Download Atomic Structure and Periodicity

Atomic Structure and Periodicity
A. Atomic structure and periodicity
1 m=109 nm
or 10-9 m 1 nm
1 J = 1 kg x m2/s2
1. Electromagnetic radiation
a. a form of radiant energy
b. three properties of waves
- wavelength ( - Greek lambda) (shorter wavelength = higher energy)
- frequency (v - Greek nu) (Hz - 1/s or s-1)
- speed (c)- equal for all kinds of e-m radiation ( = 2.9979 x 108 m/s)
c. v = c (wavelength and frequency are inversely related)
Sample problem : Calculate the wavelength of a 101.5 mHz frequency given off by a radio
station.
Solution :  = c/ v (101.5 mHz = 101.5 x 106 Hz or 101.5 x 106 s-1).
 = 2.9979 x 108 m/s / 101.5 x 106 s-1
= 2.954 m
2. Nature of matter
a. Planck - energy is quantized E = nhv (n = integer, h = Planck's constant 6.626 x 10-34 Js)
b. quantum - a small "packet" of energy - a discrete, yet varying amount (not all quanta are
the same)
c. Einstein - photons (packets of light energy) are quantized (Ephoton = hv = hc/) v = c/
d. Einstein : E = mc2
e. dual nature of light/matter
- diffraction (scattering of light from a regular array of lines or points)- property
of waves ; photoelectric effect- evidence of particle nature
f. de Broglie's equation : = h/mv (where m = mass (in kg) and v = velocity in m/s)
Sample problems :
1. Calculate the energy of a single photon and a mole of photons emitted from a vapor lamp
giving off light with a wavelength of 412.0 nm.
Solution : Ephoton = hv = hc/
= (6.626 x 10-34 J·s)(2.9979 x 108 m/s) / (412.0 x 10-9 m)
= 4.821 x 10-19 J
E mol = (4.821 x 10-19 J/e-) (6.022 x 1023 e-/mol) = 290.3 kJ
2. The first ionization energy of aluminum is 577 kJ/mol. Is light with a wavelength of 215 nm
capable of ionizing an atom of aluminum in the gas phase?
Solution : kJ/mol are given, first calculate J/atom : 577 kJ/ (6.022 x 1023) = 9.58 x 10-22
kJ/atom = 9.58 x 10-19 J/atom
Second - calculate wavelength from energy needed and compare to given wavelength :
 = hc / E = (6.626 x 10-34 J·s)(2.9979 x 108 m/s) / 9.58 x 10-19 J/atom
= 2.07 x 10-7 m = 207 nm
Answer : No. The 207 nm wavelength is a higher energy wavelength than the 215 nm
wavelength.
3. The atomic spectrum of hydrogen
a. Contrast to a continuous spectrum - all wavelengths
b. emission spectrum of hydrogen -pattern of light emitted from an excited hydrogen
atom) - a line spectrum- indicates quantized nature of hydrogen (only certain
wavelengths absorbed or emitted)
4. The Bohr model - quantum model
a. energy levels available : E = -2.178 x 10-18 J(Z2/n2)
- Z = nuclear charge (number of protons)(=1 for hydrogen atom)
- n = integer- corresponds to energy levels
- neg. sign indicates negative energy (e- is bound to nucleus therefore has
lower energy than free e- where n = )
- J is joules
b. ground state lowest possible energy state
c. calculation of energy of quantization in hydrogen for change in energy : E = Ef - Ei
(e.g. change in E when electron moves from n = 6 to n = 5)
- positive sign indicates E (light) absorbed
- negative sign indicates E (light) emitted
d. calculation of change in electron energies E= -2.178 x 10-18 J(1/nfinal2 - 1/ninitial2)
e. spectral transition diagram : ("spectral" because light is usually absorbed or emitted)
f. failure of Bohr model in other elements -the Bohr model did not hold true for
polyelectronic atoms
5. Quantum Mechanical Model of atom
a. Heisenberg, Schrodinger, de Broglie - developed wave , or quantum mechanics
- bound electron is a standing wave
- only certain circumferences will allow whole numbers of half waves
- Schrodinger's equation HΨ = EΨ
- Ψ- wave function, or "orbital" (function of coordinates x, y and z)( does
not indicate pathway of electron)
- H - operator - contains mathematical terms which produce total energy
of the atom
- E - total energy of atom (kinetic and potential)
- a specific wave function is an orbital
b. Heisenberg uncertainty principle-cannot know both momentum and position of electron
at a given time
c. Physical meaning of wave function in light of uncertainty principle : probability of elocation
- probability distribution- represents square of wave function - produces an
e- density map
- radial probability distribution - total probability within spherical shells
- orbital - no definite size-defined as area in which 90% of total eprobability
6. Quantum numbers-describe orbitals
a. Principal quantum number (n = 1, 2, 3, ...)
