Question

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) \(4 \mathrm{~s}\) or \(4 \mathrm{~d}\) (c) \(4 \mathrm{f}\) or \(6 \mathrm{~s}\) (d) 1 s or \(2 s\)

Short answer

a) 3s or 2p b) 4s or 4d c) 4f or 6s d) 1s or 2s Answer: a) 3s has a higher energy than 2p b) 4d has a higher energy than 4s c) 4f has a higher energy than 6s d) 2s has a higher energy than 1s

Step-by-step solution

(a) 3s or 2p Orbital Comparison

For 3s, we have n = 3 and l = 0. Hence, (n + l) = 3. For 2p, we have n = 2 and l = 1. Hence, (n + l) = 3. The 3s and 2p orbitals have the same (n + l) value, so we compare their n values. The 3s orbital has a higher n value (n = 3), making it higher in energy compared to the 2p orbital.

(b) 4s or 4d Orbital Comparison

For 4s, we have n = 4 and l = 0. Hence, (n + l) = 4. For 4d, we have n = 4 and l = 2. Hence, (n + l) = 6. The 4s orbital has a lower (n + l) value compared to the 4d orbital. Therefore, the 4d orbital is higher in energy than the 4s orbital.

(c) 4f or 6s Orbital Comparison

For 4f, we have n = 4 and l = 3. Hence, (n + l) = 7. For 6s, we have n = 6 and l = 0. Hence, (n + l) = 6. The 4f orbital has a higher (n + l) value compared to the 6s orbital. Therefore, the 4f orbital is higher in energy than the 6s orbital.

(d) 1s or 2s Orbital Comparison

For 1s, we have n = 1 and l = 0. Hence, (n + l) = 1. For 2s, we have n = 2 and l = 0. Hence, (n + l) = 2. The 1s orbital has a lower (n + l) value compared to the 2s orbital. Therefore, the 2s orbital is higher in energy than the 1s orbital.