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3 months ago

Draw the best lewis structure for bro4- and determine the formal charge on bromine

Chemistry

2 answers:

Alekssandra [3K]3 months ago

7 0

Answer:

The Lewis-dot structure: illustrates the connections between atoms within a molecule and indicates the unpaired electrons that are found in it. Dots are used to represent the electrons.

The molecule in question is the perbromate ion.

Bromine possesses '7' valence electrons, while oxygen has '6' valence electrons.

Thus, the overall count of valence electrons in the perbromate ion is $BrO_4^-$ = 7 + 4(6) + 1 = 32

Based on the Lewis-dot structure, there are 14 bonding electrons and 18 non-bonding electrons present.

The formula for determining formal charge:

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

$\text{Formal charge on }Br=7-0-\frac{14}{2}=0$

Displayed below is the Lewis-dot structure of the perbromate ion.

castortr0y [3K]3 months ago

6 0

The formal charge on Br is $\boxed0$ . (Refer to the attached image for the structure).

More explanation:

The interactions among atoms in covalently bonded molecules can be illustrated through specific diagrams termed Lewis structures. These diagrams also include the notation for lone pairs within the molecule. Commonly, these representations are referred to as Lewis dot diagrams, electron dot diagrams, Lewis dot structures or Lewis dot formula. Lewis structures provide insight into the geometry, polarity, and reactivity of covalent compounds.

The Lewis structure for ${\mathbf{BrO}}_{\mathbf{4}}^ -$ (See the attached image for the structure):

Calculating the total number of valence electrons for ${\text{BrO}}_4^ -$ goes as follows:

Total valence electrons = [(1) (Valence electrons of Br) + (4) (Valence electrons of O) + Charge on anion]

$\begin{aligned}{\text{Total valence electrons}}\left({{\text{TVE}}} \right)&=\left[{\left({\text{1}} \right)\left({\text{7}}\right)+\left({\text{4}}\right)\left({\text{6}} \right)+1}\right]\\&=32\\\end{aligned}$

Formal charge:

This represents the charge an atom assumes in a molecule, assuming that the bonds are shared equally between the two atoms involved, regardless of their electronegativities.

The equation for calculating the formal charge on an atom is as follows:

${\mathbf{Formal\:charge=}}\left[\begin{gathered}\left[ \begin{gathered}{\mathbf{total\:number\:of\:valence\:electrons }}\hfill\\{\mathbf{in\:the\:free\:atom}}\hfill\\\end{gathered}\right]{\mathbf{}}-\\\left[{{\mathbf{total\:number\:of\:non- bonding\:electrons}}}\right]-\\\frac{{\left[ {{\mathbf{total\:number\:of\:bonding\:electrons}}}\right]}}{{\mathbf{2}}}\\\end{gathered}\right]$

Bromine engages in three double bonds with three oxygen atoms and one single bond with another oxygen atom.

A free bromine atom has a total of 7 valence electrons.

The quantity of non-bonding electrons surrounding Br is 0.

The count of bonding electrons in Br is 14.

These values should now be substituted into equation (1) to determine the formal charge on Br.

$\begin{aligned}{\text{Formal charge on Br}}&=\left[ {7 - 0 - \frac{{14}}{2}}\right]\\&=0\\\end{aligned}$

A free oxygen atom contains 6 valence electrons.

For O, the total amount of non-bonding electrons is 4.

The total number of bonding electrons in O is 4.

Substituting these quantities into equation (1) will yield the formal charge for the blue O atom.

$\begin{aligned} {\text{Formal charge on blue O}} &=\left[ {6 - 4 -\frac{4}{2}} \right]\\&= 0\\\end{aligned}$

A free oxygen atom has 6 valence electrons in total.

O presents 6 non-bonding electrons.

The number of bonding electrons in O is 2.

$\begin{aligned}{\text{Formal charge on red O}}&=\left[ {6 - 6 -\frac{2}{2}} \right]\\&=- 1\\\end{aligned}$

Bromine has 7 valence electrons while oxygen contains 6 valence electrons. Each of the three blue-colored oxygen atoms establishes one double bond with the bromine atom, and there are two lone pairs present on each of them. The red-colored oxygen atom forms a single bond with the bromine atom and has six lone pairs. Consequently, the Lewis structure of ${\mathbf{BrO}}_{\mathbf{4}}^ -$ can be found in the attached image.

Learn more:

1. The molecular shape surrounding the central atoms in the amino acid glycine:

2. How many molecules will exist upon completion of the reaction?:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Molecular structure and chemical bonding

Keywords: Lewis structure, valence electrons, BrO4-, formal charge, 0, Br, oxygen, double bonds, single bond, bonding electrons, non-bonding electrons, total valence electrons.

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