# How do you draw the lewis structure for polyatomic ions?

May 30, 2018

Well you use VESPER....

#### Explanation:

And $\text{VSEPR"-="Valence shell electron pair repulsion theory.}$

And I will give ONE bond, the VESPER treatment of nitrate ion...we gots $N {O}_{3}^{-}$...and there are ……

$3 \times {6}_{\text{oxygen valence electrons"+5_"nitrogen valence electrons"+1_"negative charge}}$ $= \text{24 electrons}$, i.e. TWELVE electron pairs to distribute over FOUR CENTRES....

And we get....$O = \stackrel{+}{N} {\left(- {O}^{-}\right)}_{2}$...THREE of the four participating atoms have a formal charge...the leftmost oxygen is NEUTRAL (it has 8 electrons), the nitrogen has SIX electrons, and therefore has a FORMAL POSITIVE charge...and two of the oxygens bear 9 electrons, and thus have a FORMAL NEGATIVE charge... The overall charge is this $+ 1 - 1 - 1 = - 1$.

Can you try for $S {O}_{4}^{2 -}$?