+5
The oxidation state of N in HNO3 is +5. Oxygen is -2 and H is +!.
The electronic configuration of Bromine in its ground state is: 1s2 2s2p6 3s2p6d10 4s2p5. Therefore the principal quantum number for the outermost electrons in a Bromine atom is 4.
5 valence electrons exist in bromine period, at ground state bromine has 3 valence electrons
An atom of fluorine has the greatest attraction among all atoms for electrons; therefore, no other atom can extract an electron from a fluorine atom, as would be required for the fluorine to have a positive oxidation state.
The whole compound is seen to be neutral (0 oxidation state) as there are no ''-/+'' which indicate the overall charge of the compound. Although within almost all situations O has a -2 oxidation state. With this being now known and we know the whole compound is neutral (0) we can then see N must have the oxidation state of +2. 0 = (-2) + (+2)
The oxidation state of an individual sulfur atom in BaSo4 is +6.
The oxidation state of an individual sulfur atom in MgSO4 is +6.
In NaBrO3 the oxidation state of sodium (Na) is 1+, the oxidation state of bromine is 5+ and the oxidation state of each oxygen atom is 2-
The oxidation state of an individual sulfur atom in (SO3)2- is 4.
5+
in CaCO3 carbon is in the 4+ oxidation state.
The oxidation state of N in HNO3 is +5. Oxygen is -2 and H is +!.
There are no nitrogen atoms in CaCO3!
There are NO carbon (C) atoms in potassium nitrate (KNO3). B.t.w. atoms always have an oxidation state to be zero.
If you mean SO3^-2 (the sulfite anion), then S has an oxidation number of 4+, since each O atom is 2-. In sulfur trioxide, SO3, the S atom has an oxidation number of 6+.
The molecular formula should be CBr4. The oxidation numbers are -1 for each Br, +4 for C.
Oxidation state is what determines the number of each atom. This is in the ionic formula.