The Periodic Table, Valence Electrons and Bonding

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Presentation transcript:

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 15 gains 3 e- Group 2 loses 2 e- Group 16 gains 2 e- Groups 13 loses 3 e- Groups 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 14 bonds in a different way – more later] Family 15 ions = -3 Family 16 ions = -2 Family 17 ions = -1 [Family 18 doesn’t form chemical bonds.] Losing electrons is called OXIDATION and make + charged ions. Gaining electrons is called REDUCTION and make - charged ions.

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 Ve- from 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 Na Cl 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-.

Sodium lost 1 Ve- and becomes a +1 cation. 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 D) CH4 1 atom of Al 1 atom of C 3 atoms Cl 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!

Do the charges add up to = 0? 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! K1Br1 = KBr

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 ) + ( -3 ) = 0 Ca 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 ) + ( -3 ) = 0 Ca P Ca3P2

PROBLEM What is the formula for an ionic compound of Magnesium and Chlorine?

Mg Cl Mg1Cl2 = MgCl2 (+2 ) + (-1 ) = 0 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) + (-2 ) = 0 Al2 O3

So the formula becomes: Al2O3