Tx_DWboEVs IJqPU11ngY

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Tx_DWboEVs
http://www.youtube.com/watch?v=5
IJqPU11ngY

Valence electrons
› electrons in the highest occupied energy level
› always in the s and p orbitals
 normally just a draw a circle to represent these two
orbitals
› determines the chemical properties of an element
› usually only the valence electrons are used in
chemical bonds
http://wps.prenhall.com/wps/media/objects/476/488316/Instructor_Resources/Chapter_09/FG09_27-06un.JPG
Use this one

electron dot structures/diagrams are used to show
valence electrons
› each dot around the element symbol represents a
valence electron

B is 1s2 2s2 2p1;
› 2 is the outermost energy level
› it contains 3 valence electrons, 2 in the
s and 1 in the p

Br is [Ar] 4s2 3d10 4p5
How many valence electrons are
present?
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
when forming compounds, atoms tend to achieve
the electron configuration of a noble gas (ns2np6)
› this means their highest energy level will be filled
with 8 electrons that requires the least amount of
energy
› cations
 atoms of metallic elements (groups 1-14) lose
electrons producing cations (positive ions)
 the term oxidized (O.I.L – Oxidized Is Losing) is used
to say it lost an electron
 example- Ca becomes Ca2+
› anions
 atoms of nonmetallic elements (groups 15,
16, 17) gain electrons producing anions
(negative ions)
 the term reduced (R.I.G.) – Reduction Is
Gaining) is used to say gaining electron(s)
 example- Cl becomes Cl1-
Na
Mg
N
O
F
Na+
Mg2+
N3-
O2-
F1-
Cations
Anions
Ne
Ionic charge
1+
2+
3+
4+
3-
2-
1-
0
...etc.
As it turns out, atoms bond together for a very
simple reason: atoms that have full valence
shells are more stable than those that don’t.
› some transitional metals can often form more
than one cation and therefore charges may
vary
 some orbitals get filled but might not get a noble
gas electron configuration
 iron can be Fe2+ or Fe3+ (you don’t have to memorize
this)

usually composed oppositely charged metallic
cations and nonmetallic anions
+ cations
- anions
› because of the electrostatic attraction between
oppositely charged particles
 electro- electricity/electrons
 static- non-moving (vs. current which is moving charges)

forms a compound with overall zero charge
› (+ and -) need to cancel out

referred to as formula units, not a molecule,
that is a term used only for covalently bonded
atoms which is something different

ions must have a difference in electronegativity of
1.7 or greater to form an ionic bond
electronegativty
difference
0.0
probable type of bond
0.1 – 1.0
covalent, nonpolar
covalent, slightly polar
1.1 – 1.7
>1.7
covalent, very polar
ionic
Na “gives” Cl
one electron and
now both atoms
have a full
valence shell are
chemically
stable (electron
configuration of
a noble gas)
.9
3
e–
1)
2)
Na
Cl
3)
Na+
Cl–

shows the kinds and numbers of atoms in
the lowest whole-number ratio of ions in the
ionic compound
› Na 1+ + Cl 1-  NaCl
› Ca 2+ + Cl 1-  CaCl2
› Mg 2+ + O 2-  MgO
Mg and N?
 Mg2+, N3 it takes two
nitrogens to
combine with
three
magnesiums
› Mg3N2
› magnesium
nitride
Na
1+
+
1F
 NaF
 cations
keep the same name with no change
› transition metals form more than one cation
with different ionic charges
› charge is determined from the # of electrons
lost
 iron can be Fe2+ or Fe3+
› naming
 Stock system
 a Roman numeral is used to show the charge of the
transition metal
 Fe2+ is iron(II) and Fe+3 is iron(III)
› anions gain valence electrons and therefore
are negatively charged
 change ending to –ide for the name
 composed
of more than one atom that
form a single unit/ion
› these have their own name
› most end with -ite or –ate
 the –ite ending indicates one less oxygen than
the –ate ending
These will be given on tests
NO3-1
nitrate
NO2-1
nitrite
OH-1
hydroxide
ClO2-1
chlorite
ClO3-1
chlorate
HCO3-1
SO42-
hydrogencarbonate
(bicarbonate)
sulfate
SO32-
sulfite
CO3 2-
carbonate
PO43-
phosphate
NH4+
ammonium
› cation is written first, followed by the
anion with and –ide ending
 Cs2O  cesium oxide
 SrF2  strontium fluoride
 transitional metals are more difficult to
name
 CuO  copper(II) oxide
 oxygen is always 2- and therefore copper will be
2+
 Cu2O  copper(I) oxide
 oxygen is always 2- and therefore needs two
copper atoms with 1+ charge
 write
symbol of cation and then anion
 add subscripts to balance the charges
› calcium bromide
 Ca2+ and Br1- is CaBr2
› potassium sulfide
 K+1 and S2- is K2S
› iron(III) oxide
 Fe+3 and O2- is Fe2O3

the ionic charge number of each ion is crossed
over and becomes the subscript for the other
ion
 Compounds
with polyatomic ions
› do same as binary compounds
 use the name of the polyatomic ion with
the mono atomic ion
 calcium nitrate  ?
 Ca is 2+ and NO3 is 1 Ca(NO3)2
 Li2CO3  ?
 lithium carbonate
Fe(NO3)3
Choose the correct name for the compound
1. Iron trinitrate
2. iron(I) nitrate
3. iron(III) nitrite
4. iron(III) nitrate
5. none of the above
Periodic
Chart
Polyatomic
Ions
next problem
sodium chlorite
Choose the correct formula for the compound
1. NaCl
2. NaClO
3. NaClO2
4. Na(ClO)2
5. none of the above
Periodic Chart
Prefixes
next problem

high melting and boiling points
› takes a lot of energy to break apart the electrostatic
forces between cations and anions

crystalline structure
› repeating arrays of cations and ions
› an ionic lattice
 volatility- how
easily a substance turns into
a gas
› very low as electrostatic forces between cations
and anions is very strong
 electrical conductivity
› can conduct electric currents when melted or dissolved
in water because cations and anions are free moving
instead in a crystal
 solubility
› will dissolve in other polar solvents such as water
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