No, the if statement of your example *will* be executed (that is: the
test x==2, and depending on its outcome also the println).

It's like
    if (x == 2){
        System.out.println("x = 2");
    }System.out.print("foo");

but without System.out.print("foo")

And this is similar to
    if (x == 2){
        System.out.println("x = 2");
    }
    System.out.print("foo");

The <if> statement will be executed (the test and optionally the
println), and also the statement <System.out.print("foo");> will executed.

Likewise your example is similar to
    if (x == 2){
        System.out.println("x = 2");
    }
    ;

The <if> statement will be executed (the test etc), and also the empty
statement <;> will executed. But the empty statement does nothing, so
nothing will change.

So far the analogy.
Because an empty statement doesn't do a thing, the Java compiler will
not generate code for it (any smart Java compiler, that is; a "dumb"
compiler might generate some no-op code for it).

"JS" wrote...

It will be evaluated, but notice the difference between the following
constructions:

-----------------------------------
 if (x == 2)
    System.out.println("x = 2");
-----------------------------------
 if (x == 2);
    System.out.println("x = 2");
-----------------------------------

They look very similar, don't they... ;-)

A misplaced empty statement can make a logic error in the code which can be
difficult to discover.

In both cases the condition is evaluated, and hence the following statement
is executed. But in the second example the following statement is the
*empty* statement. The printout in the latter example will always execute,
as it's not a part of the if-block.

// Bjorn A

As others have said, that semicolon does nothing. The code fragment is
the same as:
 if (x == 2){
   System.out.println("x = 2");
 }
 ;

Note that the following is also legal:
 if (x == 2){
   System.out.println("x = 2");
 };;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
 y = 5;;;;;;;;

It just contains a whole bunch of useless empty statements.