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Chapter 8: Problem 50

Give an example of an ionic compound where both the anion and the cation are isoelectronic with each of the following noble gases. a. \(\mathrm{Ne}\) c. \(\mathrm{Kr}\) b. Ar d. Xe

Short Answer

Expert verified
The simple ionic compounds that are isoelectronic with the given noble gases are: a. For Ne: \(\mathrm{NaF}\) b. For Kr: \(\mathrm{RbBr}\) c. For Ar: \(\mathrm{MgO}\) d. For Xe: \(\mathrm{CsI}\)

Step by step solution

01

Identify the electronic configuration of the noble gas.

The electronic configuration of neon (Ne) is \(1s^2 2s^2 2p^6\), and it has 10 electrons.
02

Find the cation and anion that would form an ionic compound isoelectronic with Ne.

For this case, we can choose the Sodium ion (Na+). When Sodium loses one electron, it would have the same electronic configuration as Ne, namely \(1s^2 2s^2 2p^6\). To balance the charge, we can choose the Fluoride ion (F-). Fluorine has 9 electrons, and when it gains one electron, it would also have an electronic configuration of \(1s^2 2s^2 2p^6\). Therefore, the simplest ionic compound is \(\mathrm{NaF}\). b. Isoelectronic with \(\mathrm{Kr}\):
03

Identify the electronic configuration of the noble gas.

The electronic configuration of krypton (Kr) is \([Ar] 3d^{10} 4s^2 4p^6\), and it has 36 electrons.
04

Find the cation and anion that would form an ionic compound isoelectronic with Kr.

We can choose the Rubidium ion (Rb+). When Rubidium loses one electron, it would have the same electronic configuration as Kr, \([Ar] 3d^{10} 4s^2 4p^6\). To balance the charge, we can choose the Bromide ion (\(\mathrm{Br^-}\)), which has 35 electrons; when it gains one electron, it would also have an electronic configuration of \([Ar] 3d^{10} 4s^2 4p^6\). Therefore, the simplest ionic compound is \(\mathrm{RbBr}\). c. Isoelectronic with \(\mathrm{Ar}\)
05

Identify the electronic configuration of the noble gas.

The electronic configuration of argon (Ar) is \(1s^2 2s^2 2p^6 3s^2 3p^6\), and it has 18 electrons.
06

Find the cation and anion that would form an ionic compound isoelectronic with Ar.

We can choose the magnesium ion (\(\mathrm{Mg^{2+}}\)). When Magnesium loses two electrons, it would have the same electronic configuration as Ar, \(1s^2 2s^2 2p^6 3s^2 3p^6\). To balance the charge, we can choose the oxide ion (\(\mathrm{O^{2-}}\)), which has 8 electrons; when it gains two electrons, it would also have an electronic configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6\). Therefore, the simplest ionic compound is \(\mathrm{MgO}\). d. Isoelectronic with \(\mathrm{Xe}\)
07

Identify the electronic configuration of the noble gas.

The electronic configuration of xenon (Xe) is \([Kr] 4d^{10} 5s^2 5p^6\), and it has 54 electrons.
08

Find the cation and anion that would form an ionic compound isoelectronic with Xe.

We can choose the Cesium ion (\(\mathrm{Cs^{+}}\)). When Cesium loses one electron, it would have the same electronic configuration as Xe, \([Kr] 4d^{10} 5s^2 5p^6\). To balance the charge, we can choose the Iodide ion (\(\mathrm{I^{-}}\)), which has 53 electrons; when it gains one electron, it would also have an electronic configuration of \([Kr] 4d^{10} 5s^2 5p^6\). Therefore, the simplest ionic compound is \(\mathrm{CsI}\).

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