What is the oxidation number of chlorine in Al(ClO4)3?
I don't understand this question, can someone help? Thank you!
Given
O3Cl- Al - ClO3
l
ClO3
Formal Charge = [Valence - (Bonded Electrons/2) - Non-bonded electrons]
FC(Al) = 3 - (6/2) - 0 = 3 - 3 = 0
FC(O) = 6 - (2/2) - 6 = -1
FC(Cl) = 7 - (8/2) - 0 = +3
Net Charge = 1Al + 3Cl = 9O
= 1(0) + 3(+3) + 9(-1)
= 0 + 9 - 9 = 0
Conforms to conservation of matter and of charge.
I don't get this one
1. Combining solutions of silver nitrate and sodium chloride
produces a silver chloride precipitate:
AgNO3(aq) 1 NaCl(aq) → AgCl(s) 1 NaNO3(aq)
A researcher discovered that 32.0 mL of 0.100 mol/L silver
nitrate is required to precipitate all the chloride ions in 25 mL
of a solution of sodium chloride.
(a) What is the amount concentration of sodium ions in the
initial sodium chloride solution?
(b) What is the concentration, in g/L, of sodium chloride in
the initial sodium chloride solution?
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.
O is always a -2 oxidation state, unless it is in a peroxide, so -2*4*3 is -24, Al is +3 because an ion's oxidation state is its charge, which leaves the overall charge right now at -21, so to make the overall charge 0, 3 times the Oxidation state of Cl must equal positive 21, so the oxidation state of Cl is 7
Of course, I can help explain how to determine the oxidation number of chlorine in the compound Al(ClO4)3.
To find the oxidation number of an element in a compound, you need to consider a few rules:
1. The sum of the oxidation numbers of all the atoms in a compound must equal the overall charge of the compound. In this case, because Al(ClO4)3 is a neutral compound, the overall charge is zero.
2. The oxidation number of an atom in an element is always zero. In this case, the oxidation number of Al is zero.
3. The oxidation number of an ion is equal to its charge. In this case, the perchlorate ion (ClO4-) carries a charge of -1.
Now, let's break down the compound Al(ClO4)3:
Al(ClO4)3 consists of one aluminum ion (Al3+) and three perchlorate ions (ClO4-).
Since the overall charge of Al(ClO4)3 is zero, we can determine the oxidation number of chlorine by assigning appropriate oxidation numbers to the aluminum and oxygen ions.
From rule #1, we know that the sum of the oxidation numbers must be zero. Since the oxidation number of aluminum is +3 (which we get from rule #2), then the sum of the oxidation numbers of chlorine and oxygen must be -3.
Since there are four oxygen atoms in each perchlorate ion, their total oxidation number is -4 (4 * -1 = -4).
Therefore, to satisfy the sum of -3, the oxidation number of chlorine in each perchlorate ion must be +5 (-4 + x = -3, where x is the oxidation number of chlorine).
So, the oxidation number of chlorine in Al(ClO4)3 is +5.
Remember to always check the overall charge of the compound and apply the rules for determining oxidation numbers to each element in the compound.