JEE Chemistry 2026 Question with Solution

Given: Figure 1 shows the $2s$ orbital with a spherical nodal surface. Figure 2 shows the wave function $\psi_{2s}(x)$ with labeled points $A$, $B$, $C$, $D$.

Find: Which point in Figure 2 represents the nodal surface shown in Figure 1.

Concept: A node is a region where the wave function becomes exactly zero. For hydrogen-like orbitals, the number of radial nodes is

$$n-l-1$$

For the $2s$ orbital,

$$n=2,\quad l=0$$

So,

$$2-0-1=1$$

Thus, the $2s$ orbital has one spherical nodal surface.

Electron probability density is proportional to $|\psi|^2$, but the nodal surface is identified by

$$\psi=0$$

not merely by a low probability.

From Figure 2:

• Point $A$: $\psi$ is positive and maximum.
• Point $B$: $\psi=0$.
• Point $C$: $\psi$ is negative and minimum.
• Point $D$: $\psi$ is negative but non-zero.

Therefore, the nodal surface corresponds to the zero-crossing of the wave function, which is point B.

The correct option is C.

Given: The question compares a spherical nodal surface in Figure 1 with the graph of $\psi_{2s}(x)$ in Figure 2.

Find: The point on the graph where the nodal surface is represented.

A spherical nodal surface in the $2s$ orbital means that at a particular radius,

$$\psi_{2s}(r)=0$$

When this same behavior is shown along a single axis, the node appears where the plotted wave function crosses the $x$-axis.

In Figure 2, only point $B$ lies on the axis where

$$\psi_{2s}=0$$

Hence, the spherical nodal surface in Figure 1 corresponds to point B in Figure 2.

Therefore, the correct option is C.