Question 6.13: Sodium chlorite, NaClO2, is a component of mouthwashes, toot......

STEP 1   Determine the arrangement of atoms. For the polyatomic ion ClO_{2}^{\space \space -}, the central atom is Cl because there is only one Cl atom. For a polyatomic ion, the atoms and electrons are placed in brackets, and the charge is written outside at the upper right.

\left[O \space \space \space Cl \space \space \space O\right] ^{-}
STEP 2   Determine the total number of valence electrons. We use the group numbers to determine the number of valence electrons for each of the atoms in the ion. Because the ion has a negative charge, one more electron is added to the valence electrons.

Table 1
STEP 3   Attach each bonded atom to the central atom with a pair of electrons. Each bonding pair can also be represented by a line, which indicates a single bond.

\left[O \space \mathrm{:}\space Cl \space \mathrm{:}\space O\right] ^{-}        or        \left[O \space — \space  Cl \space — \space O\right] ^{-}

Four electrons (2 × 2 e^{-}) are used to bond the O atoms to the central Cl atom, which leaves 16 valence electrons.
STEP 4   Use the remaining electrons to complete octets. Of the 16 remaining valence electrons, 12 are drawn as lone pairs to complete the octets of the O atoms.

\left[\mathrm {:}\overset{\mathrm {.~.}}{\underset{.~.}{O}} \mathrm {:}Cl\mathrm {:}\overset{\mathrm {.~.}}{\underset{.~.}{O}}\mathrm {:}\right] ^{-}      or      \left[\mathrm{:}\overset{\mathrm{.~.}}{\underset{\mathrm{.~.}}{O}} \space — \space Cl \space — \space \overset{\mathrm{.~.}}{\underset{\mathrm{.~.}}{O}} \mathrm{: }\right]^{-}

The remaining four electrons are placed as two lone pairs on the central Cl atom.

\left[\mathrm{:} \overset{\mathrm{.~.}}{\underset{\mathrm{.~.}}{O}} \mathrm{:} \overset{\mathrm{.~.}}{\underset{\mathrm{.~.}}{Cl}} \mathrm{:} \overset{\mathrm{.~.}}{\underset{\mathrm{.~.}}{O}} \mathrm{:} \right] ^{-}     or     \left[\mathrm{:} \overset{\mathrm{.~.}}{\underset{\mathrm{.~.}}{O}} \space — \space  \overset{\mathrm{.~.}}{\underset{\mathrm{.~.}}{Cl}} \space — \space  \overset{\mathrm{.~.}}{\underset{\mathrm{.~.}}{O}} \mathrm{:} \right] ^{-}