Chapter 8: Problem 80

Use orbital diagrams to show the distribution of electrons among the orbitals in (a) the \(4 p\) subshell of Br; (b) the \(3 d\) subshell of \(\mathrm{Co}^{2+},\) given that the two electrons lost are \(4 s ;\) (c) the \(5 d\) subshell of \(\mathrm{Pb}\).

Short Answer

Expert verified

The electron distribution in: (a) the \(4 p\) subshell of Br is \(_{4p}\uparrow\downarrow, \uparrow\downarrow, \uparrow\), (b) the \(3 d\) subshell of \(\mathrm{Co}^{2+}\) is \(_{3d}\uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow\), and (c) the \(5 d\) subshell of Pb is \(_{5d}\uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow\).

Step by step solution

Distribution in the \(4 p\) subshell of Br

Bromine (Br) has an atomic number of 35, which means it has 35 electrons in its neutral state. Thus, it has 3 electrons in its \(4p\) subshell. Using Hund's Rule which states that orbitals of equal energy are each occupied by one electron before any orbital is occupied by a second electron, and further, that all electrons in singly occupied orbitals have the same spin, the electron distribution in the \(4 p\) subshell of Br would be \(_{4p}\uparrow\downarrow, \uparrow\downarrow, \uparrow\). There are two arrows in the first two orbitals indicating the two electrons with opposite spins (up and down), and one arrow in the last orbital indicating the third electron.

Distribution in the \(3 d\) subshell of \(\mathrm{Co}^{2+}\)

Cobalt (Co) has an atomic number of 27, and it has lost two 4s electrons to become \(\mathrm{Co}^{2+}\), so, it now has 25 electrons. When filling the energy levels, the \(3 d\) subshell will have 7 electrons. Using Hund's Rule again, the \(3 d\) subshell of \(\mathrm{Co}^{2+}\) will look like this: \(_{3d}\uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow\), with five orbitals showing two paired electrons and the last two orbitals showing single electrons.

Distribution in the \(5 d\) subshell of Pb

Lead (Pb) has an atomic number of 82 and it has a total of 10 electrons in the \(5d\) subshell. Based on the Pauli Exclusion Principle (which states that no two electrons in an atom could be in the same state or configuration) and Hund's Rule, the electron distribution of \(5d\) subshell in Pb is as follows: \(_{5d}\uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow\). All five orbitals in the \(5d\) subshell are filled with two electrons each, with opposite spins.

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Use orbital diagrams to show the distribution of electrons among the orbitals in (a) the \(4 p\) subshell of Br; (b) the \(3 d\) subshell of \(\mathrm{Co}^{2+},\) given that the two electrons lost are \(4 s ;\) (c) the \(5 d\) subshell of \(\mathrm{Pb}\).

Short Answer

Step by step solution

Distribution in the \(4 p\) subshell of Br

Distribution in the \(3 d\) subshell of \(\mathrm{Co}^{2+}\)

Distribution in the \(5 d\) subshell of Pb

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