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Transcript
Chris Khan 2007
 Standard Enthalpy of Formation (ΔHf°) is defined as the heat change that results when 1 mol of a compound is formed from its
elements at 1 atm, which is standard state. Standard enthalpies of formation can be found on a chart and can be used to
determine the Standard enthalpy of reaction (ΔHrxn°), which is the enthalpy of a reaction carried out at 1 atm.
 Consider: aA + bB  cC + dD. ΔHrxn° = [cΔHf°(C) dΔHf° (D)] – [aΔHf° (A) + bΔHf° (B)]. This is just saying that the enthalpy
of the reaction is the enthalpy of the products minus the enthalpy of the reactants.
 Stable allotropic forms of an element have zero enthalpies.
 ΔHf° can be determined by Hess’ Law, which states that when reactants are converted to products, the change in enthalpy is the
same whether the reaction takes place in one step or a series of steps. In other words, we can break down the reaction to calculate
enthalpies of formation for each part. (i.e. Calculate the standard enthalpy of formation of C 2H2 from its elements: 2C + H2 
C2H2. First, we look at each equation: C + O2  CO2, H2 + ½ O2  H2O. 2C2H2 + 5O2  4CO2 + 2H2O where the ΔHrxn°s are –
393.5kJ/mol, -285.8kJ/mol and –2598.8kJ/mol, respectively. First, we need to know what we want, and what that is, is 2C’s and
an H2 on the reactants side and a C2H2 on the products side. So, balance equations by multiplying by coefficients throughout (also
the enthalpies!) to get what you want. Make sure the product you want is on the product side. If it is not, just reverse the reaction
and reverse the sign of the enthalpy so the product you want (C 2H2) is on the products side. Multiply the first equation by 2 to get
2C + 2O2  2CO with a ΔHrxn° of –787.0kJ/mol (after x2) and multiply the last one by ½ to get only 1 mol of C 2H2. BUT, it is
on the reactants side, so after halving the ΔHrxn°, reverse the reactants and the products and change the sign of the ΔH rxn° so you
get 2CO2 + H2O  C2H2 + 2 ½ O2. Add the two new equations and the untouched one to get that, after canceling spectators, 2C
+ H2  C2H2 with a ΔHrxn° of 226.6kJ/mol.) (i.e. Calculate kJ released per gram of B 5H9 reacted with O2 when the balanced
equation is 2B5H9 + 12O2  5B2O3 + 9H2O. Use 73.2kJ/mol for B5H9. Use the chart to find ΔHrxn°s and plug in the values to
get: ΔHrxn° = [5 ΔHf° (B2O3) + 9 ΔHf° (H2O)] – [2 ΔHf° (B5H9) + 12 ΔHf° (O2)]. This comes out to –9036.6kJ/mol after solving
algebraically but we are not finished! We divide this by 2 because the B 5H9 has a coefficient of 2 in the equation. We then
multiply by 1 mol / 63.12g B5H9 to get that the heat released per gram of B 5H9 is –71.58kJ.)
 Enthalpy of Solution (ΔHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of
solvent. ΔHsoln = Hsoln – Hcomponents. The energy required to completely separate one mole of a solid ionic compound into gaseous
ions is called lattice energy (U). The enthalpy change associated with the hydration process is called the heat of hydration
(ΔHhydr). The heat of dilution is the heat change associated with the dilution process. If a process is endothermic and the
solution is diluted, more heat will be absorbed by the same solution from the surroundings.
 Max Planck discovered that atoms and molecules emit energy only in certain discrete quantities, or quanta. A wave is a
vibrating disturbance by which energy is transmitted. Waves are characterized by length and height and by the number of waves
that pass through a certain point in one second. Wavelength (𝛌) is the distance between identical points on successive waves. It
is usually expressed in Hertz (Hz), which are equal to cycles per second. Frequency (𝝂) is the number of waves that pass through
a particular point in 1 second. Amplitude is the vertical distance from the midline of a wave to the peak or trough. Speed (u) =
Wavelength (𝛌) x Frequency (𝝂). (i.e. What is the speed of a wave with a wavelength of 17.4 and a frequency of 87.4 Hz? u =
𝛌𝝂 = 17.4 cm x 84.7 Hz = 17.4 cm x 87.4/s = 1.52 x 103 cm/s.)
