Answer:
- <u>two bonding and two unshared pair of electrons.</u>
Explanation:
<em>Chlorite anion, ClO₂⁻ </em>has one atom of chlorine and two atoms of oxygen.
Chlorine atom has 7 valence electrons (group 17 of the periodic table) and oxygen has 6 valence electrons (group 16 of the periodic table).
Thus, in total the anion has 7 + 6×2 electrons, plus an additional electron indicated by the negative charge of the anion: 7 + 12 + 1 = 20 valence electrons, that must be distributed among the three atoms.
Being chlorine the most electronegative atom, you place it in the center. surrounded by the two oxygen atoms.
Thus, so far:
O Cl O (pending to place the valence electrons.
You can put two electrons between each chlorine and oxygen atoms to form the bonds:
O : Cl : O
Also, complete the octets:
. . . . . .
: O : Cl : O :
. . . . . .
That structure works because:
- there are 20 electrons
- every atom has 8 valence electrons
But you need to calculate the formal charges on each atom:
For chlorine:
- 7 valence electrons - 4 nonbonding valence electrons - 4/2 bonding electrons = 7 - 4 - 2 = + 1.
For each oxygen:
- 6 valence electrons - 6 nonbonding valence electrons - 2/2 bonding electrons = 6 - 6 - 1 = -1
The total charge of the anion is +1 -1 - 1 = -1.
And the complete structure with the charge is:
. . . . . .
[ : O : Cl : O : ] ⁻
. . . . . .
The square bracketts comprise the entire anion and the negative charge is for the entire structure.
In this structure, <em>the central atom (Cl) is surrounded by two bonding and two unshared pairs of electrons. </em>(this is the answer to the question).
There is another structure which minimizes the formal charge, which will be preferred. It is by making a double bond between chlorine and one of the oxygen atoms, by moving two of the atoms beside one chlorine.
That structure is:
. . . . . .
: O : Cl :: O
. . . . . .
There you have 20 valence electrons and complete octets too, but the formal charges are:
Chlorine:
- 7valence electrons - 4 nonbonding electrons - 6/2 bonding electrons = 7 - 4 - 3 = 0
The oxygen on the left is equal to the previous structure, thus - 1 forma charge.
The oxygen on the right:
- 6 valence electrons - 4 nonbonding electrons - 4/2 bonding electrons = 6 - 4 - 2 = 0
Hence, this las structure minimizes the formal charges and is preferred.
This is:
. . . . . .
: O : Cl :: O
. . . . . .
But you must add the negative charge which belongs to the complete anion:
. . . . . .
[ : O : Cl :: O ] ⁻
. . . . . .
The square brackets comprise the entire anion and the negative charge is for it.
Hence, in this other structue the central atom, chlorine, is surrounded by six bonding electrons and four nonbonding electrons.
