# What is the geometry of the "manganate ion", MnO_4^(2-)?

Aug 9, 2017

Tetrahedral.......

#### Explanation:

We have $M n \left(V I\right)$, i.e. ${\left(O =\right)}_{2} M n {O}_{2}^{2 -}$, and thus manganate is an analogue of sulfate ion, for which we can use VSEPR directly.....

Aug 9, 2017

The geometrical shape is tetrahedral.

#### Explanation:

1. Decide which is the central atom in the structure.

That is normally the least electronegative atom ($\text{Mn}$).

2. Draw a skeleton structure in which the other atoms are single-bonded to the central atom:

$\textcolor{w h i t e}{m m l} \text{O}$
$\textcolor{w h i t e}{m m l l} |$
$\text{O—Mn—O}$
$\textcolor{w h i t e}{m m l l} |$
$\textcolor{w h i t e}{m m l} \text{O}$

3. Put electron pairs around every atom until each gets an octet.

$\textcolor{w h i t e}{m m} \text{:Ö:}$
$\textcolor{w h i t e}{m m l l} |$
$\text{:Ö—Mn—Ö:}$
color(white)(m)" ̈ color(white)(ml)|color(white)(mml) ̈
$\textcolor{w h i t e}{m m} \text{:O:}$
color(white)(mmm) ̈

4. Count the valence electrons in your trial structure (32).

5. Now count the valence electrons you actually have available.

$\text{1 Mn + 4 O + 2e"^"-" = 1×2 + 4×6 + 2 = 28}$.

The trial structure is missing four electrons.

6. Draw a new structure, this time inserting one double bond for each missing pair of electrons and giving each $\text{O}$ atom an octet:

$\textcolor{w h i t e}{m m} \text{:Ö:}$
$\textcolor{w h i t e}{m m l l} |$
$\text{:Ö=Mn=Ö:}$
$\textcolor{w h i t e}{m m l l} |$
$\textcolor{w h i t e}{m m} \text{:O:}$
color(white)(mmm) ̈

Note: The $\text{Mn}$ atom can have an "expanded" octet.

7. Use VSEPR theory to determine the molecular geometry.

There are four bonding electron domains and no lone pairs about the $\text{Mn}$ atom.

This is an ${\text{AX}}_{4}$ molecule. The molecular geometry is tetrahedral. 