 Possible Essay: Max Karl Ernst Ludwig Planck developed the concept of the quantum, or fundamental increment of energy, basic
to quantum mechanics, and a cornerstone of modern physics. Planck began studying black body radiation in 1897 and discovered
that at very short wavelengths it did not obey the distribution laws given by Wilhelm Wien. In trying to understand this spectrum
using Wien’s laws, Planek ran smack into an "ultraviolet catastrophe". A black body would give off an infinite amount of energy
at the ultraviolet end of the spectrum. That was impossible. After working on this problem unsuccessfully for several years. he
took a fateful step in late 1900 that he eventually called "an act of desperation." Planck cautiously proposed that, contrary to
classical wave theory, matter emits and absorbs radiation in tiny discrete packets or bundles called "quanta" — not continuously,
as everyone had assumed. Initially Planck was not comfortable with this explanation and fully expected the idea to be disproved
by further research. Instead quantum theory, which gained Planck the Nobel Prize for physics in 1918, was used by Albert
Einstein to explain (1905) the photoelectric effect and by Niels Bohr to propose (1913) a model of the atom with quantified
electronic states; the theory was later developed into quantum mechanics. Understanding the implications of quantum mechanics
lead directly to the development of atomic energy and the semiconductor industry. Planck mastered every aspect of physics from
thermodynamics and electrodynamics to relativity and also wrote extensively on the philosophy of science. 1
 James Clerk Maxwell proposed that visible light consists of electromagnetic waves. An electromagnetic wave (speed of 3.00 x
108 m/s or 186,000 mi/s in a vacuum) has an electric field component and a magnetic field component. They have same speed
and travel in perpendicular planes. Electromagnetic radiation is the emission and transmission of energy in the form of
electromagnetic waves. (i.e. What is the frequency of a radiation with a wavelength of 522 nm? 𝝂 = c / 𝛌 and 𝛌 = 522 nm x 1 x 109
m / 1 nm = 522 x 10-9 m = 5.22 x 10-7 m. c = 3.00 x 108 (the speed of light). 𝝂 = c / 𝛌 = 3.00 x 108 / 5.22 x 10-7 = 5.75 x 1014/s =
5.75 x 1014 Hz.)
 Energy comes in packets. With wave nature, 1) amplitude  brightness and 2) wavelength  color. In particle nature, the
number of particles  brightness of wave and the amount of energy in quantum  color. Light is a stream of particles. Brighter
1
http://www.electro-optical.com/bb_rad/mplanck.htm
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light ejects more electrons but does not change energy. Light on  photons hit metal and collide with electrons and let them fly
off. Higher energy  harder collision. Brighter light  more electron collisions.
When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths. Classical physics assumed that
atoms and molecules could emit (or absorb) any arbitrary amount of radiant energy. Planck said that atoms and molecules could
emit (or absorb) energy in discrete qualities, like small packages or bundles. Planck gave the name quantum to the smallest
quantity of energy that could be emitted (or absorbed) in the form of electromagnetic radiation. The energy (E) of a single
quantum of energy is given by E = h𝝂 where h is given as Planck’s constant, which is 6.63 x 10 -34 J • s. According to quantum
theory, energy is always emitted in multiples of hv.
