The correct statement about probability density (except at infinite distance from nucleus) is:

We begin with the definition of probability density in quantum mechanics. For any hydrogen-like orbital, the probability density at a point in space is given by the square of the magnitude of the wave-function:

$$\text{Probability density}=|\Psi_{n,l,m}(r,\theta,\phi)|^{2}.$$

Because we are squaring an absolute value, we immediately obtain the inequality

$$|\Psi_{n,l,m}|^{2}\;\ge\;0,$$

so the probability density can never be negative at any finite point in space.

Next we recall the nodal rules for hydrogen-like orbitals. The total number of nodes possessed by an orbital with quantum numbers $$n$$ and $$l$$ is $$n-1.$$ These nodes are of two kinds:

• Radial (spherical) nodes, counted by the well-known formula

$$\text{Number of radial nodes}=n-l-1,$$

• Angular (planar/conical) nodes, whose number equals $$l.$$ At every node the wave-function $$\Psi$$ changes sign, so its value is exactly zero; consequently the probability density, being $$|\Psi|^{2},$$ is also zero at that position.

We now examine each option one by one, using the above facts.

Option A: 1s orbital has $$n=1,\;l=0.$$ The number of radial nodes is $$1-0-1 = 0,$$ and the number of angular nodes is $$0.$$ Thus, except at $$r\to\infty,$$ the wave-function is never zero, so the probability density is never zero within finite space. Hence Option A is incorrect.

Option B: We have already proved that $$|\Psi|^{2}$$ can never be negative, because it is a squared modulus. Therefore probability density cannot be negative for a 2p orbital or for any other orbital. Option B is incorrect.

Option C: 3p orbital has $$n=3,\;l=1.$$ First we compute the radial nodes: $$n-l-1 = 3-1-1 = 1.$$ In addition, because $$l=1,$$ there is $$1$$ angular node (a nodal plane). Both the radial node and the angular node are genuine points (or surfaces) where $$\Psi=0,$$ so the probability density $$|\Psi|^{2}$$ is zero there. Thus, for a 3p orbital, the probability density can indeed be zero at finite distances from the nucleus. Option C is correct.

Option D: 2s orbital has $$n=2,\;l=0.$$ The number of radial nodes is $$2-0-1 = 1,$$ so there exists a spherical surface inside which $$\Psi=0,$$ making the probability density zero on that surface. Therefore the statement that it can “never be zero” is false, and Option D is incorrect.

Out of the four alternatives, only Option C survives careful scrutiny.

Hence, the correct answer is Option C.