Atomic Structure and Periodic Trends

Atomic Structure and Periodic Trends

ATOMIC STRUCTURE AND PERIODIC TRENDS Chapter 7 ELECTROMAGNETIC RADIATION CHARACTERISTICS wavelength () ) ) (lambda) length between 2 successive crests. frequency () ) (nu in chemistry; f in physicseither is OK),

number of cycles per second that pass a certain point in space (Hz cycles per second) amplitude maximum height of a wave as measured from the axis of propagation. nodes points of zero amplitude (equilibrium position); always

occur at /2 for sinusoidal waves velocity speed of the wave, equals ) speed of light, c 2.99792458 108 m/s; we will use 3.0 x 108 m/ s (on reference sheet). All electromagnetic radiation travels at this speed. If light is involved, it travels at 3 10 8 m/s. Notice that and are inversely proportional. When one is large, the other is small. CHARACTERISTICS

PRACTICE ONE Frequency of Electromagnetic Radiation The brilliant red colors seen in fireworks are due to the emission of light with wavelengths around 650 nm when strontium salts such as Sr(NO3)2 and SrCO3 are heated. Calculate the frequency of red light of wavelength 6.50 102 nm. You must be able to convert nm to m because the speed of light constant is in meters per second. 1 nm = 1 x10-9m

THE NATURE OF MATTER In the early 20th century, certain experiments showed that matter could not absorb or emit any quantity of energy this did not hold with the generally accepted notion.

1900 - Max Planck solved the problem. He made an incredible assumption: There is a minimum amount of energy that can be gained or lost by an atom, and all energy gained or lost must be some integer multiple, n, of that minimum. There is no such thing as a transfer of energy in fractions of quanta, only in whole numbers of quanta. CALCUALTING ENERGY Energy = Energy = n() h)

where h is a proportionality constant, Planck's constant, h = 6.6260755 10-34 joule seconds. We will use 6.626 x 10-34 Js (on reference sheet). This is the lowest frequency that can be absorbed or emitted by the atom, and the minimum energy change, h, is called a quantum of energy. Think of it as a packet of energy equal to h. PRACTICE TWO

The Energy of a Photon The blue color in fireworks is often achieved by heating copper(I) chloride to about 1200C. Then the compound emits blue light having a wavelength of 450 nm. What is the increment of energy (the quantum) that is emitted at 4.50 102 nm by CuCl? Planck's formula requires frequency so you must find that first before solving for energy. THE PHOTOELECTRIC EFFECT

In 1900 Albert Einstein was working as a clerk in the patent office in Bern, Switzerland and he proposed that electromagnetic, EM, radiation itself was quantized. He proposed that EM radiation could be viewed as a stream of particles called photons. photoelectric effect light bombards the surface of a metal and electrons are ejected.

A Demonstration: http://www.youtube.com/watch?v=0qKrOF-gJZ4 Stop at 3:37 PHOTOELECTRIC EFFECT frequency a minimum must be met or alas, no action! Once minimum is met, intensity increases the rate of ejection.

photon massless particles of light. Use Plancks formula and constant Ephoton = h = CALCULATING MASS Rearrange Einsteins famous formula: m=E c2 Therefore

c2 m = E = hc/ = h c2 cc So mass = Plancks constant divided by wavelength and the speed of light. SUMMARY

Energy is quantized. It can occur only in discrete units called quanta [h]. EM radiation exhibits wave and particle properties. This phenomenon is known as the dual nature of light. MORE Since light which was thought to be wavelike now has certain characteristics of particulate matter, is the converse also true?

French physicist Louis de Broglie (1923): If m = h/c, cc, substitute v [velocity] for c for any object NOT traveling at the speed of light, then rearrange and solve for lambda (wavelength). This is called the de Broglie equation: = PRACTICE THREE Calculations of Wavelength Compare the wavelength for an electron (mass = 9.11 10-31 kg) traveling at a speed of 1.0 107 m/s with that for

a ball (mass = 0.10 kg) traveling at 35 m/s. FUN FACTS The more massive the object, the smaller its associated wavelength and vice versa. Davisson and Germer @ Bell labs found that a beam of electrons was diffracted like light waves by the atoms of a thin sheet of metal foil and that de Broglie's relation was followed quantitatively.