Albert Einstein solved the photoelectric effect, in which electrons are ejected from the surface of certain metals exposed to light
of at least a certain minimum frequency, called the threshold frequency. Higher threshold frequency  electrons move faster or
have more kinetic energy. Below threshold frequency, kinetic energy = 0. The number of e- ejected was proportional to the
brightness of the light, but the energies of the electrons were not. Einstein said that a beam of light is really a stream of particles,
called photons, each of which possesses energy E where E = hv where v is the frequency of the light. (i.e. Calculate the energy of
a photon with a 5.00 x 104 nm wavelength (infrared) and a photon with a wavelength of 5.00 x 10 -2 nm (x-ray). Since E = hv and
v = c/𝛌, E = hc/𝛌 = (6.63 x 10-34 J • S)(3.00 x 108 m/s) / (5.00 x 104 nm)(1 x 10-9 m / 1 nm) = 3.98 x 10-21 J. For the second one,
we can show that the energy of the photon that has a wavelength of 5.00 x 10 -2 nm is 3.98 x 10-15 J.)
Possible Essay: Einstein’s paper proposed the simple description of "light quanta" (later called "photons") and showed how they
could be used to explain such phenomena as the photoelectric effect. The simple explanation by Einstein in terms of absorption of
single quanta of light explained the features of the phenomenon and helped explain the characteristic frequency. Einstein's
explanation of the photoelectric effect won him the Nobel Prize of 1921. The idea of light quanta was motivated by Max Planck's
published law of blackbody radiation. Einstein, by assuming that light actually consisted of discrete energy packets, wrote an
equation for the photoelectric effect that fit experiments (it explained why the energy of the photoelectrons was dependent only on
the frequency of the incident light and not on its intensity: a low intensity, high frequency source could supply a few high energy
photons, whereas a high intensity, low frequency source would supply no photons of sufficient individual energy to dislodge any
electrons). This was an enormous theoretical leap and the reality of the light quanta was strongly resisted. Even after experiments
showed that Einstein's equations for the photoelectric effect were accurate there was resistance to the idea of photons, since it
appeared to contradict Maxwell's equations, which were believed to be well understood and well verified. Einstein's work
predicted that the energy of the ejected electrons would increase linearly with the frequency of the light. Perhaps surprisingly, that
had not yet been tested. In 1905 it was known that the energy of the photoelectrons increased with increasing frequency of
incident light -- and independent of the intensity of the light -- but the manner of the increase was not experimentally determined
to be linear until 1915 when Robert Andrews Millikan showed that Einstein was correct.2 The Photoelectric Effect is the
phenomenon of electrons being emitted from a metal when struck by incident electromagnetic radiation. Explanation of the
photoelectric effect was one of the first triumphs of quantum mechanics. In his famous paper of 1905, Einstein extended Planck's
quantum hypothesis by postulating that quantization was not a property of the emission mechanism, but rather an intrinsic
property of the electromagnetic field. Using this hypothesis, Einstein was able to explain the observed phenomenon that the
maximum kinetic energy K of emitted electrons varied with frequency of incident radiation. This is exactly the result expected
if photons are quantized with energies, which is exactly what Einstein proposed.3
Einstein paved the way to the solution of the emission spectra of atoms. Emission spectra is either continuous or line spectra of
radiation emitted by substances. These line spectra are the light emission only at specific wavelengths. Niels Bohr explained
the emission spectrum of the hydrogen atom. His model included the idea of electrons moving in circular orbits, but with a severe
restriction: the single electron in the hydrogen atom could be located only in certain orbits. The energies associated with electron
motion in the permitted orbits must be fixed in value, or quantized. Light is also quantized. n is an integer called the principal
quantum number. The – sign is a convention signifying that the energy of the electron in the atom is lower than the energy of a
free electron. The ground state, or ground level, refers to the lowest energy state of a system. An excited state, or an excited
level, is higher in energy than the ground state. Radiant energy absorbed by the atom causes the electron to move from a lowerenergy state to a higher-energy state. Radiant energy is emitted when the electron moves from a higher-energy state to a lowerenergy state.
Absorption Spectrum—the spectrums you see when you shine white light through gas and then through a prism and find some
colors are missing. Emission Spectrum—start out with gas and heat it to glowing and pass that light through the prism; now the
same frequencies that were missing are now visible. Emission and Absorption spectrums have the same frequencies.
