Sn4+
Tin(IV) = Sn4+Chloride = Cl-Formula = SnCl4
The symbol of lead IV ion is Pb^4+.
Sn4+ is the symbol for Tin(IV), that is, the element tin with a oxidation state of 4.
The formula for tin(IV) hydroxide is Sn(OH)4. Tin(IV) has a 4+ oxidation state, leading to the "IV" in its name. Hydroxide is a polyatomic ion consisting of one oxygen and one hydrogen atom with a 1- charge, hence the formula Sn(OH)4.
Tin (IV) phosphide
The formula for the tin four ion is Sn+4
Tin(IV) = Sn4+Chloride = Cl-Formula = SnCl4
The symbol of lead IV ion is Pb^4+.
The ion symbol for tin with 2 electrons lost is Sn2+.
A Stannic ion is the ion of Tin(IV). Its formula is Sn4+ . Here tin is in its highest oxidation state of +4. The other ion is Stannous(Sn2+).
Sn4+ is the symbol for Tin(IV), that is, the element tin with a oxidation state of 4.
Tin(IV) acetate has the chemical formula (not symbol) Sn(C2H3O2)4.
Sn(CrO4)2 Tin 4 means that the tin is a cation with a +4 charge. Chromate is a polyatomic ion with the formula (CrO4)-2. Since the tin ion has a +4 charge in this case, and the chromate ion has a -2 charge, there is a 1:2 ratio of tin ions to chromate ions.
Tin(III) Phosphate
The tin IV ion has 50 protons and 48 electrons. Tin has an atomic number of 50, so it normally has 50 electrons. However, since it is in the +4 oxidation state, it loses 4 electrons during ionization, leaving it with 50 protons and 46 electrons. The charge of +4 means it has 4 more protons than electrons, making the total number of protons and electrons in the tin IV ion 50 and 48, respectively.
If you mean Sn2+ it is known as Tin(II) ion in the stock system or stannous ion in the old naming system.
The name for SnS2 ionic compound is tin (IV) sulfide. It is formed when the tin ion (Sn^4+) combines with the sulfide ion (S^2-).