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

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.....

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.