According to De Broglie, an electron bound to the nucleus behaves like a standing wave. Some points on a string, called nodes,
do not move at all; that is, the amplitude of the wave at these points is zero. De Broglie argued that if an electron does behave
like a standing wave in the hydrogen atom, the length of the wave must fit the circumference of the orbit exactly. Otherwise, the
wave would partially cancel itself on each successive orbit. Eventually, the amplitude of the wave would be reduced to 0 and the
wave would not exist. De Broglie deduced that waves can behave like particles and particles can exhibit wavelike properties.
Clinton Davisson, Lester Germer, and G.P. Thomson demonstrated that electrons do possess wavelike properties.
http://en.wikipedia.org/wiki/Photoelectric_effect
http://scienceworld.wolfram.com/physics/PhotoelectricEffect.html
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The Heisenberg uncertainty principle says that it is impossible to know the momentum (p) and the position of a particle with
certainty. TThe Schrödinger Equation describes the behavior and energies of submicroscopic particles in general. Electron
Density gives the probability that an electron will be found in a particular region of an atom. An atomic orbital can be thought
of as the wave function of an electron in an atom. Many-electron atoms are atoms that contain 2+ electrons. Three quantum
numbers are required to describe the distribution of electrons in hydrogen and other atoms. The principal quantum number is n,
which can have integral values of 1, 2, 3… and determines the energy of an orbital. The larger n is, the greater the average
distance of an electron in the orbital from the nucleus and therefore the larger the orbital. The angular momentum quantum
number is l, which tells us the shape of the orbitals. l’s values depend on the value of n and range from 0 to (n-1). (i.e. If n=3,
l={0, 1, 2}) If l=0, we have an s orbital, 1, we have a p orbital, 2, we have a d orbital, 3, f orbital, 4, g orbital, 5, h orbital. When
n=2, the hell has two subshells, which are the 2s and 2p subshells where 2 denotes the value of n, and s and p denote the values of
l. The Magnetic quantum number (ml) describes the orientation of the orbital in space. ml depends on l; its value varies from –l
to +l. (i.e. If l=-1, m={-1, 0, 1}) (i.e. If n=2 and l=1, we have a 2p subshell and three 2p orbitals because there are 3 values of m l,
given my -1, 0 and 1) The electron spin quantum number is ms and has a value of -1/2 or +1/2. (i.e. Give the values of the
quantum numbers associated with the orbitals in the 3p subshell. We can automatically determine that n = 3 and p = 1, so m l = -1,
0 or 1 and ms = ±½. Now, just pair up all possible combinations.) (i.e. What is the total number of orbitals associated with n = 4?
Since n = 4, l = 0, 1, 2, or 3 and therefore, we have an s, p, d, and an f orbital. There is one orbital with the s, 3 with the p, 5 with
the d and 8 with the f. Therefore, we have 16 orbitals, and with 2 electrons each, we can hold 32 electrons. But, the question only
asks for the 16 orbitals.) (i.e. Write 4 quantum numbers for an electron in a 5p orbital. We know n = 5 and l = 1, so m l = -1, 0, or
1 and ms = ±½, so our pairs are {5, 1, -1, ±½}, {5, 1, 0, ±½}, {5, 1, 1, ±½}.)
The relation between quantum numbers and atomic orbitals are given as 2l + 1 = the number of orbitals. The p orbitals start with
the principal quantum number. The angular momentum quantum number can then tell us in what orbital the electron is located.
The energies of hydrogen orbitals increase by the principal quantum number. Electrons come off the highest energy state.
Orbitals with the same quantum number all have the same energy. We should not write electron configurations in the order from
the textbook, but rewrite them in order of principal quantum numbers.
