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{{nav|CH 101|Keys |{{CH 101num_exerc}}}}
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Back to [[CH 101]]
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Hover your mouse over the ionic formula to see the name of the ion or the common oxidation states of the elements. <br>
<big>'''Make sure you memorize these.'''</big>

{| style="float:left; margin-left:0.5em; margin-bottom:0.5em; border: 1px solid #AAAAAA; border-collapse:collapse; text-align:center;" cellpadding="3" rules="all"
!colspan="5" bgcolor="lightgreen"|common anions
|-
|{{E|NO<sub>3</sub><sup>-</sup>|nitrate}}
|{{E|NO<sub>2</sub><sup>-</sup>|nitrite}}
|{{E|PO<sub>3</sub><sup>3-</sup>|phosphite}}
|{{E|N<sup>3-</sup>|nitride}}
|{{E|CrO<sub>4</sub><sup>2-</sup>|chromate}}
|-
|{{E|SO<sub>4</sub><sup>2-</sup>|sulfate}}
|{{E|SO<sub>3</sub><sup>2-</sup>|sulfite}}
|{{E|S<sub>2</sub>O<sub>3</sub><sup>2-</sup>|thiosulfate}}
|{{E|S<sup>2-</sup>|sulfide}}
|{{E|MnO<sub>4</sub><sup>-</sup>|permanganate}}
|-
|{{E|ClO<sub>4</sub><sup>-</sup>|perchlorate}}
|{{E|ClO<sub>3</sub><sup>-</sup>|chlorate}}
|{{E|ClO<sub>2</sub><sup>-</sup>|chlorite}}
|{{E|Cl<sup>-</sup>|chloride}}
|{{E|SnO<sub>3</sub><sup>2-</sup>|stannate}}
|-
|{{E|BrO<sub>3</sub><sup>-</sup>|bromate}}
|{{E|BrO<sub>2</sub><sup>-</sup>|bromite}}
|{{E|BrO<sup>-</sup>|hypobromite}}
|{{E|Br<sup>-</sup>|bromide}}
|{{E|SnCl<sub>6</sub><sup>2-</sup>|hexachlorostannate}}
|-
|{{E|IO<sub>4</sub><sup>-</sup>|periodate}}
|{{E|IO<sub>3</sub><sup>-</sup>|iodate}}
|{{E|IO<sub>2</sub><sup>-</sup>|iodite}}
|{{E|I<sup>-</sup>|iodide}}
|{{E|VO<sub>4</sub><sup>3-</sup>|(ortho)vanadate}}
|-
|{{E|CO<sub>3</sub><sup>2-</sup>|carbonate}}
|{{E|HCO<sub>3</sub><sup>-</sup>|"hydrogen carbonate (bicarbonate)"}}
|{{E|PO<sub>4</sub><sup>3-</sup>|phosphate}}
|{{E|H<sub>2</sub>PO<sub>4</sub><sup>-</sup>|"dihydrogen phosphate"}}
|{{E|AuCl<sub>4</sub><sup>-</sup>|tetrachloroaurate}}
|-
|{{E|SeO<sub>3</sub><sup>2-</sup>|selenite}}
|{{E|HSO<sub>4</sub><sup>-</sup>|"hydrogen sulfate (disulfate)"}}
|{{E|PS<sub>4</sub><sup>3-</sup>|thiophosphate}}
|{{E|CN<sup>-</sup>|cyanide}}
|{{E|FeO<sub>4</sub><sup>2-</sup>|ferrate}}
|-
|{{E|BO<sub>3</sub><sup>3-</sup>|borate}}
|{{E|S<sub>2</sub>O<sub>8</sub><sup>2-</sup>|"peroxydisulfate persulfate"}}
|{{E|AsO<sub>4</sub><sup>3-</sup>|arsenate}}
|{{E|As<sup>3-</sup>|arsenide}}
|{{E|WO<sub>4</sub><sup>2-</sup>|tungstate}}
|-
|{{E|BF<sub>4</sub><sup>-</sup>|fluoroborate}}
|{{E|O<sub>2</sub><sup>2-</sup>|"peroxide"}}
|{{E|S<sub>2</sub><sup>2-</sup>|disulfide}}
|{{E|H<sup>-</sup>|hydride}}
|{{E|OCN<sup>-</sup>|cyanate}}
|}

{| style="float:left; margin-left:0.5em; margin-bottom:0.5em; border: 1px solid #AAAAAA; border-collapse:collapse; text-align:center;" cellpadding="3" rules="all"
!colspan="5" bgcolor="lightgreen"|common oxidation states
|-
|{{E|Al|"3+"}}
|{{E|C|"4+"}}
|{{E|Co|"2+,3+"}}
|{{E|Fe|"2+,3+ occasionaly 6+"}}
|{{E|Au|"1+, 3+"}}
|-
|{{E|Ag|"1+"}}
|{{E|Cu|"2+ with O,F,Cl 1+ others"}}
|{{E|Zn|"2+"}}
|{{E|Ga|"3+"}}
|{{E|In|"3+, somtimes 1+"}}
|-
|{{E|Tl|"1+, sometimes 3+"}}
|{{E|Sn|"2+, 4+"}}
|{{E|Ge|"4+"}}
|{{E|F|"1-"}}
|{{E|Cl|"7+,5+,3+,1+,1-"}}
|-
|{{E|I|"7+,5+,3+,1+,1-"}}
|{{E|Br|"7+,5+,3+,1+,1-"}}
|{{E|S|"6+,4+,2+,1-,2-"}}
|{{E|Se|"6+,4+,2+,1-,2-"}}
|{{E|O|"2-, 1-"}}
|-
|{{E|Na|"1+"}}
|{{E|Mn|"7+,6+,4+,3+,2+"}}
|{{E|Rb|"1+"}}
|{{E|Cr|"6+,4+,3+,2+"}}
|{{E|Cs|"1+"}}
|-
|{{E|N|"5+,3+,1+,-3"}}
|{{E|Ce|"4+,3+"}}
|{{E|Sc|"3+"}}
|{{E|Ca|"2+"}}
|{{E|Ba|"2+"}}
|-
|{{E|Ni|"2+"}}
|{{E|Zr|"4+"}}
|{{E|Os|"8+,3+,2+"}}
|{{E|Be|"2+"}}
|{{E|U|"2+,3+,6+"}}
|}
<br clear="all">

