In compounds, barium (Ba) typically has an oxidation state of +2, as it is an alkaline earth metal. In a chemical reaction where barium is a reactant and remains a part of the product, its oxidation state generally stays at +2 for both the reactant and product. However, it's essential to examine the specific compounds involved to confirm this, as complex interactions might alter its state in rare cases.
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.
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.
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.