I have a question which asks...
What is the Oxidation number of Nitrogen in Cu(NO3)2
I know the answer is 5+ but how do you get that??
Cu= 2+
O= 2- (Theres 3 of them so -2x3=-6...then, theres a two out of the brackets so 6- x 2 =-12)
+2 - (-12)= -10? And then what?
I think my method is wrong. :/ It's really frustrating and i've been at the same question for over an hour now.
Cu=2+
O=2-
Put x for N
Then ,
2(ie cu)+2×((x+3×-2))=0 (3 times O is there so 3×-2 , also both N and and O are there two times..so multiply them with two)
2+2x-12=0
2x=10
X=+5
Hope you understood🙂
To determine the oxidation number of nitrogen (N) in Cu(NO3)2, you need to consider the overall charge of the compound and the known oxidation numbers of the other elements present.
In this case, the compound Cu(NO3)2 can be broken down as follows:
Cu - Nitrate (NO3) is a polyatomic ion, and the sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion. In this case, nitrate (NO3) has a charge of -1. Since the compound contains two nitrate ions, the total charge from the nitrate ions is -2 (-1 x 2).
Cu has an oxidation number of +2. This is known because Cu(NO3)2 is a neutral compound, and the sum of the oxidation numbers of all the elements must equal zero.
To find the oxidation number for nitrogen (N), you can use the formula:
Total charge of compound = charge of cation + (charge of anion x number of anions)
In this case, the total charge of the compound is zero, the charge of the cation is +2, and the charge of the anion is -1. The number of anions is two.
So, 0 = +2 + (-1 x 2)
Solving this equation gives:
0 = +2 - 2
Therefore, the oxidation number of nitrogen (N) in Cu(NO3)2 is +5.
To determine the oxidation number of nitrogen (N) in Cu(NO3)2, you can assign oxidation numbers to the other elements in the compound and then use the known charge of the compound to find the oxidation number of nitrogen.
In this case, you correctly identified the oxidation state of copper (Cu) as 2+.
Next, consider the nitrate (NO3) ion. Nitrate is a polyatomic ion, which means it has an overall charge. The known charge of the copper nitrate compound (Cu(NO3)2) is neutral, which means the sum of the oxidation numbers of all the elements must add up to zero.
Since oxygen (O) is usually assigned an oxidation number of -2 in compounds, and there are three oxygen atoms in nitrate, the total oxidation number contribution from the oxygen atoms will be -2 x 3 = -6.
Given that the compound is neutral and the copper ion contributes +2 to the overall charge, we can calculate the oxidation number of nitrogen as follows:
Total oxidation number contribution from all elements in the compound = 0
Oxidation number of copper (Cu) = +2
Oxidation number of nitrogen (N) = X
Total oxidation number contribution from oxygen in nitrate = -6
Therefore, the equation becomes: (+2) + X + (-6) = 0
Simplifying the equation, we have: X - 4 = 0
Solving for X, we find that the oxidation number of nitrogen is +4.
Therefore, in the compound Cu(NO3)2, the oxidation number of nitrogen (N) is +4, not +5 as you initially stated. It seems there was an error in your reasoning, possibly in calculating the total oxidation number contribution from the oxygen atoms.
Remember to double-check your calculations and keep in mind the overall charge of the compound to arrive at the correct oxidation number.