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Text Elements
octet
2
4
4
6
2
Duet
metal
nonmetal
nonmetals
metal atoms
metal atoms
nonmetals
metals lose electrons to become cations nonmetals gain electrons to become anions
Electron configuration of ions: write the configuration for the neutral atom then adjust electrons positive ions lose the number of electrons of the charge form the valence shell ( outermost energy level ) negative ions gain the number of electrons of the charge into the valence shell ( outermost energy level )
Ions (cation) : metals lose e ions ( anion ) metals gain e
examples:
Na
1s2 2s2 2p6 es1
Na 1+ positively charged ( metals lose electrons )
1s2 2s2 2p6
Cl 1s2 2s2 2p6 3s2 3p5 Cl 1- ( nonmetals always gain ) 1s2 2s2 2p6 3s2 3p6
ionic - m + nm + - transfer
covalent - sharing nm + nm
picture non polar covalent bonds like 2 twins sharing equally
a polar bond is unequal sharing two different nonmetals, one is stronger than the other.
metallic bonding - m + m sea of electrons pool of electrons know this means metallic.
polar
nonpolar
Cl + Cl - > Cl Cl
H1
F7
H2F7
electrons
goes on one pager
ionic
non polar covalent
polar covalent
polar covalent
| electronegativity of first element - electronegativity of second element | = electronegativity difference 0.9 - 3.0 2.1 4.0 - 4.0 = 0 2.1 - 3.5 1.4 2.5 - 2.1 .4
2.1
m/nm
nm/nm
nm/nm
nm/nm
0
ionic
nonpolar covalent
1.4
polar covalent
.4
nonpolar covalent
npc
npc
solids
liquids or gasses
metals
water
lattice
molecules
electrons
protons
Type
Elements involved
valence electrons are
Lewis
1 ) count valence electrons 2 ) find center atom ( always C, never H, 1st one ) 3 ) put single bonds to terminal atoms ( subtract from valence electrons ) 4 ) add lone pairs to terminal atoms to make 8 (subtract) 5 ) add lone pairs to center atom 6 ) if no valence electrons, make double + triple bonds if needed.
No2
C1H4 (1) ( 4 * 1 ) + ( 1 * 4 ) = 8 (2) C (3) H | H - C - H | H each line is 2 electrons. total of 8 8 - 8 = 0 great! i want to use all my valence electrons. check that we have met the octet rule or the duet rule H | H - C - H | H
terminal atoms are outside atoms.
each hydrogen is happy because they each have 2 electrons
CH3I count valence electrons VE = (41) + (13) + (7*1) = 14 C
H
H C I
H
8
6
-6
0
PCI3: VE ( 5 * 1 ) + ( 7 * 3 ) = 26 you will always get even for valence electrons Cl
Cl P
Cl
-6
20 -18
2
double bonds are when 2 pairs of electrons are shared between the same two atoms lone pairs are pairs of electrons not shared only one atom counts them a triple bond occurs when two atoms share 3 pairs of electrons
H - C - C - H
H = 1 valence electron ( 1 * 1 ) C = 4 valence electrons ( 4 * 1 ) N = 5 valence electrons ( 1 * 1 ) 10 total valence electrons -4 6 -6 0
C
H
N
C
H
N
C
O
O
6 + 6 + 4 = 16 -4 = 12 -6 = 6 -6 = 0
single bonds are longer and weaker triple bonds are shorter and stronger
Add or subtract electrons from the valence electrons based on the charge if charge is negative, add electrons if positive, subtract electrons only other thing you do is you put the lewis dot diagram in hard brackets [] with a charge on the outside
Co3 -2 VE = (4 * 1 ) + ( 6 * 3 ) +2 = 24 -6 = 18 - 18 = 0
O — C — O | O
[
[
-2
Valence Shell Electron Pair Repulsion Theory
Valence Shell - outer shell involved in bonding Electron Pair - bonds are made of electron pairs Repulsion Theory
Linear
Linear
2 bonds 0 lone pairs
any 2 atom molecule has no central atom
Tetrahedral
4 bonds 0 lone pairs
2 bonds 1 lone pair is bent 2 bonds and 2 lone pairs is bent
bent
2 bonds 1 lone pairs
make sure that you know you have a good lewis dot drawing before you figure out the shape when you’re looking for the shape, you need to look at the center atom ONLY if there is no center atom, and you only have 2, its always linera. if there are only 2 atoms, it is always linear. you do not have to count bonding and nonbonding, because it is linear. when you have more than 2 you are counting bonding domains and bonding groups. you are looking to see where the electrons are hanging out. you dont care if it is a double bond or a single bond or a triple bond, it only counts as one bonding domain. 1 single bond is 1 bonding domain 1 double bond is 1 bonding domain 1 triple bond is 1 bonding domain
H C N
^ 1
^ 1
total of 2 bonding domains.
HCN
VE = ( 1 * 1 ) + ( 4 * 1 ) + ( 5 * 1 ) = 10. 10 - 4 = 6 ( from the 2 bonds ) . 6 - 6 ( from the getting the nitrogen to 8 ) = 0. nitrogen still not happy. switch to double bond. now carbon has 6. switch to triple bond. now carbon has 8. 10 valence electrons center is carbon
C
H
N
boron only needs 6 valence electrons hydrogen only needs 2 beryllium only needs 4
positive
Negative
2 poles
δ+
δ-
H - F
δ-
δ+
: N ::: N :
nonpolar ^
S
H
H
it has lone pairs on it - it is bent so it is a polar molecule
Cl
B
Cl
Cl
if the terminal atoms ( outside atoms ) are all the same, and there are no lone pairs on the center atom, then it is NON POLAR otherwise it is polar.
H
C
H
O
N
H
H
H
polar
polar
H - O - H
polar
bent
trigonal pyramidal
4+ 6 + 6 = 16-4 = 12
-12
=0
O - C - O
nonpolar
linear
C
4 + 7*4 = 32 -8
Cl
Cl
Cl
Cl
tetrahedral
nonpolar
2h2O - > 2h2 + o2
2h2O(i) - > 2h2o(g)
larger
positive
more electrons
negative
weird+ and weird-
network
ionic, metallic, and covalent
Hydrogen, dipole, london dispersion
intramolecular is ALWAYS stronger than intermolecular chemical bonds are always stronger than physical bonds.
ionic, metallic, and covalent
Hydrogen, dipole, london dispersion
ionic covalent or metallic
Ionic IMF
Is the molecule polar?
yes
does it have an H bonded to an F, O, or N
yes
H bonding IMF
no
Dipole Dipole IMF
no
LDF
Metallic IMF
nm nm
2 different things with london dispersion forces the stronger of the two can be found by counting electrons the more electrons the stronger the london dispersion forces if you have 2 things and they both have london dispersion forces, figuring out which ones stronger is just counting electrons.
Shes probably gonna move the test - gonna be moved to thursday tuesday is tutoring
the number and strength of the forces affect the properties of the substance
nm m
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