Using the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), the diagonal of the square $\sqrt{(1.5)^2+(1.5)^2}=\sqrt{4.5}>2$. Because this is longer than $2$( length of the sides of two adjacent squares), the card can be placed like so, covering $12$ squares. $\rightarrow \boxed{\text{(E)}\ 12\ \text{or more}}$.
[asy] for (int a = -2; a <= 2; ++a) {     draw((-2,a)--(2,a)); draw((a,-2)--(a,2)); } pair A,B,C,D; A=(0,sqrt(2.25)); B=(sqrt(2.25),0); C=(0,-sqrt(2.25)); D=(-sqrt(2.25),0); draw(A--B--C--D--cycle); fill(A--B--C--D--cycle,lightgray); [/asy]
A checkerboard consists of one-inch squares.  A square card, $1.5$ inches on a side, is placed on the board so that it covers part or all of the area of each of $n$ squares.  The maximum possible value of $n$ is
$\text{(A)}\ 4\text{ or }5 \qquad \text{(B)}\ 6\text{ or }7\qquad \text{(C)}\ 8\text{ or }9 \qquad \text{(D)}\ 10\text{ or }11 \qquad \text{(E)}\ 12\text{ or more}$