Multiple Choice
Which of the following wavelengths corresponds to the energy required to excite a hydrogen atom from the n=2 state to the n=5 state?
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10. Periodic Properties of the Elements
The Electron Configuration
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In a multi-electron atom, which of the following orbitals will have the highest energy?
A
3p
B
4s
C
4p
D
3d
Verified step by step guidance1
Understand that in multi-electron atoms, the energy of orbitals depends on both the principal quantum number $n$ and the azimuthal quantum number $l$. The principal quantum number $n$ indicates the main energy level, while $l$ (subshell type) affects the shape and energy due to electron-electron repulsions and penetration effects.
Recall the order of orbital energies in multi-electron atoms generally follows the $(n + l)$ rule, where orbitals with lower $(n + l)$ values have lower energy. If two orbitals have the same $(n + l)$ value, the one with the lower $n$ has lower energy.
Calculate the $(n + l)$ values for each orbital: For 3p, $n=3$, $l=1$ so $(n + l) = 4$; for 4s, $n=4$, $l=0$ so $(n + l) = 4$; for 4p, $n=4$, $l=1$ so $(n + l) = 5$; for 3d, $n=3$, $l=2$ so $(n + l) = 5$.
Compare the orbitals with the same $(n + l)$ values: between 3p and 4s (both 4), 3p has lower energy because it has lower $n$. Between 4p and 3d (both 5), 4p has higher $n$ so 3d has lower energy than 4p. However, 3d has a higher $l$ value, which generally increases energy within the same principal level.
Conclude that among the given orbitals, 3d has the highest energy because it has a higher $l$ value and a relatively low $n$, making it higher in energy than 3p, 4s, and 4p orbitals in a multi-electron atom.
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