ANY moving particle has an associated wavelength. We now know that Energy is really a form of matter, and ALL matter shows the same types of properties. That is, all matter exhibits both particulate and wave properties. ATOMIC LINE () EMISSION) SPECTRA emission spectrum the spectrum of bright lines, bands, or continuous radiation that is provided by a specific emitting substance as it loses energy and returns

to its ground state OR the collection of frequencies of light given off by an "excited" electron absorption spectrum a graph or display relating how a substance absorbs electromagnetic radiation as a function of wavelength line spectrum isolate a thin beam by passing through a slit then a prism or a diffraction grating which sorts into discrete frequencies or lines HYDROGEN ATOMIC EMISSION AND

ABSORPTION SPECTRA HYDROGEN ATOMIC EMISSION AND ABSORPTION SPECTRA HYDROGEN ATOMIC EMISSION SPECTRA Niels Bohr connected spectra and the quantum ideas of Einstein and Planck: the single electron of the hydrogen

atom could occupy only certain energy states, stationary states An electron in an atom would remain in its lowest E state unless otherwise disturbed. Energy is absorbed or emitted by a change from this ground state. AN EXPLANATION an electron with n = 1 has the most negative energy and

is thus the most strongly attracted to the positive nucleus. [Higher states have less negative values and are not as strongly attracted to the positive nucleus.] ground state--n = 1 for hydrogen Ephoton = h = AN EXPLANTION To move from ground to n = 2 the electron/atom must

absorb no more or no less than a prescribed amount of energy What goes up must come down. Energy absorbed must eventually be emitted. The origin or atomic line spectra is the movement of electrons between quantized energy states. IF an electron moves from higher to lower E states, a photon is emitted and an emission line is observed. CALCULATING ENERGY OF ENERGY LEVELS IN

HYDROGEN ATOM Bohrs equation for calculating the energy of the E levels available to the electron in the hydrogen atom: E = -2.178 x 10-18J Z 2 n2 where n is an integer (larger n means larger orbit radius, farther from nucleus), Z is the nuclear charge, the (-) sign simply means that the E

of the electron bound to the nucleus is lower than it would be if the electron were at an infinite distance [n = ] from the nucleus where there is NO interaction and the energy is zero. CALCULATING ENERGY OF ENERGY LEVELS IN HYDROGEN ATOM This equation can be used to calculate the change in energy of an electron when it changes orbits. This happens when electrons gain energy (get excited) and

jump to a higher energy level or lose this energy and drop to a lower energy level. E = difference between the energies for the two levels. If E is (-), energy was lost. If E is (+), then energy was gained. CALCULATING ENERGY OF ENERGY LEVELS IN HYDROGEN ATOM You

can also calculate the wavelength of this emitted photon using the equation: = hc E Use the absolute value of E. Energy flow is indicated by saying the photon was emitted from the atom.

PRACTICE FOUR Energy Quantization in Hydrogen Calculate the energy required to excite the hydrogen electron from level n = 1 to level n = 2. Also calculate the wavelength of light that must be absorbed by a hydrogen atom in its ground state to reach this excited state. Two important points about Bohr's Model 1. The model correctly fits the quantized energy

levels of the hydrogen atom and postulates only certain allowed circular orbits for the electron. 2. As the electron becomes more tightly bound, its energy becomes more negative relative to the zero energy reference state (corresponding to the electron being at infinite distance from the nucleus). As the electron is brought closer to the nucleus, energy is released into the system. Two shortcomings about Bohr's Model

1. Bohr's model does not work for atoms other than hydrogen. 2. Electron's do not move in circular orbits. PRACTICE FIVE Calculate the energy required to remove the electron from a hydrogen atom in its ground state. Removing the electron from a hydrogen in its ground state corresponds to taking the electron from ninitial = 1 to nfinal = .

THE QUANTUM MECHANICAL MODEL OF THE ATOM Bohr model would not work. de Broglie opened a can of worms among physicists by suggesting the electron had wave properties because that meant that the electron has dual properties. Werner Heisenberg and Max Born provided the uncertainty principle - if you want to define the momentum of an electron,

then you have to forego knowledge of its exact position at the time of the measurement. Max Born suggested: if we choose to know the energy of an electron in an atom with only a small uncertainty, then we must accept a correspondingly large uncertainty about its position in the space about the atom's nucleus. This means that we can only calculate the probability of finding an electron within a given space. THE WAVE MECHANICAL VIEW OF THE ATOM

Schrodinger equation: solutions are called wave functions - chemically important. The electron is characterized as a matter-wave sort of standing waves - only certain allowed wave functions (symbolized by the Greek letter, , pronounced sigh) Each for the electron in the H atom corresponds to an allowed energy. For each integer, n, there is an atomic state characterized by its own and energy En.

