To solve this problem, we need to determine which digit could not be at the units place of the square of a number given that it is 2 more than the units digit of the original number. Let's analyze the possibilities for the units digit of a number and their corresponding squares:

1. Units digit = 0: \[0^2 = 0\]. The result is 0, not 2 units more than 0.
2. Units digit = 1: \[1^2 = 1\]. The result is also not 2 more than 1.
3. Units digit = 2: \[2^2 = 4\]. 4 is indeed 2 more than 2.
4. Units digit = 3: \[3^2 = 9\]. The result is not 2 more than 3.
5. Units digit = 4: \[4^2 = 16\]. The result (6) is not 2 more than 4.
6. Units digit = 5: \[5^2 = 25\]. Here, the units digit is 5, which matches one of the choices, but 5 is not 2 more than the units place of the original number (5 itself).
7. Units digit = 6: \[6^2 = 36\]. 6 is indeed 2 more than 4.
8. Units digit = 7: \[7^2 = 49\]. The result is not 2 more than 7.
9. Units digit = 8: \[8^2 = 64\]. 4 is indeed 2 more than 2.
10. Units digit = 9: \[9^2 = 81\]. The result is not 2 more than 9.

From the analysis above, it is evident that when the units digit of a number is 2, 6, or 8, the units digit of the square can be 4 (2+2), 6 (4+2), or 4 (2+2), respectively. However, the digit 5 does not satisfy the condition as there is no number whose square ends in 5 with a units digit 2 less than it (since 5 is only 5 when squared, not modified).

Thus, the digit that cannot be the units digit of the square of a number, under the given condition, is 5.