For each of the following sets of atoms and ions, arrange the members in order of increasing size.

Mg2+ F- and Na+

It is somewhat like this as a trend:

Metal ions get smaller as you go to the right on the periodic table, and non-metals get larger.

Now, those are trends. In this case, look at the actual sizes.
http://www.chemguide.co.uk/atoms/properties/atradius.html

thanks:)

To arrange the given atoms and ions in order of increasing size, we need to consider their atomic numbers and charges.

First, let's look at the neutral magnesium (Mg) atom. The atomic number of magnesium is 12, indicating that it has 12 protons and 12 electrons. Since it is neutral, the number of protons is equal to the number of electrons, so the size of the neutral magnesium atom is determined by its atomic radius.

Next, let's consider the magnesium ion with a +2 charge (Mg2+). Due to the loss of 2 electrons, the number of protons remains unchanged at 12, but now there are only 10 electrons. The decrease in the number of electrons results in a smaller ionic radius compared to the neutral magnesium atom.

Moving to the fluoride ion (F-), fluorine (F) has an atomic number of 9 and normally has 9 electrons. However, with a -1 charge, the fluoride ion gains one extra electron, resulting in a total of 10 electrons. This increase in the number of electrons leads to a larger ionic radius compared to the neutral fluorine atom.

Lastly, let's consider the sodium ion with a +1 charge (Na+). Sodium (Na) has an atomic number of 11 and normally has 11 electrons. With the loss of one electron, the number of electrons decreases to 10. Similar to the magnesium ion, the decrease in the number of electrons results in a smaller ionic radius compared to the neutral sodium atom.

Now, let's arrange the atoms and ions in order of increasing size based on the explanation above:

1. F- (Fluoride ion) - largest ionic radius
2. Mg (Magnesium atom)
3. Na+ (Sodium ion)
4. Mg2+ (Magnesium ion) - smallest ionic radius