QJEEM25Ch6144: JEE Main 2025 Chemistry - Class 11 JEEM25Ch6144 Solution: Which of the following is/are not correct with respect to…

First, recall the key fact that for a hydrogen atom (and hydrogen‐like ions), all orbitals having the same principal quantum number $n$ have the same energy. In other words,

$E_{n\ell} \text{ depends only on }n\text{, and not on }\ell.$

Hence, the energy ordering for hydrogen‐like atoms follows

1s \;<\; 2s = 2p \;<\; 3s = 3p = 3d \;<\; 4s = 4p = 4d = 4f \;<\;\dots

Analyzing Each Statement

We check each statement against the true ordering for the hydrogen atom.

(A) \;1s < 2p < 3d < 4s

Because $2p$ is an $n=2$ orbital, $3d$ is $n=3$ , and $4s$ is $n=4$ , this partial ordering

E(1s) \;<\; E(2p) \;<\; E(3d)\;<\; E(4s)

is consistent with the fact that n=1 < n=2 < n=3 < n=4 .

Even though hydrogen also has 2s degenerate with 2p, and 3s and 3p degenerate with 3d, nothing in (A) contradicts the correct order E_1 < E_2 < E_3 < E_4 .

So (A) is correct for hydrogen’s energy levels as stated.

(B) \;1s < 2s = 2p < 3s = 3p

For hydrogen, indeed $2s=2p$ in energy, and also $3s=3p=3d$ .

The statement does not mention 3d, but it does correctly say $3s=3p$ . That part is still true for hydrogen.

So for the orbitals it listed, (B) is also correct.

(C) \;1s < 2s < 2p < 3s < 3p

This is the kind of ordering that appears in many‐electron atoms (where subshell splitting occurs).

But for a hydrogen atom, $2s$ and $2p$ have exactly the same energy, so 2s<2p is not correct.

Hence (C) is not correct for hydrogen.

(D) \;1s < 2s < 4s < 3d

In multi‐electron atoms (e.g. the usual “(n+ $\ell$ ) rule”), often $4s$ is lower than $3d$ .

But for hydrogen, all orbitals with $n=3$ lie below all orbitals with $n=4$ . That is, $3d$ is lower in energy than $4s$ .

So (D) is not correct for hydrogen.

Conclusion

(C) and (D) are not correct for the hydrogen atom.

(A) and (B) are correct.

Hence the final answer matches the choice stating:

(C) and (D) only are not correct.

Looking at the given options, that is:

$\boxed{\text{Option (D): (C) and (D) only.}}$