Basically: the energy of electrons is quantized. THE WAVE MECHANICAL VIEW OF THE ATOM the square of gives the intensity of the electron wave or the probability of finding the electron at the point P in space about the nucleus. electron density map, electron density and electron probability ALL mean the same thing!

matter-waves for allowed energy states are also called ORBITALS. THE WAVE MECHANICAL VIEW OF THE ATOM To solve Schrodinger's equation in a 3-dimensional world, we need the quantum numbers n, , and m The amplitude of the electron wave at a point depends on the distance of the point from the nucleus.

Imagine that the space around a Hydrogen nucleus is made up of a series of thin shells like the layers of an onion, but these shells as squishy. THE WAVE MECHANICAL VIEW OF THE ATOM Plot the total probability of finding the electron in each shell versus the distance from the nucleus and you get the radial probability graph you see in (b). Darker color

means more probability of finding the ethere. THE WAVE MECHANICAL VIEW OF THE ATOM The maximum in the curve occurs because of two opposing effects. the probability of finding an electron is greatest near the nucleus [electrons just cant resist the attraction of a

proton!]. BUT - the volume of the spherical shell increases with distance from the nucleus, so we are summing more positions of possibility, so the TOTAL probability increases to a certain radius and then decreases as the electron probability at EACH position becomes very small. QUANTUM NUMBERS n principal energy level 1 to infinity. Determines the total energy of the electron.

It indicates the most probable distance of the electron from the nucleus within 90%. A measure of the orbital size or diameter. 2n2 electrons may be assigned to a shell. Translation: the energy level that electron is in. If its a 3s electron, n = 3, if its a 4d electron, n = 4, etc. QUANTUM NUMBERS - angular momentum 0,1,2,....(n-1) electrons within each shell may be grouped into subshells

or sublevels each characterized by its certain wave shape. Each is a different orbital shape or orbital type. n limits the values of to no larger than n-1. Thus, the number of possibilities for is equal to n. s,p,d,f sublevels 0,1,2,3 -values respectively spdf stand for - Sharp, principle, diffuse, fundamental. Translation: 3 sublevels for 3rd energy level (s, p, d), 4 for 4th energy level (s, p, d, f), etc.

QUANTUM NUMBERS m, magnetic this designates the orientation of an orbital in a given sublevel s = 1 sublevel, p = three sublevels, d = five sublevels, f = seven sublevels. Assign the slots in orbital diagrams with zero on the middle blank and then - through zero to +; that means that the range of orbitals is from + to -

- The down arrow in the 3p, -1 slot is the last electron placed, a valence electron. So its quantum numbers are 3 (n), 1 (), -1 (m) PRACTICE SIX For principal quantum level n = 5, determine the number of allowed subshells (different values of ), and give the designation of each.

2014 AP Chemistry test The 2014 AP Chemistry test will not require you to determine the quantum numbers for any given electron, but you still need to know what they are and how they are found. QUANTUM NUMBERS AND ORBITALS ORBITALS SHAPES AND ENERGIES s sharp

Spherical The size increases with n. The nodes you see at right represent ZERO probability of finding the electron in that region of space. The number of nodes equals n-1 for s orbitals. ORBITALS SHAPES AND ENERGIES

p principle has one plane that slices through the nucleus and divides the region of electron density into 2 halves. Nodal plane -the electron can never be found there!! 3 orientations: px, py, and pz. ORBITALS SHAPES AND ENERGIES d diffuse has two nodal planes slicing through the nucleus to create

four sections 5 orbitals. ORBITALS SHAPES AND ENERGIES f fundamental has three nodal planes slicing through the nucleus eight sections 7 orbitals ELECTRON SPIN AND THE PAULI PRINCIPLE

If electrons interact with a magnetic field, there must be one more concept to explain the behavior of electrons in atoms. ms - the 4th quantum number; accounts for the reaction of electrons in a magnetic field QUANTUM NUMBERS

ms, electron spin Each electron has a magnetic field with N and S poles Electron spin is quantized such that, in an external magnetic field, only two orientations of the electron magnet and its spin are possible + and MAGNETISM magnetite - Fe3O4, natural magnetic oxide of iron 1600 -William Gilbert concluded the earth is also a large

spherical magnet with magnetic south at the geographic North Pole. NEVER FORGET: opposites attract & likes repel whether its opposite poles of a magnet or opposite electrical charges PARAMAGNETISM AND UNPAIRED ELECTRONS diamagnetic - not magnetic [magnetism dies]; in fact they

are slightly repelled by a magnetic field. Occurs when all electrons are PAIRED. paramagnetic - attracted to a magnetic field; lose their magnetism when removed from the magnetic field; has one or more UNPAIRED electrons. ferromagnetic - retain magnetism upon introduction to and then removal from a magnetic field All of these are explained by electron spins

