Question

Arrange the following in decreasing order of ionic radius:
N3-,Mg2+,Na+,O2-​

Answer 1

Answer:

The decreasing order of the ionic radii of the following species is N3->O2->Na+>Mg2+

Explanation

We know that the atomic size of a particular species depends on the effective nuclear charge and the shielding effect due to the surrounding electrons. The greater the effective nuclear charge on the species, the smaller is the species and the greater the shielding effect of the surrounding electrons on the species, the larger is the atom. When a neutral atom gains an electron, an anion is formed and when a neutral atom loses an electron, a cation is formed. Also, when an atom gains an electron it increases in size because the effective nuclear charge on the atom decreases and the size of the atom increases but when a neutral atom loses an electron the effective nuclear charge of the species increases but the size decreases.

This the correct decreasing order of the ionic radii of the following species is N3->O2->Na+>MG2+

Answer 2

Answer: The decreasing order of ionic radius is: $N^{3-}>O^{2-}>Na^+>Mg^{2+}$.

Explanation:

Now, let us look at the question, $N^{3-}, Mg^{2+}, Na^+, O^{2-}$.

All the four ions $N^{3-}, Mg^{2+}, Na^+, O^{2-}$ are having 10 electrons and have only two shells, but the number of protons is different.

In $N^{3-}$, 7 protons are holding 10 electrons. The effective nuclear charge will be less, and hence size increase.

In $Mg^{2+}$, 12 protons are holding 10 electrons. The effective nuclear charge will be high, and hence size decreases.

In $Na^{+}$, 11 protons are holding 10 electrons. The effective nuclear charge will be high, and hence size decreases a little.

In $O^{2-}$, 8 protons are holding 10 electrons. The effective nuclear charge will be less, and hence size increases but the increase is less in comparison to  $N^{3-}$.

Thus, the order is $N^{3-}>O^{2-}>Na^+>Mg^{2+}$.