Question

Which set of the orbitals is listed in the sequential order of filling in a multi electron atom?
A. \[3s,3p,3d\]
B. \[3d,4s,4p\]
C. \[3d,4p,5s\]
D. \[4p,4d,5s\]

Answer 1

Hint: The aufbau principle, states that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. We can correlate the energy of the orbital with \[(n+l)\]rule. According to\[(n+l)\] rule, the orbital which has the least value of \[(n+l)\] will be filled first to the electrons. In cases where \[(n+l)\] is the same for two orbitals, the\[(n+l)\] rule says that the orbital with lower \[n\] has lower energy.

Complete answer:
An atom is composed of a nucleus containing neutrons and protons with electrons dispersed throughout the remaining space. Electrons, however, are not simply floating within the atom; instead, they are fixed within electronic orbitals. Electronic orbitals are regions within the atom in which electrons have the highest probability of being found.The energy of atomic orbitals increases as the principal quantum number, , increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values differ so that the energy of the orbitals increases within a shell in the order \[s < p < d < f\].
As the principal quantum number, \[n\], increases, the size of the orbital increases and the electrons spend more time farther from the nucleus. Thus, the attraction to the nucleus is weaker and the energy associated with the orbital is higher (less stabilized). Hence, we can apply the\[n+l\]rule to find out which orbital gets filled first.

Consider the options:
A. \[3s,3p,3d\]: Incorrect. Because after \[3p\] is completely filled we will have 4s orbital and not \[3d\] orbital.
B. \[3d,4s,4p\]: Incorrect. Because \[4s\] orbital is filled before \[3d\] orbital.
C. \[3d,4p,5s\]: Correct. They all have the same \[n+l\] and they are in their increasing order of \[n\].
D. \[4p,4d,5s\]: Incorrect. Because \[5s\] orbital fills before \[4d\] orbital.
Hence, the correct option is option C which is \[3d,4p,5s\].

Note: Although, up to Z = 20 (calcium), the \[(n+l)\] rule (and the Aufbau diagram) correctly predicts: Orbital energy levels, The order of occupancy of the orbitals. The physical meaning of the \[(n+l)\] rule is related to the size (n) and shape (l) of a given orbital. For Z > 20 (starting at the transition metals): The \[(n+l)\] rule fails to correctly predict orbital energy levels. Even when we know the orbital energies, this particular knowledge is yet not sufficient to predict the order of filling. Hence, this rule fails for the heavier atoms.