What is the oxidation number of chlorine in Al(ClO4)3?

Question

I don't understand this question, can someone help? Thank you!

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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?