Meaning of probability density

For an atomic orbital, the probability density is $\rho = |\psi|^2$ where $\psi$ is the wave function.

Since it is a square modulus:

• $\rho \ge 0$ always
• It cannot be negative
• It can be zero wherever the wave function $\psi = 0$

Check each option

Option A: It can be zero for $1s$ orbital

For a $1s$ orbital, the wave function has no node at any finite distance from the nucleus.

In fact, for hydrogen-like atoms, $\psi_{1s} \propto e^{-r/a_0}$ which is never zero for any finite $r$.

So probability density is not zero at any finite distance.

A is false.

Option B: It can be zero for $3p$ orbital

For $3p$ orbital:

• $n=3$
• $l=1$

Number of radial nodes: $n-l-1 = 3-1-1 = 1$ So the wave function becomes zero at one finite radial distance. Hence probability density can be zero there.

Also, $p$ orbitals have angular nodal planes where $\psi=0$. Therefore probability density can certainly be zero.

B is true.

Option C: It can never be zero for $2s$ orbital

For $2s$ orbital:

• $n=2$
• $l=0$

Number of radial nodes: $n-l-1 = 2-0-1 = 1$ So $2s$ has one radial node where $\psi=0$. Hence probability density is zero at that finite distance.

C is false.

Option D: It can negative for $2p$ orbital

Probability density is $|\psi|^2$, so it can never be negative for any orbital.

D is false.

Conclusion

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