The electron configuration of an element is 1s22s22p63s1. Describe what most likely happens when an atom of this element comes near an atom having seven valence electrons.

According to the octet rule, there must be 2, 8, 18, 32 and so on number of electrons present in K, L, M, N and so on shells.

Since, there is there is 1 valence electron present in the valence shell of given electronic configuration as there are 2, 8, 1 electrons. Therefore, it will donate its one extra electron to another atom.

For example, an atom with 7 valence electrons will most likely take up this extra electron in order to complete its octet to attain stability.

Hence, an ionic bond will be formed between both the atoms.

What is likely to happens when an atom of with 1s2 2s2 2p6 3s1 electron configuration comes near an atom having seven valence electrons is that

An ionic bond will be formed between element with electronic configuration of 1s2 2s2 2p6 3s1 and atom with seven valence electrons.

Explanation

Ionic bond is formed when one atom loses valence electron while the other atom gain electron to fill the valence electron.

Atom with 1s2 2s2 2p6 3s1 electron configuration loses one electron while atom having seven valence electrons gain one electron to fill its valence electrons that is obtain octet electron configuration.