Match List I with List II:
|
List I (Nitrogen oxides) |
List II (Oxidation number of nitrogen) |
| A. \(NO\) | I. \(+4\) |
| B. \(N_2O_4\) | II. \(+2\) |
| C. \(N_2O_3\) | III. \(+5\) |
| D. \(N_2O_5\) | IV. \(+3\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is option 3. A-II, B-I, C-IV, D-III.
|
List I (Nitrogen oxides) |
List II (Oxidation number of nitrogen) |
| A. \(NO\) | II. \(+2\) |
| B. \(N_2O_4\) | I. \(+4\) |
| C. \(N_2O_3\) | IV. \(+3\) |
| D. \(N_2O_5\) | III. \(+5\) |
Let us determine the oxidation numbers of nitrogen in each of the nitrogen oxides listed:
1. Nitric oxide (\(NO\):
In \(NO\), the oxidation state of nitrogen can be found by setting up the equation based on the known oxidation state of oxygen (\(-2\)).
Let \(x\) be the oxidation state of nitrogen in \(NO\). The total oxidation state should equal the charge of the molecule, which is 0.
\(x + (-2) = 0\)
\(x = +2\)
2. Dinitrogen tetroxide (\(N_2O_4\)):
In \(N_2O_4\), let \(x\) be the oxidation state of nitrogen.
Each oxygen has an oxidation state of \(-2\). So, for \(N_2O_4\), the equation is
\(2x + 4(-2) = 0\).
\(2x - 8 = 0\)
\(2x = +8\)
\(x = +4\)
3. Dinitrogen trioxide (\(N_2O_3\)):
In \(N_2O_3\), let \(x\) be the oxidation state of nitrogen. The equation is
\(2x + 3(-2) = 0\).
\(2x - 6 = 0\)
\(2x = +6\)
\(x = +3\)
4. Dinitrogen pentoxide (\(N_2O_5\)):
In \(N_2O_5\), let \(x\) be the oxidation state of nitrogen. The equation is
\(2x + 5(-2) = 0\).
\(2x - 10 = 0\)
\(2x = +10\)
\(x = +5\)
So, the correct matches are:
A → II (\(NO\) has oxidation state \(+2\))
B → I (\(N_2O_4\) has oxidation state \(+4\))
C → IV (\(N_2O_3\) has oxidation state \(+3\))
D → III (\(N_2O_5\) has oxidation state \(+5\))
The correct answer is: 3. A-II, B-I, C-IV, D-III