.. ..

: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:

..

.. ..

: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:

..

has three bonding pairs and one non-bonding pair, which is the geometry for trigonal pyramidal, not trigonal planar.

There are six electron domains in . has five bonding and one nonbonding pairs of electrons. The electron domain geometry of is octahedral. When a molecule has octahedral electron domain geometry and has five bonding and one nonbonding pairs, the molecular geometry is square pyramidal.

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.

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.

Use the molar mass and percent by mass of sodium to find the moles of sodium in the compound.

Since the compound consists of only sodium and element X, we can subtract to find the percent by mass of element X.

Use the percent by mass to find the mass per mole of compound of element X.

Because there is a 1:1 ratio, between sodium and element X, they must have an equal number of moles in the compound. Since there are 2 moles of sodium, there must also be 2 moles of element X. Divide the mass of element X by the moles of element X to find its atomic mass and its identity.

By consulting a periodic table, we can find that element X is oxygen.

Therefore element X is Oxygen.

Therefore, the molecular formula is and the empirical formula is .

BCl3 only has six electrons around boron, while NO2 (with an odd number of electrons) would have only 7 electrons around the central nitrogen.

BCl3 only has six electrons around boron, while NO2 (with an odd number of electrons) would have only 7 electrons around the central nitrogen.

Use the molar mass and percent by mass of sodium to find the moles of sodium in the compound.

Since the compound consists of only sodium and element X, we can subtract to find the percent by mass of element X.

Use the percent by mass to find the mass per mole of compound of element X.

Because there is a 1:1 ratio, between sodium and element X, they must have an equal number of moles in the compound. Since there are 2 moles of sodium, there must also be 2 moles of element X. Divide the mass of element X by the moles of element X to find its atomic mass and its identity.

By consulting a periodic table, we can find that element X is oxygen.

Therefore element X is Oxygen.

Therefore, the molecular formula is and the empirical formula is .