Draw a lewis structure for BrO4- in which all atoms have the lowest formal changes. Indicate the values of nonzero formal charges and include lonepair electrons.

To draw the Lewis structure for the BrO4- ion and determine the lowest formal charges on the atoms, we need to follow a step-by-step process:

Step 1: Determine the total number of valence electrons.
The atomic number of bromine (Br) is 35, and oxygen (O) is 8. Since there are four oxygen atoms, the total number of valence electrons in BrO4- is:

(1 Bromine atom x 7 valence electrons) + (4 Oxygen atoms x 6 valence electrons) + 1 extra electron for the negative charge = 7 + 24 + 1 = 32 valence electrons

Step 2: Identify the central atom.
In the BrO4- ion, the central atom is bromine (Br) since it is less electronegative than oxygen.

Step 3: Connect the atoms using single bonds.
Place the bromine atom in the center and connect it to the four oxygen atoms with single bonds.

O
||
Br--O
||
O
||
O

Step 4: Arrange the remaining electrons.
Distribute the remaining valence electrons around the atoms, placing them as lone pairs on the oxygen atoms. Initially, each oxygen atom should have six lone-pair electrons (since they all want to achieve an octet), and the bromine atom should have two lone pairs of electrons.

:O:
||||
O-Br-O
||||
:O:
||||
:O:

Step 5: Check for octets and adjust electron pairs.
After arranging the electrons, check if all the atoms have achieved an octet, except for hydrogen that can only accommodate two electrons. In this case, the oxygen atoms have an octet, but the bromine atom has ten electrons.

Step 6: Convert lone pairs into bonding pairs.
To accommodate the extra two electrons on bromine, you can convert one of the lone pairs on an oxygen atom into a bonding pair by creating a double bond between the bromine and one of the oxygen atoms.

:O:
||||
O=Br-O
||||
:O:
||||
:O:

Step 7: Calculate formal charges.
To determine the formal charges, compare the number of electrons assigned to each atom with the number of valence electrons it should have.

Formal charge = Valence electrons - (Lone pair electrons + 0.5 x Bonding electrons)

Formal charge (Br) = 7 - (2 + 6) = -1

Formal charge (each oxygen) = 6 - (6 + 0.5 x 2) = -1

So, all the atoms in the BrO4- ion have the lowest formal charges (-1).

To draw the Lewis structure for BrO4-, we need to consider the valence electrons of each atom involved.

Step 1: Determine the total number of valence electrons.
Bromine (Br) is in group 7A, so it has 7 valence electrons.
Oxygen (O) is in group 6A, so it has 6 valence electrons.
Since there are four oxygen atoms in BrO4-, the total number of valence electrons is calculated as follows:
Total valence electrons = Valence electrons of Br + (Valence electrons of O × Number of O atoms)
Total valence electrons = 7 + (6 × 4) = 31 valence electrons

Step 2: Start by connecting the atoms.
Place the bromine atom (Br) in the center, as it is the least electronegative. Connect the four oxygen atoms (O) around the bromine atom using single bonds (one oxygen on each side).

O
|
O -- Br -- O -- O

Step 3: Distribute the remaining electrons.
Distribute the remaining valence electrons around the atoms, beginning with the oxygen atoms. Place lone pairs of electrons around each oxygen atom, completing their octets. Since there are 4 oxygen atoms, we have used 4 × 6 = 24 valence electrons so far.

O
lone pairs |
O -- Br -- O -- O
|
O lone pairs

Step 4: Assign formal charges.
To determine whether formal charges are necessary, examine the formal charge of each atom. Recall that formal charge is calculated as the group number of the atom minus the number of bonds to that atom minus the number of lone pair electrons.

For each oxygen atom (O), the formal charge calculation is as follows:
Formal charge = Group number – Lone pair electrons – Number of bonds
Formal charge = 6 – 6 – 1 = -1

For the bromine atom (Br), the formal charge calculation is as follows:
Formal charge = Group number – Lone pair electrons – Number of bonds
Formal charge = 7 – 0 – 4 = +3

Step 5: Check if the overall formal charge matches the charge of the ion.
The overall charge of BrO4- is -1. Currently, the sum of the formal charges is +104. To balance the overall charge, we can take one of the lone pairs on an oxygen atom and convert it into a bonding pair.

Final Lewis structure for BrO4- with all atoms having the lowest formal charges:

O
lone pairs |
O -- Br -- O = O
|
O lone pairs

The nonzero formal charges are: Bromine (Br) has a formal charge of +3, and two oxygen atoms (O) have formal charges of -1.

It is almost impossible to draw lewis structures on these boards due to the spacing problem. Here is a structure for BrO3^- from Purdue.

http://www.chem.purdue.edu/gchelp/molecules/bro3m.html
To the ball and stick model drawn there, if you will change each line to a pair of electrons, and add a pair of electrons to the top, left, and bottom of the left and right oxygen atom, add a pair of electrons to the lowest oxygen at the left, right, and bottom, then add the final two electons to the top of the Br atom, you will have the structure. Finally, add a negative charge to the ion. You should have 25 electrons total. I hope this helps.