Question

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

Answer 1

Tetrahedral.......

Explanation:

We have `Mn(VI)`, i.e. `(O=)_2MnO_2^(2-)`, and thus manganate is an analogue of sulfate ion, for which we can use VSEPR directly.....

Answer 2

The geometrical shape is tetrahedral.

Explanation:

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

That is normally the least electronegative atom (`"Mn"`).

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

`color(white)(mml)"O"`
`color(white)(mmll)|`
`"O—Mn—O"`
`color(white)(mmll)|`
`color(white)(mml)"O"`

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

`color(white)(mm)":Ö:"`
`color(white)(mmll)|`
`":Ö—Mn—Ö:"`
`color(white)(m)" ̈ color(white)(ml)|color(white)(mml) ̈`
`color(white)(mm)":O:"`
`color(white)(mmm) ̈`

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

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

`"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 `"O"` atom an octet:

`color(white)(mm)":Ö:"`
`color(white)(mmll)|`
`":Ö=Mn=Ö:"`
`color(white)(mmll)|`
`color(white)(mm)":O:"`
`color(white)(mmm) ̈`

Note: The `"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 `"Mn"` atom.

This is an `"AX"_4` molecule.

The molecular geometry is tetrahedral.