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structure diagram of tetrafluoroborate anion

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tetrafluoroborate anion | structure diagram
tetrafluoroborate anion | structure diagram

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Draw the Lewis structure of tetrafluoroborate anion. Start by drawing the overall structure of the molecule:  Count the total valence electrons of the boron (n_B, val = 3) and fluorine (n_F, val = 7) atoms, including the net charge: n_B, val + 4 n_F, val - n_charge = 32 Calculate the number of electrons needed to completely fill the valence shells for boron (n_B, full = 8) and fluorine (n_F, full = 8). Note that boron has three valence electrons but four bonds, implying that the net charge was given to boron, allowing boron to make the fourth bond and fully fill its valence shell: n_B, full + 4 n_F, full = 40 Subtracting these two numbers shows that 40 - 32 = 8 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 4 bonds and hence 8 bonding electrons in the diagram. Fill in the remaining unbonded electrons on each atom. In total, there remain 32 - 8 = 24 electrons left to draw. Lastly, fill in the formal charges: Answer: |   |
Draw the Lewis structure of tetrafluoroborate anion. Start by drawing the overall structure of the molecule: Count the total valence electrons of the boron (n_B, val = 3) and fluorine (n_F, val = 7) atoms, including the net charge: n_B, val + 4 n_F, val - n_charge = 32 Calculate the number of electrons needed to completely fill the valence shells for boron (n_B, full = 8) and fluorine (n_F, full = 8). Note that boron has three valence electrons but four bonds, implying that the net charge was given to boron, allowing boron to make the fourth bond and fully fill its valence shell: n_B, full + 4 n_F, full = 40 Subtracting these two numbers shows that 40 - 32 = 8 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 4 bonds and hence 8 bonding electrons in the diagram. Fill in the remaining unbonded electrons on each atom. In total, there remain 32 - 8 = 24 electrons left to draw. Lastly, fill in the formal charges: Answer: | |