Q1.
If the valence shell of an atom has less than four electrons, it would generally donate its valence electrons to achieve a stable electronic configuration. Identify four such elements among the first 18 elements by using their electronic configurations given in Table 8.4 of the Chapter 8, Journey Inside the Atom.
Answer
Lithium, beryllium, sodium, magnesium and aluminium — any four of these.
| Element | Z | Electronic configuration | Valence electrons | Ion formed on donating them |
|---|---|---|---|---|
| Lithium (Li) | 3 | 2, 1 | 1 | Li⁺ |
| Beryllium (Be) | 4 | 2, 2 | 2 | Be²⁺ |
| Sodium (Na) | 11 | 2, 8, 1 | 1 | Na⁺ |
| Magnesium (Mg) | 12 | 2, 8, 2 | 2 | Mg²⁺ |
| Aluminium (Al) | 13 | 2, 8, 3 | 3 | Al³⁺ |
Why it happens: giving away 1, 2 or 3 electrons is a short journey; gaining 5, 6 or 7 to complete an octet is a long one. When sodium (2, 8, 1) loses its single M-shell electron, the shell underneath is already a full 2, 8 — the same stable arrangement as neon. So one small loss buys a complete octet. That is why elements with fewer than four valence electrons are metals and form cations.
Tip: boron (2, 3) also has fewer than four valence electrons, but with three electrons to give and a small, tightly holding atom it prefers to share. The “less than four → donate” rule is a good guide, not a law.