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Chapter 8: Problem 55

Give three ions that are isoelectronic with krypton. Place these ions in order of increasing size.

Short Answer

Expert verified
The three ions that are isoelectronic with krypton are \(Sr^{2+}\), \(Rb^{+}\), and \(Br^-\). They are placed in order of increasing size as follows: \(Sr^{2+} < Rb^{+} < Br^{-}\).

Step by step solution

01

Identify Elements Close to Krypton

First, we need to find elements surrounding Krypton in the periodic table, as they are most likely to have ions isoelectronic to Krypton. Some examples are Rubidium (Rb, atomic number 37), Bromine (Br, atomic number 35), and Strontium (Sr, atomic number 38).
02

Determine the Ions

Rubidium (Rb) can lose one electron to form Rb+, Bromine (Br) can gain one electron to form Br-, and Strontium (Sr) can lose two electrons to form Sr2+. All of these ions have 36 electrons, which is the same as Krypton, so they are isoelectronic.
03

Determine the Size Order

Since the number of protons in the nucleus determines the attraction between electrons and the nucleus, larger atomic numbers lead to a stronger attraction and, hence, smaller ion size. Therefore, the order of increasing size would be as follows: \[Sr^{2+} < Rb^{+} < Br^{-}\] These three ions are isoelectronic with Krypton and are placed in order of increasing size: Sr2+ < Rb+ < Br-.

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