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(A) $1$. The energy of an orbital in a multielectron system is determined by the $(n+\ell)$ rule. Lower $(n+\ell)$ value means lower energy. If $(n+\e

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$(C) < (E) < (D) < (A) < (B)$

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(A) $1$. The energy of an orbital in a multielectron system is determined by the $(n+\ell)$ rule. Lower $(n+\ell)$ value means lower energy. If $(n+\ell)$ values are equal, the orbital with the lower $n$ value has lower energy.$2$. Calculate $(n+\ell)$ for each set:$A: 4+1 = 5$$B: 4+2 = 6$$C: 3+1 = 4$$D: 3+2 = 5$$E: 4+0 = 4$$3$. Group by $(n+\ell)$ values:$(n+\ell) = 4: C (n=3), E (n=4)$. Since $n_C $(n+\ell) = 5: D (n=3), A (n=4)$. Since $n_D $(n+\ell) = 6: B (n=4)$.$4$. Combining these, the order of increasing energy is: $(C) < (E) < (D) < (A) < (B)$.

Compare the energies of the following sets of quantum numbers for a multielectron system:
$A. n=4, \ell=1$
$B. n=4, \ell=2$
$C. n=3, \ell=1$
$D. n=3, \ell=2$
$E. n=4, \ell=0$
Choose the correct order of increasing energy from the options given below:

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Compare the energies of the following sets of quantum numbers for a multielectron system:$A. n=4, \ell=1$$B. n=4, \ell=2$$C. n=3, \ell=1$$D. n=3, \ell=2$$E. n=4, \ell=0$Choose the correct order of increasing energy from the options given below: