Q.Consider the isoelectronic species, Na^+, Mg^2+, F^- and O^2-. The correct order of increasing length of their radii is _________.

The key idea here is the trend of ionic radii in isoelectronic species.

For isoelectronic species (those having the same number of electrons), the ionic radius is inversely proportional to the nuclear charge (atomic number, $Z$). A higher nuclear charge exerts a stronger pull on the electron cloud, drawing it closer to the nucleus and resulting in a smaller ionic radius.

1. All the given species ($Na^+$, $Mg^{2+}$, $F^-$, $O^{2-}$) are isoelectronic, each possessing 10 electrons.
2. Determine the atomic number ($Z$) for each element:
   • Oxygen ($O$): $Z = 8$
   • Fluorine ($F$): $Z = 9$
   • Sodium ($Na$): $Z = 11$
   • Magnesium ($Mg$): $Z = 12$
3. The order of increasing nuclear charge is $O < F < Na < Mg$.
4. Since ionic radius decreases with increasing nuclear charge for isoelectronic species, the order of increasing ionic radii will be the reverse of the nuclear charge order.

When comparing the radii of different ions, especially when they are isoelectronic, the key factor to consider is the effective nuclear charge ($Z_{eff}$). Isoelectronic species are atoms or ions that have the same number of electrons.

The effective nuclear charge is the net positive charge experienced by an electron in a multi-electron atom. It is influenced by the actual nuclear charge (number of protons, $Z$) and the shielding effect of inner electrons.

For a series of isoelectronic species, the number of electrons is constant. This means the electron-electron repulsion and the shielding effect are roughly comparable across the series. In such a scenario, the dominant factor determining the size of the ion is the magnitude of the nuclear charge ($Z$).

A higher nuclear charge means a stronger attractive force exerted by the nucleus on the electron cloud. This stronger pull draws the electrons closer to the nucleus, resulting in a smaller ionic radius. Conversely, a lower nuclear charge leads to a weaker attraction, allowing the electron cloud to expand and resulting in a larger ionic radius.

Let's apply this concept to the given species:

1. Identify the number of electrons in each species.

   • Sodium (Na) has an atomic number of 11, meaning 11 protons and 11 electrons. Na$^{+}$ has lost one electron, so it has $11 - 1 = 10$ electrons.
   • Magnesium (Mg) has an atomic number of 12, meaning 12 protons and 12 electrons. Mg$^{2+}$ has lost two electrons, so it has $12 - 2 = 10$ electrons.
   • Fluorine (F) has an atomic number of 9, meaning 9 protons and 9 electrons. F$^{-}$ has gained one electron, so it has $9 + 1 = 10$ electrons.
   • Oxygen (O) has an atomic number of 8, meaning 8 protons and 8 electrons. O$^{2-}$ has gained two electrons, so it has $8 + 2 = 10$ electrons. All four species (Na$^{+}$, Mg$^{2+}$, F$^{-}$, O$^{2-}$) are indeed isoelectronic, each possessing 10 electrons.

2. Determine the nuclear charge ($Z$) for each species.

   The nuclear charge is simply the atomic number of the element.

   • O$^{2-}$: $Z = 8$ (Oxygen)
   • F$^{-}$: $Z = 9$ (Fluorine)
   • Na$^{+}$: $Z = 11$ (Sodium)
   • Mg$^{2+}$: $Z = 12$ (Magnesium)

3. Order the species based on their nuclear charge.

   Arranging them in increasing order of nuclear charge:

   $Z(\text{O}^{2-}) < Z(\text{F}^{-}) < Z(\text{Na}^{+}) < Z(\text{Mg}^{2+})$

   $8 < 9 < 11 < 12$

4. Determine the order of increasing ionic radii.

   Since ionic radius is inversely proportional to nuclear charge for isoelectronic species, the species with the highest nuclear charge will have the smallest radius, and the species with the lowest nuclear charge will have the largest radius.

   Therefore, the order of increasing ionic radii will be the reverse of the order of increasing nuclear charge:

   $\text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} < \text{O}^{2-}$

   Let's summarize this in a table:

The correct order of increasing length of their radii is Mg$^{2+}$ < Na$^{+}$ < F$^{-}$ < O$^{2-}$.