Q.The radius of Na^+ cation is less than that of Na atom. Give reason.

The key idea here is Ionic Radii Trends.

1. A neutral sodium atom ($\text{Na}$) has 11 protons and 11 electrons, with its outermost electron in the $3s$ orbital.
2. When $\text{Na}$ forms a cation ($\text{Na}^+$), it loses its single valence electron from the $3s$ orbital. This results in the complete removal of the outermost (3rd) electron shell.
3. The $\text{Na}^+$ ion now has 11 protons but only 10 electrons. The remaining 10 electrons are in the 2nd shell. …

When an atom forms a cation, it loses one or more electrons. This process fundamentally alters the balance between the attractive forces from the nucleus and the repulsive forces among electrons, leading to a change in size. For sodium, the formation of a cation involves a significant structural change that reduces its radius.

Here's a step-by-step explanation:

1. Electron Configuration of Sodium Atom ($\text{Na}$):

   • A neutral sodium atom ($\text{Na}$) has an atomic number of 11, meaning it has 11 protons in its nucleus and 11 electrons orbiting the nucleus.
   • Its electron configuration is $1s^2 2s^2 2p^6 3s^1$.
   • The outermost electron, also known as the valence electron, is in the $3s$ orbital, which belongs to the third electron shell. This $3s$ electron is relatively far from the nucleus and is shielded by the inner 10 electrons.

2. Formation of Sodium Cation ($\text{Na}^+$):

   • Sodium readily loses its single $3s$ valence electron to achieve a stable electron configuration, identical to that of the noble gas Neon ($\text{Ne}$).
   • The $\text{Na}^+$ cation still has 11 protons in its nucleus, but now it has only 10 electrons.
   • Its electron configuration becomes $1s^2 2s^2 2p^6$.

3. Loss of an Entire Electron Shell:

   • In the neutral $\text{Na}$ atom, electrons occupy the first, second, and third electron shells. The atomic radius is determined by the extent of the third shell.
   • When $\text{Na}$ loses its $3s^1$ electron, the entire third electron shell is removed. The $\text{Na}^+$ cation now only has electrons in the first and second shells.
   • This reduction in the number of principal electron shells is the most significant factor contributing to the decrease in size.

4. Increased Effective Nuclear Charge ($Z_{eff}$):

   • The nuclear charge (number of protons) remains constant at $+11$ for both $\text{Na}$ and $\text{Na}^+$.
   • However, in $\text{Na}$, 11 protons attract 11 electrons, while in $\text{Na}^+$, the same 11 protons attract only 10 electrons.
   • With fewer electrons to attract, the positive charge of the nucleus is distributed among a smaller number of electrons. This means each of the remaining 10 electrons in $\text{Na}^+$ experiences a stronger attractive pull from the nucleus. This increased attraction is quantified as an increase in the effective nuclear charge ($Z_{eff}$) experienced by each electron. …