unit 6 ch 7 quantum theory and the electronic structure … · chem. 1 hons. unit 6 ch. 7 quantum...

CHEM. 1 HONS. UNIT 6 CH. 7 Quantum Theory 1

UNIT 6 CH. 7: QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS CH. 8: RELATIONSHIPS AMONG ELEMENTS

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READ P. 244 - 277 ASSIGNMENTS: #1 P.5 notes #1-12 Light #2 P.7 notes #1-3 Particle-Wave Duality #3 P. 11-12 notes #1-11 Apartment Block Analogy #4 P. 17 notes #1-16 Quantum Numbers #5 P.21 notes #1-9 More Quantum Number Practice #6 P.23 notes #1-11 Electron Configurations #7 P.27notes #1-10 Ion Formation #8 P.36 notes #1-18 Trends

H. 7: QUANTUM THEORY AND THE ELECTRON STRUCTURE OF THE ATOM • What is the structure of the atom? • How can atomic structure account for the periodic properties observed? • What is quantum mechanics?

ISTORY OF THE DEVELOPMENT OF MODELS OF THE ATOM hese drawings show how the model of the atom has changed as physicists have learned more about its structure.

LECTROMAGNETIC RADIATION AND THE ELECTROMAGNETIC SPECTRUM

n this chapter, the coverage of the electromagnetic spectrum leads into the electronic structure of atoms. From this information, e can rationalize, and predict, such properties of atoms as size, ionization energy, and the way in which they will form bonds.

lectromagnetic radiation is one of the ways that energy travels through space. Electromagnetic radiation includes waves such s light waves from the sun, X-rays used by a dentist, ultraviolet waves, microwaves from a microwave oven, and radiant heat rom a fireplace. Electromagnetic radiation is the only source of information from celestial objects other than the Moon, Mars, nd Venus. We can’t directly touch the stars, but we can receive their radiation and learn about their composition and formation rom this. Although these forms of radiant energy seem quite different, they all exhibit the same type of wavelike behavior and ravel at the speed of light in a vacuum. ll types of electromagnetic radiation travel at the speed of light (3.00 x 108 m/sec).

November 30 Test: 07/01/2006


THE QUANTUM MECHANICAL MODEL OF THE ATOM

The Quantum Mechanical Model of the atom is the one we will study in this chapter. This model of the atom is different than all previous models. It is not a “picture” model but a mathematical description of the energy of an electron within an atom. The quantum mechanical model describes the probability of finding a particular electron in a region of space around the nucleus.

In 1926 Austrian physicist Erwin Schrödinger took atomic models one step further. He used the new quantum theory to write and solve a mathematical equation describing the location and energy of the one electron in a hydrogen atom. The modern description of the electrons in atoms, (the quantum mechanical model) is derived from the mathematical solution to the Schrödinger equation. Previous models were physical models based on the extension of knowledge of the motion of large objects. In contrast, the quantum mechanical model is strictly mathematical.

Like the Bohr model, the quantum mechanical model of the atom leads to quantized energy levels for an electron. Quantized means that there are whole-number multiples of the energy levels; the energy levels are not fractional. Examples of quantized quantities are: matter is quantized because the number of electrons, protons, and neutrons and the numbers of atoms are all integers (1, 2, 5, 18……etc.). Eggs laid by hens are quantized to an integral number – 1, 3, 5 eggs, not ¼ or ½ of an egg.

The quantum mechanical model is based on the likelihood of finding an electron in a certain position. (Analogy – bees in and around a hive.) This probability can be portrayed as a blurry cloud of negative charge. The cloud is most dense where the probability of finding the electron is large. It is less dense where the probability of finding the electron is small. Hence it is difficult to say where an electron cloud ends. There is at least a slight chance of finding the electron a considerable distance from the nucleus. A model is needed, however to show where the electron probably is. By convention, a surface is drawn around the model of an electron cloud so that the electron is inside the surface 90% of the time. The surface is shaped so that the probability for finding the electron inside is the same for all points on the surface. The shape of the surface can now give us a useful picture of the shape of the cloud in which an electron is found.

This is analogous to locating the position of a rapidly rotating fan blade. At any instant it may be in any one of a number of possible positions. Over a period of time these positions all blend to form a round blur. We know with high probability that the fan blade is somewhere within this blur. Similarly the electron cloud gives us a time-averaged view of the electron. The electron is somewhere in that electron cloud.

The modern quantum mechanical model of the atom arose out of the study of light and waves. Light travels in waves. Light consists of electromagnetic waves that travel in a vacuum at the speed of light 3.00 x 108 meters/sec. Light is described as energy in the form of radiation going through space as varying, vibrating electric and magnetic fields. This variation takes place in a regular, repeating fashion. If the strength of the variation is plotted against time, the graph would show the “waves” of energy.

This Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Electromagnetic radiation (EMR) includes: radio waves, visible light, infrared light, X-rays, gamma rays, microwaves, ultraviolet waves, etc. They all come from the sun and travel at the speed of light.

In the diagram above: gamma rays have the shortest wavelength and highest frequency; radio waves have the longest wavelength and the lowest frequency. Each type of radiation is spread over a specific range of wavelengths (and frequencies). Visible light ranges from a wavelength of 400 nm (violet) to 700 nm (red). You should know the order of wavelength covered by each region of the E/M spectrum.



PROPERTIES OF WAVES • Wavelength (λ) – distance from peak to peak (or crest of the wave). The units of wavelength are usually meters or

nanometers (1 m = 1 x 109 nm) • Cycle – is one complete wavelength. • Frequency (ν) – number of wave peaks (cycles) that occur in a unit of time.) There are 2 units of frequency: 1/sec and Hertz

(Hz). ****NOTE: Use the unit of frequency 1/sec when doing calculations so that the units cancel. Use Hz to report an answer for frequency

• Amplitude – is the vertical distance from the midline of a wave to the peak or trough. • Hertz (Hz) is the unit of frequency – one hertz = 1 sec−1. Different waves travel at different speeds (Ocean waves, sound waves and light waves). All electromagnetic waves travel at the same speed which is the speed of light designated as “c”. Speed of light c = 3.00 x 108 m/sec.

Wavelength and frequency are related by the equation: c = λν c = speed of light, ν = frequency,

λ = wavelength. As the wavelength of light increases, the frequency decreases. Hence the product of frequency and wavelength always equals a constant, c, the speed of light. The SI unit of frequency is hertz (Hz). Ex. 1: What is the wavelength of light having a frequency of 7.5 x 1014 Hertz? (4.0 x 10−7 m)



Ex. 2: What is the frequency of light with a wavelength of 5.00 x 10−10 m? (6.00 x 1017 Hz) Sunlight consists of light with a continuous range of wavelengths and frequencies. The wavelength and frequency of yellow light is different from that of red. When sunlight is passed through a prism the light separated into a spectrum of colors. Each color blends into the next in order: red, orange, yellow, green, blue, indigo, and violet. These are the colors of the rainbow. In the visible spectrum, red light has the longest wavelength and the shortest frequency. Violet light has the shortest wavelength and the highest frequency. Wavelength (m)

Visible light ranges from a wavelength of 400 nm (violet) to 700 nm (red). THE NATURE OF MATTER - PLANCK’S HYPOTHESIS The spectrum of light includes all electromagnetic radiation emitted from the sun. In an attempt to explain the spectrum, Planck proposed that energy (EMR) is not given off as a continuous flow of energy. He said it was radiated or given off in little packets or quanta. If a quanta of energy is viewed as a particle, the quanta is called a PHOTON. Photons are considered to be beams of light existing as a stream of particles. This leads to the hypothesis of the DUAL NATURE OF LIGHT in which light is considered have the properties of both waves and particles. The amount of radiation energy emitted by an object at a certain temperature depends on its frequency. The energy E of a single quantum (photon) of energy is given by the equation:

E= hν where E = energy of a photon in Joules, h = 6.626 x 10−34 J⋅sec. (Planck’s constant) NOTE: J⋅sec not J/sec ν = frequency in Hz (Hz = sec−1)

Ex. 1: Find the energy of red light that has a frequency of 4.60 x 1014 Hz. (Planck’s constant h = 6.626 x 10−34 J sec) (3.05 x 10-19 J)



