1. Write the relevant facts for the hydrogen $1s$ orbital

For the hydrogen atom in the $1s$ state:

• The wavefunction is spherically symmetric.
• The probability density is $|\\psi_{1s}|^2 = \\frac{1}{\\pi a_0^3} e^{-2r/a_0}$
• The radial probability distribution is $P(r) = 4\\pi r^2 |\\psi_{1s}|^2 = \\frac{4r^2}{a_0^3} e^{-2r/a_0}$

The distinction is important:

• Probability density means $|\\psi|^2$
• Radial probability means probability of finding electron between $r$ and $r+dr$, i.e. $P(r)dr$

2. Check option A

Option A says: The probability density of finding the electron is maximum at the nucleus.

Since $|\\psi_{1s}|^2 = \\frac{1}{\\pi a_0^3} e^{-2r/a_0}$ it is largest at $r=0$.

So, A is correct.

3. Check option B

Option B says: The total energy of the electron is maximum when it is at a distance $a_0$ from the nucleus.

This is incorrect because in the Bohr model / quantum mechanical stationary state, the electron in the hydrogen $1s$ orbital has a fixed total energy: $E_1 = -13.6\,\text{eV}$ It does not depend on distance $r$.

The radius $a_0$ is the distance where the radial probability distribution is maximum for the $1s$ orbital, not where total energy is maximum.

So, B is incorrect.

4. Check option C

Option C says: The electron can be found at a distance $2a_0$ from the nucleus.

For the $1s$ orbital, the wavefunction is nonzero for all finite $r$: $\\psi_{1s}(r) \propto e^{-r/a_0}$ Thus the probability of finding the electron at $r=2a_0$ is nonzero.

So, C is correct.

5. Check option D

Option D says: The $1s$ orbital is spherically symmetrical.

This is a standard property of all $s$ orbitals, especially the hydrogen $1s$ orbital.

So, D is correct.

6. Conclusion

The only incorrect statement is: $\boxed{\text{B}}$

7. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

They match.