The electron configuration of an atom is how the electrons are distributed among the various atomic orbitals. (i.e. With 1s1, the
1 denotes n, the s denotes l and the superscript of 1 denotes the number of electrons in the orbital or subshell) The Pauli
Exclusion Principle states that no two electrons in an atom can have the same 4 quantum numbers. If they have the same n, l,
and ml values, the ms values must be different. (i.e. Helium can be written as 1s 2 with two up arrows on top, two down arrow on
top, or one up and one down arrow on top of the line above the orbital name [1s]) Paramagnetic Substances are those that
contain net unpaired spins and are attracted by a magnet. Diamagnetic Substances do not contain net unpaired spins and are
slightly repelled by a magnet. Hund’s Rule states that the most stable arrangement of electrons in subshells is the one with the
greatest number of parallel spins. Only two electrons (max) can be placed in each orbital, but remember that p’s have three
orbitals, holding 6 electrons and d’s have 5 orbitals, holding 10 electrons. When writing electron configurations, we can just use
noble gas cores to make it easier and shorter. This is written as the noble gas PRECEDING the element you are trying to write
the configuration of in brackets and followed by the rest of the electron configuration. (i.e. Cl is written as [Ne]3s 23p5) (i.e.
Write the electron configuration for Cr {trick question}. Since Cr is in the 4 th row and Argon precedes it, we start with the Argon
core ([Ar]). Now we write 4s2 since the s-block is full. Then we write 3d4 because the element is 4 over in the d-block. BUT, we
notice that Cr is one of the exceptions we should memorize! (see colored bullet point below) We have to write what we have so
far as [Ar]4s23d4, but to make it easier, change it to [Ar]3d 44s2. Since elements with half-filled subshells are pretty stable, we
should take an electron from the 4s, since it’s of higher energy, and give it to the 3d to make the d half full, making the
configuration [Ar]3d54s1.)
You can either read the electron configuration right off the chart by looking at the d-block and the p-block or you can use this
handy chart (on the left) to know in which order the subshells are filled. For example, the 4s subshell is filled before the 3d
subshell because the 4th arrow hits 4s and the 5th arrow hits 3d. HOWEVER, when writing the electron configuration, reorder
the configuration so it is in order of principal quantum numbers so it is easier to take electrons off the highest energy level. The
chart above (middle) tells you in what order to write electron configuration. Note that, when writing s or p orbital configuration,
the number corresponds to the row number (principal quantum number). But when writing configuration of the d block, the
number is the row number – 1 and when doing the f block, the number is the row number – 2.
Isoelectronic means that two elements have the same number of electrons and are in the same orbital. Ions can easily be
Isoelectronic. Fluoride, Oxide, and Nitride ions are Isoelectronic with Neon where the Nitride ion is largest because it is N3- and
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has three more electrons (than the regular N atom). Excited Elements are elements in which their electrons jump to another level
when heated. (i.e. 1s12s1 is an excited He atom)
Exceptions You Should Know Are: Cr, Mo, Cu, Ag and Au.
In chapter 8, all we learn is that size matters!
All elements from 1A to 7A are representative elements, which do not have completely filled subshells of the highest quantum
number. Except He, all of the noble gases have completely filled p subshells. (i.e. What is the ground state configuration of an
element with 20 electrons? Is it diamagnetic or paramagnetic? Since we know we have 20 electrons, the configuration must be
1s22s22p63s23p64s2, which makes it a representative element, with no filled 4p subshell. Since all electrons are paired, the
element is diamagnetic.)