Ex. 2: Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00 x 104 nm (infrared region) and (b) a photon with a wavelength of 5.00 x 10−2 nm (X-ray region). From this solution, you will be able to see that an X-ray photon is one million times more energetic than an infrared photon. ((a) 3.98 x 10−21 (b) 3.98 x 10−15) Ex. 3: The energy of a photon is 5.87 x 10−20 J. What is its wavelength? (in nanometers) (3.39 x 103)

DO ASSIGNMENT #1 P. 5 NOTES #1-12

ASSIGNMENT #1 LIGHT

1. What is the frequency of light with a wavelength of 450 nm? (6.7 x 1014) 2. What is the energy of a quantum of light of frequency 4.31 x 1014 Hz? (2.86 x 10−19) 3. What is the energy of light with wavelength 662 nm? (3.00 x 10−19) 4. A certain green light has a frequency of 6.26 x 1014 Hz. What is its wavelength? (4.79 x 10−7) 5. What is the energy content of one quantum of light in problem # 4? (4.15 x 10−19) 6. A certain violet light has a wavelength of 412 nm. What is its frequency? (7.28 x 1014) 7. How does the frequency of a wave change as the wavelength of the wave increases? 8. If the frequency goes up, what happens to the wavelength and energy? 9. The average distance between Mars and Earth is about 1.30 x 108 miles. How long would it take TV pictures transmitted

from the Viking space vehicle on Mars” surface to reach Earth? (1 mile = 1.61 km) (698) 10. The SI unit of time is the second, which is defined as 9,192,631,770 cycles of radiation associated with a certain emission

process in the cesium atom. Calculate the wavelength of this radiation (to three significant figures). In which region of the electromagnetic spectrums is this wavelength found? (3.26 x 10−2)

11. The blue color of the sky results from the scattering of sunlight by air molecules. The blue light has a frequency of about

7.5 x 1014 Hz. (a) Calculate the wavelength, in nm, associated with this radiation, and (4.0 x 102 (b) Calculate the energy, in joules, of a single photon associated with this frequency. (5.0 x 10−19)

12. What is the wavelength, in nm, of radiation that has an energy content of 1.0 x 103 kJ/mol? In which region of the

electromagnetic spectrum is this radiation found? (1.2 x 102)



THE DUAL NATURE OF THE ELECTRON (CALCULATING THE WAVELENGTHS OF PARTICLES)THE DUAL NATURE OF THE ELECTRON (CALCULATING THE WAVELENGTHS OF PARTICLES) Albert Einstein showed that light (electromagnetic radiation) can possess particle like properties. DeBroglie reasoned that if waves can behave like particles, then particles can exhibit wave properties, like a moving baseball that is producing waves as it travels through the air. He said that the wavelength of a particle depends on its mass and velocity. DeBroglie deduced that the particle and wave properties are related by the expression:

)

This equation implies that a particle in motion can be treated as a wave and a wave can exhibit the properties of a particle. As the mass of a particle reaches macroscopic size (large) (say > 10−12 grams) its wavelength becomes extremely short and wave properties cannot be observed. SUMMARY: • EMR was found to possess particle-like properties. • Particles, like electrons were found to have an associated wavelength. • Matter and energy are not distinct. • Energy is really a form of matter. • Large pieces of matter are predominately particulate. • Very small pieces of matter (ex. photons) are predominately wavelike (but can exhibit particulate properties). • Intermediate pieces of matter (ex. electrons) are wavelike as well as particulate. NOTE: ENERGY EQUATION OF PLANCK’S HYPOTHESIS E= hν.: “ν” IS A GREEK LETTER AND MEANS FREQUENCY IN HERTZ.

THE DE BROGLIE EQUATION λ = mvh “v” is lower case v of the alphabet and means velocity of the particle in m/s.

EXAMPLES: 1. The fastest serve in tennis is about 140 miles per hour or 62 m/s. Calculate the wavelength associated with a 6.0 x 10−2

kg tennis ball traveling at this velocity. (1.8 x 10−34) 2. Calculate the wavelength (in nanometers) of a H atom (mass = 1.674 x 10−24 grams) moving at 7.00 x 102 cm/s.

(5.66 x 10−8) Y

DO ASSIGNMENT #2 P. 7 notes #1-3 PARTICLE-WAVE DUALIT

THE de BROGLIE EQUATION

λ =mvh where λ = wavelength of the moving particle

m =mass of the particle (in kg) v = velocity of the particle (because matter does not travel at the speed of light



ASSIGNMENT #2 PARTICLE-WAVE DUALITY 1. Thermal neutrons are neutrons that move at speeds comparable to those of air molecules at room temperature. These neutrons

are most effective in initiating a nuclear chain reaction among U isotopes. Calculate the wavelength (in nm) associated with a beam of neutrons moving at 7.00 x 10

2352 m/s. (Mass of a neutron = 1.675 x 10−27 kg.)

2. Protons can be accelerated to speeds near that of light in particle accelerators. Estimate the wavelength (in nm) of such a

proton moving at 2.90 x 108 m/s. (Mass of a proton = 1.673 x 10−27 kg.) (1.37 x 10−6) 3. What is the de Broglie wavelength, in cm, of a 12.4-g hummingbird flying at 1.20 x 102 mph? (1 mile = 1.61 km)



SPECTROSCOPY Bohr proposed the planetary model of the atom. He used SPECTROSCOPY (the study of wavelengths) to improve his

theory. This led to the discovery of energy levels of atoms and then finally to the new quantum mechanical model of the atom. Spectroscopy is the study of substances that are exposed to some continuous exciting energy such as heat or light. It is the

process of producing and analyzing waves of a particular wavelength called spectra. Every element emits light if it is heated by passing an electric discharge through its gas or vapor. The atoms absorb energy and

move to a higher energy level farther from the nucleus of the atom. They then lose the energy and emit it as light as they drop back a lower energy level. The spectrum that is created by exciting atoms and then observing the light that they emit as they return to lower energy is called an emission spectrum. The spectrum appears as colored lines.

The spectrum of a substance is the set of wavelengths absorbed or emitted by that substance. In spectroscopy, the wavelength of light absorbed is characteristic of the substance being excited. This set would be emitted by all excited particles of the same substance.

These unique line spectra are extremely useful for the identification of an unknown substance. Each spectrum is a unique “fingerprint” for each element. Each element has its own pattern. This is especially useful in identifying the substances for example in crime labs (illegal drugs).

Bohr used this principle to improve his model of the atom. The unique energy change that occurred as an atom absorbed energy was used to determine the energy levels in the model of the atom.. An energy level corresponds to a certain distance from the nucleus and a specific amount of energy. Electrons position themselves at certain distances from the nucleus according to their energy levels. Low energy electrons stay closer to the nucleus and higher energy electrons can move further away from the nucleus.

When the spectral lines produced by an excited atom are studied, it was found that each frequency of light corresponds to an exact amount of energy. Each spectral line represents an amount of energy (quanta) that can be calculated. From these energy calculations the positions of the electron around the nucleus were determined. This led to the mathematical model of the atom called the quantum mechanical mode

Figure: (a) An experimental arrangement for studying the emission spectra of atoms and molecules. The gas under study is in a discharge tube containing two electrodes. As electrons flow from the negative electrode to the positive electrode, they collide with the gas. This collision process eventually leads to the emission of light by the atoms (or molecules). The emitted light is separated into its components by a prism. Each component color is focused at a definite position, according to its wavelength and forms a colored image of the slit on the photographic plate. The colored images are called spectral lines. (b) The line emission spectrum of hydrogen atoms.



FLAME TESTS Flame tests illustrate emission of light as excited electrons return to lower energy levels. When exposed to a flame, the ions listed below emit colored light that is characteristic of the metal element in the compound.

FLAME TESTS COLOR OF METAL ELEMENTS ELEMENT COLOR

Na+ Yellow K+ Violet

Ca2+ Yellow-red Sr2+ Deep red Li+ Crimson

Ba2+ Green-yellow Cu2+ Blue-green

OBJECT: To detect the presence of an element by the use of flame tests. OBSERVATIONS:

SUBSTANCE COLOR OF FLAME calcium chloride

cupric chloride

lithium chloride

potassium chloride

sodium chloride

strontium chloride

CONCLUSION: 1. The POSITIVE ion of the salt gives the characteristic flame test for the metallic element involved.

