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Chapter 7: Problem 67

For each of the following pairs of hydrogen orbitals, indicate which is higher in energy: (a) \(1 s, 2 s ;\) (b) \(2 p\), \(3 p ;\) (c) \(3 d_{x y}, 3 d_{y z};\) (d) \(3 s, 3 d ;\) (e) 4f, \(5 s.\)

Short Answer

Expert verified
The higher energy orbitals in each pair are: (a) 2s, (b) 3p, (c) Both are equal, (d) 3d, (e) 5s.

Step by step solution

01

Match (a)

In a hydrogen atom, energy levels increase with the principal quantum number \(n\), which represents the shell number. Thus, a \(2s\) orbital has higher energy than a \(1s\) orbital.
02

Match (b)

Again, using the rule that energy increases with the principal quantum number, a \(3p\) orbital has higher energy than a \(2p\) orbital.
03

Match (c)

Orbitals with the same value of \(n\) and \(l\) (the azimuthal quantum number) will have the same energy in a hydrogen atom. Therefore, in this case, the \(3 d_{xy}\) and the \(3d_{yz}\) orbitals have the same energy.
04

Match (d)

In a given energy level, s orbitals are lower in energy than p orbitals, which are lower in energy than d orbitals. Thus, in this case, a \(3s\) orbital has lower energy than a \(3d\) orbital.
05

Match (e)

When comparing orbitals of different energy levels, those with larger principal quantum numbers are higher in energy. Hence, a \(5s\) orbital has higher energy than a \(4f\) orbital.

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