Lect 12 3D Molecular Structure - Santiago Canyon College
MOLECULAR STRUCTURE
Objective
To be able to figure out the shape, 3D structure of a molecule, starting from the chemical formula.
For instance: to predict that, & explain why, H2O molecule is bent & polar, though CO2 molecule is linear & non-polar.
Why important?
From the shape of molecule, one can predict physical properties of material (say, that water is a strong dielectric).
Chemical properties are also determined by the shape of molecule:
how accessible are its electrons to attack by another agent?
Say, why CH4 molecule is rather non-reactive, though NH3 violently reacts with HCl, & C2H4 forms polymers?
What molecules can exist, & what can’t?
What do we already know?
Electrical charges of the same sign repel each other.
Especially important is that electrons repel each other.
There are valence (outermost) & core (inner-shell) electrons in atoms.
Only valence electrons are shared by bonded atoms.
Covalent bond is a shared electron pair, occupying a MOLECULAR ORBITAL.
The physical reason of pairing electrons is that each electron is a small magnet. 2 electrons may cancel each other’s magnetic field (( (we say they have antiparallel SPINS, thi is a more stable situation:
an electron pair at the same molecular orbital) or sum up their magnetic field (( (unstable situation: both electrons cannot belong to the same orbital)
Electron octet configuration is especially stable.
Lewis dot formulas of molecules are compiled to satisfy
the OCTET RULE: Chemical compounds form so that each atom, by gaining, losing or sharing electrons has an octet (8) of its valence electrons (2e- for H)
H:H H2 H – H
Cl2 Cl – Cl
HF H – F
When more than 2 atoms bonded, there is a quest: which atoms are bound to each other & which are not?
Methane, CH4
H
|
H – C – H
|
H
In SF2 F – S – F there are two S – F bonds, but none of F to F
Structural formula shows the order of connectivity of atoms.
Ammonium NH4+ H +
ion |
H - N - H
|
H
Sulfate ion, SO42- :Ö: 2-
|
:Ö - S - Ö:
|
:O:
Nitrate ion, NO3- |Ö| -
N = O|
| Ö
VALENCE-SHELL ELECTRON-PAIR REPULSION THEORY
VSEPR
predicts the shape of molecules.
1. Only outermost, or valence electrons, count.
Core electrons do not contribute to bonding, ( do not affect the shape of a molecule.
2. Valence electrons in molecules are arranged in pairs, when possible.
3. Lewis electron-dot formula is valid, with bonding & lone electron pairs.
(H-Ö-H Oxygen in water has 2 bonding & 2 non-bonding electron pairs)
4. All electron pairs (both bonding & non-bonding) repel each other & occupy molecular orbitals (MO) located as far apart as possible.
5. Formation of double bonds do not affect the geometry.
Major Cases
Two electron pairs around central atom.
BeH2 total 2+ 2= 4 val. e-, 2 el. pairs & 2 bonds H-Be-H, no lone el. pairs, linear molecule
Bond angle 180o
2 double bonds & no lone el. pairs around central atom give the same linear geometry, as in CO2
O=C=O
Bond multiplicity does not affect the basic geometry!
3 electron pairs around central atom
BH3 3+3=6e, 3 bonds are formed, no lone pairs,
trigonal planar H
|
B
H H
4 electron pairs around central atom*
The most important case
8 valence electrons (4 pairs) & 4 bonds around central atom.
The most remote location of electron pairs – tetrahedral.
CH4 tetrahedron
All bond angles 109.5o
4 electron pairs – 1 lone pair around central atom
:NH3 5+3=8e-, 4 MO, 3 bonds N-H, one lone e- pair on N
Lewis electron dot formula is
Electron configuration is still tetrahedral,
but the molecule geometry is different:
4 electron pairs – 2 lone pairs around central atom
H2O 8e-, 4 MO, 2 bonds, 2 lone pairs on O,
:O:
bent molecule: H H
while electron configuration is still tetrahedral
Electronic structures
with more than 8 e- around central atom
Octet rule is strictly valid for the elements of 2nd Period. Starting with 3th period octet rule may be violated & more than 8 electrons, ( more than 4 el. pairs surround the central atom.
For the surrounding atoms – ligands –
octet rule is still valid
5 electron pairs around central atom
Trigonal bipyramid
Bond angles: 90 & 120º.
PF5 (but not NF5), SnCl5-
Lone pair, if present, repels other pairs stronger than bonding pairs. Therefore, lone pair occupies an equatorial position (bond angle (120o), not an axial position ((90o)
6 electron pairs around central atom
Octahedron
Review of the step-by-step procedure:
POLARITY OF MULTIATOMIC MOLECULES
Covalent bond between atoms of different electronegativity is polar:
δ+ δ− δ+ δ− δ+ δ−
H – O C – O H – N
However the net polarity of a multiatomic molecule depends on the bond angle between polar bonds.
δ− δ+ δ−
O = C = O or
-----------------------
(( ((
[pic]
Be
B
H †††††††††††††††††††††††ഠ††⁈††††††††††††††††††††䠠††††ㄠ〸൯⁈††敂††ൈ䄍汬
H H
180o
H Be H
All bond angles are 120º
H
H
H
H
C
SF4 6+7(4= 34e-
|F – S – F |
|F| |F|
1 lone pair
“see-saw” molecule
Lone pair
2 lone pairs
2 lone pairs
XeF4 8+7(4 = 36e-
IF5 7+7x5= 42e-
|F – I – F|
|F| |F|
|F|
1 lone pair. El. configuration: octahedral
Molecule geometry: square pyramid
(109.5o
O C O
O
H H
Total polarity:
Two dipoles add to each other to a total molecular dipole
H
NH3
N
H
H
Three dipoles add to each other to a total zero molecular dipole
H2O
..
..
H : N : H
H
attract electron pair ((
repel
N N N
S S
S N
N S
N S
S
S
N
3 equatorial positions
structural formula shows that there are
4 C – H bonds, but none
of H to H
Group V: PCl3 & PCl5 both exist, but only NF3
1) In the molecular formula, PCl3
identify the central atom.
2) Count total valence e- P + 3Cl = 5+3(7 = 26 e-
3) Assign 8e- (or 4 el pairs) 3Cl take 3(8 = 24e-
to each ligand.
4) Assign one pair as bonding for each ligand [pic]
5) Assign remaining e-, if any, to the central 1 lone pair
atom as lone pair(s).
6) Count total (bonding & lone) pairs
around central atom – identify el. configuration 4 total pairs -
tetrahedron
[pic]
7) Out of the ligand positions only,
identify the geometry of the molecule trigonal pyramid
2 axial positions
Here 2 lone pairs are present.
They repel stronger, ( go to opposite positions.
The molecule geometry is square planar.
All bond angles 90º. SF6, SiF62-
S
P
IF3 28e-
I
|F F|
|F| “T-shape”
in CCl4 (tetrahedron)
4 C-Cl dipoles cancel each other to net zero.
F
B
F F
In BF3, 3 bond dipoles cancel each other to add up to zero total polarity
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