2. The NEGATIVE ion of the salt has no bearing on the flame test and cannot be used to detect the element.

THE QUANTUM ATOMIC MODEL The most modern model of the atom is not one you can draw because the location of the electrons is not known for certain. It

has been proven that electrons do not spin around the nucleus in fixed paths called orbits. Instead, they occupy a region in space around the nucleus. When trying to locate a particular electron in an atom, the best guess as to where it may be found is based on a prediction of where it is most likely to be found 90% of the time. So the quantum mechanical model of the atom does not give a specific location of an electron. It describes a region in space where an electron probably can be found 90% of the time. This cannot be drawn on paper because it would just look like a big blur. The new model just describes mathematically where the electron may be. This is not something you can indicate by a drawing.

The quantum mechanical model is not a drawn model of an atom but a mathematical representation. To describe the problem of trying to locate an electron, we must consider The Heisenberg Uncertainty Principle. It states that it is impossible to know the exact velocity and exact position of an electron at any time. We can only describe the probability of the electron being at a certain point in space.

The three-dimensional region in space around the nucleus where an electron can be found is called an orbital. (Note: An orbital is not the same as an orbit.) The farther an electron is away from the nucleus, the more energy it has.

Three quantum numbers n, l, and m will be required to describe the location of an electron in space. These three quantum numbers are derived from the mathematical solution of the Schrödinger equation for the hydrogen atom. These three quantum numbers will describe the probability or position of an electron in an atom at a particular time A fourth quantum number s will describe the behavior of a specific electron in an orbital and only serves to identify the atom once you have found its location..

Electrons will be located by first finding what shell (or level) they occupy. Each shell is then made of subshells, and finally each subshell contains orbitals where the electron is probably going to be found.

shell (n) subshell (l) orbital (m→ → l)



QUANTUM NUMBERS

In three-dimensional space, three numbers are required to describe the location of an object in space. In quantum mechanics, three quantum numbers are required to describe the location of an electron in an atom. Each electron occupies an orbital which describes where the electron can be found most of the time. An orbital has both a characteristic energy and a characteristic shape.

The three numbers used to describe an orbital are called the principle quantum number, the angular momentum quantum number, and the magnetic quantum number. These quantum numbers will be used to describe atomic orbitals and to label electrons that reside in them. A fourth quantum number – the spin quantum number – describes the behavior of a specific electron and completes the description of electrons in atoms.

The analogy of an apartment block will be used to help you determine what the quantum numbers mean.

7TH FLOOR7

6TH FLOOR6

5TH FLOOR5

4TH FLOOR4

3RD FLOOR3

2ND FLOOR2

1ST FLOOR1

DO ASSIGNME

floors apartments rooms → → LEVEL SUBLEVEL ORBITAL

ATOMIC ORBITALS - APARTMENT BLOCK ANALOGY

s 7p 7d 7f

s 6p 6d 6f

s 5p 5d 5f

s 4p 4d 4f

s 3p 3d

s 2p

s

NT #3 P. 11-12 notes #1-11 THE APARTMENT BLOCK ANALOGY



ASSIGNMENT #3 THE APARTMENT BLOCK ANALOGY

Keep in mind that we are studying the basic model of a very complex theory. A good way to look at the model is to compare it to an apartment building. An apartment building has different floors, different apartments on each floor, and different rooms in each apartment.

We can look upon the electrons of an atom as rather peculiar apartment dwellers. Electrons prefer the floor closest to the ground and the smallest apartments. Electrons also prefer to live one to a room until each room in an apartment has one occupant. The electrons will then pair up until each room has two. Each room in the apartment can only hold two electrons.

NOTE: A “shell” is also called a “level.”A “subshell” is also called a “sublevel”.

Apartment buildings may have several floors. The model we discuss has several floors but only the first seven floors will be occupied. All the electrons of the elements known today will fit within seven floors of the building. Additional floors are available but will only be occupied in special cases. The floors in the apartment building are called shells or levels in the

electron model and are numbered 1 through 7. According to what you

have just read, what shell (floor) will be occupied first by electrons? _______

Each shell (or floor) in the model has one or more apartments which are called subshells or sublevels. These subshells are apartments of different sizes: s, p, d, f, g, h, i, j, k, l…etc. After the “f” apartment, the apartment sizes are named alphabetically: g, h, i, j, k….etc. All the electrons of the elements known today will fit within seven floors of the building and will need only occupy apartments s, p, d, and f. There are not enough electrons available yet in the elements known to need apartments “g” and higher. The subshells (apartments) 1. The subshells are apartments of different sizes:

• An s subshell (apartment) has only a single room. • A p subshell (apartment) has 3 rooms. • A d subshell (apartment) has 5 rooms. • An f subshell (apartment) has 7 rooms.

(a) A g subshell will have _______ rooms;

(b) A h subshell will have _______ rooms,

(c) A i subshell will have _______ rooms, etc.

2. Each room in an apartment can only hold two electrons. An s subshell (apartment) then will hold a maximum of 2 electrons according to the model.

(a) A p subshell will hold a maximum of how many electrons? _______

(b) How many electrons will a d subshell hold? _______

(c) An f subshell will hold ______ electrons; a g subshell will hold _______ electrons, a h subshell will hold _______

electrons, etc.

3. Each room in a subshell (apartment) is called an orbital. Then an s subshell (apartment) will consist of one orbital (room)

with a capacity of 2 electrons.

(a) A p subshell will consist of 3 orbitals (room) with a total subshell capacity of _______ electrons.

(b) A d subshell will have _______ orbitals and a total capacity of _______ electrons.

(c) An f subshell will have _______ orbitals and hold _______ electrons.

4. The first shell (floor) has only one subshell (apartment) which is an s subshell. Because of its location on the first shell

(floor), it is called a 1s subshell.

(a) How many orbitals (rooms) are there in this 1s subshell? _______

(b) How many electrons will the subshell hold? _______



5. The second shell (floor) only has an s subshell (apartment) and a p subshell.

(a) If the s subshell is call 2s, what do you call the p subshell on the second shell? _______

(b) How many orbitals (rooms) are in that p subshell (apartment)? _______

(c) How many subshells (apartments) are in the second shell? _______

(d) How many orbitals (rooms) are there on the second shell (floor)? _______

(e) How many electrons can occupy the second shell (floor)? _______

6. The third shell has three subshells: s, p, and d.

(a) What are they called? _______, _______, _______

(b) How many subshells are in the third shell? _______

(c) How many orbitals are in the third shell? _______

(d) How many electrons can be in the third shell? _______

7. The fourth shell has four subshells; s, p, d, and f.

What would you call the subshells in the fourth shell? _______, _______, _______, _______

8. The sixth shell can have six subshells.

(a) What would you call the subshells in the sixth shell? ______, _______, _______ _______ _______, _______

(b) How many subshells can there be in the fifth shell? _______ seventh shell? _______

(c) How many orbitals are there in the fourth shell? _______

(d) How many electrons will the fourth shell hold? _______

As mentioned previously, electrons prefer the lower shells (floors) and the smallest subshells (apartment). Electrons prefer the smaller subshells to such a degree that they will sometimes occua smaller subshell on the next higher shell rather than the larger subshell on the lower shell.

py

s

By experiment, it has been determined that electrons will fill the 1s subshell (apartment) first. They will then fill the 2s subshell, and then the 2p subshell. Next, they will fill the 3s subshell and then the 3p subshell. However, before going into the large five-orbital 3d subshell, electrons will first fill the 4s subshell. After filling the 4s subshell, electrons will then proceed to fill the 3d subshell. The 4p subshell is filled next. The electrons prefer to fill the small 5s subshell before filling the larger 4d subshell. The 4d ifilled after 5s. Next, the electrons fill the 5p subshell. Then the small 6s subshell is filled. The very large 4f subshell is occupied only after 6s is filled. After 4f comes 5d. Next is 6p, then 7s and then 5f. A diagram to show you the order of filling is shown on the right.

