1. We evaluate each statement one by one.

A. For $1s$ orbital, the probability density is maximum at the nucleus

Probability density is given by

For the $1s$ orbital, the wave function is maximum at $r=0$ (the nucleus), so $|\psi|^2$ is also maximum at the nucleus.

So, A is correct.

B. For $2s$ orbital, the probability density first increases to maximum and then decreases sharply to zero

For a $2s$ orbital, the probability density does not simply rise to one maximum and then fall to zero. It has a radial node.

That means as we move away from the nucleus:

• probability density starts from a finite value,
• changes with distance,
• becomes zero at the radial node,
• then increases again and decreases later.

So the given description is incomplete/incorrect.

Thus, B is incorrect.

C. Boundary surface diagrams of the orbitals encloses a region of $100\%$ probability of finding the electron

Boundary surface diagrams do not represent $100\%$ probability. They usually enclose about 90–95% probability of finding the electron.

So, C is incorrect.

D. $p$ and $d$-orbitals have 1 and 2 angular nodes respectively

Number of angular nodes is equal to azimuthal quantum number $l$.

• For $p$ orbital: $l=1$ $\Rightarrow$ angular nodes $=1$
• For $d$ orbital: $l=2$ $\Rightarrow$ angular nodes $=2$

Hence, D is correct.

E. probability density of $p$-orbital is zero at the nucleus

For a $p$ orbital, the wave function vanishes at the nucleus, so

at the nucleus.

So, E is correct.

2. Count the correct statements

Correct statements are:

• A
• D
• E

So, total number of correct statements is

3. Comparison with stored answer

Stored correct answer = $3$

Our derived answer = $3$

Hence, the answer agrees with the stored answer.