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Chapter 9: Problem 34

Which of the following species has the greatest number of unpaired electrons (a) \(\mathrm{Ge} ;\) (b) \(\mathrm{Cl} ;\) (c) \(\mathrm{Cr}^{3+}\) (d) Br -?

Short Answer

Expert verified
The species with the greatest number of unpaired electrons is \(Cr^{3+}\) as it has 3 unpaired electrons.

Step by step solution

01

Identify Atomic Numbers

The atomic numbers for each species are as follows: Ge (32), Cl (17), Cr (24), Br (35). These atomic numbers are found on the periodic table.
02

Determine the electron configuration

Write down the electron configurations according to the Aufbau principle, Hund's rule, and Pauli's exclusion principle.\n For Ge: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^2\]\n For Cl: \[1s^2 2s^2 2p^6 3s^2 3p^5\]\n For Cr3+: Chromium has a special electron configuration. Its ground-state electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\] however in case of Cr3+, three electrons are lost, two from 4s orbital and one from 3d, hence the configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\]\n For Br-: When Bromine gains an electron, become Br-. Its electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\]
03

Identify Number of Unpaired Electrons

Now, count the number of unpaired electrons in the last shell for each species. Ge has 2, Cl has 1, Cr3+ has 3 and Br- has 0 unpaired electron.

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