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Chapter 7: Problem 169

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 identified elements are: a. Tellurium (Te) - Ground state electron configuration b. Germanium (Ge) - Ground state electron configuration c. Fluorine (F) - Excited state electron configuration

Step by step solution

01

Identify element for configuration a

For configuration a, \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\), we first need to determine the atomic number of Kr. Kr has an atomic number of 36. Now, we add the electrons from the additional subshells: 36 (from Kr) + 2 (from \(5s^2\)) + 10 (from \(4d^{10}\)) + 4 (from \(5p^4\)) = 52 The element with atomic number 52 is Tellurium (Te).
02

Identify element for configuration b

For configuration b, \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{2}\), we first need to determine the atomic number of Ar. Ar has an atomic number of 18. Now, we add the electrons from the additional subshells: 18 (from Ar) + 2 (from \(4s^2\)) + 10 (from \(3d^{10}\)) + 2 (from \(4p^2\)) = 32 The element with atomic number 32 is Germanium (Ge).
03

Identify element for excited state configuration c

For the excited state configuration c, \(1s^{2} 2s^{2} 2p^{4} 3s^{1}\), we sum up the number of electrons: 2 (from \(1s^2\)) + 2 (from \(2s^2\)) + 4 (from \(2p^4\)) + 1 (from \(3s^1\)) = 9 The element with atomic number 9 is Fluorine (F). We can conclude that: a. Tellurium (Te) - Ground state electron configuration b. Germanium (Ge) - Ground state electron configuration c. Fluorine (F) - Excited state electron configuration

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Most popular questions from this chapter

Cesium was discovered in natural mineral waters in 1860 by R. W. Bunsen and G. R. Kirchhoff using the spectroscope they invented in 1859. The name came from the Latin caesius (“sky blue”) because of the prominent blue line observed for this element at 455.5 nm. Calculate the frequency and energy of a photon of this light.

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electromagnetic radiation is emitted in each transition? a. \(n=4 \rightarrow n=3\) b. \(n=5 \rightarrow n=4\) c. \(n=5 \rightarrow n=3\)

Does a photon of visible light $(\lambda \approx 400 \text { to } 700 \mathrm{nm})$ have sufficient energy to excite an electron in a hydrogen atom from the \(n=1\) to the \(n=5\) energy state? from the \(n=2\) to the \(n=6\) energy state?

An electron is excited from the \(n=1\) ground state to the \(n=\) 3 state in a hydrogen atom. Which of the following statements are true? Correct the false statements to make them true. a. It takes more energy to ionize (completely remove) the electron from \(n=3\) than from the ground state. b. The electron is farther from the nucleus on average in the \(n=3\) state than in the \(n=1\) state. c. The wavelength of light emitted if the electron drops from \(n=3\) to \(n=2\) will be shorter than the wavelength of light emitted if the electron falls from \(n=3\) to \(n=1 .\) d. The wavelength of light emitted when the electron returns to the ground state from \(n=3\) will be the same as the wavelength of light absorbed to go from \(n=1\) to \(n=3\) e. For \(n=3,\) the electron is in the first excited state.

Give a possible set of values of the four quantum numbers for the 4s and 3d electrons in titanium.
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