Question

The following electron configurations represent excited states. Identify the element, and write its ground-state condensed electron configuration. (a) \(1 s^{2} 2 s^{2} 3 p^{2} 4 p^{1},\) (b) \([\mathrm{Ar}] 3 d^{10} 4 s^{1} 4 p^{4} 5 s^{1}\), (c) \([\mathrm{Kr}] 4 d^{6} 5 s^{2} 5 p^{1}\) (a) Determine which elements emit radiation in the visible part of the spectrum. (b) Which element emits photons of highest energy? Of lowest energy? (c) When burned, a sample of an unknown substance is found to emit light of frequency \(6.59 \times 10^{14} \mathrm{~s}^{-1}\). Which of these elements is probably in the sample?

Short answer

The elements represented by the given excited electron configurations are nitrogen (N), selenium (Se), and rhodium (Rh). Their ground-state electron configurations are \(1s^2 2s^2 2p^3\) for N, \([\mathrm{Ar}] 3d^{10} 4s^2 4p^4\) for Se, and \([\mathrm{Kr}] 4d^6 5s^2\) for Rh. Among these elements, nitrogen is most likely to emit radiation in the visible part of the spectrum. Rhodium is expected to emit the highest energy photons, while nitrogen emits the lowest energy photons. Given the frequency of emitted light, \(6.59 \times 10^{14}\) Hz, nitrogen is probably in the sample.

Step-by-step solution

Identify the elements given their excited electron configurations

To identify the element, we just need to count the total number of electrons: (a) \(1s^2 2s^2 3p^2 4p^1\) -> Total electrons = 2 + 2 + 2 + 1 = 7; This corresponds to nitrogen (N). (b) \([\mathrm{Ar}] 3d^{10} 4s^1 4p^4 5s^1\) -> Total electrons = 18 (from Ar) + 10 + 1 + 4 + 1 = 34; This corresponds to selenium (Se). (c) \([\mathrm{Kr}] 4d^6 5s^2 5p^1\) -> Total electrons = 36 (from Kr) + 6 + 2 + 1 = 45; This corresponds to rhodium (Rh).

Write the ground-state electron configurations

Each element's ground-state configuration fills the sublevels in increasing order: (a) Nitrogen (N) ground-state configuration: \(1s^2 2s^2 2p^3\) (b) Selenium (Se) ground-state configuration: \([\mathrm{Ar}] 3d^{10} 4s^2 4p^4\) (c) Rhodium (Rh) ground-state configuration: \([\mathrm{Kr}] 4d^6 5s^2\)

Determine which elements emit radiation in the visible spectrum

Since the problem asks about the "elements" that emit radiation in the visible part of the spectum, we can omit the specific configurations. Elements with bigger energy spacing between energy levels tend to emit higher energy photons when electrons jump between levels, corresponding to shorter wavelengths. Therefore, it is expected that nitrogen - as a light element - is most likely to emit radiation in the visible part of the spectrum.

Determine the highest and lowest energy photons

As the elements' atomic number increases, the energy difference between the higher energy levels will be larger. Larger energy differences correspond to larger energy emitted photons: - Highest energy photon: Rhodium (Rh) - Lowest energy photon: Nitrogen (N)

Find which element's emitted frequency matches the given value when burned

Given the frequency of emitted light, \(6.59 \times 10^{14}\) Hz, we have to find the corresponding element emitting in the visible part of the spectrum. Since nitrogen is the only element in this group that likely emits visible radiation, it supports the hypothesis that nitrogen (N) is probably in the sample.

Key concepts

Ground-State Electron Configuration

The ground-state electron configuration of an atom is the arrangement of electrons in the lowest available energy levels, according to the Aufbau principle, Pauli exclusion principle, and Hund's rule. In the ground state, the electron configuration will fill sublevels in increasing order of energy. For example, in nitrogen (N), the ground-state configuration is represented as \(1s^2 2s^2 2p^3\), where the superscripts denote the number of electrons occupying each sublevel.

Understanding this formative concept is crucial in chemistry, as it underlies the prediction of an element's chemical behavior. The ground-state electron configuration can also show us the quantum numbers associated with an atom's electrons, which are essential in delineating the energies and shapes of orbitals. For instance, in the given exercise, determining the correct ground-state configuration enabled us to identify elements based on their excited states.

Visible Spectrum Radiation

When excited electrons return to lower energy levels, they emit energy in the form of photons. In some cases, this energy falls within the visible spectrum, producing light that can be seen by the human eye. This spectrum spans from approximately 380 nm to 750 nm in wavelength, which corresponds to frequencies ranging from about \(7.89 \times 10^{14}\) Hz for violet light to \(3.99 \times 10^{14}\) Hz for red light.

Elements emit different colors based on the specific energy level transitions of their electrons. In our exercise, we inferred that among the given elements, nitrogen is most likely to emit radiation within the visible spectrum. This determination ties into the concept that smaller atoms with simpler electron configurations tend to have larger spacing between energy levels available for electronic transitions that fall within the visible spectrum.

Energy Levels in Atoms

Atoms consist of quantized energy levels, where electrons can reside. The energy levels are arranged hierarchically, with the ground state being the lowest energy level. Each subsequent level requires more energy, and when an electron transitions between these levels, it must absorb or release quantities of energy equal to the differences between the levels.

Moreover, these energy differences influence the emission spectra of elements. High-energy photons are emitted when electrons drop from high energy levels to significantly lower ones. In the exercise, we discussed that the highest energy photon would be emitted by rhodium (Rh), due to its larger energy differences between higher energy levels. On the other hand, nitrogen (N), a much lighter atom, emits lower energy photons corresponding to the smaller energy spacing in its atomic structure.