The anion in SnBr4 is bromide (Br⁻). In this compound, tin (Sn) has a +4 oxidation state, and each bromine atom contributes a -1 charge, resulting in a total of four bromide anions to balance the charge of the tin cation. Therefore, SnBr4 consists of one tin cation and four bromide anions.
The cation for SnBr4 is Sn^4+, which is the tin(IV) cation.
SnBr4 + 2MgO -> SnO2 + 2MgBr2
The correct stock name for SnBr4 is tin(IV) bromide. In this compound, tin has a +4 oxidation state, indicated by the Roman numeral IV, while bromide ions carry a -1 charge. Therefore, four bromide ions are needed to balance the +4 charge of the tin ion.
Tin can be either an anion or a cation depending on the compound it forms. Tin can exhibit a valence of +2 or +4, making it capable of forming both tin (II) cations (Sn2+) and tin (IV) cations (Sn4+).
The name of the compound SnBr is tin (II) bromide.
The old name of the formula SnBr4 is tin(IV) bromide.
The ionic compound SnBr4 is called tin (IV) bromide. It consists of tin cations with a +4 charge and bromide anions with a -1 charge.
The anion in SnBr4 is bromide (Br⁻). In this compound, tin (Sn) has a +4 oxidation state, and each bromine atom contributes a -1 charge, resulting in a total of four bromide anions to balance the charge of the tin cation. Therefore, SnBr4 consists of one tin cation and four bromide anions.
The cation for SnBr4 is Sn^4+, which is the tin(IV) cation.
SnBr4 + 2MgO -> SnO2 + 2MgBr2
First of all, the correct formula is SnO2, and the Stock name is tin(IV) oxide.
The correct stock name for SnBr4 is tin(IV) bromide. In this compound, tin has a +4 oxidation state, indicated by the Roman numeral IV, while bromide ions carry a -1 charge. Therefore, four bromide ions are needed to balance the +4 charge of the tin ion.
It's a mystery.
Tin can be either an anion or a cation depending on the compound it forms. Tin can exhibit a valence of +2 or +4, making it capable of forming both tin (II) cations (Sn2+) and tin (IV) cations (Sn4+).
The ionic compound SnBr4 is named tin(IV) bromide.
Tin(IV) oxide is primarily considered an ionic compound, as it is composed of a metal cation (tin) and a non-metal anion (oxygen) that form an ionic bond.