What is the oxidation number of chlorine in Al(ClO4)3?
You can read about it here.
http://www.chemteam.info/Redox/Redox-Rules.html
Please read the above but here is how you do this one.
All compounds are zero.
Al is in group III (or 13 depending onh the system used) so the oxidation state is 3.
You have 12 O at -2 each for -24. That leaves -24+3 = -21. So 3 Cl atoms must be +21 to balance the -21 there which makes +21/3 or +7 for each Cl.
To determine the oxidation number of chlorine in Al(ClO4)3, we need to apply some rules and calculations.
1. The oxidation number of aluminum (Al) is +3 since it is a Group 3A metal.
2. The overall charge of any compound is zero since it is neutral.
3. The compound includes three chlorate ions (ClO4-). The overall charge of these ions combined is (-1) * 3 = -3.
Now, let's assign "x" as the oxidation number of chlorine (Cl).
To calculate the oxidation number of Cl, multiply the oxidation number of Cl by the number of Cl atoms present in the compound (which is 1) and then add it to the total charge of the compound (which is -3).
So, we can set up the equation:
x * 1 + (+3) = -3
Simplifying the equation:
x + 3 = -3
Subtracting 3 from both sides:
x = -6
Therefore, the oxidation number of chlorine in Al(ClO4)3 is -6.
To find the oxidation number of chlorine in Al(ClO4)3, let's break down the compound:
- ClO4 is the perchlorate ion, which has a charge of -1.
- Al(ClO4)3 implies that there are three perchlorate ions attached to one aluminum ion.
Since the overall charge of Al(ClO4)3 is neutral, we can set up an equation:
(negative charge of ClO4) * (number of ClO4 ions) + (charge of Al) = 0
-1 * 3 + x = 0 (where x is the oxidation number of Cl)
Simplifying the equation:
-3 + x = 0
x = +3
Therefore, the oxidation number of chlorine in Al(ClO4)3 is +3.