(a) Why does Li have a larger first ionization energy than Na? (b) The difference between the third and fourth ionization energies of scandium is much larger than the difference between the third and fourth ionization energies of titanium. Why? (c) Why does Li have a much larger second ionization energy than Be?

The first ionization energy is the energy required to remove one electron from the outermost shell of an atom. It depends mainly on the effective nuclear charge and atomic size of the atom. Li (lithium) has an atomic number of 3 (1s²2s¹), while Na (sodium) has an atomic number of 11 (1s²2s²2p⁶3s¹). Both elements are in the same group (alkali metals), so the main difference will be in their atomic size and the effective nuclear charge they have. Li has a smaller atomic size and a higher effective nuclear charge, hence more energy is needed to remove its outer electron than Na, which has a larger atomic size and a lower effective nuclear charge.

Scandium (Sc) has an atomic number of 21 (1s²2s²2p⁶3s²3p⁶4s²3d¹), and Titanium (Ti) has an atomic number of 22 (1s²2s²2p⁶3s²3p⁶4s²3d²). Now let's analyze each ionization step. For the first three ionizations, electrons are removed from the 4s and 3d orbitals in both atoms. However, significant energy is needed to remove the fourth electron from Sc as it has to come out of the 3d orbital. In Ti, after the third ionization, one electron is still left in its 3d orbital, making the fourth ionization less difficult. This is why the difference between the third and fourth ionization energies of Sc is much larger than that of Ti. The 3d electrons in Sc are closer to the nucleus than the 4s electrons, providing additional shielding effect and making the ionization process harder.

Lithium (Li) has an electron configuration of 1s²2s¹, and Beryllium (Be) has an electron configuration of 1s²2s². When the first ionization occurs in Li, an electron from the outermost shell (2s¹) is removed. After this, the electron configuration becomes 1s², which corresponds to the noble gas configuration of helium. Hence, removing the second electron from Li will involve breaking a stable noble gas configuration, requiring a significantly larger amount of energy. In the case of Be, after the first ionization, the electron configuration becomes 1s²2s¹, which doesn't correspond to a noble gas configuration. Therefore, removing the second electron from Be will require less energy compared to Li. In summary, Li has a much larger second ionization energy than Be due to the stability of the noble gas configuration formed after the first ionization in Li.