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 $

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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.

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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.}} $