The octet rule requires that each atom in the molecule has eight valence electrons, fully filling thesandpsubshells. Atoms that have a full octet tend to be more stable and lower in energy.

In the compound, each fluorine has seven initial valence electrons and boron has three initial valence electrons. Upon forming the compound bonds, boron shares its three electrons with each fluorine through covalent bonds. This givens each fluorine a full octet, but leave boron with only three covalent bonds, resulting in only six valence electrons. Boron does not satisfy the octet rule in this compound.

Each atom in the other listed compounds forms bonds to generate a full octet. Sodium chloride achieves this through an ionic bond. Methanal and diatomic oxygen both use double bonds to help achieve octets.

.. ..

:F — B — F: Boron can have a violated octet (6 e^-) and each F has 3 lone pairs

.. | .. for a total of 9 pairs of unpaired electrons

:F:

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BCl₃only has six electrons around boron, while NO₂(with an odd number of electrons) would have only 7 electrons around the central nitrogen.

Lewis dot diagrams can help us keep track on how the valence electrons will disperse themselves among orbitals in the atom. When drawing these diagrams, it is important to keep two things in mind.

1. Thesorbital will be filled before theporbital begins to be filled.

2. Electrons will never pair in the same orbital if an orbital of equal energy is unoccupied.

Knowing these two facts, we can predict which of the following elements will have two unpaired electrons.

Fluorine will have seven valence electrons, meaning that only one orbital will not be completely filled.

Beryllium only has two valence electrons, but they will both be found in the2sorbital, because it must be filled before theporbitals can receive electrons.

Nitrogen will have five valence electrons, so two will be found in the2sorbital. The other three will be present in their ownporbitals, meaning nitrogen has three unpaired electrons.

Neon will have eight valence electrons, fully filling the2sand2p轨道。It will have no unpaires electrons.

Carbon has one less valence electron than nitrogen, meaning that it will have two unpaired electrons in thep轨道。

Isoelectronic means having the same electron configuration (i.e. the same number of electrons).andboth have the same number of electrons (18), while sulfur has 16, phosphorous has 15, neon has 10, and the sodium cation has 10.

When drawing a Lewis dot structure, we are always trying to reach an electron count where all atoms involved are stable and (usually) have full octets. We are also trying to estabilsh a structure in which we have the smallest formal charge possible. The general rule is first to draw out all of the elements involved and their valence electrons, then start piecing them together trying to reduce the formal charge and get all elements involved to an octet. There are a couple exceptions to the octet rule.

钠和氯形成离子键,th意义at one atom will donate an electron and the other will receive it. This gives each atom a charge. Chlorine has seven valence electrons, while sodium has one valence electron. For each atom to arrive at an octet, sodium will need to lose one electron and chlorine will need to gain one electron. This would give chlorine a negative charge, and sodium a positive charge.

Thus, the answer is a sodium with a positive charge (due to one lost electron) and a chlorine with eight electrons and a negative charge (due to one electron gained).

When drawing a Lewis dot structure, we are always trying to reach an electron count where all atoms involved are stable and (usually) have full octets. We are also trying to estabilsh a structure in which we have the smallest formal charge possible. The general rule is first to draw out all of the elements involved and their valence electrons, then start piecing them together trying to reduce the formal charge and get all elements involved to an octet. There are a couple exceptions to the octet rule.

In this case, boron actually has an incomplete octet. Though there are resonance forms in which boron has a full octet, when you calculate the formal charge of these configurations it will not be zero.