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Chapter 7: Problem 159

Identify the following three elements. a. The ground-state electron configuration is \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\). b. The ground-state electron configuration is \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{2} .\) c. An excited state of this element has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{4} 3 s^{1}\).

Short Answer

Expert verified
The elements corresponding to the given electron configurations are: a. Tellurium (Te) with the ground-state electron configuration \([\mathrm{Kr}] 5s^2 4d^{10} 5p^4\). b. Germanium (Ge) with the ground-state electron configuration \([\mathrm{Ar}] 4s^2 3d^{10} 4p^2\). c. Fluorine (F) in an excited state with the electron configuration \(1s^2 2s^2 2p^4 3s^1\).

Step by step solution

01

Calculate the atomic number of each element

For each given configuration, we will count the total number of electrons. a. \([\mathrm{Kr}] 5s^2 4d^{10} 5p^4\) b. \([\mathrm{Ar}] 4s^2 3d^{10} 4p^2\) c. Excited State: \(1s^2 2s^2 2p^4 3s^1\)
02

Identify the elements corresponding to the atomic numbers

Match the calculated atomic numbers to their corresponding elements on the periodic table. a. The ground-state electron configuration is \([\mathrm{Kr}] 5s^2 4d^{10} 5p^4\): The total number of electrons is \(36+2+10+4=52\). The element with the atomic number of 52 is Tellurium (Te). b. The ground-state electron configuration is \([\mathrm{Ar}] 4s^2 3d^{10} 4p^2\): The total number of electrons is \(18+2+10+2=32\). The element with the atomic number of 32 is Germanium (Ge). c. An excited state of this element has the electron configuration \(1s^2 2s^2 2p^4 3s^1\): The total number of electrons is \(2+2+4+1=9\). The element with the atomic number of 9 is Fluorine (F). Note that this configuration is an excited state, meaning that at least one electron has moved to a higher energy level temporarily. Answers: a. Tellurium (Te) b. Germanium (Ge) c. Fluorine (F) in an excited state

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atomic Number
The atomic number is a fundamental property of an element and it indicates the number of protons found in the nucleus of an atom. Since atoms are electrically neutral, this number also reflects the number of electrons orbiting the nucleus under normal circumstances.

For example, the atomic number of Tellurium (Te) is 52, which means it has 52 protons and, when in the ground state, an equal number of electrons. Atomic numbers are sequential and are assigned based on increasing number of protons in the elements, making it a unique identifier for each element.
Periodic Table
The periodic table is an organized chart of elements, arrayed by increasing atomic number, electron configurations, and recurring chemical properties. Elements are placed in specific rows (periods) and columns (groups) which are indicative of the element's electron configuration and the behavior of its electrons.

Different blocks within the periodic table (s, p, d, and f) correspond to the highest energy level of electron orbitals that are being filled. For instance, Germanium (Ge) resides in the p-block, period 4 of the table, which correlates with its valence electron configuration of \(4s^2 3d^{10} 4p^2\).
Ground-State Configuration
Ground-state configuration of an element refers to the most stable, low energy arrangement of electrons in an atom when it is not excited. This configuration can be predicted using principles like the Aufbau principle, Pauli exclusion principle, and Hund's rule.

The ground-state electron configurations provided in the exercise for Tellurium is \(\left[\mathrm{Kr}\right] 5s^2 4d^{10} 5p^4\), and for Germanium is \(\left[\mathrm{Ar}\right] 4s^2 3d^{10} 4p^2\). In these notations, the part in brackets, \(\left[\mathrm{Kr}\right]\) and \(\left[\mathrm{Ar}\right]\), represents the electron configuration of the noble gas from the previous period, signifying a stable electron configuration that the elements’ electrons are building upon.
Excited State
An excited state occurs when one or more electrons in an atom absorb energy and move to a higher energy orbital, which is not the case in ground-state configurations. These excited configurations are usually unstable and the excited electrons will eventually return to their ground-state configurations, releasing energy in the process.

In the exercise, Fluorine (F) \(1s^2 2s^2 2p^4 3s^1\) is in an excited state because one electron has moved to a 3s orbital, which is a higher energy level than expected for its ground-state. The ground-state configuration for Fluorine would be \(1s^2 2s^2 2p^5\), without the \(3s^1\) electron.

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Most popular questions from this chapter

Predict some of the properties of element 117 (the symbol is Uus, following conventions proposed by the International Union of Pure and Applied Chemistry, or IUPAC). a. What will be its electron configuration? b. What element will it most resemble chemically? c. What will be the formula of the neutral binary compounds it forms with sodium, magnesium, carbon, and oxygen? d. What oxyanions would you expect Uus to form?

Assume that a hydrogen atom's electron has been excited to the \(n=5\) level. How many different wavelengths of light can be emitted as this excited atom loses energy?

The ionization energy for a \(1 s\) electron in a silver atom is \(2.462 \times 10^{6} \mathrm{~kJ} / \mathrm{mol}\) a. Determine an approximate value for \(Z_{\text {eff }}\) for the Ag \(1 s\) electron. Assume the Bohr model applies to the \(1 s\) electron. \(Z_{\text {eff }}\) is the apparent nuclear charge experienced by the electrons. b. How does \(Z_{\text {eff }}\) from part a compare to \(Z\) for Ag? Rationalize the relative numbers.

Consider the following ionization energies for aluminum: $$ \begin{aligned} \mathrm{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} & I_{1}=580 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} & I_{2}=1815 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} & I_{3}=2740 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} & I_{4}=11,600 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\).

For hydrogen atoms, the wave function for the state \(n=3, \ell\) \(=0, m_{\ell}=0\) is $$ \psi_{300}=\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta} $$ where \(\sigma=r / a_{0}\) and \(a_{0}\) is the Bohr radius \(\left(5.29 \times 10^{-11} \mathrm{~m}\right)\). Calculate the position of the nodes for this wave function.
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