For the following pairs of orbitals, indicate which is higher in energy in a many-electron atom. (a) \(3 \mathrm{~s}\) or \(2 \mathrm{p}\) (b) 4 s or \(4 \mathrm{~d}\) (c) \(4 \mathrm{f}\) or \(6 \mathrm{~s}\) (d) 1 s or \(2 \mathrm{~s}\)

We have the following orbitals: (a) 3s: Principal quantum number (n) = 3, azimuthal quantum number (l) = 0 2p: Principal quantum number (n) = 2, azimuthal quantum number (l) = 1 (b) 4s: Principal quantum number (n) = 4, azimuthal quantum number (l) = 0 4d: Principal quantum number (n) = 4, azimuthal quantum number (l) = 2 (c) 4f: Principal quantum number (n) = 4, azimuthal quantum number (l) = 3 6s: Principal quantum number (n) = 6, azimuthal quantum number (l) = 0 (d) 1s: Principal quantum number (n) = 1, azimuthal quantum number (l) = 0 2s: Principal quantum number (n) = 2, azimuthal quantum number (l) = 0

Now we'll apply the n + l rule to determine which orbital in each pair has higher energy: (a) 3s: n + l = 3 + 0 = 3 2p: n + l = 2 + 1 = 3 Since both orbitals have the same n + l value, we compare their n values. The 2p orbital has a lower n value, so it has lower energy. Therefore, the 3s orbital has higher energy. (b) 4s: n + l = 4 + 0 = 4 4d: n + l = 4 + 2 = 6 The 4s orbital has a lower n + l value, so it has lower energy. Therefore, the 4d orbital has higher energy. (c) 4f: n + l = 4 + 3 = 7 6s: n + l = 6 + 0 = 6 The 6s orbital has a lower n + l value, so it has lower energy. Therefore, the 4f orbital has higher energy. (d) 1s: n + l = 1 + 0 = 1 2s: n + l = 2 + 0 = 2 The 1s orbital has a lower n + l value, so it has lower energy. Therefore, the 2s orbital has higher energy.

Based on our comparisons, the higher-energy orbitals in the given pairs are as follows: (a) 3s (b) 4d (c) 4f (d) 2s