-3.
+3
The ammonium ion has net charge of +1. The central nitrogen atom is bonded to four hydrogen atoms which have oxidation status +1. The oxidation number of nitrogen is -3 after balancing the charges.
The oxidation number of ammonium is 1+. The oxidation number of ammonium is 1+.
The oxidation number of the ammonium ion is +I.
NO2 is a acidic gas. Nitrogen shows +4 oxidation number.
+3
The ammonium ion has net charge of +1. The central nitrogen atom is bonded to four hydrogen atoms which have oxidation status +1. The oxidation number of nitrogen is -3 after balancing the charges.
The oxidation number of ammonium is 1+. The oxidation number of ammonium is 1+.
The oxidation number of the ammonium ion is +I.
Ammonium, NH4, forms a +1 ion.
NO2 is a acidic gas. Nitrogen shows +4 oxidation number.
Oxidation number of N is +1. Oxidation number of O is -2.
+4 for nitrogen
Nitrogen's oxidation number is -4.Carbon's oxidation number is +3.The cyanide ion has -1 charge. Nitrogen is in -3 state. By balancing the charges: the oxidation number of carbon is +4.
The oxidation number of NO, nitrogen oxide, is +3.
The answer is +5 for the Nitrate anion and -3 for the Ammonium cation.Here's how I came up with that answer:NH4NO3 = (NH4)+ and (NO3)-So we start with the known oxidation numbers, such as Hydrogen which is +1 and Oxygen which is -2.Ammonium has 1 Nitrogen atom and 4 Hydrogen atoms with an overall Oxidation number of +1, so the algebraic equation is:X + 4(+1) = (+1)X + 4 = 1X = -3Nitrate has 1 Nitrogen atom and 3 Oxygen atoms with an overall Oxidation number of -1, so the algebraic equation is:X + 3(-2) = (-1)X - 6 = -1X = 5
The assumed oxidation number of nitrogen in ammonia (3+) in this question, is wrongly signed:The correct oxidation number if nitrogen in NITRIDES (like in ammonia NH3, ammonium NH4+ and amino groups -NH2) is minus 3, so hydrogen has oxidation value of plus 1(one, like in H+) which is in fact the only possible form when attached to nonmetals.