The electron configuration you've provided is \( \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^1 \). This corresponds to the element sodium (Na), which is in group 1 of the periodic table. Sodium has one valence electron in the 3s orbital.
You mentioned the other atom has two valence electrons. An atom with two valence electrons could be an alkaline earth metal, like magnesium (Mg) or calcium (Ca).
When sodium approaches an atom with two valence electrons, it is likely to form a bond based on the tendency to achieve a more stable electronic configuration. Sodium, having one valence electron, tends to lose this electron and form a cation (\( \text{Na}^+ \)). On the other hand, an atom with two valence electrons (e.g., Mg) can either share or transfer electrons to achieve a full outer shell.
In this scenario, one possible interaction might be:
- Sodium losing its one valence electron: Sodium would become positively charged (\( \text{Na}^+ \)).
- Other atom (e.g., Mg) could gain an electron: If the other atom is Mg, it could then lose one of its valence electrons to achieve a stable configuration.
However, since sodium can only lose one electron and does not have the capacity to share more than that in simple ionic bonding, an ionic bond could form if it reacts with a suitable element to facilitate electron transfer.
If this atom with two valence electrons is also compatible for sharing electrons (thereby forming a covalent bond), then a bond can form. However, it would primarily depend on the specific behavior of the two atoms in the context of their electronegativities, the type of bond (ionic or covalent), and other factors.
In summary, yes, a bond can form; the specific type of bond (ionic or covalent) will depend on the individual characteristics of the interacting atoms.