Use principles of atomic structure to answer each of the following: \(^{[4]}\) (a) The radius of the Ca atom is 197 pm; the radius of the \(C a^{2+}\) ion is 99 pm. Account for the difference. (b) The lattice energy of \(\mathrm{CaO}(s)\) is \(-3460 \mathrm{kJ} / \mathrm{mol}\); the lattice energy of \(\mathrm{K}_{2} \mathrm{O}\) is \(-2240 \mathrm{kJ} / \mathrm{mol}\). Account for the difference. (c) Given these ionization values, explain the difference between \(\mathrm{Ca}\) and \(\mathrm{K}\) with regard to their first and second ionization energies. $$\begin{array}{|c|c|c|} \hline \text { Element } & \text { First lonkation anerey (caluol) } & \text { Second lonkation anary (Calmo) } \\ \hline \mathrm{K} & 419 & 3050 \\ \hline \mathrm{Ca} & 590 & 1140 \\ \hline \end{array}$$ (d) The first ionization energy of \(\mathrm{Mg}\) is \(738 \mathrm{kJ} / \mathrm{mol}\) and that of \(\mathrm{Al}\) is \(578 \mathrm{kJ} / \mathrm{mol}\). Account for this difference.

Understanding the difference between atomic and ionic radii is fundamental in grasping concepts of atomic structure. For instance, a neutral calcium atom (Ca) has a significantly larger radius than its ion (Ca2+). This occurs because when an atom loses electrons to form a cation, the decreased electron cloud size leads to less electron-electron repulsion. Consequently, the remaining electrons are drawn closer to the nucleus by its positive charge. In the provided example, where a Ca atom becomes Ca2+, it is shedding its entire outer energy level, causing the radius to halve from 197 pm to just 99 pm.

In the atomic radius comparison, we observe that as atoms lose electrons and become more positively charged, their radii decrease due to the increased nuclear pull on the fewer remaining electrons. It's a bit like a group of friends holding hands in a circle; if some let go, the circle gets smaller as they're pulled closer together.

Lattice energy is a term that explains the energy released when ions bond to form a solid. It's a great indicator of a compound's stability - the higher the lattice energy, the more stable the compound. For example, the lattice energy of calcium oxide (CaO) is far greater than that of potassium oxide (K2O). This disparity stems from two main factors: the charges on the ions and their radii.

Calcium ions (Ca2+) and oxygen ions (O2-) have higher charges than potassium ions (K+) and oxygen ions (O2-). A simple rule of thumb is that the larger the charge, the stronger the force of attraction between the ions, and thus, the higher the lattice energy. Also, CaO has potentially smaller ions compared to K2O, leading to a stronger interaction between them due to a shorter distance, reinforcing the lattice strength and energy. Imagine two magnets: stronger magnets (larger charges) that are closer together (smaller radii) stick together more firmly, a bit like the ions in these compounds.

Ionization energy refers to the amount of energy needed to remove an electron from an atom or ion. It's how we measure an atom's hold on its electrons. Different elements have varying ionization energies. For instance, calcium has higher first and second ionization energies than potassium. The first ionization energy of potassium (K) is lower because its valence electron is further away from the nucleus and less tightly held. Conversely, calcium's (Ca) first valence electron is closer to the nucleus and more tightly bound, costing more energy to remove.

When comparing second ionization energies, it's a bit like going through a series of security doors. The first door isn't too tough for either element, but the second door for potassium doesn't just open willingly. For potassium, this second door represents breaking into a very stable electron configuration, like a security vault, needing lots more energy. Meanwhile, calcium's second electron removal is just another regular door, easier to open as it's just removing another somewhat loosely held valence electron. This imaginative analogy helps to conceptualize the relative ease or difficulty in removing successive electrons from different elements.

The ionic radius measures the size of an ion in a crystal lattice. Unlike the atomic radius, ions can be larger or smaller than their corresponding atoms depending on whether they gain or lose electrons. When an atom loses an electron and becomes a positively charged cation, its ionic radius decreases as seen with calcium transforming into Ca2+. This decrease is due to the reduction in electron repulsion and increased nuclear pull. Conversely, when an atom gains electrons to become a negatively charged anion, its ionic radius increases because the added electrons increase repulsion forces, allowing them to occupy a larger volume.

Looking back at calcium, its smaller ionic radius as Ca2+ compared to the neutral atom exemplifies the cation effect. It is essential to note, as it does in the lattice energy discussion earlier, that the ionic radius plays a crucial role in the strength of ionic bonds: smaller ions can pack more tightly, leading to stronger bonds. So, whether an ion is snug or loose in its lattice quarters tells us a lot about the compound's properties and reactivity.