How can the bonds between B and N be polar while the compound B3N3H6(inorganic benzene) is non polar?
In: Chemistry
A single B-N bond is polar because the significant difference in their electronegativity causes the compound to form the positive-negative poles on the molecule. However B3N3H6 is a ring structure where the electronegativity is evenly distributed around the molecular structure so there is no polarity.
Your title needs to be amended in that *single* bonds between B-N are polar. Also the question is really “how polar”?
In benzene there’s sufficient polarization that it is more water soluble than most other hydrocarbons. It dissolves slightly in water.
The bonding structure in borazine is analogous to that of benzene, indeed the two compounds have almost the same molar mass and density.
Like benzene, borazine contains delocalized Pi bonds. In benzene these form a pair of donut-shaped, electron-dense bonding regions above and below the plane of the molecule. The relatively higher electron density in the pi-bonding regions explains the slight water solubility. However the bonding differs in borazine due to the greater positive charge and greater electronegativity of nitrogen. The pi bonds take on a roughly triangular shape because the nitrogens gain more negative charge by withdrawing electrons. For this reason the nitrogen atoms are also smaller and the bond angles differ slightly from the 120° in benzene. Nitrogen also has a greater tendency to withdraw electrons from their respective associated hydrogens. Therefore in borazine the nitrogen associated hydrogens have a greater positive charge. Overall this makes borazine slightly more polar than benzene. It is yet more water soluble.
can someone explain the question to me as though I were five? bc *woosh* over my head lol
So the individual B-N bonds are polar but because of the ring shape of the molecules all of the dipoles cancel out the fields from the others. If you draw vector lines and sum them you will find they all add to zero.
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