Download Introduction to Ionic Bonding 11-4-16

The Periodic Table,
Valence Electrons
and Bonding
How Elements
Form Compounds
• I CAN describe chemical
bonding and determine a
binary ionic chemical
formula.
Chemical Bonding
• Chemical Bonding is the joining of two
or more atoms to form a compound.
• Bonding occurs when atoms transfer
electrons between them, or, in some
cases, move close together and share
electrons between them.
Valence Electrons
• During bonding, the only electrons
involved are those at the outer edge of
the atom’s electron cloud.
• These are called VALENCE ELECTRONS.
• Atoms can have between 1 and 8 Valence
Electrons [ Ve- ].
• When atoms come into close contact with
each other, VALENCE ELECTRONS can
be TRANSFERRED from one atom to
another.
• Typically, atoms on the LEFT SIDE of the
periodic table [the METALS] LOSE
electrons.
• Atoms on the RIGHT SIDE of the periodic
table [the NON-METALS] gain electrons.
• The number of electrons GAINED or LOST
depends on the GROUP NUMBER the atom
is in:
• LOSE ELECTRONS GAIN ELECTRONS
• Group 1 loses 1e• Group 2 loses 2 e• Groups 13 loses 3 e-
Group 15 gains 3 eGroup 16 gains 2 eGroups 17 gains 1 e-
Atoms To Ions
• When an atom gains or loses electrons
and there is a change in its overall
charge, the atom is transformed into
another form called an
ION
• POSITIVE IONS are called CATIONS and
are formed from METALS.
• NEGATIVE IONS are called ANIONS and
are formed from NON-METALS.
Charges of IONS
• Family 1 ions = +1
• Family 2 ions = +2
• Family 13 ions = +3
Losing electrons is
called OXIDATION
and make + charged
ions.
[Family 14 bonds in a different way – more later]
• Family 15 ions = -3
• Family 16 ions = -2
• Family 17 ions = -1
Gaining electrons is
called REDUCTION
and make - charged
ions.
• [Family 18 doesn’t form chemical bonds.]
Label the charges of the groups on your
period table as indicated below:
IONIC BONDING
• IONIC BONDING occurs between two
IONS with OPPOSITE CHARGES.
• Remember OPPOSITES ATTRACT!
Characteristics of IONIC Compounds
• Generally consist of a METAL CATION and
a NON-METAL ANION.
• High melting points
• Almost always solid at room temperature
• Brittle
• Generally not very flammable
• Dissolve in water [electrolytes]
Forming an IONIC Compound
• One method to illustrate the formation of
an ionic bond is to use LEWIS
STRUCTURES to show the transfer of Vefrom one atom to another.
PRACTICE PROBLEM
Use Lewis Structures to show how an atom
of Sodium and an atom of Chlorine would
form an ionic bond.
STEPS
• 1. Draw the LEWIS STRUCTURE for
ccceach element.
– Since we wish to show electron transfer, use
dots for one atom and Xs for the other.
•
Na
Cl
• 2. Note that chlorine has an empty space
that can accept an electron from sodium.
– Sodium has only 1 Ve- and is too far from 8 to
try to gain electrons, so it loses one Ve- to
chlorine:
•
Na
Cl
• 3. Now redraw the Lewis Structures with the
Ve- transferred from sodium to chlorine:
Na Cl
• Chlorine now has 8 Ve-, and so does
sodium.
– Since sodium lost its only Ve-, the next PEL
below becomes the new valence level and has 8
Ve-.
• 4. How has the charge changed for both
atoms?
Sodium lost 1 Ve- and becomes a +1 cation.
Chlorine gained 1 Ve- and becomes a -1 anion:
•
Na Cl
• Opposites Attract! The atoms stick together
due to their charges being opposite but equal!
Practice Problems
• Use Lewis Structures to show how ionic
bonds would form between the following
elements.
• A) K and I
B) Ca and Br
Determining Ionic Compound
Formulas Using Criss-Cross
(Charge Balance Method)
Forming an IONIC Compound
• The simplest IONIC COMPOUNDS are called
BINARY IONIC compounds because they
consist of only TWO DIFFERENT kinds of
atoms…..one a POSITIVE ION from the metal
side of the PT and the other a NEGATIVE ION
from the non-metal side of the PT.
• Together their CHARGES MUST add up to
equal ZERO.
• Ionic compounds are represented by an
EMPIRICAL FORMULA. This type formula
shows the least whole number ratio of the
elements in the compound.
Empirical Formulas
• Empirical Formulas, like all chemical
formulas, have two components:
– Chemical Symbols representing the elements
in the compound.
– Subscripts – numbers written AFTER a
symbol and LOWER than the symbol to
represent HOW MANY of each atom in the
formula.
– The subscript ALWAYS applies only to the
element it follows!
EXAMPLE
• What elements and how many atoms of
each are represented in these empirical
formulas?
• A) H2O
B) Mg2P3
•
2 atoms of H
2 atoms of Mg
•
1 atom of O
3 atoms of P
• C) AlCl3
•
1 atom of Al
•
3 atoms Cl
D) CH4
1 atom of C
4 atoms of H
PROBLEM
What would be the formula for a compound resulting from
ionic bonding between Potassium and Bromine?
• STEPS
• 1. Find the elements in the Periodic Table and write
down their chemical symbols.
• 2. Use the element’s location on the Periodic Table to
predict the charge of each atom as an ION.
• 3. Criss-cross the charge numbers (IGNORING + and SIGNS)…these become the SUBSCRIPTS in the
formula.
• 4. Write the FORMULA…positive ion always goes first!
• Write the symbols and charges in a
parenthesis above the symbols in this
manner:
( +1 ) + ( -1 ) = 0
K
Br
Do the charges add up to = 0?
If YES, the formula contains only 1 atom of
each element.
• So the formula is:
K1Br1
•
•
Since the symbol itself represents 1
atom, a subscript is not needed for 1!
K Br = KBr
1
1
What If the Charges
Do Not Add to Zero?
• BALANCE the charges by criss-crossing
the numbers from the charges (ignore the
+ or – sign).
(+2 )
Ca
+
( -3 ) = 0
P
To write the EMPIRICAL FORMULA for a binary
ionic compound:
• Write the symbol of the POSITIVE ION first, the
NEGATIVE ION second.
– Place the number you criss-crossed AFTER and
slightly BELOW the symbol.
– This number is called a SUBSCRIPT.
( +2 )
Ca
+
( -3 )
P
Ca3P2
= 0
PROBLEM
What is the formula for an ionic compound
of Magnesium and Chlorine?
• Determine the symbol and charge of both
elements and cross the numbers:
•
(+2 ) +
(-1 ) = 0
• Mg
Cl
• Mg1Cl2 = MgCl2
• Remember ONE is not usually written as a subscript!
Problem
What is the formula of an ionic compound of
Aluminum and Oxygen?
•
(+3) + (-2) = 0
Al
O
•
(+3) +
•
Al2
(-2 ) = 0
O3
• So the formula becomes:
• Al2O3