MX2(s)<--> M2+(aq)+ 2X-(aq)
ksp=[M2+][X-]2
the question asks.. why is [MX2] not used in the expression for the equilibrium?
is it because they aren't used in the expressions?
Your explanation is not an explanation of anything. I will give you two answers.
a. This answer makes good sense but isn't the easy way to go with this. However, if MX2 is a solid and it is slightly soluble, then it will have a certain solubility which is a constant.
So we can write the equation as
Keq = (M^2+)(X^-)^2/(MX2). But if (MX2) = a constant, then we can rewrite the equation as
Keq*(MX2) = ((M^2+)(X^-)^2. But Keq*k = another constant which we call Ksp and not Keq.
#2 explanation.
By definition, materials that are in their normal state at room T and P as pure SOLIDS or LIQUIDS have an activity of 1 are not included in Keq expressions. (Of course you could argue that Keq*1 = Keq which doesn't change anything anyway.)