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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