Which of the following order is correct for the size of \(\mathrm{Fe}^{3+}, \mathrm{Fe}\) and \(\mathrm{Fe}^{2+} ?\) (a) \(\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}<\mathrm{Fe}\) (b) \(\mathrm{Fe}<\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}\) (c) \(\mathrm{Fe}<\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}\) (d) \(\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}<\mathrm{Fe}\)

When we talk about atomic and ionic sizes, we're referring to how large an atom or ion is, which directly influences their physical and chemical properties. Atomic size is typically determined by the distance from the nucleus to the outermost shell of electrons.

For neutral atoms, as we move down a group in the periodic table, atomic size increases because additional electron shells are added. However, when atoms form ions, their size changes as well depending on the gain or loss of electrons.

Impact of Electron Loss or Gain on Size

• Cations: Atoms that lose one or more electrons to become positively charged cations generally decrease in size. This is due to the loss of an entire electron shell in some cases and also because the electrons which remain experience a stronger effective nuclear charge, are drawn closer to the nucleus, and reduce the radius of the ion.
• Anions: Atoms that gain electrons to become negatively charged anions increase in size. This is because the addition of extra electrons causes increased electron-electron repulsion and can increase the size of the electron cloud.

The concept of effective nuclear charge (often abbreviated as Z_eff) is integral to understanding atomic and ionic sizes. It refers to the net positive charge experienced by an electron in a multi-electron atom. The higher the effective nuclear charge, the more strongly electrons are attracted to the nucleus, leading to a smaller atomic or ionic radius.

Effective nuclear charge increases across a period on the periodic table due to the addition of protons in the nucleus, leading to a greater positive charge and a stronger pull on the electrons. It's important to note that despite the additional electrons also being added, they don't completely cancel out the effect of the extra protons because of their shielding effect.

Understanding Shielding

Electrons in inner shells partially shield the outer electrons from the pull of the nucleus. As a result, electrons in the outermost shell do not experience the full charge of the nucleus. However, the shielding effect is not strong enough to completely offset the increase in nuclear charge. Thus, in ions, when electrons are removed, the effective nuclear charge felt by the remaining electrons increases, pulling them closer to the nucleus and consequently decreasing the ionic size.

Comparing sizes of cations involves understanding how the removal of electrons affects the remaining electron cloud. As mentioned earlier, cations are formed when an atom loses one or more electrons, which results in a decrease of the ionic radius.

For a specific element, the more electrons that are removed, the smaller the ion becomes. The increase in effective nuclear charge due to the loss of electrons causes a stronger attraction toward the nucleus, squeezing the electron cloud more tightly and leading to smaller ionic sizes.

Size Comparison of Iron Cations

Taking iron as an example, when comparing Fe, Fe^{2+}, and Fe^{3+}, the removal of electrons (two for Fe^{2+} and three for Fe^{3+}) result in progressively smaller ions. The Fe atom is the largest because it has the most electrons, which generate enough repulsion to maintain a larger size, and has a lower effective nuclear charge relative to its own cations.

• Fe is larger than Fe^{2+} because Fe^{2+} has lost two electrons, increasing the effective nuclear charge and reducing its size.
• Fe^{2+} is larger than Fe^{3+} because Fe^{3+} has lost an additional electron, further increasing the effective nuclear charge and making it even smaller.

This results in the size order: Fe > Fe^{2+} > Fe^{3+}, which matches option (a) as the correct answer to the exercise.