On a $8 \times 8$ chessboard, the squares are colored alternately white and black. Thus there are white squares and black squares. Each row/column of the chessboard has squares.
It is thus possible to tile this chessboard with dominoes (1 x 2 pieces) by laying say 4 dominoes per column. (tile means lay the dominoes, so that they cover the chessboard, no two dominoes overlapping.)

Now suppose we remove two squares from the chessboard, from DIAGONALLY opposite corners. Suppose one of the squares we remove is white. Now there are white squares left and black squares left.

Q: Is it possible to cover the modified chessboard (with the two diagonally opposite corners removed) with dominoes? Why?

A. Yes. Since there are an equal number of white and black squares remaining on the modified chessboard, one can tile the modified chessboard with dominoes each covering one white and one black square.
B. Yes. It is possible to tile the modified chessboard by placing dominoes, alternating between horizontal and vertical placements in a suitable way.
C. No. Since the total number of remaining squares on the chessboard is odd and every domino covers 2 squares and hence can only be used to tile a region with an even number of squares.
D. No. Since every time we lay down a domino it covers one white square and one black square. Thus since the number of white squares is not equal to the number of black squares on the modified chessboard, it is impossible.