Note that as we fill consecutive subshells, the energy of the electrons increases. Electrons in the 2s subshell have a higher energy than electrons in the 1s subshell; 2p electrons have a higher energy than 2s electrons, and so on. 9. Neon has 10 electrons. The order of filling its subshells is first 1s, then 2s, and finally 2p. What is the order of filling the

subshells in an atom of magnesium (Mg #12)? _____________. 10. The following notation is used to indicate the number of electrons in each subshell of an atom. For example, neon has 10

electrons, therefore its subshells are written as 1s2 2s2 2p6. The numbers to the upper right of each subshell indicate the number of electrons in each subshell. If we add these numbers (2 + 2 + 6 = 10) we get the number of electrons in a neon atom.

11. How would you use this notation for the magnesium (Mg) atom? ___________________________



THE 4 QUANTUM NUMBERS ( n, l, ml, ms) THE PRINCIPAL QUANTUM NUMBER (n) • Designates the main energy level (floor) or shell. • Values: 1, 2, 3, …….∞. n = 7 is the highest number of shells needed for the

number of electrons in the atoms known. • Describes the CLOUD SIZE. The larger the value of n, the larger the cloud size. • Energy levels closer to the nucleus have lower energy. As n increases, the orbital

becomes larger and the electron spends more time farther from the nucleus. An increase in n also means that the electron has a higher energy and is therefore less tightly bound to the nucleus.

• The larger n is, the greater the average distance of an electron in the orbital from the nucleus and therefore the larger (and less stable) the orbital.

• The maximum number of electrons possible in a given shell is 2n2.

LEVEL

Maximum no. of

electrons 2n2

Actual no. of e−‘s (for existing

elements) n = 1 2 2 n = 2 8 8 n = 3 18 18 n = 4 32 32 n = 5 50 32 n = 6 72 18 n = 7 98 ≈ 6 n = 10 200 0

• In reality there are only 32 electrons in level 5. Levels 6 and 7 have a lot fewer than the maximum possible because the number of levels of the existing elements discovered or man-made does not exceed n = 7 (element #118). Levels greater than 7 have not yet been necessary. There are no elements existing or man-made that would have electrons in levels greater than 7.

ANGULAR MOMENTUM QUANTUM NUMBER (l) • Designates the sub-level (apartment) where the electron can be found. • Gives the SHAPE OF THE ORBITALS. • Values of l: from 0 to (n-1) for each value of n. • The value of l for a particular orbital is generally designated by the letters s, p, d, and f corresponding to l values of 0, 1, 2,

and 3. For the known atoms, only s, p, d, and f exist.

value of l 0 1 2 3 4 5 sub-level s p d f g* h*

*These sublevels are not used in the ground state of any known element. After sublevel designation f, the sublevels are named in alphabetical order: g, h, i, j, k, l, etc…Because only 108 elements

are known, the last designation necessary for the 108th electron is the 7p sublevel • Each sublevel is given a letter designation (like s) and a number designation (s = 0). This allows you to put these two

quantum numbers (level and sublevel) together to identify the shape and location of the atomic orbital .(1s, 4f, 3d etc.) MAGNETIC QUANTUM NUMBER (ml) • Designates the orbital (room) where the electron can be found. • Gives the DIRECTION IN SPACE that the orbital takes. • ml specifies to which orbital within a subshell the electron is assigned. Orbitals in a given subshell differ only in their

direction in space, not in their shape. • Values of ml: from –l, ….0, ….+l. • The middle orbital of a subshell has a value of 0. Orbitals to the left of the middle orbital have negative numbers; to the

right, they have + numbers. ml = 0 −1 0 +1 −2 −1 0 +1 +2 −3 −2 −1 0 +1 +2 +3

l = 0 1 2 3



SPIN QUANTUM NUMBER (ms) • Designates the SPIN OF THE ELECTRON and describes the behavior of the electron, not the location. • Once the location of an electron has been found by quantum numbers n, l, and ml, quantum number ms now is used to

identify the atom (recognize it). • Values: +½ , −½. • Arrow up ↑ is +½, (referred to as “spin up”); arrow down ↓ is –½ (referred to as “spin down”.)



SHAPES OF ATOMIC ORBITALS

Boundary surface diagrams of the five 3d orbitals. Although the 3dz2 orbital looks different, it is equivalent to the other four orbitals in all other respects. The d orbitals of higher principal quantum numbers have similar shapes.

Figure: Representation of the 4f orbitals in terms of their boundary surfaces.



SUBLEVEL VALUE OF l ORBITAL SHAPE s 0 spherical (size of s increases as n increases) p 1 dumbbell shape; 2 regions of electron density d 2 4 regions of electron density f 3 8 regions of electron density

DO ASSIGNMENT #4 P. 17 Notes #1-16 QUANTUM NUMBERS



NAME ________________________________ DATE __________________ PERIOD _________ ASSIGNMENT #4 QUANTUM NUMBERS 1. List the values of n, l, and ml for orbitals in the 4d subshell. 2. Give the values of the quantum numbers associated with the orbitals in the 3p subshell. 3. What is the total number of orbitals associated with the principal quantum number n = 3?_____

4. What is the total number of orbitals associated with the principal quantum number n = 4?_____

5. What is the maximum number of electrons that can occupy the fourth energy level.. ________

6. When n = 5 what are the possible values for l? ________

7. When l = 2 what are the possible values of ml? __________

8. How many sublevels are possible in the third energy level? ___________

9. What is the maximum number of electrons that can occupy a d sublevel? _______

10. What is the maximum number of electrons that can occupy a g sublevel? __________

11. When n = 2, the values of l can be ____ and _____.

12. When l = 1, the values of ml can be ______________________and the subshell has the letter designation of ___________.

13. When a subshell is labeled s, the value of l is _____ and ml has value(s) of ________________.

14. When a subshell is labeled p, _________ orbitals occur within the subshell.

15. When a subshell is labeled f, there are _____ values of ml and ______ orbitals occur within the subshell.

16. Suppose an electron in an atom has the following set of quantum numbers: n = 2, l = 1, ml = +1, ms = +½ and it is the

last electron of an atom. What is the name of the atom?



THE ENERGIES OF ORBITALS (ORDER OF FILLING ORBITALS WITH ELECTRONS) The relative energy of sublevels will now determine the arrangement of electron in atoms (electrons moving into the apartment block). The ground state of an atom is the state in which the electrons are located in the lowest energy levels and sublevels possible. They can move to higher levels or an exited state if excited by some form of energy (light). Sublevels of some energy levels overlap. A 4s subshell has lower energy than a 3d subshell. The effect of the overlap is that the atom is more stable when the 4s sublevel is of lower energy than the 3d sublevel. When drawing orbital diagrams (placing electrons in orbitals of increasing energy), electrons occupy sublevels of the lowest energy. In drawing orbit filling diagrams the following guidelines must be followed: • THE AUFBAU PRINCIPLE (Building up principle) ELECTRONS ENTER ORBITALS OF

LOWEST ENERGY FIRST. Electrons are added to atomic orbitals starting with the lowest energy orbital and building up to higher energy orbitals.

• PAULI EXCLUSION PRINCIPLE – ONLY 2 ELECTRONS CAN OCCUPY AN ORBITAL

AND THEY MUST HAVE OPPOSITE SPINS. No two electrons in an atom can have the same four quantum numbers. This means that electrons occupying the same orbital cannot have the same spin. (No 2 electrons can be in the same place at the same time).

↑↓ ↑↑ THIS NOT THIS:

• HUND’S RULE: WHEN ELECTRONS OCCUPY ORBITALS OF EQUAL ENERGY, ONE ELECTRON ENTERS

EACH ORBITAL UNTIL ALL THE ORBITALS CONTAIN ONE ELECTRON WITH SPINS PARALLEL. The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins, without violating the Pauli Exclusion Principle. (Electrons occupy orbitals in a sublevel one at a time until each orbital has 1 electron. They will not double up until each orbital has at least one electron.)

