oxidation numbers?

Question

Oxidation numbers?
can you please help me with my homework?
i am having trouble in the chemical formula and oxidation numbers.
it states here that i should compute for the oxidation number of calcium in calcium nitrate, oxygen in barium iodide, nickel in Ni(NO3)2, sulfur in sulfate ion and phosphorous in phosphate ion.?.please i need help!

Answer 1

firstly, u should write down the chemical formula for the compounds given.

for calcium nitrate, tat'll be Ca(NO3)2. calcium usually exists in oxidation state +2 only, so tis question is easy to answer.

barium iodide will be BaI2, cos barium, like calcium is from Grp II in the periodic table, and the predominant form of oxidation state is +2. there is no oxygen in this compound, so i think u better double check if u copied the question wrongly.

already Ni(NO3)2, the chemical formula is given. NO3 is nitrate ion, with a charge of -1 (oxygen is -2, nitrogen is +5). Ni(NO3)2 is a salt, its neutral, so u can calculate (oreven see) that Ni is +2 here (since there are 2 nitrate ions bonded to one Ni ion)

sulfate is (SO4)2-. let x be the oxidation number of S.
x + (4)(-2) = -2 (oxygen is -2, and u have 4 of them)
hence, x = -2 - (-8) = +6

phosphate ion is written as (PO4)3-. let y be the oxidation number of phosphate, P:
y + (4)(-2) = -3
y = -3 - (-8) = +5
note that phosphate is in the same grp as nitrogen, Grp V

Answer 2

Well, I don't have a periodic table in front of me now so here is the short. You simply need to account for ALL oxygen with in the compounds on the oxidated solution. For example:

NO3^2=N^2/O^6 if all oxygen is lost from the compound.

Hope this helps.

Answer 3

Ni(NO3)2

(NO3)2 HAS the O.N -1.

SO,

let x be the o.n of NI.

x+(2X-1)=0
x-2=0
x=2.