source : yahoo.com
If the boron nitride molecule, BN, were to form, what would its structure look like?
Add bonds and lone pairs as necessary.?
Boron nitride ….
Boron has three valence electrons and nitrogen has five, for a total of 8 electrons (what do you know, an octet). But that’s not enough electrons for both atoms to have an octet.
In its standard form, boron nitride is a network solid and has a very high melting point. It would not exist as discrete molecules of B≡N:, and yes, that is the best arrangement if it did exist as discrete molecules. It will deviate from the octet rule, but remember, the octet rule isn’t all its cracked up to be. A better indicator of structure is formal charge. And in B≡N: the formal charges on B and N are both zero, indicating that the triple bond and lone pair on nitrogen is the best arrangement.
Formal charge is not the actual charge on an atom in a compound, nor is it the oxidation state of an element in a compound. In BN, nitrogen has an oxidation state of -3 and boron has an oxidation state of +3, but those are not actual charges (nor are they formal charges). Formal charge (FC) is computed for each element in a compound using the formula below. The closer the formal charges are to zero, the more likely is the structure.
FC = VE – NBE – ½BE … VE=valence electrons, NBE=nonbonding electrons, BE=bonding electrons
Boron nitride is thought to exist in three different crystalline forms. More about that can be found here:
Boron and nitrogen together have 8 electrons, just like two carbon atoms. Boron and nitrogen can form structures which are analogous to carbon structures, like graphite and diamond. In the graphite-like structure, each B and each N has three bonds to boron atoms and each B has three bonds to nitrogen atoms, flat sheets of hexagons. The flat sheets are held together by London dispersion forces.
Does your program what you to draw that?
In BN(hex) there are single bonds, so draw BN with a single bond and two lone pairs on N and one lone pair on B. Now nothing, not even N has an octet of electrons, but the formal charges will be zero on each.
: B−N :
Actually you could draw it with two unpaired electrons on B to indicate two more bonds and two unpaired electrons on N to indicate two more bonds, and N will have a lone pair.
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