In barium chlorite, Ba(ClO2)2, the oxidation state of barium (Ba) is +2. Each chlorite ion (ClO2-) has a charge of -1. Within the chlorite ion, the oxidation state of oxygen is -2, so for two oxygen atoms, that totals -4. To balance this, the oxidation state of chlorine must be +3. Therefore, the oxidation state of Cl in Ba(ClO2)2 is +3.
Barium phosphide is represented as Ba₃P₂ because it consists of three barium (Ba) ions for every two phosphide (P³⁻) ions. Barium has a +2 oxidation state, while phosphorus in phosphide form has a -3 oxidation state. To achieve electrical neutrality, the ratio of barium to phosphide must be three to two, resulting in the formula Ba₃P₂. This ratio balances the positive and negative charges in the compound.
Ba2+ is the ion of the Ba atom. That means it has given away two electrons to other atoms in order to reach a stable state with 8 electrons in its outer shell.
The chemical reaction is:BaO + H2O = Ba(OH)2
The abbreviation for barium is Ba.
Barium in group 2 has just two oxidation numbers, 0 in the metal, +2 in its compounds.
In barium chlorite, Ba(ClO2)2, the oxidation state of barium (Ba) is +2. Each chlorite ion (ClO2-) has a charge of -1. Within the chlorite ion, the oxidation state of oxygen is -2, so for two oxygen atoms, that totals -4. To balance this, the oxidation state of chlorine must be +3. Therefore, the oxidation state of Cl in Ba(ClO2)2 is +3.
BA is in the group2. It generally shows +2 oxidation state.
The oxidation number of Ba in Ba²⁺ is +2. This is because in ionic compounds, like BaCl₂ where Ba²⁺ is formed, the atom loses electrons to achieve a stable electron configuration, resulting in a positive oxidation state.
The oxidation state of an individual sulfur atom in BaSo4 is +6.
The oxidation number of Ba is +2. In the ionic compound Ba2+, the oxidation number of Ba is still +2.
The oxidation number for BaSO4 is 6. It goes as follows: +2 for Ba +6 for S -2 for O
Barium is +2 valenced ion, so its oxidation state is +2 in ion form (Ba2+) and 0 in elemental (Ba) form
The oxidation number for Ba in BaO2 is +2, as the oxidation number for oxygen is typically -2. Since there are two oxygen atoms in BaO2, the overall charge must be balanced out by Ba having a +2 oxidation number.
Well, I'm not entirely sure what you're asking. In this reaction equation: Ba + ZnSO4 --> BaSO4 + Zn, barium is a reactant. The equation is balanced as is, so everything is in a nice even 1-to-1 molar ratio, making stoichiometric calculations pretty easy. That's about all I can tell you though, based on the limited information present.
In BaF2, Ba has an oxidation number of +2 and F has an oxidation number of -1. This is because Ba typically forms ionic compounds where it loses 2 electrons to achieve a stable electron configuration, while F gains one electron.
Barium Manganate O.S. of Ba = +2 O.S. of Mn = +6 O.S. of O = -2