Download Chem 1a Review

Chem 1a Review
Absorption an demission Spectra of Atomic Hydrogen:
hc
1.
E  h 
 hc

2

Rydberg equations
a)
b)
c)


 1
1 
, RH  13.6 eV = 2.18 1018 J
2
2 
n
n
 f
i 
2 
R Z 1
1 
E
(preferred units cm-1)
  H 

 Hz and  
2
2 

h n f n i 
hc
For emission: Lyman mf = 1; Balmer nf = 2; Paschen nf = 3
E  RH Z 2 


Bohr Theory

 only for 1 electron atoms and
1 Works
ions
2. Bohr used the classical equations for a negatively charged particle revolving around positive
heavier particle. The electrostatic force on the electron equals the centripetal force on the
electron in an orbital:
Ze 2
mev 2
Fele 
 Fcent 
40 r 2
r
where me is the mass of the electron, v the velocity of the electron and r the radius of the orbit.
3) Bohr postulated that only specific values of the angular momentum are allow
h
, where n is an integer
 mevr  nh  n
2
n2
h2 4 0
This gives rn  a0 ; and a0 
Z
mee
4) The orbital energy is given by the classical equations
1
Ze 2
Z2
E  KE  PE  mv 2 
  2 RH
2
4 0 r
n

Quantum Mechanics
h
h


1. DeBroglie Wave:
mv p
2. Heisenberg Uncertainty Principle:
h
h
px  ; or Et 
2
2

h
h
p  mv  p = mv ; and  
    2 p
p
p
3. Wave 1D



Standing wave
with one node
Not a standing wave
ends not tied down not allowed
Boundary conditions: Ends tied down standing wave
More nodes  higher energy
4. 2D waves for a square or round drum have degenerate sets (i.e. two or more waves have
the same energy. They are the same wave only rotate in space.
Schrödinger Equations: 3D wavefunctions (r) or Orbitals








Boundary conditions:  (r)  0 as r   (this is same as standing wave and makes
probability finite) and  (r)  ; at all times
 (r)2 dV is probability of finding an electron in small volume dV around r
  (r)2 dV is the total probability of finding the electron anywhere and must = 1
with specific values of the quantum numbers, n, l, ml (and ms added later)
Only solutions
are allowed.
nlml (r,,  )  Rml (r)Ylml (,  )

  (r)
2
dV 


 R
(r)Ylml (,  ) r 2 sin(  )drdd 
2
ml
 R
ml
(r) r 2 dr  Ylml (,  ) sin(  )dd  1
2
2
 R (r) r dr  1
 Y (, ) sin(  )dd  1
2 2


ml
This gives
2
lml

Probability of finding a particle at a distance r from the nucleus for any angles  and is
2
P(r)    (r) 2 r 2 sin(  )dd   Rml (r)Ylml (,  ) r 2 sin(  )dd

given as
 r 2 Rml (r)
2
 Y
lml
(,  ) sin(  )dd
2
 r 2 Rml (r)
2
Quantum Numbers:
n  principle = total number of nodes + 1, mostly determines the energy and orbital size
l  orbital shape quantum number = number of angular nodes has some effect on energy and
size in
multielectron atoms
ml  orientation quantum number (degeneracy of the l orbital is 2l + 1) values –l,
–l +1,… ,0, 1, 2, …, l determines the orientation of the orbital in space
ms spin quantum number + ½ and – ½
Orbit type
l
ml
degeneracy
# of electrons
in orbital
s
0
0
1
2
p
1
-1,0,1
3
6
d
2
-2,-1,0,1,2
5
10
f
3
-3,-2,7
14
1,0,1,2,3
Orbitals Sizes, Shapes, and graphs:
1s
2s
3s
3p
-
2p
+
+
-
No nodes;
1 node
2 nodes
3d(x2-y2)
3d(z2)
+
-
+
+
-
-
y
-
y
+
x
z
y
x
3d(xy)
1s
z
x
3d(xz)
y
z
3d(yz)
n=2 Shell
n=3 shell
n=5 shell
Look at pictures of the hydrogen orbitals at http://www.shef.ac.uk/chemistry/orbitron/ . Note
particularly the shape, nodes and sign of the orbitals. Note that the s and d orbitals are
symmetric to inversion through the origin while the p is anti-symmetric toward inversions.
Orbital energy:


one electron atoms the energy only dependents on n
multi-electron atoms:
o Penetration for orbitals of same n gives energy order of ns < np < nd < nf < …
o General orbital energy order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 5p
 Exceptions: Neutral atoms: Cr, Cu, Nb, Mo,
 Ions: all transition metals when they ionize the first two electrons that are
lost are from the ns shell not the (n-1)d shell.
Filling orbitals
1. Pauli Principle: Every electron must have a unique set of 4 quantum numbers
2. Aufbau principle: Fill lowest energy orbitals first
3. Hund's Rule: In a degenerate orbital aligned spins have lower energy than paired spins
4. When an electron is removed from a transition metal the result is that one s electron is
lost and the other electron moves into a d orbital. When 2 electrons are removed from a
transition metal the result is that two s electrons are lost.
Periodic Trends:
 Atomic size increases as you move down the periodic table (PT); decreases as you move
across PT

When you add 1 electron (to a partially filled orbital) and 1 proton the added electrons do
not completely shield the added positive charge. Thus the effective nuclear charge goes
up and the electrons are held more tightly. The ionization energy, Iz, also goes up, and
the atom becomes smaller. When you add an electron to a completely empty orbital of a
larger n the size of the atom gets much larger and the ionization energy will go down.