• ↑ ↓ ↑ ↑ ↑ ↑ ↑↓ ↑ THIS NOT THIS NOT THIS **NOTE: Orbitals in the same subshell are of equal energy so an electron may occupy any of the orbitals in that subshell with equal probability. This means that the first orbital on the left is not necessarily the first one filled. By convention, we use this as the first orbital filled. Also the first electron in a sublevel may be spin up ↑ or spin down ↓ with equal probability. We designate the first electron to be spin up ↑ by convention. Once the first one in has a spin, ↑ or ↓ the remaining 1st electrons in the remaining orbitals must have the same spin according to Hund’s Rule.

↑ ↑ ↑ ↑ ↑ ↑ All these are of equal probability

↓ ↓ ↓ ↓ ↓ ↓ These are not correct.

↑ ↓ ↑ ↓ ↓ ↑

↓ ↓ ↑ ↑ ↑ ↓ ↑ ↓ ↑



ORDER OF FILLING SUBLEVELS • The Diagonal Rule can be used to determine which order the sublevels fill. (shown below). A better way to do it is to use

the periodic table. In using the periodic table: • Period numbers are the level number. • Go from left to write. Designate each period and sublevel ex. 2p, 3f • Sublevel d is one level lower than the main energy level. • Sublevel f is two levels lower than the main energy level. • ORDER 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p



Use the following energy diagrams to draw the orbital filling diagrams for the astatine atom (#85):

**USE YOUR TEST COPY OF THE PERIODIC TABLE. Exercise: An oxygen atom has a total of eight electrons. Write the 4 quantum numbers for each of the eight electrons in the ground state. Note the configuration of 2p4 which means the 2p sublevel has 4 electrons. The orbital diagram is:

↑↓ ↑↓ ↑↓ ↑ ↑

1s2 2s2 2p4

Electron n l ml ms orbital 1 1 0 0 +½ 1s 2 3 4 5 2 1 −1 +½ 2p 6 7 8

DO ASSIGNMENT #5 P. 21 Notes #1 – 9 MORE QUANTUM NUMBER PRACTICE



ASSIGNMENT #5: – MORE QUANTUM NUMBER PRACTICE

1. For the following sets of quantum numbers for electrons, indicate which quantum numbers n, l, ml could not occur and

state why. (a) 3, 2, 2 (b) 2, 2, 2 (c) 2, 0, -1

2. Which of the following sets of quantum numbers are not allowed for describing an electron in an orbital?

n l ml ms a. 3 2 −3 +½ b. 2 3 0 −½ c. 2 1 0 −½

3. Which choice is a possible set of quantum numbers for the last electron added to make up an atom of gallium Ga in its

ground state? n l ml ms

a. 4 2 0 −½ b. 4 1 0 +½ c. 4 2 −2 −½ d. 3 1 +1 +½ e. 3 0 0 −½

4. Which of the following are incorrect designations for an atomic orbital?

(a) 3f (b) 4s (c) 2d (d) 4f 5. How many orbitals in an atom can have the following designations?

(a) 2s (b) 3d (c) 4p (d) n = 3 6. For each of the following give the subshell designation, the ml values, and the number of possible orbitals.

(a) n = 3 l = 2 (b) n = 4 l = 3 (c) n = 5 l = 1 7. Identify the atoms whose last electron is represented by the set of quantum numbers below:

n l ml ms a. 3 1 0 +½ b. 5 2 +2 −½ c. 2 0 0 +½ d. 3 2 −2 −½ e. 4 1 +1 −½

8. What is the subshell designation for each of the following cases

(a) n = 2, l = 0 (b) n = 4, l = 2 (c) n = 5, l = 1

(d) n = 3, l = 2 (e) n = 4, l = 3

9. The quantum numbers listed below are for four different electrons in the same atom. Arrange them in order of increasing

energy. Indicate whether any two have the same energy. (a) n = 4, l = 0 ml = 0 ms = +½ (b) n = 3, l = 2 ml = 1 ms = +½ (c) n = 3, l = 2 ml = −2 ms = −½ (d) n = 3, l = 1 ml = 1 ms = −½


CHEM. 1 HONS UNIT 6 CH. 7 Quantum Theory 22

ELECTRON CONFIGURATIONS We will now consider the arrangements of electron in atoms. The way in which the electrons are distributed among the various orbitals is called the electron configuration. The most stable, or ground state electron configuration of an atom is that in which the electrons are in the lowest possible energy states. The electron configuration of an element is a shorthand way of writing the orbital filling diagram of an atom. The sublevels are written in order of filling and the number of electrons in each sublevel is written as a superscripted number. The sum of the superscripts is equal to the atomic number. For example, the electron configuration of sodium is: Na: 1s2 2s2 2p6 3s1 (read as “one s two, two s two.) A more simple form of the electron configuration is THE NOBLE GAS CORE CONFIGURATION. This configuration shows in brackets the noble gas element that most nearly precedes the element being considered, followed by the symbol for the highest filled subshells in the outmost shells beyond the noble gas configuration. The noble gas in brackets represents the noble gas configuration. This method is especially useful in writing configurations of the higher atomic number elements. The noble gas core of sodium is:

Na: [Ne] 3s1 where [Ne] = 1s2 2s2 2p6.

Another representation of electrons in an atom is called an orbital diagram. Instead of drawing boxes for each orbital, draw only a line and place the arrows over the lines. Make sure you separate the sublevels. Use the noble gas core to represent the rest of the atom. Orbital diagram of chlorine : [Ne] ↑↓ ↑↓ ↑↓ ↑ 3s 3p PROPERTIES OF FAMILIES OF ELEMENTS The properties of an atom or group of atoms are determined by the configuration of the outmost shell. The outmost shell consists of the subshells that are filled last. This configuration explains the similar chemical properties within the same group. The alkaline earth metals all end in ns2. it is this configuration which is responsible for the similar properties of the family of elements. PARAMAGNETIC AND DIAMAGNETIC SUBSTANCES If 2 electrons in the 1s orbital of a helium atom had the same or parallel spins (↑↑ or ↓↓), their net magnetic fields would reinforce each other. Such an arrangement would make the helium atom paramagnetic. (Fig. a). Paramagnetic substances are those that are attracted by a magnet. On the other hand, if the electron spins are paired, or antiparallel to each other (↑↓ or ↓↑) the magnetic effects cancel out and the atom is diamagnetic. (Fig b). By experiment, helium is diamagnetic in its ground state. RULE: Any atom with at least one unpaired electron is paramagnetic. An atom is diamagnetic if all electrons are paired.

EXAMPLE: Give the electron configuration, the noble gas core configuration, and the orbital filling notation of nickel (28). Label as paramagnetic or diamagnetic. Nickel: Electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d8 Noble gas core configuration: [Ar] 4s2 3d8 Orbital filling notation: [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ PARAMAGNETIC 4s 3d

DO ASSIGNMENT #6 P. 23 notes #1-11 ELECTRON CONFIGURATIONS


CHEM. 1 HONS UNIT 6 CH. 7 Quantum Theory 23

ASSIGNMENT #6: ELECTRON CONFIGURATIONS

1. Write the full electron configuration and the noble gas core configuration, and the orbital diagram of the following elements: hydrogen, nitrogen, potassium, nickel, bromine, iron, silver, molybdenum, barium, osmium, radon,

2. Write the noble gas core configuration of the alkali metals. What is the similarity in the configurations that account for the

similar properties of the alkali metals? 3. Complete the following table. An occupied shell, subshell or orbital contains at least one electron, but is not necessarily

filled.

