C=4
n= -3
oxygen is -2
OCN has net charge of -1 so here is my expalnation
we will take C as unknown
-2+x-3=-1
x=4
now N is the unknown :
-2+4+x=-1
x=-3
and Oxygen is always -2 except peroxides and others cases...
hope i helped !
Wiki User
∙ 12y agoThe oxidation number of nitrogen in HCN (hydrogen cyanide) is -3. This is because hydrogen has an oxidation number of +1 and carbon has an oxidation number of -1, leaving nitrogen with an oxidation number of -3 to balance out the charge of the molecule.
Wiki User
∙ 12y agoas H has oxd'n 1 so oxd'n of N is -1
Wiki User
∙ 11y agoH +1
c +2
n -3
Wiki User
∙ 13y ago-3.
Wiki User
∙ 12y ago-1
In both HCN and HNC molecules, the oxidation number of carbon (C) is -3. This is because hydrogen (H) is almost always assigned an oxidation number of +1, and nitrogen (N) is usually assigned an oxidation number of -3. By assigning the oxidation numbers of H and N, we can then determine the oxidation number of C that makes the overall charge of the molecule neutral.
The minimum oxidation number for nitrogen is -3.
The oxidation number of nitrogen in ammonium nitrite (NH4NO2) is +3. In the ammonium ion (NH4+), nitrogen has an oxidation number of -3 and in the nitrite ion (NO2-), nitrogen has an oxidation number of +3.
The oxidation number of nitrogen in N2 is 0 since it is in its elemental form where the oxidation number is always 0.
The oxidation number of hydrogen in NH3 is +1, and the oxidation number of nitrogen is -3. This is because hydrogen typically has an oxidation number of +1 and in compounds, nitrogen usually has an oxidation number of -3.
In both HCN and HNC molecules, the oxidation number of carbon (C) is -3. This is because hydrogen (H) is almost always assigned an oxidation number of +1, and nitrogen (N) is usually assigned an oxidation number of -3. By assigning the oxidation numbers of H and N, we can then determine the oxidation number of C that makes the overall charge of the molecule neutral.
The minimum oxidation number for nitrogen is -3.
The oxidation number of nitrogen in ammonium nitrite (NH4NO2) is +3. In the ammonium ion (NH4+), nitrogen has an oxidation number of -3 and in the nitrite ion (NO2-), nitrogen has an oxidation number of +3.
The oxidation number of nitrogen in N2 is 0 since it is in its elemental form where the oxidation number is always 0.
The oxidation number of hydrogen in NH3 is +1, and the oxidation number of nitrogen is -3. This is because hydrogen typically has an oxidation number of +1 and in compounds, nitrogen usually has an oxidation number of -3.
The oxidation number of nitrosyl (NO) is +1. Nitrogen typically has an oxidation number of -3, and oxygen typically has an oxidation number of -2. In NO, nitrogen has a -3 oxidation number and oxygen has a -2 oxidation number, leading to an overall oxidation number of +1 for the nitrosyl ion.
The oxidation number of nitrogen (N) in nitric oxide (NO) is +2.
The oxidation number of nitrogen in NO is +2. This is because oxygen has an oxidation number of -2, and since the overall charge of NO (nitrogen monoxide) is 0, the nitrogen atom must have an oxidation number of +2 to balance the equation.
The oxidation number of nitrogen in NO2 is +4. Each oxygen atom has an oxidation number of -2, and since the overall charge of NO2 is 0, the nitrogen atom must have an oxidation number of +4 to balance the charges.
The oxidation number of nitrogen in hydrazine (N2H4) is -2. This is because hydrogen usually has an oxidation number of +1, and in this case, since there are two hydrogens bonded to each nitrogen, the total oxidation number for nitrogen must be -2 to balance it out.
The oxidation number of nitrogen can vary depending on the compound it is in. In most cases, nitrogen has an oxidation number of -3 when it is in its elemental form or in compounds like ammonia (NH3). However, in compounds like nitrate (NO3-), nitrogen has an oxidation number of +5.
The oxidation number of nitrogen in nitrogen dioxide (NO2) is +4. Each oxygen atom in NO2 has an oxidation number of -2, and since the molecule is neutral, the sum of the oxidation numbers must equal zero. Therefore, the oxidation number of nitrogen is calculated as follows: 2(-2) + x = 0, where x represents the oxidation number of nitrogen. Solving for x gives x = +4.