Problem 61 - Chemistry and Chemical Reactivity

Question

Rank the following orbitals in the \(\mathrm{H}\) atom in order of increasing energy: \(3 s, 2 s, 2 p, 4 s, 3 p, 1 s,\) and \(3 d\)

Step-by-Step Solution

Answer

The order is: \(1s, 2s, 2p, 3s, 3p, 3d, 4s\).

Step 1: Understanding the Problem

We need to rank these orbitals in an atom based on their energy levels. We are given orbitals: \(3s, 2s, 2p, 4s, 3p, 1s,\) and \(3d\), that belong to the hydrogen atom. Our task is to arrange them starting from the lowest energy to the highest.

Step 2: Using Orbital Energy Order for Hydrogen Atom

In a hydrogen atom, the energy of an orbital depends solely on the principal quantum number \(n\). Hence, energy increases with increasing \(n\). The order from lowest to highest is: \(1s\), \(2s, 2p\), \(3s, 3p, 3d\), \(4s\).

Step 3: Ranking the Orbitals by Principal Quantum Number

Identify the principal quantum number \(n\) for each orbital: \(1s (n=1)\), \(2s\) and \(2p (n=2)\), \(3s\), \(3p\), and \(3d (n=3)\), \(4s (n=4)\). Arrange them sequentially based on increasing \(n\): \(1s \), \(2s\), \(2p\), \(3s\), \(3p\), \(3d\), \(4s\).

Step 4: Conclusion

The ranking of the given orbitals in order of increasing energy for the hydrogen atom is: \(1s, 2s, 2p, 3s, 3p, 3d, 4s\).

Key Concepts

Principal Quantum Number

The principal quantum number, denoted by \(n\), is a crucial number in the study of atomic orbitals, especially when examining hydrogen atom orbitals. It is an integer that determines the energy level and the size of the orbital.

As \(n\) increases, the electron is found farther from the nucleus, meaning larger orbitals. Therefore, an electron at \(n=1\) is closest to the nucleus and has lower energy compared to higher \(n\) values.

• \(n=1\): First energy level.
• \(n=2\): Second energy level.
• \(n=3\): Third energy level.
• \(n=4\): Fourth energy level.

This principle is essential in understanding how to rank orbitals based on energy. For the hydrogen atom, the principal quantum number directly influences the energy of the orbital—higher \(n\) signifies a higher energy level.

Orbital Energy Levels

The concept of orbital energy levels becomes simpler in a hydrogen atom compared to many-electron atoms. In hydrogen, the energy of an orbital is solely reliant on its principal quantum number \(n\). This means the various orbitals for a given \(n\) level (like \(2s\) and \(2p\)) have identical energy levels, an occurrence described as degenerate.

In more complex atoms, energy levels rely on both \(n\) and the azimuthal quantum number \(l\), making things more complicated. However, in the case of hydrogen:

• For \(n=1\), only \(1s\) is present.
• For \(n=2\), the orbitals \(2s\) and \(2p\) have the same energy.
• For \(n=3\) you will encounter \(3s\), \(3p\), and \(3d\) which are also the same in energy.

Thus, as the principal quantum number increases, the energy levels also increase, and this forms the basis for ranking orbitals in terms of energy.

Atomic Orbital Ranking

Ranking atomic orbitals by their energy is a straightforward task when dealing with hydrogen due to its simple electron configuration.

In the hydrogen atom, as discussed, the principal quantum number \(n\) dictates the energy level—higher \(n\) suggests higher energy. This means the ranking starts with orbitals having the lower \(n\) and progresses to higher ones.

• \(1s\) is lowest because it has \(n=1\).
• Next, \(2s\) and \(2p\) are at \(n=2\).
• Then, \(3s\), \(3p\), and \(3d\) correspond to \(n=3\).
• Finally, \(4s\) is at \(n=4\) and has the highest energy.

Through understanding of the energy level order, we can confidently position orbitals from lowest to highest energy: \(1s, 2s, 2p, 3s, 3p, 3d, 4s\). This hierarchy helps in predicting electron configurations and understanding atomic behavior.