What is the smallest number of queens that can be placed on an $8\times8$ chess board so that every square is either occupied or can be reached in one move? (A queen can be moved any number of unoccupied squares in a straight line vertically, horizontally, or diagonally.)

$\text{(A) }4\qquad\text{(B) }5\qquad\text{(C) }6\qquad\text{(D) }7\qquad\text{(E) }8$
To determine the smallest number of queens required to cover an $8 \times 8$ chessboard such that every square is either occupied or can be reached in one move, we need to analyze the coverage capabilities of the queens.

1. **Understanding Queen's Movement**:
   A queen can move any number of squares vertically, horizontally, or diagonally. Therefore, a queen placed on any square can control all squares in its row, column, and both diagonals passing through it.

2. **Maximum Coverage by One Queen**:
   A single queen can cover up to $15$ additional squares (7 in its row, 7 in its column, and 1 in each diagonal, minus the overlap at the queen's position).

3. **Coverage by Multiple Queens**:
   We need to determine the minimum number of queens such that every square on the board is either occupied by a queen or can be attacked by at least one queen.

4. **Known Result**:
   It is known that 5 queens can cover an $8 \times 8$ chessboard. We need to show that it is not possible to cover the board with fewer than 5 queens.

5. **Proof by Contradiction**:
   Assume that it is possible to cover the board with 4 queens. We will show that this leads to a contradiction.

6. **Maximum Coverage by 4 Queens**:
   The maximum number of squares covered by $q$ queens is given by:
   \[
   22q - 4 \binom{q}{2}
   \]
   For $q = 4$:
   \[
   22 \cdot 4 - 4 \binom{4}{2} = 88 - 4 \cdot 6 = 88 - 24 = 64
   \]
   This suggests that 4 queens could theoretically cover all 64 squares. However, we need to show that this maximum is unattainable due to overlapping coverage.

7. **Overlapping Coverage**:
   Consider two queens $Q_1$ and $Q_2$ placed such that they cover the maximum number of unique squares. The number of squares covered by both queens is at least 4. If $Q_1$ and $Q_2$ are placed such that they only overlap in 4 squares, the distance between them horizontally ($h$) and vertically ($v$) must satisfy $h + v > 7$.

8. **Adding a Third Queen**:
   Place a third queen $Q_3$ such that it is $h'$ horizontally and $v'$ vertically away from the closest other queen (say $Q_2$). We have:
   \[
   h' \leq 3 \quad \text{and} \quad v' \leq 7 - v < h
   \]
   If $v' = h'$, the intersections $J_1$ or $J_2$ must exist. If $v' < h'$, intersections $J_1$ or $J_2$ must exist since the vertical distance is less than 6. If $v' > h'$, then $K_1$ must exist as the horizontal distance between it and $Q_2$ is less than $h$.

9. **Conclusion**:
   Therefore, there exists a pair of queens that both cover at least 5 squares, making it impossible to cover all squares with only 4 queens. Hence, the minimum number of queens required to cover the entire $8 \times 8$ chessboard is 5.

The final answer is $\boxed{5}$