Number of Occupied

Symbol

Total # of electrons

Ground state electron configuration

shells

subshells

orbitals

diamagnetic or

paramagnetic

F

1s2 2s2 2p6 3s2 3p6 4s1

V

Cu

[Kr] 5s2 4d10 5p2

41

4. What is the electron configuration of phosphorus (full and noble gas core)? 5. What quantum numbers in phosphorus describe:

(a) The 1st electron (b) The 5th electron (c) The 11th electron

6. Identify the atoms with the following quantum numbers of the last electron in that atom:

(a) (b) (c) (d) (e) (f) n = 3 n = 3 n = 5 n = 4 n = 4 n = 2 l = 1 l = 2 l = 0 l = 2 l = 1 l = 1 ml = 0 ml = +1 ml = 0 ml = −1 ml = +1 ml = +1 ms = +½ ms = −½ ms = +½ ms = −½ ms = −½ ms = −½

7. What is the maximum number of electrons in a single atom that can have a set of quantum numbers containing the

following? (a) n = 3, l = 2 (b) n = 3, l = 1, ml = −1 (c) n = 4 (d) n = 4, l = 3

(e) n = 5, l = 1 (f) n = 5, l = 3 (g) n = 5, l = 3, ml = 0

8. In the ground state of krypton (36) how many electrons have ml = +1 as one of their quantum numbers? 9. How many electrons have l = 1 as one of their quantum numbers? 10. Describe how the ground state electron configurations of the elements of a period compare to one another and how the

ground state electronic configurations of the elements of a group are related. 11. Identify the atoms that have the following ground state electron configurations in their outer shell or shells. (Be careful –

you must notice something first) (a) 5s2 5p2 (b) 3s2 3p6 3d5 4s2 (c) 4s2 4p6 4d10 5s2

(d) 4s2 4p6 5s2 (e) 5s2 5p5


CHEM. 1 HONS UNIT 6 CH. 8 Periodic Relationships Among the Elements 24

ELECTRON CONFIGURATION EXCEPTIONS

• Sometimes the electron configuration of an element is not what you expect it to be. This is due to the need for stability. An atom will change its configuration to become more stable. An electron may move from one sublevel to another if it makes the atom more stable.

• IN ORDER OF STABILITY: 8 e−’s in outer level > full subshell > ½ filled subshell > no arrangement (no stability) • These exceptions occur in the transition metals where the d subshells are incompletely filled.

Ex: chromium #24: Expected configuration: [Ar] 4s2 3d4 Actual configuration [Ar] 4s1 3d5

• 1 electron from 4s is shifted between 2 very closely spaced sublevels 4s and 3d. The atom assumes the new configuration

because 2 half-filled subshells are more stable than 1 full subshell and one subshell with no special arrangement. These exceptions normally occur only in the 3d and 4d sublevels because the 4s and 3d and the 5s and 4d sublevels are very close in energy level. Exceptions do not occur in the 5d and 6d sublevels (except for Au #79) because 6s and 5d are not close in energy levels. There is a 4f sublevel between them which shields the 6s electrons and prevents them from shifting to a much higher energy level.

• Only other exceptions in the transition elements are Cu (29), Mo (42), Ag (47) and Au (79). (Others will not be

studied.) • Look at the periodic table below. You will notice a lot of exceptions but you only need to know the 5 that I gave you.

November 30 Test. 07/03/2006


EXERCISE: 1. Determine the expected and actual configuration of the copper (#29) the molybdenum atom (#42), and the gold atom (79).

Explain why and how each configuration occurs. 2. Without adding up the number of electrons in the configurations below, give the group number for the elements that have

the electron configurations: Identify the element. (Be careful – look at each configuration carefully!) a) 1s2 2s2 2p1 b) 1s2 2s2 2p6 3s2 3p5 c) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 d) [Xe] 6s2 4f14 5d8

CHAPTER 8: RELATIONSHIPS AMONG ELEMENTS IMPORTANT DEFINITIONS AND POINTS: Valence electrons – These are the outermost level electrons of an atom which are the ones involved in chemical bonding. The group number for A GROUP elements determines the number of valence electrons in an atom. All the transition elements have 2 valence electrons determined by the outermost level electrons. The valence electrons are the s and p sublevel electrons only.

GROUP EXAMPLE e− CONFIGURATION # VALENCE e−‘s 1A Alkali metals

Na #11

[Ne] 3s1

1 (from 3s)

2A Alkaline Earth metals

Sr #38

[Kr] 5s2

2 (from 5s)

3A

In #49

[Kr] 5s2 4d10 5p1

3 (from 5s and 5p)

7A Halogens

I #53

[Kr] 5s2 4d10 5p5

7 (from 5s and 5p)

3B

Sc #21

[Ar] 4s2 3d1

2 (from 4s)

7B

Re #75

[Xe] 6s2 4f14 5d5

2 (from 6s)

EXERCISE: How many valence electrons are there in each of the following atoms? N (#7), Nb (#41), Sb (#51), Hg (#80), V (#23), W (#74), Y (39).



FORMING IONS FROM ATOMS IMPORTANT: METALS LOSE ELECTRONS TO FORM CATIONS; NON-METAL ATOMS GAIN ELECTRONS TO FORM ANIONS. We will now use the electron configuration and number of valence electrons to determine the charge of the cations and anions formed by different elements depending on their location in the periodic table. You will use the electron configuration of the atom to write the electron configuration of the cation or anion.

RULES FOR DETERMINING CHARGES OF CATIONS AND ANIONS

• Atoms can combine by gaining or losing electrons in order to gain an inert gas configuration. • FOR THE A GROUP ELEMENTS: The electrons involved in bonding are the HIGHEST OUTER LEVEL

ELECTRONS (THE VALENCE ELECTRONS) (not the highest energy sublevel e−’s) • Valence electrons are the s and p sublevel electrons only). • The order of electron filling does not determine or predict the order of electron removal for transition metals. • The transition metals lose the highest energy outer level electrons first (s electrons), and then they can lose d

sublevel electrons one at a time. • s and p electrons cannot be lost one at a time − it is all or none.

IONS DERIVED FROM REPRESENTATIVE ELEMENTS (Group 1A to 7A) In forming cations, one or more electrons are lost from the highest level (not sub-level):

Na: [Ne] 3s1 Na+: [He] 2s2 2p6 Na loses the 1 VE in the 3s sublevel. Ca: [Ar] 4s2 Ca2+: [Ne] 3s2 3p6 Ca loses the 2 VE’s in the 4s sublevel. Al: [Ne] 3s2 3p1 Al3+: [He] 2s2 2p6 Al loses the 3 VE’s in the 3rd level (3s and 3p).

Pb: [Xe] 6s2 4f14 5d10 6p2 Pb4+: [Xe] 4f14 5d10 Pb loses the 4 VE’s: 2 in the 6p sublevel and 2 in the 6s sublevel. Level 6 is the highest level and the s and p electrons are lost all at once, not 1 at a time.

In forming anions, one or more electrons are added to the highest partially filled n shell. H: 1s1 H−: 1s2 or [He] F: 1s2 2s2 2p5 F−: 1s2 2s2 2p6 or [Ne] O: 1s2 2s2 2p4 O2−: 1s2 2s2 2p6 or [Ne] N: 1s2 2s2 2p3 N3−: 1s2 2s2 2p6 or [Ne]

All of these anions have stable noble gas configurations. Notice that F−, Na+, and Ne (and Al3+, O2−, and N3−) have the same electron configuration. They are said to be isoelectronic because they have the same number of electrons, and hence the same ground state electron configuration. H- and He are also isoelectronic. F- is isoelectronic to Ne. CATIONS FORMED BY THE TRANSITION METALS You have seen that the 4s orbital is filled before the 3d. Consider the Mn2+ ion. Mn atom configuration: [Ar] 4s2 5d5. Mn2+ ion configuration: [Ar] 5d5. When the Mn2+ ion is formed, you might expect the 2 electrons to be removed from the 3d orbitals to explain the 2+. This would give a configuration of [Ar]4s2 3d3. The true configuration of Mn2+ is [Ar] 3d5. Where the 4s is always filled before the 3d orbital in Mn, electrons are removed from the 4s orbital in forming Mn2+ because the 3d orbital is more stable than the 4s orbital in transition metal ions. Therefore, when a cation is formed from an atom of a transition metal, electrons are always removed first from the highest energy s orbitals first and then from the d orbitals one at a time. As a result, Mn can form cations with charges of +2, +3, +4, +5, +6, and +7.

DO ASSIGNMENT #7 P. 27 Notes #1-10 ION FORMATION



NAME ________________________________ DATE __________________ PERIOD _________

ASSIGNMENT #7: ION FORMATION Explain how the following ions are formed. In your explanation, you must write the configuration of both the atom and the ion so I know what you are talking about when you refer to the configurations. Which sublevel electrons are gained or lost and explain fully why. 1. Na: +1 2. Ba: +2 3. Ga: +3 4. Sc: +3 5. Sulfur: −2. 6. Vanadium: +2, +3, +4, +5. 7. Iron (26) forms 2 cations, +2 and +3.

a) Explain how and why each is formed.

b) Which cation do you expect to be formed most often or will both be formed equally? Give reasons for your answer.

c) Why does Fe4+ not form? 8. How would you explain the ionic charges +2 and +4 of tin (#50)? 9. What ions do you expect bismuth (#83) to form? 10. Which bismuth ion would you expect to be more favorable?