- indicates size and energy of orbital (higher n, higher energy-less bound,
less neg. energy)
b. Angular momentum number (azimuthal quantum number) (l)
- integral values from 0 to n-1
- indicates shape of orbital (l = 0 = s; l = 1 = p; l = 2 = d; l = 3 = f )
c. magnetic quantum number (ml)
- integral values from l to - l including 0
- indicates position of orbital in space relative to other orbitals
d. Spin quantum number (ms) - indicates rotation of electron
- -½ or -½
Table 1 : Quantum numbers for first four energy levels
n
ml
Orbital
l
(n = 1,
(ml = -l …0…+l)
designati
(l = 0 to n2,
3…)
on
1)
No. of
orbitals
(= n2)
Total number
of electrons
(= 2n2)
(2)
1s
1
0
0
1
2
2s
2
0
0
1
2 (8)
2p
1
1, 0, -1
3*
6
3s
3
0
0
1
2
(18)
3p
1
1, 0, -1
3
6
3d
2
2, 1, 0, -1, -2
5**
10
4s
4
0
0
1
2
4p
1
1, 0, -1
3
6 (32)
4d
2
2, 1, 0, -1, -2
5
10
4f
3
3, 2, 1, 0, -1, -2, -3
7
14
* The 3 p orbitals are x, y, and z (e.g. 2px, 2py, 2pz)
** The 5 d orbitals are xz, yz, xy, x2-y2, and the z2 (e.g. 3dxz, 3dyz, 3dxy, 3dx2-y2,
and the 3dz2)
Sample problems :
1. Which of the following sets are incorrect?
a. n = 2, l = 2, ml = -1
b. n = 3, l = 2, ml = -3
c. n = 4, l = 2, ml = -1
Answer : sets a (l cannot exceed n-1)and b (ml must be between + l and - l)
2. How many electrons can have the quantum numbers n = 5, ml = +1?
Answer : 8 In n = 5, l will equal 0, 1, 2, 3, 4. For l = 1, 2, 3, 4 there will be one orbital in which ml = +1, and
each orbital will contain two electrons.
7. Orbital shapes and energies
a. nodal surface (nodes)- regions of zero probability of finding an eb. hydrogen atom - orbitals in energy levels (same n value) are degenerate - have the
same energy
8. Electron Spin and the Pauli Principle
a. Pauli exclusion principle - in a given atom, no two electrons can have the same set of four
quantum numbers (i.e. -only two electrons per orbital)
9. Polyelectronic atoms
a. three forces affecting distribution of electrons
- kinetic energy of electrons (outward)
- potential energy of attraction of nucleus (inward)
- potential energy of repulsion of electrons
b. electron correlation problem-since paths of electrons cannot be known exactly,
repulsions cannot be calculated exactly- must make approximations
c. shielding - inner electrons shield outer electrons from nuclear charge (held less
tightly)
d. orbitals in polyelectronic atoms are not degenerate (s < p < d < f)
e. penetration effect -ex. In general, 2p electrons are closer to the nucleus and
therefore have a lower energy than 2s, however, 2s electrons spend a small, but
significant amount of time closer to the nucleus and are said to "penetrate" and
thus are considered to have a lower energy state and are filled first
d. Electron configuration problems
Sample problems :
1. What is the electron configuration of neon?
Answer : Neon has ten electrons. In the ground state these will be in the
lowest energy levels possible giving the following :
(1s22s22p6)
2. What is the electron configuration of chromium ?
Answer : [Ar] 4s13d5 (note this is an exception to the normal rules)
3. What is the identity of the element with the electron configuration
1s22s22p63s13p1 in an excited state ?
Answer : Magnesium - 12 e-, with one 3s1 e- moved up into the 3p1 orbital.
4. Write the electron configuration for the following orbital notation for an
atom with 8 protons, identify the species (atom, ion etc) and write its
formula .
Answer : 1s22s22p6, oxide ion , O2-
5. How many unpaired electrons are found in nitrogen in its ground state?
Answer : Three.
6. Give a set of values for the four quantum numbers of the valence electrons of phosphorus
in its ground state.
Answer : The valence electrons of phosphorus are the 3s23p3 electrons :
Electrons
n
ms
l
ml
3s
3
0
0
+½
3s
3
0
0
-½
3p
3
1
-1
3p
3
1
0
3p
3
1
+1
+½
(or -½)
+½
(or -½)
+½
(or -½)
10. History of periodic table
a. Mendeleev given most of the credit because of his emphasis on using it as a tool to
predict yet unknown elements.