PARAMAGNETISM AND UNPAIRED ELECTRONS H is paramagnetic; He is diamagnetic, WHY? H ____ H has one unpaired electron He ____ He has NO unpaired electrons; all spins offset and cancel each other out

FERROMAGNETIC clusters of atoms have their unpaired electrons aligned within a cluster, clusters are more or less aligned and substance acts as a magnet. Don't drop it!! When all of the domains, represented by the arrows, are aligned, it behaves as a magnet. If you drop it: The domains go in all different

directions and it no longer operates as a magnet. PAULIS EXCLUSION PRINCIPLE No two electrons in an atom can have the same set of four quantum numbers. This means no atomic orbital can contain more than 2 electrons When two electrons occupy the same orbital, they must be of opposite spin!!

ATOMIC ORBITAL ENERGIES AND ELECTRON ASSIGNMENTS Order of orbital energies - the value of n determines the

energy of an electron. Use the diagonal rule. subshell orbitals contract toward the nucleus as the value of increases. contraction is partially into the volume of space occupied by the core electrons. the energy of the electrons in these subshells is raised due to the repulsions between the subshell electrons and the core electrons.

a subshell's energy rises as its quantum number increases when inner electrons are present Order of Orbital Assignments each electron is lazy and occupies the lowest energy space available based on the assumption inner electrons have no effect on which orbitals are assigned to outer or valence electrons

You should always use the energy order for filling when doing electron configuration though you may see it written in quantum number order. ORBITAL ENERGY ORDER DIAGONAL RULE PERIODIC TABLE

ELECTRON CONFIGURATION Need energy level, orbital, and number of electrons in orbital Number of electrons in orbital Principle quantum number

2s 2 orbital Maximum electrons allowed: s = 2e-, p = 6e-, d = 10e-, f = 14e- ORBITAL DIAGRAMS

Using the electron configuration, draw arrows for the electrons in each orbital. Use lines or boxes to designate each orbital. ORBITAL DIAGRAMS http://intro.chem.okstate.edu/WorkshopFolder/

Electronconf.html Click on Electron Configuration Animation. Once the animation comes up, click on the screen to advance from Hydrogen on up by atomic number. HUNDS RULE The most stable arrangement of electrons is that with the

maximum number of unpaired electrons WHY?? it minimizes electron-electron repulsions (everyone gets their own room) all single electrons also have parallel spins to reduce electron-electron repulsions. When two electrons occupy the same orbital they must have opposite spins (Pauli exclusion principle); this also helps to minimize electron-electron repulsions PRACTICE SEVEN

Write the electron configuration for the following elements. The first one is done for you. S 1s22s22p63s23p4 Cd La Hf Ra Ac PRACTICE SEVEN

Write the electron configuration for the following elements. The first one is done for you. Cd 1s22s22p63s23p64s23d104p65s24d10 La 1s22s22p63s23p64s23d104p65s24d105p66s25d1 Hf 1s22s22p63s23p64s23d104p65s24d105p66s24f145d2 Ra 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s2 Ac 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s26d1 PRACTICE EIGHT Write the orbital diagrams for the following elements

Cd La Hf Ra Ac

PRACTICE EIGHT Write the orbital diagrams for the following elements Cd (Z = 48) / / / / / / / / / La (Z = 57) / / / / / / / / / / / Hf (Z = 72) / / / / / / / / / / / / Ra (Z = 88)

/ / / / / / / / / / / / / / / Ac (Z = 89) / / / / / / / / / / / / / / / ELECTRON CONFIGURATION EXCEPTIONS and ION ORBITAL ENERGIES Elements in the transition and inner transition areas do

not always follow the appropriate filling. This is sometimes referred to as the dfs overlay. Exceptions do not occur in ion configurations since the valence electrons are the first to go. The shell energy ranges separate more widely as electrons are removed. Transition metals with +2 or higher have no ns electrons Fe+2 is paramagnetic to the extent of 4 unpaired electrons and Fe+3 is paramagnetic to the extent of 5 unpaired electrons