TRENDS OR PATTERNS IN THE PERIODIC TABLE Many of the chemical properties of the elements can be understood in terms of their electron configurations. Because electrons fill atomic orbitals in a fairly regular fashion, it is not surprising that elements with similar electron configurations, such as sodium and potassium, behave similarly in many respects and that, in general, the properties of the elements exhibit observable trends. Chemist in the nineteenth century recognized periodic trends in the physical and chemical properties of elements long before quantum theory came onto the scene. The electron configurations show a periodic variation with increasing atomic number. Periodic means that there is a trend or pattern that becomes predictable. A periodic trend is: “I come to Chemistry class once a day.” The atomic size (radius) of an atom is a periodic trend that has a direct bearing on its ability and tendency to form ions. ATOMIC RADIUS (The size of the atom) AS YOU GO LEFT TO RIGHT ACROSS A PERIOD, THE RADIUS OF THE ATOM DECREASES. The reason for this is that electrons are being added to the same principal energy level. The nuclear charge increases and pulls the outer electrons closer. AS YOU GO FROM TOP TO BOTTOM IN A GROUP, THE RADIUS OF THE ATOM INCREASES. Electrons are being added to the next principal energy level. The nuclear charge increases but the outermost electrons are farther away. Nuclear attraction for farther electrons is shielded from the outer electrons by occupied levels between the nucleus and the outermost electrons. The inner level electrons shield the outer level electrons from the nuclear attraction (a book between a magnet and a nail).



ION SIZE VE There is no general trend with respect to ion size as you go awhen compared to the size of the metal atom, and anion size AN INCREASED NUMBER OF ELECTRONS ARE H A DECREASED NUMBER OF ELECTRONS ARE H

COMPARING CATIONS AND THE METAL ATOMS F POSITIVE IONS ARE SMALLER THAN THE NEUTRRADIUS!!) By removing one or more electrons from an atomthe same so the electron cloud shrinks. The cation is smaller on fewer electrons. A Li+ ion is smaller than the Li atom. The most metallic elemhave the highest metallic character. Where are the most metallic elements located? COMPARING ANIONS AND THE NON-METAL ATOM NEGATIVE IONS ARE LARGER THAN THEIR NEUTelectrons are being added. The nuclear charge remains the sadispersed. There is additional repulsion of electrons resultingelectron cloud. As a result, the anion is larger than its atom. Oones that gain electrons most easily and have the least metall

EXERCISES: 1. Arrange the following atoms in order of increasing radius: P, Si, N. _____________________________________ 2. Arrange the following atoms in order of decreasing radius: C, Li, Be. _____________________________________ 3. Arrange the following atoms in order of increasing radius: N, O, P, S. _____________________________________

RSUS ATOM SIZE.

cross and down the periodic table. There is a trend in cation size when compared to the size of the non-metal atom.

ELD MORE LOOSELY FORMING A LARGER ION. ELD MORE TIGHTLY FORMING A SMALLER ION.

ROM WHICH THEY ARE FORMED:

AL ATOMS FROM WHICH THEY ARE FORMED. (THINK reduces electron-electron repulsion but the nuclear charge remains

than the atom because there is the same number of protons pulling

ents are the ones that lose electrons most easily. They are said to

S FROM WHICH THEY ARE FORMED.

RAL ATOMS. (THINK RADIUS!!) When anions are formed, me but there are more electrons to pull on and the attraction is being from the additional electrons and this enlarges the size of the

2− is larger than the O atom. The most non-metallic elements are the ic character. Where are these elements located?




Questions: 1. Which has the larger radius:

a. sulfur atom or sulfur ion?

b. Rb+ ion or Sr2+ ion?

For each of the following pairs, indicate which one of

the two species is larger:

a. N3− or F−

b. Mg2+ or Ca2+

c. Fe2+ or Fe3+

Select the smaller ion in each of the following pairs:

a. K+, Li+

b. Au+, Au3+

c. P3−, N3−


IONIZATION ENERGY Ionization energy is the minimum energy (in kJ/mole) required to remove an electron from a gaseous atom in its ground state, i.e. it is the amount of energy in kilojoules needed to strip one mole of electrons from one mole of gaseous atoms (gaseous atoms because they are not influenced by other atoms around it). The amount of ionization energy used is a measure of how tightly the electron is held in the atom. The higher the ionization energy, the more difficult it is to remove the electron. TRENDS IN IONIZATION ENERGY IN THE PERIODIC TABLE First ionization energy is the energy required to remove the outermost electron from an atom. Think radius!! An atom with a smaller radius holds the electrons more tightly because they are closer to the nucleus. Consider the boron atom which has five electrons − two in an inner core (1s2) and three valence electrons (2s2 2p1). The five steps and their successive ionization energies I1 to I5 are:

B(g) B→ +(g) + e− I1 = 801 kJ/mol

B+(g) B→ 2+(g) + e− I2 = 2,427 kJ/mol

B2+(g) B→ 3+(g) + e− I3 = 3,660 kJ/mol

B3+(g) B→ 4+(g) + e− I4 = 25,025 kJ/mol

B4+(g) B→ 5+(g) + e− I5 = 32,822 kJ/mol Notice that each successive removal of an electron takes more energy. I4 is a lot higher than I3. This is because B3+ has an inert gas configuration and does not want to give up its stable configuration easily. Therefore, the 4th electron of boron is very difficult to remove.

EXERCISE: 1. Write equations showing the 1st ionization energy for oxygen.

2. Write and equation showing the 4th ionization energy for oxygen.

THE FIRST IONIZATION ENERGY INCREASES AS YOU GO FROM LEFT TO RIGHT ACROSS THE PERIODIC TABLE (RADIUS GETS SMALLER – MORE TIGHTLY HELD – HARDER TO PULL AWAY).

THE FIRST IONIZATION ENERGY INCREASES AS YOU GO FROM LEFT TO RIGHT ACROSS THE PERIODIC TABLE (RADIUS GETS SMALLER – MORE TIGHTLY HELD – HARDER TO PULL AWAY). THE FIRST IONIZATION ENERGY DECREASES AS YOU GO DOWN THE PERIODIC TABLE (LARGER RADIUS – ELECTRONS FARTHER FROM THE NUCLEUS – LESS TIGHTLY HELD).

THE FIRST IONIZATION ENERGY DECREASES AS YOU GO DOWN THE PERIODIC TABLE (LARGER RADIUS – ELECTRONS FARTHER FROM THE NUCLEUS – LESS TIGHTLY HELD). Metals are characterized by low ionization energy (very little energy needed to remove the electron). Nonmetals have high ionization energies (smaller radius – electrons are tightly held).

Metals are characterized by low ionization energy (very little energy needed to remove the electron). Nonmetals have high ionization energies (smaller radius – electrons are tightly held). Increased distance of the outer electrons from the nucleus and the shielding effect of the inner electrons tend to lower the ionization energy. Though it appears that the increased nuclear charge of an element with greater atomic number tends to increase ionization energy, the lowering tendency is greater. The number of electrons in the outermost sublevel is the same for all elements in a column or group.

Increased distance of the outer electrons from the nucleus and the shielding effect of the inner electrons tend to lower the ionization energy. Though it appears that the increased nuclear charge of an element with greater atomic number tends to increase ionization energy, the lowering tendency is greater. The number of electrons in the outermost sublevel is the same for all elements in a column or group. FACTORS AFFECTING IONIZATION ENERGY.FACTORS AFFECTING IONIZATION ENERGY.

1. Nuclear charge – The larger the nuclear charge, the greater the ionization energy. 2. Shielding effect – The greater the shielding effect the less the ionization energy. 3. Radius – The greater the distance between the nucleus and the outer electrons of an atom, the less the ionization energy. 4. Sublevel – An electron from a full or half-full sublevel requires additional energy to be removed.



