Answer :
It would be created like this
":" means lone pair of electrons, ignore the "~"
H---O---Br===O:
The formal charge would be as follows:
H: 1-1/2(2)-0=0
O: 6-1/2(4)-4=0
Br: 7-1/2(6)-4=0 O: 6-1/2(4)-4=0
Br can hold extra electrons because it can leak those electrons over to the D orbital.
":" means lone pair of electrons, ignore the "~"
H---O---Br===O:
The formal charge would be as follows:
H: 1-1/2(2)-0=0
O: 6-1/2(4)-4=0
Br: 7-1/2(6)-4=0 O: 6-1/2(4)-4=0
Br can hold extra electrons because it can leak those electrons over to the D orbital.
The formal charge on Br, H and O is [tex]\boxed0[/tex]. The Lewis structure of [tex]{\mathbf{HBr}}{{\mathbf{O}}_{\mathbf{2}}}[/tex] is attached in the image
Further Explanation:
The bonding between the different atoms in covalent molecules is shown by some diagrams known as the Lewis structures. These also show the presence of lone pairs in the molecule. These are also known as Lewis dot diagrams, electron dot diagrams, Lewis dot structures or Lewis dot formula. In covalent compounds, the geometry, polarity, and reactivity are predicted by these structures.
The total number of valence electrons of [tex]{\text{HBr}}{{\text{O}}_{\text{2}}}[/tex] is calculated as,
Total valence electrons = [(1) (Valence electrons of H) + (1) (Valence electrons of Br) + (2) (Valence electrons of O)]
[tex]\begin{aligned}{\text{Total valence electrons}}\left({{\text{TVE}}}\right)&=\left[{\left({\text{1}} \right)\left({\text{1}}\right)+\left({\text{1}}\right)\left({\text{7}}\right)+\left( 2 \right)\left(6\right)}\right]\\&=20\\\end{gathered}[/tex]
Formal charge:
It is the charge that an atom acquires in a molecule by considering that the chemical bonds are shared equally between the two atoms, irrespective of their electronegativities.
The formula to calculate the formal charge on an atom is as follows:
[tex]\text{Formal Charge}=\left[\begin{aligned}\left[\text{total number of valence electrons in the free atom}\right]\\-[\text{total number of non-bonding electrons}]\\-\left[\frac{\text{total number of bonding electrons}}{2}\right]\end{aligned}\right][/tex] ......(1)
H forms one single bond with an oxygen atom and no lone pair is present on it.
The total number of valence electrons in the free hydrogen atom is 1.
The total number of non-bonding electrons in H is 0.
The total number of bonding electrons in H is 2.
Substitute these values in equation (1) to find the formal charge on H.
[tex]\begin{aligned}{\text{Formal charge on H}}&=\left[{1-0-\frac{2}{2}}\right]\\&=0\\\end{aligned}[/tex]
Br forms one single bond with O-2 and one double bond with O-1 and two lone pairs are present on it.
The total number of valence electrons in the free bromine atom is 7.
The total number of non-bonding electrons in Br is 4.
The total number of bonding electrons in Br is 6.
Substitute these values in equation (1) to find the formal charge on Br.
[tex]\begin{aligned}{\text{Formal charge on Br}}&=\left[ {7-4-\frac{6}{2}}\right] \\&=0\\\end{aligned}[/tex]
O-1 forms one double bond with one bromine atom and has two lone pairs on it.
The total number of valence electrons in the free oxygen atom is 6.
The total number of non-bonding electrons in O is 4.
The total number of bonding electrons in O is 4.
Substitute these values in equation (1) to find the formal charge on O-1.
[tex]\begin{aligned}{\text{Formal charge on O-1}}&=\left[{6-4-\frac{4}{2}}\right]\\&=0\\\end{aligned}[/tex]
O-2 forms one single bond with one bromine atom and one single bond with one hydrogen atom and two pair on it.
The total number of valence electrons in the free oxygen atom is 6.
The total number of non-bonding electrons in O is 4.
The total number of bonding electrons in O is 4.
Substitute these values in equation (1) to find the formal charge on O-2.
[tex]\begin{aligned}{\text{Formal charge on O-2}}&=\left[{6-4-\frac{4}{2}}\right]\\&=0\\\end{aligned}[/tex]
Bromine atom has 7 valence electrons, oxygen atom has 6 valence electrons and oxygen atom has 6 valence electrons. Therefore, the total numbers of valence electrons in [tex]{\text{HBr}}{{\text{O}}_{\text{2}}}[/tex] are 20. In the Lewis structure of [tex]{\text{HBr}}{{\text{O}}_{\text{2}}}[/tex] , Br forms one single bond with O-2 and one double bond with O-1 and two lone pairs are present on it. Therefore, Lewis structure of [tex]{\mathbf{HBr}}{{\mathbf{O}}_{\mathbf{2}}}[/tex] is attached in the image.
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Molecular structure and chemical bonding
Keywords:
Lewis structure, valence electrons, HBrO2, formal charge, 0, Br, oxygen, double bonds, single bond, bonding electrons, non-bonding electrons, total valence electrons, H, O-1, O-2.
