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Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory
Chapter 6, Problem 38b

How many protons and electrons are in each of the following ions? (b) Rb+

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Step 1: Identify the atomic number of the element. The atomic number of an element is equal to the number of protons in its nucleus. For Rb (Rubidium), the atomic number is 37. Therefore, there are 37 protons in a Rb atom.
Step 2: Determine the charge of the ion. The charge of the Rb+ ion is +1. This means it has lost one electron.
Step 3: Calculate the number of electrons. The number of electrons in a neutral atom is equal to the number of protons. However, since Rb+ has a +1 charge, it means it has one less electron than protons. Therefore, the number of electrons in Rb+ is 37 (number of protons) - 1 (charge of the ion) = 36 electrons.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Atomic Structure

Atoms consist of protons, neutrons, and electrons. Protons are positively charged particles found in the nucleus, while electrons are negatively charged and orbit the nucleus. The number of protons defines the element, and in a neutral atom, the number of electrons equals the number of protons.
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Ions

Ions are atoms or molecules that have gained or lost one or more electrons, resulting in a net charge. Cations are positively charged ions formed by losing electrons, while anions are negatively charged ions formed by gaining electrons. The charge of an ion indicates the difference between the number of protons and electrons.
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Rubidium Ion (Rb+)

Rubidium (Rb) is an alkali metal with an atomic number of 37, meaning it has 37 protons and, in its neutral state, 37 electrons. The Rb+ ion indicates that it has lost one electron, resulting in a total of 37 protons and 36 electrons. This loss of an electron gives Rb+ its positive charge.
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