The correct order of increasing ionization energy for the given species is:

A $\mathrm{Ar} < \mathrm{P}^{3-} < \mathrm{S}^{2-} < \mathrm{Cl}^{-}$ B $\mathrm{Ar} < \mathrm{Cl}^{-} < \mathrm{S}^{2-} < \mathrm{P}^{3-}$ C $\mathrm{P}^{3-} < \mathrm{S}^{2-} < \mathrm{Cl}^{-} < \mathrm{Ar}$ (D) $\mathrm{Cl}^{-} < \mathrm{S}^{2-} < \mathrm{P}^{3-} < \mathrm{Ar}$

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The correct option is C: $\mathrm{P}^{3-} < \mathrm{S}^{2-} < \mathrm{Cl}^{-} < \mathrm{Ar}$.

To understand why, let's analyze the reasoning behind this order:

  1. Electron Configuration and Nuclear Charge:

    • $\mathrm{P}^{3-}, \mathrm{S}^{2-}, \mathrm{Cl}^{-}$ all have the same number of electrons, 18. However, they have different nuclear charges because the number of protons varies.
    • The greater the negative charge on an anion, the larger its radius. This is due to the increased net repulsion among electrons, which outweighs the nuclear attraction, causing the ion to expand.
  2. Order of Ionic Radii:

    • The ionic radii follow the order: $\mathbf{\mathrm{P}^{3-} > \mathrm{S}^{2-} > \mathrm{Cl}^-}$.
  3. Ionization Energy (IE):

    • Argon (Ar), being a noble gas, has a very stable electronic configuration, resulting in the highest ionization energy.
    • Since ionization energy is inversely related to ionic radius, the order of increasing ionization energy is:

$$ \mathbf{\mathrm{P}^{3-} < \mathrm{S}^{2-} < \mathrm{Cl}^{-} < \mathrm{Ar}} $$

Thus, the option C correctly represents the increasing order of ionization energy for the species given.

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