ELECTRON CONFIGURATION EXPLANING THE EXCEPTIONS The sublevels had differences in energies long ago. The increasing order of energies: s < p < d < f < g Now you have to be able to EXPLAIN it on the AP test. Throughout this discussion keep some fundamental scientific principles close at hand electrons repel each other

electrons are negative and thus, attracted by the positive nucleus forces () attractive & repulsive) dissipate with increasing distance. EXPLAINING THE EXCEPTIONS penetrates closest to the nucleus

mighty close to the nucleus; energy is very low EXPLAINING THE EXCEPTIONS MORAL: The greater the penetration, the less energy that orbital has. Since you already knew the order with respect to energy, s < p

< d < f , the degree of penetration is ss penetrate most and fs penetrate least. THE HISTORY OF THE PERIODIC TABLE 1800ish 1864 1870 1913

Johann Dobereiner, triads John Newlands octaves Dmitrii Mendeleev & Julius Lothar Meyer, by mass Mosley, by number of protons. BASIC FAMILY FACTS Group 1, ALKALI METALS the most reactive metal family; must be stored under oil; react with water violently!

Group 2, ALKALINE EARTH METALS except for Be(OH)2, the metal hydroxides formed by this group provide basic solutions in water; pastes of these used in batteries Group 16, CHALCOGENS many found combined with metal ores Group 17, HALOGENS known as the salt-formers ; used in modern lighting BASIC FAMILY FACTS Groups 18, NOBLE GASES known for their disinterest in other

elements; once thought to never react; neon used to make bright red signs Groups 3-12, TRANSITION METALS fill the d orbitals; Anomalies occur at Chromium and Copper to minimize electron/c, electron repulsions - it is about lowering energy by minimizing electron/electron repulsions. RARE EARTH ELEMENTS () Inner Transition) fill the f sublevels. Lanthanides and Actinides. These sometimes put an electron in d [just one or two electrons] before filling f. This is that dsf overlay referred to earlierthe energies of the sublevels are

very similar. PERIODIC TRENDS This is an important sectionthere is almost always an essay involving this topic on the AP exam.

There are several arguments you will evoke to EXPLAIN a periodic trend. Remember opposites attract and likes repel. The trick is learning which argument to use when explaining a certain trend! A TREND IS NOT AN EXPLANATION THE ARGUMENTS

1. Effective nuclear charge, Zeff essentially equal to the group number. Think of the Alkali Metals (Group 1) having a Zeff of one while the Halogens (Group 7) have a Zeff of 7. The idea is that the higher the Zeff, the more positive the nucleus, the more attractive force there is emanating from the nucleus drawing electrons in or holding them in place. Relate this to ENERGY whenever possible.

THE ARGUMENTS 2. Distance Attractive forces dissipate with increased distance. Distant electrons are held loosely and thus easily removed. Relate this to ENERGY whenever possible. 3. Shielding Electrons in the core effectively shield the nucleus

attractive force for the valence electrons. Use this ONLY when going up and down the table, NOT across. There is ineffective shielding within a sublevel or energy level. Relate this to ENERGY whenever possible. THE ARGUMENTS 4. Minimize electron/electron repulsions This puts the atom at a lower energy state and makes it more stable. Relate this to ENERGY whenever possible.

THE ARGUMENTS Trends across a period Effective Nuclear Charge, Zeff Minimize electron/electron repulsions Trends down a group Distance Shielding

ATOMIC RADIUS Often called covalent atomic radii because of how they are determined. Use diatomic molecules and determine radius then react with other atoms to determine the radius of those atoms TREND: Atomic radii decreases () ) moving across a period AND increases () ) moving down

a group (family). ATOMIC RADIUS WHY across? The effective nuclear charge (Zeff) increases (more protons for the same number of energy levels) as we move from left to right across the periodic table, so the nucleus has a greater positive charge, thus the entire electron cloud is more strongly attracted and shrinks. This shrinks the electron cloud untilthe point

at which electron/electron repulsions overcome the nuclear attraction and stop the contraction of the electron cloud. ATOMIC RADIUS WHY down? The principal level, n, determines the size of an atomadd another principal level and the atoms get MUCH larger radii. As we move down a family, the attractive force the nucleus exerts on the valence

electrons dissipates. Shielding is only a valid argument when comparing elements from period to period (up and down the table) since shielding is incomplete within a perioduse this argument with extreme caution! It should NOT be your favorite! IONIZATION ENERGY, IE Energy required to remove an electron from the atom IN THE GAS PHASE. Costs Energy.