As nuclear charge increases, atoms get smaller because p-orbitals are held together more tightly. Nuclear charge, however, stays
the same with the loss and gain of electrons. The size of an atom depends on its outermost shell. The f-block atoms have smaller
size changes because the f-block goes down two subshells. We define the size of an atom in terms of its atomic radius, or ½ the
distance between the two nuclei in two adjacent metal atoms, which is measured in picometers. We can determine the size of an
atom’s atomic radius by simply looking at the periodic table. From left to right in a row, the atomic radius gets smaller. Going
from up to down in a column, the radius gets bigger. (i.e. Arrange Li, Be and C in order of atomic radius size from largest to
smallest. Since they are all in the same row, we just go from left to right to see what is smaller. Li is before Be, which is before
C, so Li > Be > C.) The electron configuration of F is [He]2s22p5, but the configuration of F- is [He]2s22p6 because it gains an
electron. N3- is larger than O2- and F- ions even though they are Isoelectronic with Ne. Smaller electron clouds are more tightly
held. Li+ is smaller than He even though they are Isoelectronic. K + is Isoelectronic with Ar, but K + has a greater positive charge
and therefore holds orbitals more closely. Nuclear charge stays the same even with loss or gain of electrons.
The ionic radius is the radius of a cation or anion measured by X-ray diffraction. If an atom forms an anion, the size gets larger
because the negative charge increases the number of electrons. If we examine Isoelectronic ions, we can see that cations are
smaller than anions. Na+, for example, is larger the F- even though they are both Isoelectronic to Ne. Both have the same number
of electrons, but Na has more protons than F, so the larger effective nuclear charge of Na + results in a smaller radius. (i.e. Which
ions are smaller in each pair: K+ and Li+, Au+ and Au3+, P3- and N3-. Li+ is smaller than K+ because Li is two rows up from K.
Au3+ is smaller than Au+ because Au3+ has two less electrons than Au+. N3- is smaller than P3- because it is one row up.) When
two ions are in the same row and are Isoelectronic to the same noble gas, the one towards the left more is usually larger (i.e. N3- is
larger than F-). When two ions are in the same column but different rows, the one towards the bottom in a row with a higher
principal quantum number is usually larger (i.e. Ca2+ is larger than Mg2+). When two ions are of the same element but the
element has different charges, the one with more electrons is usually bigger (i.e. Fe 2+ is larger than Fe3+.) (i.e. F- is bigger than
Ne, Na+, and Mg2+ because they are all isoelectronic and F- has the most protons.)
Ionization Energy is the energy needed to remove an electron from a gaseous atom in its ground state. The energy required to
remove the first electron from an atom is first ionization energy. Ionization is always an endothermic process. Ionization
energy increases as you go from left to right in a row. A lone electron in a p-subshell has lower ionization energy. 1A has very
low ionization energies while noble gases have the highest ionization energies because it takes their electron away so they don’t
have a full p subshell. N has a ½ filled subshell and therefore, higher ionization energy. Throughout the transitional metals,
ionization energy barely changes even though nuclear energy increases because we fill inner shells (i.e. 3d shell in 4 th row). For
first ionization energy, you can look at the chart on the previous page (on the right) to see where ionization energy increases. (i.e.
Does N or P have a larger first ionization energy? Since they are in the same column and N is higher, N has a larger first
ionization energy according to the chart on the previous page.) For 2 nd ionization energy, you need to see how easy it would be to
take an electron off of an ion. (i.e. If there is a Mg and a Na atom, Mg would have a smaller second ionization energy because
after taking off one electron, Na has the same electron configuration as Neon [a noble gas] and Mg is one electron away from it,
so it would take less energy to take off an electron from the Mg ion because then it would have a full 2p shell and it wants that
subshell filled.) Ionization energy increases on the amount of atoms of an element in a molecule. (i.e. I < I 2 < I3 < I4) As with
everything, there are exceptions. For some reason, O has a lower ionization energy than N, just like S has lower than P.
Ionization never breaks a noble gas core.
Electron Affinity is for – ions and is the amount of energy given off when we add one electron to an atom. The electron
affinities for halogens are high while the electron affinities for noble gases are less than 0. 2A’s electron affinities are lower than
that of 1A’s. The electron affinity of the halogens are high while the noble gas electron affinities are less than zero.
Electronegativity is the ability of an atom to attract electrons to itself in a chemical bond. It is also the “greediness in electron
sharing.”