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 fluoride ion (F⁻), sodium ion (Na⁺), and magnesium ion (Mg²⁺). The order of increasing size for these ions is F⁻ > Na⁺ > Mg²⁺.

Step-by-step solution

Find the electron configuration of neon

Neon is a noble gas with atomic number 10, which means it has 10 electrons in its neutral state – two in the first shell and 8 in the second shell. Therefore, its electron configuration is 1s²2s²2p⁶.

Identify ions with the same number of electrons

We need to find ions that have the same number of electrons (10) as neon. To do this, we can identify ions that have gained or lost electrons compared to their neutral state. Three such ions are: 1. Fluoride ion (F⁻): Gained 1 electron; atomic number 9 2. Sodium ion (Na⁺): Lost 1 electron; atomic number 11 3. Magnesium ion (Mg²⁺): Lost 2 electrons; atomic number 12 All three ions have 10 electrons, just like neon.

Arrange the ions in order of increasing size

Since all three ions have the same number of electrons, their size will depend on the effective nuclear charge experienced by the outermost electrons. Here, atomic number plays a role: 1. F⁻: 9 protons, 10 electrons 2. Na⁺: 11 protons, 10 electrons 3. Mg²⁺: 12 protons, 10 electrons Effective nuclear charge is calculated by subtracting the shielding effect due to inner shells from the total number of protons. Since all three ions have the same number of inner-shell electrons, we can use the atomic numbers to determine the order of increasing effective nuclear charge: F⁻ (9 protons) < Na⁺ (11 protons) < Mg²⁺ (12 protons) Outer electrons are attracted more strongly to the nucleus as the effective nuclear charge increases. As the attractive force increases, the atomic size (ion radii) decreases. Therefore, the order of increasing size is: F⁻ > Na⁺ > Mg²⁺