Question

Because fluorine has seven valence electrons \(\left(2 s^{2} 2 p^{5}\right),\) seven covalent bonds in principle could form around the atom. Such a compound might be \(\mathrm{FH}_{7}\) or \(\mathrm{FCl}_{7}\). These compounds have never been prepared. Why?

Short answer

Compounds \(FH_{7}\) or \(FCl_{7}\) are not formed due to the octet rule. Fluorine only needs one additional electron to form a stable octet, so it can only form one covalent bond. Making seven bonds, as in \(FH_{7}\) and \(FCl_{7}\), would violate the stability provided by the octet rule.

Step-by-step solution

Understanding Fluorine's Electron Configuration

The electron configuration of fluorine is \(2 s^{2} 2 p^{5}\). This means does not have the capability to form seven covalent bonds. There are only five available places in the p-orbital for bonding, two electrons filling the s-orbital already.

Applying the Octet Rule

The Octet Rule states that atoms will gain, lose, or share electrons in order to achieve a stable configuration of 8 electrons in their outermost shell (an octet). For Fluorine, the octet rule is fulfilled when it is surrounded by eight electrons, gaining just one more electron to fill the 2p sub-level of its second energy level.

Explaining why \(FH_{7}\) or \(FCl_{7}\) are not Formed

As Fluorine has fulfilled the octet rule by having 7 electrons and only needing one more for stability, it can only form one covalent bond. It is thus impossible for Fluorine to form compounds such as \(FH_{7}\) or \(FCl_{7}\), as these would require Fluorine to have seven bonds, which doesn't obey the Octet rule. Indeed, Fluorine only forms one bond in compounds.