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(A) The energy of an orbital in a multielectron system is determined by the $(n+\ell)$ rule.$1$. Calculate $(n+\ell)$ for each set:$A: 4+1 = 5$$B: 4+2

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

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(A) The energy of an orbital in a multielectron system is determined by the $(n+\ell)$ rule.$1$. 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$$2$. Compare the $(n+\ell)$ values: $6 (B) > 5 (A, D) > 4 (C, E)$.$3$. For orbitals with the same $(n+\ell)$ value,the orbital with the higher $n$ value has higher energy.Comparing $A (n=4)$ and $D (n=3)$: $A > D$.Comparing $E (n=4)$ and $C (n=3)$: $E > C$.$4$. Combining these,the order of energy is: $(B) > (A) > (D) > (E) > (C)$.

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 answer 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 answer from the options given below: