Question

Which of the following electron configurations correspond to an excited state? Identify the atoms and write the ground-state electron configuration where appropriate. a. \(1 s^{2} 2 s^{2} 3 p^{1}\) b. \(1 s^{2} 2 s^{2} 2 p^{6}\) c. \(1 s^{2} 2 s^{2} 2 p^{4} 3 s^{1}\) d. \([\mathrm{Ar}] 4 s^{2} 3 d^{5} 4 p^{1}\) How many unpaired electrons are present in each of these species?

Short answer

The excited-state electron configurations are for options c and d. Option c is an excited-state configuration for Fluorine (F) with 2 unpaired electrons, and its ground-state configuration is \(1s^2 2s^2 2p^5\). Option d is an excited-state configuration for Iron (Fe) with 6 unpaired electrons, and its ground-state configuration is \([\mathrm{Ar}] 4s^{2} 3d^{6}\).

Step-by-step solution

Option a

\(1s^{2} 2s^{2} 3p^{1}\) This configuration already looks like a ground-state configuration. To find which element it is, we just count the total number of electrons: \(2 + 2 + 1 = 5\). This corresponds to the atomic number \(5\), which is Boron (B). The configuration is a ground-state configuration for Boron. The number of unpaired electrons in this configuration is 1 (the electron in the 3p orbital).

Option b

\(1s^{2} 2s^{2} 2p^{6}\) This configuration also appears to be a ground-state configuration. The total number of electrons is \(2 + 2 + 6 = 10\), which corresponds to the atomic number \(10\) for Neon (Ne). This configuration is the ground-state configuration for Neon. There are no unpaired electrons here.

Option c

\(1s^{2} 2s^{2} 2p^{4} 3s^{1}\) This configuration doesn't seem to be in a ground state since we have an electron in the 3s orbital while the 2p orbital is not completely filled. The total number of electrons is \(2 + 2 + 4 + 1 = 9\), which corresponds to the atomic number \(9\) for Fluorine (F). The ground-state electron configuration for Fluorine should be \(1s^2 2s^2 2p^5\). The number of unpaired electrons in this excited configuration is 2 (one electron in the 2p orbital and one electron in the 3s orbital).

Option d

\([\mathrm{Ar}] 4s^{2} 3d^{5} 4p^{1}\) This configuration indicates an excited state because there is an electron present in the 4p orbital while the 3d orbital is not completely filled. The total number of electrons is the electron sum plus the atomic number of Argon (\(18\)). We have: \(18 + 2 + 5 + 1 = 26\), corresponding to the atomic number \(26\) for Iron (Fe). The ground-state configuration for Iron should be \([\mathrm{Ar}] 4s^{2} 3d^{6}\). The number of unpaired electrons in this excited configuration is 6 (five in the 3d orbital and one in the 4p orbital). In summary: a. Ground-state configuration for Boron (B) with 1 unpaired electron b. Ground-state configuration for Neon (Ne) with 0 unpaired electrons c. Excited-state configuration for Fluorine (F) with 2 unpaired electrons and a ground-state configuration of \(1s^2 2s^2 2p^5\) d. Excited-state configuration for Iron (Fe) with 6 unpaired electrons and a ground-state configuration of \([\mathrm{Ar}] 4s^{2} 3d^{6}\)

Key concepts

Excited State

An excited state is when an electron in an atom moves to a higher energy level, higher than its typical ground state. Imagine this like getting more energy and jumping to a more spacious area. This can occur when an electron absorbs energy from an external source, such as photons or electrical energy.

Let's delve into the excitement using our examples from the problem. In option c, we find an excited state for Fluorine because we see an electron jumping to the 3s orbital even as the 2p orbital isn't filled completely. Similarly, in option d, an electron is excited into the 4p orbital while the 3d should still have more electrons if it were in the ground state.

To recognize excited states, look for configurations that seem to skip a step or have incomplete orbitals below filled ones.

Ground-State Configuration

The ground-state configuration is the "normal" or "default" electronic arrangement of an atom's electrons. It's like their comfort zone, where they naturally reside without extra energy involved. It's crucial to know the correct order of filling for electrons in different orbitals based on their energy levels.

Based on the exercise, we can identify ground-state configurations by ensuring all orbitals in one level are completely filled before moving to the next, reflecting minimum energy.

In option a, we identified the ground-state configuration of Boron. All the electrons are in their expected minimal energy states, without any unnecessary jumps. Option b, similarly, gives us the ground-state configuration for Neon, as all electrons occupy fully filled orbitals before moving to higher energy levels.

• For Boron, we have: \(1s^2 2s^2 2p^1\).
• For Neon, it's \(1s^2 2s^2 2p^6\). This shows no skipped steps, ensuring it stays low on energy, which is the soul of the ground-state.

Unpaired Electrons

Unpaired electrons are those that do not share their orbital with another electron. Imagine having a seat all to yourself instead of sitting next to someone else. The presence of unpaired electrons explains many properties of elements, such as magnetism.

In our examples, each configuration gives insight into the number of unpaired electrons. Counting these involves looking at the last filled orbitals or any with only one electron.

• For Boron, we noted 1 unpaired electron in the 2p orbital.
• Neon had none because all its orbitals were filled with paired electrons.
• In the excited state of Fluorine, there are two unpaired electrons (found in 2p and 3s).
• Iron's excited state presented 6 unpaired electrons primarily due to its 3d and 4p arrangement.

This simple count can reveal a lot about an element's reactivity and other chemical properties.