Question

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

Short answer

The three ions that are isoelectronic with neon are Na^+, F^-, and Al^3+. In order of increasing size, these ions are: Al^3+ < Na^+ < F^-.

Step-by-step solution

Determine the number of electrons in neon

Neon (Ne) is a noble gas, and its atomic number is 10. This means that it has 10 protons and 10 electrons.

Identify ions with the same number of electrons as neon

To find ions that are isoelectronic with neon, we should look for ions that have 10 electrons. Some examples are: 1. Sodium ion (Na^+): Sodium has 11 electrons, but when it loses 1 electron, it will have 10 electrons (Na^+). 2. Fluoride ion (F^-): Fluorine has 9 electrons, but when it gains 1 electron, it will have 10 electrons (F^-). 3. Aluminum ion (Al^3+): Aluminum has 13 electrons, but when it loses 3 electrons, it will have 10 electrons (Al^3+).

Order the ions based on their size

Let's compare the sizes of these ions based on their nuclear charge and electron configuration. 1. Na^+: Sodium ion has 11 protons and 10 electrons. It has lost its valence electron, making its outermost electron shell closer to the nucleus. 2. F^-: Fluoride ion has 9 protons and 10 electrons. It has gained an extra electron in its outermost shell, making it a little larger than neon. 3. Al^3+: Aluminum ion has 13 protons and 10 electrons. It has lost 3 electrons, making its outermost electron shell much closer to the nucleus. When an ion has more protons, its nucleus attracts electrons more strongly, making the ion smaller. Based on this information, we can order these ions in terms of increasing size: Al^3+ < Na^+ < F^-