==Determining oxidation states==

This is a little puzzle in which you make all the charges balance. The best way to figure oxidation states is to do a little calculation, e.g for the dichromate ion Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup>, you would start by putting the charge of the entity (anion in this case) behind the equal sign:

:.... = 2-

Before the equal sign you start to put in what you know for sure. Unfortunately both chromium and oxygen can have more than one oxidation state, but for O the 2- state is the most common one, while chromium can be a lot of things. So your best bet is to make O equal to 2- for the moment and compute their contribution to the charge:

:  ... +7*(-2) = -2

We now have to make an assumption: that both chromium atoms are in the ''same'' oxidation state. That is commonly the case, although mixed valent compounds do exist where an element occurs in two different states. If both chromiums are the same we can write their charge as 2x and complete the equation:

: 2x + 7*(-2) = -2

Now do a little simple algebra:

: 2x - 14 = -2

Bring the 14 to the other side (sign changes!).

: 2x = +14 -2 = +12

Dividing by two gives:

: x = +6

So we can make the charges balance by assuming that Cr is 6+ and O is 2-. This confirms that we do not need to look at a possible -1 state for oxygen.

I do admit that this procedure is ''not'' airtight. There are cases where the stoichiometry does not lead to a single set of possible oxidation states. In difficult cases the only way out is to determine them experimentally. Using XPS that is possible because the energies of the core levels of say Cu<sup>2+</sup> and Cu<sup>+</sup> are a bit different and you can see that in the kinetic energies of the photoelectrons you liberate from the compound. We will not deal with those difficult cases in CH 101 though.

==Sour and bitter oxides==

Many oxides can react with water to form either an '''oxo-acid''' or a '''hydroxide'''. These oxides are know as '''acidic''' oxides and '''basic''' oxides respectively. The oxidation number and the electronegativity of the central atom determine what kind of oxide you get.

;Low ox / lox EN  ====> Basic
;High ox / high EN ====> Acidic

So elements is the lower left of the periodic table give basic oxides:

:Na<sub>2</sub>O + H<sub>2</sub>O ==> 2NaOH   (sodium hydroxide)
:BaO + H<sub>2</sub>O ==> Ba(OH)<sub>2</sub>   (sodium hydroxide)

Notice that we write OH as a ''group'', the '''''hydroxyl group''''', because it is this group that tends to react.

In the top right we get acidic oxides, particularly at high oxidation numbers:

:SO<sub>3</sub> + H<sub>2</sub>O ==>  H<sub>2</sub>SO<sub>4</sub> (sulfuric acid)

Notice that we do not write  SO<sub>4</sub>H<sub>2</sub> or SO<sub>2</sub>(OH)<sub>2</sub>. In principle, we could do that but it is the 'acidic' hydrogens that tend to react (not -OH groups) and we express that by showing them first.  

Lower oxidation states usually give ''weaker acids'', e.g. 

:SO<sub>2</sub> + H<sub>2</sub>O ==>  H<sub>2</sub>SO<sub>3</sub> (sulfur''ous'' acid)

Some of the transition metals also have high oxidation numbers and moderate electronegativities and this leads to acidic oxides, e.g. CrO<sub>3</sub> is acidic but Cr<sub>2</sub>O<sub>3</sub> is a basic oxide.

==Neutralization and salt formation==

Hydroxides and oxo-acids react with each other in a scheme called '''neutralization'''. Their active -OH and -H groups will then combine to form back water, and what is elft is called a '''salt''':
{|
|Na'''OH'''|| +|| '''H'''NO<sub>3</sub>|| ==>|| '''H<sub>2</sub>O'''||+|| NaNO<sub>3</sub>
|-
|hydroxide ||+|| oxo-acid||==>||water||+||salt
|-
|<font color = blue>sodium</font> hydroxide||+||<font color = red>nitric</font> acid||==>||water||+||<font color = blue>sodium</font> <font color = red>nitrate</font>
|}

Notice that the salt simply inherits the metal name <font color = blue>sodium</font>, but the acid name <font color = red>nitric</font> turns into <font color = red>nitrate</font> in the salt. Nitrates are said to be "the salts of nitric acid"

We can always think of a salt as having been formed from an acid and a base, even though it may not be possible to do the reaction in practice, e.g. because one of the two do not really exist.

The acids and bases mentioned here are known as Arrhenius acids and bases. Later we will see a considerable expansion of the definition.

==Exercises==

Write the chemical formula for the following compounds and determine the oxidation states for all elements. Apart from the hydrides, all the compounds are salts, so you could imagine that they were prepared by reacting an appropriate hydroxide and acid. Write out the formation reaction (balance equation please!) for each of them.

* lanthanum sulfate
* sodium bicarbonate
* calcium peroxide
* lithium aluminium hydride
* rubidium ferrate
* scandium arsenide
* barium phosphate
* calcium borate
* potassium chlorite
* lithium fluoroborate
* cesium hydride
* vanadium(II)sulfate
* iron(II) disulfide
* erbium selenate
* sodium hypochlorite
* calcium hydrogen sulfate
* indium chloride
* thallium bromide

(See key page for solutions)
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