Tin(IV) has a positive four charge.
Selenide usually forms a 2- ion.
Ionic compounds need a neutral, balanced charge.
With the above information, the rest is pretty much just math.
Tin - Sn(+4)
Selenide - Se(-2)
So you need two of Se to balance the +4 with a -4.
= SnSe2
The answer above is OK, however if you are familiar with the idea of oxidation states SnIV is tin in its 4th oxidation state. Selenium has a common oxidation stae of -2 and once again you can do the math.
The bonding in SnSe2 is not ionic so using oxidation states is preferable.
The systematic name of this ionic compound is Tin(IV) Selenide.
Tin(IV) selenide
MnSe2 is manganese IV selenide.
SnO2, if you meant (tin(IV) oxide) or tin dioxide that is
The manganese(IV) selenide has the chemical formula MnSe2.
The systematic name of this ionic compound is Tin(IV) Selenide.
Tin has two oxidation states (II and IV), and exhibits approximately equal stability in both its II and IV oxidation state. The chemical formula Tin (II) Iodide is SnI2. The chemical formula for Tin (IV) Iodide is SnI4.
Tin(IV) selenide
Cesium selenide
lead (IV) selenide
Almost certainly you are looking for Titanium(IV) Selenide, TiSe2, though the Titanium(III) compound, Ti2Se3 may exist as well. You should be more specific :)
MnSe2 is manganese IV selenide.
The Answer To Your Question Is.... tin (IV) oxide
Tin (IV) phosphide
SnO2, if you meant (tin(IV) oxide) or tin dioxide that is
Sn4+ is the symbol for Tin(IV), that is, the element tin with a oxidation state of 4.
The manganese(IV) selenide has the chemical formula MnSe2.