EXCEPTIONS TO THE TREND


General trend – left to right, Ionization energy increases. (See chart). Exception #1: Be 899.5 kJ/mole, and B = 800.7 kJ/mole. I.E. should be higher for boron than for beryllium but it isn’t. Why? Be 1s2 2s2 B 1s2 2s2 2p1

Be has a full 2s subshell. It is stable and wants to stay that way. It is easier to remove the 2p electron from the B atom than an electron from the full 2s subshell of Be. Exception #2: N: 1402.4 kJ/mole, O: 1314.0 kJ/mole. The I.E. of O should be greater than N because of its smaller radius but it is not. Why? N 1s2 2s2 2p3 O 1s2 2s2 2p4 The ½ full subshell of N is stable so the 4th 2p (2p4)electron of O is easier to remove than the electron from nitrogen(2p3 – half-filled subshell) A useful generalization is that the noble gas structure is particularly stable and therefore difficult to “break”. As electrons are removed from an atom one by one there is a steady rise iionization energy until the structure of a nobgas is reached. Then there is a dramatic rise in ionization energy.

n le


MULTIPLE IONIZATION ENERGIES It is possible to measure other (second, third, and so on) ionization energies of an atom. These measurements provide the same

evidence for atomic structure as first ionization energies. Consider the aluminum atom 1s2 2s2 2p6 3s2 3p1

1st 2nd 3rd 4th 5th Ionization Energies for aluminum (kJ/mole)

577.4 1,816 2,744 11,600 14,800

Example: The second ionization energy of aluminum (1s2 2s2 2p6 3s2 3p1) is about 3 times as large as the first. The difference

can be explained by the fact that the first ionization energy removes a p electron and the second removes an s electron from a full s sublevel. The third ionization energy is about one and two thirds times as large as the second. The second and third electrons are in the same sublevel. Yet the third electron’s I.E. is greater because the nuclear charge remains constant as we remove electrons. As a result, the remaining electrons are more tightly held.

The 4th I.E. is about 4 times as large as the third. The jump in energy between the 3rd and 4th I.E. is so large due to the location of the 4th electron removed. The 4th electron would come from the full second energy level that is closer to the nucleus. The 2s2 2p6 level with eight electrons is stable. Thus, a large amount of energy will be required to remove that 4th electron.

ELECTRON AFFINITY The word “affinity” means have a want or love of something ex. “Butchie has an affinity for chocolate.” The attraction of an

atom for an electron to form a negative ion is called electron affinity. An atom with a high electron affinity loves to gain electrons and has a high negative energy value. Electron affinity values were obtained by atoms in the gaseous state gaining an electron to become a negative ion.

Ex. The electron affinity of fluorine is –328 kJ/mole. The electron affinity reaction is:

F(g) + 1e− F→ −(g) E.A. = −328 kJ/mole Other electron affinity reactions:

Al(g) + 1e− → Al−(g) E.A. = −42.6 kJ/mole When an electron is removed from an atom it takes energy (+ sign: endothermic) to remove that electron. When an atom forms

an anion by gaining an electron, that same amount of energy is given off therefore the sign for electron affinity is negative. The EA for fluorine is –328 kJ/mole; for iodine –295 kJ/mole. This means that fluorine has a higher electron affinity. The negative sign just means that more energy is given off when the fluorine anion if formed. The negative sign is not a mathematical sign that indicates a negative number. All the electron affinity values are positive. The negative sign indicates an exothermic reaction.

The same factors that affect ionization energy will also affect electron affinity. As electron affinity increases, an increase in ionization energy can be expected.

The greater the affinity of an atom for an electron, the more negative the value of EA. Metals have low electron affinity (they don’t want electrons). Nonmetals have a high electron affinity (they love and

want electrons) The halogens have the highest EA of all the elements. Atoms with a value of 0 have no affinity for electrons and will not form anions.



TRENDS IN THE PERIODIC TABLE: THINK RADIUS!!! Left to right: radius decrease, electron affinity increases. The electron can get closer to the nuclear charge and is pulled in more easily. Top to bottom, electron affinity decreases. The electron is farther from the nucleus and is not attracted very strongly to the nucleus which is farther away.

November 30 Test. 07/03

/2006



Examples: 1. How do you account for the exceptions in electron affinity of Be, N, and Ne in Period 2? Explain each fully. 2. Why are the electron affinities of the alkaline earth metals shown in the table 0? 3. Is it likely that Ar will form the anion Ar−? Explain. 4. Which element Al or Si is expected to have the less negative electron affinity value?

REVIEW OF TRENDS

Periodic properties are those that vary as you move across and down the periodic table. These properties include the atomic radius, electron affinity, ionization energy, nuclear charge, shielding effect, and electronegativity of the elements.

**Electronegativity will be studied in the next unit but it is included here so that you see that it follows the trends of the other periodic properties.

DO ASSIGNMENT #8 P. 36 notes 1-18 TRENDS


ASSIGNMENT #8 TRENDS 1. Arrange the following atoms in order of decreasing atomic radius: Na, Al, P, Cl, Mg. 2. Why is the radius of the lithium atom considerably larger than the radius of the hydrogen atom? 3. Indicate which one of the two species in each of the following pairs is smaller:

(a) Cl or Cl− (b) Na or Na+ (c) O2− or S2−

(d) Mg2+ or Al3+ (e) Au+ or Au3+

4. List the following ions in order of increasing ionic radius: N3−, Na+, F−, Mg2+, O2− 5. Define ionization energy.

(a) Ionization energy measurements are usually made when atoms are in the gaseous state. Why? (b) Why is the second ionization energy always greater than the first ionization energy for any element?

6. Use the third period of the periodic table as an example to illustrate the change in first ionization energies of the elements as we move from left to right. Explain the trend.

7. In general, ionization energy increases from left to right cross a given period. Aluminum, however, has a lower ionization energy than magnesium. Explain.

8. Two atoms have the electron configurations 1s2 2s2 2p6 and 1s2 2s2 2p6 3s1. The first ionization energy of one is 2080 kJ/mol, and that of the other is 496 kJ/mol. Match each ionization energy with one of the given electron configurations. Justify your choice.

9. Arrange the elements in each of the following groups in increasing order of the most positive electron affinity: (a) Li, Na, K (b) F, Cl, Br, I

10. Specify which of the following elements you would expect to have the greatest electron affinity: He, K, Co, S, Cl. 11. Considering their electron affinities, do you think it is possible for the alkali metals to form an anion like M−, where M

represents an alkali metal? 12. Explain why alkali metals have a greater affinity for electrons than alkaline earth metals. 13. State whether each of the following properties of representative elements generally increases or decreases

(a) From left to right across a period and (b) From top to bottom within a group, The properties are: metallic character, atomic size, ionization energy.

14. Write equations representing the following processes: (a) The electron affinity of S−. (b) The third ionization energy of titanium.

(c) The electron affinity of Mg2+. (d) The ionization energy of O2−.

15. March each of the elements on the right with its description on the left: (a) A dark-red liquid (b) A colorless gas that burns in oxygen gas. (c) A reactive metal that attacks water. (d) A shiny metal that is used in jewelry. (e) A totally inert gas.

Calcium (Ca) Gold (Au) Hydrogen (H2) Argon (Ar) Bromine (Br2)

16. Arrange the following species in isoelectronic pairs: O+, Ar, S2−, Ne, Zn, Cs+, N3−, As3+, N, Xe

17. You are given four substances: a fuming red liquid, a dark metallic-looking solid, a pale-yellow gas, and a yellow-green gas that attacks glass. You are told that these substances are the first four members of group 7A. Name each one.

18. The H− ion and the He atom have two 1s electrons each. Which of the two species is larger? Explain. THE NEXT 4 PAGES CONTAIN GOOD TEST REVIEW QUESTIONS. YOU CANNOT DO THEM ALL BUT TRY SOME.



TEST REVIEW SET #1 (Answers at end)




TET REVIEW PROBLEM SET #2


TEST REVIEW PROBLEM SET #3 (Zumdahl) Omit #22 and 23.


unit 6 ch 7 quantum theory and the electronic structure … · chem. 1 hons. unit 6 ch. 7 quantum...

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