Ionization Energy Decreases as you move down PT, increases as you move across left
to right of PT.
Note:
A: Ionization energies are very high at the
noble gases because they have the highest
number of protons (positive charge) for
that quantum number n.
B: Exception to general trend of increase Iz
with increase Z. Due to going from filling
1s shell to 1p shell and since p penetrates
less well then s it is easier to remove.
C: Exception to general trend due to
forcing two electrons into same p
suborbital causes electron-electron
repulsion to go up (2 electrons in same
region of space).
D: General trend but increase is slow. Due
to filling of inner orbital, (n-1)d, on the transition metals. So this makes outer orbital ns see only
a small change in positive charge.
E: Exception to general trend due to filling the 4p orbital and now the electron that is lost is a 4p
and no longer a 4s.
Exception to general trends:



closed shell, closed subshell (2p6. 3p6, 3d10) and half-filled subshell (2p3, 3p3, 3d5) are
stable and have lower ionization energy.
P electrons are better shielded compared to s (p electrons do not penetrate as close to the
nucleus as do the s) thus the atom with [He]2s2 2p1 is easier to ionize that one with
[He]2s2
When you have one more that a half filled subshell the electrons must pair and thus two
electrons share the same suborbital. Due to electron-electron repulsion this electron is
easier to ionize.
Lewis structures:
Rules:
1. Total number of valence electrons is sum of valence electrons of each atom minus the
overall charge
2. Arrange the atoms in a structure and distribute the electrons so that each atom has 8
electrons around it (exceptions, H has 2, B can have 6 and third row and lower atoms can
have more than 8).
3. Try to maximize the number of bonds that each atom has.
4. Calculated the partial charge on each atom (it must add up to the total charge).
5. See if you can optimize the structure to reduce partial charges, put negative partial
charges on electronegative atoms and positive on electropositive atoms and maximize the
bonds to all atoms.
6. Look for resonance structures.
Resonance Structures:
1. To form resonance structures you can move electrons but not atoms.
2. Structures with higher partial charges are less stable
3. Structures with negative changes on electropositive atoms are not favorable
4. Structures with positive charges on electronegative atoms are not favorable
Oxiacids:
The four rules of Oxiacids are:
1. As electronegativity of central atom increases the acidity increases
2. As the positive formal charge (no expanded octet) on the central atom increases the
acidity increases
3. As the number of resonance structures of the dissociated acid increase the acidity
increases.
4. As the number of O atoms that do not have an attached H increases acidity increases.
VSEPR:
 Assign steric number
 Assign electron pair geometry of from steric number
Steric #
Electron Pair
# lone pair
Shape around
geometry
atom
6
Oh
0
Oh
1
square pyramidal
l5
Trig bipyramidal
4
Tetrahedral
3
trigonal planar
2
linear
2
3
4
0
1
2
3
0
1
2
0
1
0
square planar
bent T
linear
Trig bipy
saw horse
T shape
linear
tetrahedral
pyramidal
bent
trigonal planar
bent
linear
o
Assign shape around atom remembering that lone pairs are not seen
Lone pairs are fat.
Larger molecules
o assign steric number and shape around each atom.
o Assign shape to whole molecule
MO Theory
 Plot Orbital energy vs r for H2
 Linear Combination of Atomic Orbitals
o b = 1sa + 1sb
o * = 1sa - 1sb
o extra electron density between nuclei in bonding less electron density in
antibonding
 MO diagrams
o adding electrons (lowest energy first
o Hund's rule (electrons in different suborbitals have lower energy then electrons in
the same suborbital and unpaired electrons have lower energy then paired)
o bond order
o Paramagnetic/ Magnetism
o Excited states
o P orbitals
  bonds
 bonds
 b Node on line between nucleus
 * 2 nodes one as in bonding and one perpendicular to line
between nucleus
 MO diagram for Homonuclear molecules
C
C
o bond order for Li to N (see
right side picture
2p
2p
2p
o Bond order for O – F (and S 2p
– Cl) see left side picture



