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1) First write the bonding sequence. Usually the first atom, other than H is the central atom. In this example Br is the central atom. Write Br first and draw two bonds to connect the 2 extra Br atoms in the Br3-.

Br-Br-Br

2) Count the total number of valence electrons : 3 X 7 + 1 = 22 (since bromine belongs to VII group, each Br has 7 valence electrons and add 1 for each - ve charge)

3) Since there are two bonds in the structure drawn in rule 1, subtract 4 bonding electrons (2 electrons for each bond) from the total valence electrons, 22 (22-4 =18). This gives 18 electrons.

4) Distribute these electrons to the atoms in the Br3-. First satisfy the surrounding atoms, making sure that each atom has an octet around it (H will have doublet). Each surrounding Br atom needs 6 electrons to attain octet. Place three pairs of electrons on each of the surrounding Br atom.

.. ..

:Br - Br - Br: (Note: lone pairs are not placed correctly, and I am not able to edit it)

.. ..

12 electrons (2 x 6=12) are used. Still 6 (18-6= 12) electrons are remaining from rule 3. Place them as three pairs around the central Br atom.

.. .. ..

:Br - Br - Br:

.. .. .. ..

In this example, the central atom Br has 10 electrons around it, in other words it exceeded the octet. Elements from the third period and below can exceed the octet since they have the d orbitals to accommodate the excess electrons.

Since this is an ionic species, draw a square bracket and show the -ve charge of the ion as superscript.

.. .. .. _

[:Br - Br - Br:]

.. .. .. ..

(Note: This editor doesn't let me put the lone pairs on the atoms correctly. I tried to edit and correct it, but goes back to the same way as given above. You please redraw it with three electron pairs around each Br with the -ve charge superscripted.)

Pushpa Padmanabhan

Lansing Community College

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13y ago

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