Which of the following is/are not correct with respect to energy of atomic orbitals of hydrogen atom?

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

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

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

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

Choose the correct answer from the options given below :

<p>First, recall the key fact that <strong>for a hydrogen atom (and hydrogen‐like ions)</strong>, all orbitals having the same principal quantum number $n$ have the <strong>same energy</strong>. In other words,</p> <p>$ E_{n\ell} \text{ depends only on }n\text{, and not on }\ell. $</p> <p>Hence, the energy ordering for hydrogen‐like atoms follows</p> <p>$ 1s \;<\; 2s = 2p \;<\; 3s = 3p = 3d \;<\; 4s = 4p = 4d = 4f \;<\;\dots $</p> <hr /> <h2>Analyzing Each Statement</h2> <p>We check each statement against the <strong>true</strong> ordering for the hydrogen atom.</p> <p><strong>(A) $\;1s < 2p < 3d < 4s$</strong> </p> <p><p>Because $2p$ is an $n=2$ orbital, $3d$ is $n=3$, and $4s$ is $n=4$, this partial ordering </p> <p>$ E(1s) \;<\; E(2p) \;<\; E(3d)\;<\; E(4s) $</p> <p>is consistent with the fact that $n=1 < n=2 < n=3 < n=4$. </p></p> <p><p>Even though hydrogen also has 2s degenerate with 2p, and 3s and 3p degenerate with 3d, <strong>nothing in (A) contradicts</strong> the correct order $E_1 < E_2 < E_3 < E_4$. </p></p> <p><p>So (A) is <strong>correct</strong> for hydrogen’s energy levels as stated.</p></p> <p><strong>(B) $\;1s < 2s = 2p < 3s = 3p$</strong> </p> <p><p>For hydrogen, indeed $2s=2p$ in energy, and also $3s=3p=3d$. </p></p> <p><p>The statement does <em>not</em> mention 3d, but it does correctly say $3s=3p$. That part is still true for hydrogen. </p></p> <p><p>So for the orbitals it listed, (B) is also <strong>correct</strong>.</p></p> <p><strong>(C) $\;1s < 2s < 2p < 3s < 3p$</strong> </p> <p><p>This is the kind of ordering that appears in many‐electron atoms (where subshell splitting occurs). </p></p> <p><p><strong>But</strong> for a hydrogen atom, $2s$ and $2p$ have exactly the <strong>same</strong> energy, so $2s<2p$ is <strong>not</strong> correct. </p></p> <p><p>Hence (C) is <strong>not</strong> correct for hydrogen.</p></p> <p><strong>(D) $\;1s < 2s < 4s < 3d$</strong> </p> <p><p>In multi‐electron atoms (e.g. the usual “(n+$\ell$) rule”), often $4s$ is lower than $3d$. </p></p> <p><p><strong>But</strong> for hydrogen, <em>all</em> orbitals with $n=3$ lie below <em>all</em> orbitals with $n=4$. That is, $3d$ is lower in energy than $4s$. </p></p> <p><p>So (D) is <strong>not</strong> correct for hydrogen.</p></p> <hr /> <h2>Conclusion</h2> <p><p>(C) and (D) are <strong>not correct</strong> for the hydrogen atom. </p></p> <p><p>(A) and (B) are correct.</p></p> <p>Hence the final answer matches the choice stating:</p> <blockquote> <p><strong>(C) and (D) only</strong> are not correct.</p> </blockquote> <p>Looking at the given options, that is:</p> <p>$ \boxed{\text{Option (D): (C) and (D) only.}} $</p>