the number of moles of gas present (or grams since moles = grams/molecular weight). P (atm) x V (liters) = n (moles) x R x T(K) R is a numerical constant with the value of 0.0821 at all times. This value of R has the units of (liters * atm) / ( K * mol) This equation makes it possible to calculate the number of moles (or grams) of gas present if the pressure, volume and temperature are known. This equation should be used to solve any gas problem in which grams or moles are given or asked for.

Constant ν x λ = c

ν x λ = c = 2.998 x 108 m/s

EX: 28Ni

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58Ce

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