If an element has gained or lost electrons in a reaction, this would be reflected by a change in its oxidation number, i.e the oxidation number expresses the oxidation state of an element.
RULES FOR DETERMINING OXIDATION NUMBERS.
The set of rules used for the determination of oxidation numbers are as follows :1. The oxidation number of an element in his uncombined or free state is zero(0).That is if oxygen is uncombined(has not reacted with anything its oxidation state is zero).
2. The oxidation state of a simple ion(an ion consisting of a single element) has the same size and sign as the charge of the ion. The oxidation number of Na+, Mg²+,Al³+,Cl- and S²- are +1,+2,+3,-1 and -2 respectively. Note that the negative or positive sign for an oxidation number always comes before the number.
Where the ion consists more than one element, its oxidation number is the algebraic sum of all the oxidation numbers of all elements in the ion.
3. The algebraic sum of the oxidation numbers of all the elements in a compound is zero.
4. In most compounds, the oxidation number of oxygen is - 2 (except in peroxides where its oxidation number is -1,e.g Hydrogen peroxide(H2O2),while that of hydrogen is +1 (except in metallic hydrides {e.g Lithium hydride(LiH)} where its oxidation is - 1).
How to know the oxidation number of first 20 elements.
Now we are going to use our electronic configuration to find it out, remember the oxidation number of uncombined element is zero, we are calculating the oxidation numbers so that we can use the numbers in calculating the oxidation number of compounds.
H, Z(symbol for atomic number) =1,now hydrogen we give away its 1 to become stable, so when it gives away 1,it's short of 1 so lets say (-1) but electrons are always negative so (-1)×electron(e) = +1,so the oxidation number of hydrogen is +1.
If you learnt electronic configuration elements with valence electrons of 1 or 2 or 3 or 4 tend to give their electrons away, but from 5-7 tend to borrow electrons.
Nitrogen, Z:7=2,5
Now it will borrow 3 to become octet/stable so borrowing 3 is +3×electron(e) = - 3,so its oxidation number is - 3.
Magnesium, Z:12=2,8,2, it will give the two electrons to become octet so giving away two is - 2×e=+2.
Chlorine, Z:17=2,8,7
it needs to borrow 1 to be octet, so borrowed 1 is +1×e=-1.
So you can do as many as you want so now lets solve examples on Oxidation numbers.
Before we start solving if you were to find the oxidation number of an element in a two-element compound, the element at the left is the predominant element, while in three-element compound the one in the middle is the predominant element, and you use a letter to represent the oxidation number of the predominant element so that you will ill find the value for the letter.
Examples
1. Find the oxidation number of sulphur in H2SO4.Solution
Hydrogen, Z=1=-1×e=+1
Sulphur, predominant element
Oxygen, Z:8=2,6= +2×e=-2
Let the oxidation number of sulphur be n
(H¹)2(Sⁿ)(O²-)4 = 1×2+1×n+4×-2=0
2+n-8=0
n-6=0,n=6
So the oxidation number of sulphur in H2SO4 is 6,thats why its called hydrogen tetraoxosulphate (vi).
2. Find the oxidation number of PO4³-.
Now the sum of oxidation number in an ion is equal to the charge which in this case is -3.
P= predominant element
Oxygen, Z:8=2,6=+2×e=-2
Let the oxidation number of phosphorous be n.
(Pⁿ)(O²-)4 =-3(charge of the ion)
1×n+4×-2=-3
n-8=-3,n=8-3=5.
So the oxidation number of phosphorous in PO4³- is 5.
3. Find the oxidation number of Chromium in K2Cr2O7.
Potassium, Z=19=-1×e=+1
Chromium= predominant element
Oxygen, Z=8=+2×e=-2
2(K¹){(Crⁿ)2}(O²-)7
2×1+2×n+7×-2=0
2+2n-14=0
2n-12=0,2n=12,n=12/2=6.
So the oxidation number of Chromium in K2Cr2O7 is 6.
4. Find the oxidation number of Lithium in LiH.
Lithium= predominant element
Hydrogen =-1(hydride mostly when hydrogen reacts with alkali metals)
(Liⁿ)(H-¹)=0
1×n+1×-1=0
n-1=0,n=1,so the oxidation number of Lithium in LiH is 1,
You can drop a comment and can also read my post on how to name inorganic and organic compounds.
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