11. Aufbau ("building up") principle and the periodic table
a. Aufbau principle - as protons are added to successive atoms of elements, so are
electrons
b. Hund's rule - the lowest energy configuration for an atom is the one having the most
unpaired electrons allowed by the Pauli principle in degenerate orbitals
c. Valence electrons - electrons in the outermost principal quantum energy level of an
atom
d. core electrons - "inner" electrons
e. configurations of representative elements vs. transition metals and inner transition
metals (lanthanides and actinides)
f. exceptions to expected configurations - chromium [Ar] 4s13d5 and copper [Ar] 4s13d10
g. arrangement of periodic table- representative elements, transition metals,
lanthanides, actinides, s block, p block, d block and f block
h. IUPAC periodic table- numbers Groups 1-18 instead of 1A-8A and "B" designation for
transition metals
i. determination of electron configurations
12. Periodic trends in atomic properties
a. major factors affecting trends
- nuclear charge - increases across a period causing a greater attraction for
electrons
- shielding - additional energy levels down a group causes a decrease in the
attraction of the nucleus for the outermost electrons
- electron configuration - most important - groups have similar electron
configurations giving them similar chemical and physical properties
b. ionization energy- the energy needed to remove an electron from a gaseous atom (X(g)
-->X+(g) + e-) in its ground state
- first ionization energy (I1)- energy needed to remove the first electron
- second ionization energy (I2) - energy needed to remove second electron
- note pattern in aluminum [Ne]3s23p1
I1- removes p1electron = 580 kJ/mol
I2 removes one 3s electron = 1815 kJ/mol - higher due to positive charge on
+
Al ion
I3 removes remaining 3s electron = 2740 kJ/mol (Al2+ )
I4 removes core 2p electron = 11,600kJ/mol - core electrons come from
stable noble gas configuration
- periodic trend is for ionization energy to increase across a period - increased
nuclear charge with no increase in shielding
-exceptions-due to electron repulsion i.e. nitrogen and oxygen
N:
O:
- easier to remove outer
2p electron from oxygen than nitrogen because of doubly occupied
2porbital
-group trend is to decrease down a group - increased shielding, outer electrons
are farther from nucleus
c. electron affinity - the energy change associated with the addition of an electron to a
gaseous atom : X(g) + e- ---> X (g)
- sign of energy change- negative if exothermic, positive if endothermic
- periodic trend - negative electron affinities increase across a period (more
energy released)
- exceptions : carbon has a negative electron affinity (becomes more stable) and
nitrogen has a positive electron affinity (becomes unstable- higher energy)
C:
N:
- an electron added to carbon
goes into an empty orbital whereas an electron added to nitrogen goes into
an occupied orbital resulting in electron repulsion
- Group trend - electron affinities become more positive (less negative) because
less energy is released due to shielding and increased distance from
nucleus (less attraction for e-)
Note relationship between ionization energy ( energy change for X(g) -->X+(g) + e-) and electron affinity
( energy change for X(g) + e- ---> X (g)). Since the changes are opposite their energies are opposite.
Sample problem : What is the electron affinity for the ionization of the Al2+ ion (The ionization energy
for Al+(g) + e-  Al2+(g) = 1850 kJ/mol)?
Answer : -1850 kJ/mol : Electron affinity is the energy change associated with the addition of an
electron to a gaseous atom making the reaction : Al2+(g) + e-  Al+(g). The electron affinity is
therefore opposite (EA = -IE) and equals -1850 kJ/mol.
c. Atomic radii - defined as half the distance between the nuclei of diatomic molecules
of the same element (e.g. O2)
- periodic trend - atomic radii decrease across a period - due to increased nuclear
charge
- Group trend - atomic radii increase down a group - large n value, increased
shielding
d. ion size
- metallic ions lose electrons and become smaller due to loss of volume occupied
by lost electrons and by the increased ratio of positive to negative charge
- nonmetallic ions gain electrons and increase in size for reasons opposite of
metallic ions
- trend for metallic ions is to decrease in size (consider 3rd period metallic ions
Na+, Mg2+, Al3+)
- trend for nonmetallic ions is to also decrease in size (consider the 3rd period
nonmetallic ions P3-, S2-, Cl-)
13. Information contained in the Periodic Table
a. number of valence electrons - main determiner of chemical properties - know well
b. electron configuration - know representative elements and Cr and Cu
c. know : metals, nonmetals, alkali metals, alkaline earth elements, halogens, noble gases,
transition metals, inner transition metals ( lanthanides and actinides) and
metalloids (semimetals)
d. characteristics of metals and nonmetals
- metals - low ionization energies, low (less negative) electron affinities, tend to
form cations
- nonmetals - high ionization energies, higher (more negative) electron affinities,
tend to form anions
14. The Alkali metals - Group 1A H, Li, Na, K, Rb, Cs, and Fr
a. hydrogen is a nonmetal mainly because of its small size - holds electron tightly
b. density increases down group - typical of all groups - mass increases at a faster rate
than size
c. melting points and boiling points generally decrease
d. tend to lose outer e- making them reducing agents - lower ionization energy makes
them stronger reducing agents thus order of reducing strength is Cs > Rb > K >
Na > Li for solid metals only
- in aqueous solution the reducing strength is Li > K > Na Why change in position
for lithium in aqueous solution?- Lithium's small size gives it a high charge
density which makes it more able to bond to water (hydration energy - energy
released when a substance bonds to water in solution) and form the Li+ ion (and
thus lose its valence electron) more readily than K or Na.
-If lithium is a stronger reducing agent and bonds to water more strongly than K
or Na, then why do experiments show that K and Na react more violently with
water? - Na and K have lower melting points causing them to react with water in
the liquid state giving more surface area resulting in a faster reaction. Main
point of this - thermodynamics (energy change) tells us if a reaction will be
spontaneous or not, kinetics (study of reaction mechanisms/rates) tells us how
fast a reaction will occur.