2s
2s
2s
2s
o Reason is energy distance between 2s and 2p orbitals and the interaction between
them
o Predict bond order, number of unpaired electrons,
 Heteronuclear MO
o MO diagram
o Bond order for NO.
o Placement of AO use Ionization energies for the various orbitals.
Hybridization:
 Driving force for hybridization is due to better overlap between hybrid orbitals than just
for p's or s's. Why
 Type of hybridization, shapes and orientation:
hybrid
shape
sp
linear
sp2
trigonal planar
3
sp
tetrahedral
 Use VSEPR to assign geometry and then can assign hybrid
 Lone pairs are possible in hybrids and they are still fat
Shapes of large Molecules:
 Use VSEPR on all atoms
 Assign hybridization of all C atoms (and N)
 Unused p orbitals maybe available for  bonding.
 Use MO theory to create the  bonding for unused p orbitals
o double bonds do not allow rotation
o triple bonds allow partial rotation
o single bonds allow full rotation
 Aromatic molecules and delocalized  bonds
o C6H6
MO of  bonds
 Cyclic molecules:
o Energy level diagram
 Draw molecule with point down:
 Put one orbital for each apex
 Fill in electrons
 Mark energy zero (remember it must be
such that sum of all orbitals equals zero).
 Aromatic if there are no partially filled degenerate orbitals and that only
bonding and nonbonding orbitals are used
 If non aromatic the molecule will distort so that bond angles and energies
are optimized
o Orbitals
 lowest energy orbital has no nodes
 each orbital at higher energy has one more node
 degenerate orbitals have same number of nodes
 show signs of the AO that contribute to the MO
 Linear molecules
o
o
o
o
No degenerate orbitals
Lowest energy orbitals has no nodes
each higher orbital has one more node
Assign signs on the AO in a symmetric manner
18 Electron Rule
 Electron Count on a central metal should add up to 18.
 Two ways to count electrons:
o Way one; Example (Fe(CN)64 use the number of valence electrons for the neutral atom then add(or
remove) the number of electrons to give the complex the total charge that
exists on the complex.
 Fe has 8 electrons, CN has one lone electron to donate to the Fe and then
you need to add 4 more to get the correct charge
 8 + 6(1) +4 =18;
 Example 2 Cr(CO)6
 Cr has 6 CO donates 2: 6 + 6(2) =18;
o Way two:
 Assign each atom or group in the molecule a charge such that the sum of
all the charges adds up to the total charge.
 Count electrons using the valence electrons of the charged groups
 [Fe2+(CN-)6]4- you see that the charges add up +2 + 6(1-) = -4
 Fe2+ has 6 valence electrons, while CN– has a lone pair to donate so it
gives 2 electrons; Thus 6 + 6(2) =18;
 Remember that you don't use the total charge this second method because
you have already used the partial charges and these add up to the total
charge.
 Groups that can contribute electrons to a central metal
o C5H5 (note that this is C6H6 that has lost one H). contributes 5 if it is 5 4 if it is
if it is etc
o A double bond can tribute 2
o Metals can form a metal-metal bond that can contribute 1 electron from each
metal to the other, or if it forms n bonds it can contribute n electrons from each to
the other:
o (Fe(CO)(Cp))4 is tetrahedron with a bond between each Fe; thus the electron
count at a particular Fe is 8 electros from that Fe, 1 electron from each of the
other Fe, 2 electrons from the CO and 5 from the Cp 8 + 3(1) +2 + 5 =18
Transition Metals
 Isomers PtCl2(NH3)2 square planar
 Crystal field theory
o Oh three down and two up which ones?
o Td two down and three up
o Predict how they split for a square planar arrangement of ligands
o Predict how they split for a linear arrangement of ligands
o Fill orbitals and get high field and low field complexes
o Predict magnetic properties of complexes
o Spectrochemical Series => ligand field strength
o Ligand field strength and charge on central metal ion
o Ligand field strength and larger central metal ion.
o Jahn-Teller distortions only for when a degenerate orbital is partially filled
o Ligand Field Stabilization energy
 Chelate Effect
Crystal Structures:
 Metals
o Simple crystals, simple cubic, body center cubic, face center cubic (fcc)
 Number of atoms in unit cell
 Calculate % volume occupied
 Type of holes in lattice
 Binary Compounds
o One atom forms one of the unit cells the other atom goes into the holes in the
structure
 Hole ins structure:
 Simple cubic: center hole with 8 coordination (Nate calls Body
center cubic)
 hole to atom ration 1:1 CsCl
 fcc: two type of holes Oh and Td holes
 fcc hold to atom ratio Oh 1:1 Rock Salt structure
 fcc hole to atom ration Td 2:1
o Fill half of the Td holes Zinc Blend
o Fill all the Td hole Anti-fluorite or Fluorite
 fcc file both Oh and Td hole ratio 3:1
 Radius Ratio
 Form structure for bigger atom: (usually anion) place other atom
(cation) into holes. Structure only stable if cations are large
enough to keep anions from touching:
 Calculate radius ration for all structures
 Calculate Madelung constant for any arrangement that is given
 Semiconductors
 dopants, electron (n) or hole (p) donors