Removing each subsequent electron requires more energy: Second IE, Third IE, etc. Some subsequent IEs cost more E than others!! A HUGE energy price is paid if the subsequent removal of electrons is from another sublevel or another principal E level (core). TREND: down a family and across a period IONIZATION ENERGY, IE

IONIZATION ENERGY, IE WHY down a family increased distance from the nucleus and increased shielding by full principal E levels means it requires less E to remove an electron WHY across a period due to increasing Zeff. The higher the Zeff, the stronger the nucleus attracts valence electrons, the more energy required to remove a valence electron.

IONIZATION ENERGY, IE EXCEPTIONS At a half-filled or filled sublevel. When electron pairing first occurs within an orbital, there is an increase in electron/electron repulsions which makes it require less energy [easier] to remove an electron thus the IE drops. From p3 to p4. It requires less energy to remove an electron from oxygens valence IN SPITE OF AN INCREASING Zeff because oxygens p4 electron is the first to pair within the orbital thus experiencing

increased repulsion. The increased repulsion lowers the amount of energy required to remove the newly paired electron! From any s2 to p1. This is also IN SPITE OF AN INCREASING Zeff. This drop in the energy required to remove a valence electron is due to the fact that you are removing a p electron rather than an s electron. The ps are less tightly held BECAUSE they do not penetrate the electron cloud toward the nucleus as well as an s electron. The general trend is that s is held most tightly since it penetrates more then p, d and f ELECTRON AFFINITY

An affinity or liking for electrons force feeding an element an electron. Energy associated with the addition of an electron to a gaseous atom: X(g) + e X(g) If the addition of an electron is exothermic, then youll see a negative sign on the energy change and the converse is also true. The more negative the quantity, the more E is released.

TREND: down a family and across a period ELECTRON AFFINITY ELECTRON AFFIINITY WHY down a family becomes less negative, a.k.a. more positive, giving off less energy due to increased distance from the nucleus with each increasing principal E level. The nucleus is farther from the valence level and

more shielded. WHY across a period becomes more negative, giving off more energy; the increasing Zeff more strongly attracts the electron. The interactions of electron/electron repulsions causes exceptions. ELECTRON AFFINITY ELECTRON AFFIINITY

EXCEPTIONS First the lines on the diagram above connect adjacent elements. The absence of a line indicates missing elements whose atoms do not add an electron exothermically and thus do not form stable isolated anions remember these are all 1 ions at this point. An anomaly No N1 yet there is a C-1 this is due to their electron/electron repulsions compared to their electron configurations. N is p3 while C is p2.

ELECTRON AFFIINITY Exceptions O-2 doesnt exist in isolated form for the same reason. It is p4, so adding the first electron causes a subsequent pairing BUT it has a greater Zeff than N, so it can form O 1 . BUT adding the second electron fills the ps and that increased e-/c, e- repulsion overpowers the Zeff of oxygen. F has really strong e-/c, e- repulsion since the p orbitals are really small in the second level therefore repulsions are high. In subsequent halogen orbitals, it is not as

noticeable. IONIC RADII TREND: Cations shrink big time since the nucleus is now attracting fewer electrons; Anions expand since the nucleus is now attracting MORE electrons than there are protons AND there is enhanced electron/electron repulsion to boot. Isoelectronic ions containing the same number of

electrons - consider the # of protons to determine size. Oxide vs. Fluoride. Fluoride has one more proton which further attracts the electron cloud, so it is smaller. O-2 versus F-1 IONIC RADII METALS IONIC RADII

NONMETALS ELECTRONEGATIVITY, Eneg The ability of an atom IN A MOLECULE, participating in a BOND, to attract shared electrons to itself. Higher electronegativity means it wins the tug of war for the electron(s). TREND: across a period, down a group. Why is F the most? Highest Zeff and smallest so that the

nucleus is closest to the valence e Why is Fr the least? Lowest Zeff and largest so that the nucleus is farthest from the valence e-. ELECTRONEGATIVITY PRACTICE NINE Trends in Ionization Energies The first ionization energy for phosphorus is 1060 kJ/mol, and that for sulfur is 1005 kJ/mol. Why?

PRACTICE TEN Ionization Energies Consider atoms with the following electron configurations: a. 1s22s22p6 b. 1s22s22p63s1 c. 1s22s22p63s2 Identify each atom. Which atom has the largest first ionization energy, and which one has the smallest second ionization energy? Explain your choices.

PRACTICE ELEVEN Trends in Radii Predict the trend in radius for the following ions: Be2+, Mg2+, Ca2+, and Sr2+. Justify your answer.

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