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Chapter 7: Problem 91

Use the Aufbau principle to obtain the ground-state electron configuration of selenium.

Short Answer

Expert verified
The ground-state electron configuration of selenium using the Aufbau principle is \[1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{2} 4d^{10} 5p^{4}\]. In a more simplified format, using the noble gas notation, the configuration is \[(Ar) 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^4 \].

Step by step solution

01

Recognizing the atomic number

From the periodic table, it's known that Selenium (Se) has an atomic number of 34. Thus, this denotes that a neutral atom of selenium has 34 electrons that need to be distributed in the atomic orbitals.
02

Applying the Aufbau Principle

The Aufbau principle indicates that electrons fill the lowest energy level first and then successively fill higher energy levels. Using this principle, the electron configuration process starts at hydrogen, and builds up adding protons (which makes an heavier element) and adding the corresponding electrons as well. In this way following the Aufbau principle, the electron configuration would be: \[1s ^ 2 \rightarrow 2s ^2 \rightarrow 2p ^6 \rightarrow 3s ^2 \rightarrow 3p ^6 \rightarrow 4s ^2 \rightarrow 3d ^{10} \rightarrow 4p ^6 \rightarrow 5s ^2 \rightarrow 4d ^{10} \rightarrow 5p^4 \]
03

The Electron Configuration of Selenium

Following this configuration set in the previous step, Selenium's electron configuration will be: \[1s ^2 2s ^2 2p ^6 3s ^2 3p ^6 4s ^2 3d ^{10} 4p ^6 5s ^2 4d ^{10} 5p ^4 \]
04

Simplified Configuration using Noble Gas

The electron configuration of selenium can also be written in short form. The Noble Gas that falls before Selenium is Argon (Ar). Since Argon has 18 electrons, the first 18 electrons in Selenium would have exactly the same electron configuration as Argon. Thus, the remaining electron distribution can be written next to the Argon (Ar) symbol in brackets [(Ar)]. Hence, the shorthand electron configuration for Selenium will be: \[(Ar) 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^4 \]

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Most popular questions from this chapter

How is the concept of electron density used to describe the position of an electron in the quantum mechanical treatment of an atom?

(a) What is the wavelength (in nanometers) of light having a frequency of \(8.6 \times 10^{13} \mathrm{~Hz} ?\) (b) What is the frequency (in Hz) of light having a wavelength of \(566 \mathrm{nm} ?\)

An electron in a hydrogen atom is excited from the ground state to the \(n=4\) state. Comment on the correctness of the following statements (true or false). (a) \(n=4\) is the first excited state. (b) It takes more energy to ionize (remove) the electron from \(n=4\) than from the ground state. (c) The electron is farther from the nucleus (on average) in \(n=4\) than from the ground state. (d) The wavelength of light emitted when the electron drops from \(n=4\) to \(n=1\) is longer than that from \(n=4\) to \(n=2\). (e) The wavelength the atom absorbs in going from \(n=1\) to \(n=4\) is the same as that emitted as it goes from \(n=4\) to \(n=1\).

Calculate the energies needed to remove an electron from the \(n=1\) state and the \(n=5\) state in the \(\mathrm{Li}^{2+}\) ion. What is the wavelength (in \(\mathrm{nm}\) ) of the emitted photon in a transition from \(n=5\) to \(n=1 ?\) The Rydberg constant for hydrogen-like ions is \((2.18 \times\) \(\left.10^{-18} \mathrm{~J}\right) Z^{2},\) where \(Z\) is the atomic number.

Make a chart of all allowable orbitals in the first four principal energy levels of the hydrogen atom. Designate each by type (for example, \(s, p\) ) and indicate how many orbitals of each type there are.
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