Chemistry, 20.10.2019 19:00 tristasbarker03

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

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Answer from: jjgccc

Answer :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule. The electrons are represented by dot.

The given molecule is, perbromate ion.

Bromine has '7' valence electrons and oxygen has '6' valence electron.

Therefore, the total number of valence electrons in perbromate ion, BrO_4^- = 7 + 4(6) + 1 = 32

According to Lewis-dot structure, there are 14 number of bonding electrons and 18 number of non-bonding electrons.

Formula for formal charge :

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

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

The Lewis-dot structure of perbromate ion is shown below.

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

Answer from: TristonC1680

The formal charge on Br is $\boxed0$ . (For structure, refer to the attached 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.

Lewis structure of ${\mathbf{BrO}}_{\mathbf{4}}^ -$ (Refer to the structure in the attached image):

The total number of valence electrons of ${\text{BrO}}_4^ -$ is calculated as,

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:

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:

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

Br forms three double bonds with three oxygen atoms and one single bond with one oxygen atom.

The total number of valence electrons in the free bromine atom is 7.

The total number of non-bonding electrons in Br is 0.

The total number of bonding electrons in Br is 14.

Substitute these values in equation (1) to find the formal charge on Br.

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

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 blue O.

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

Total number of valence electrons in the free oxygen atom is 6.

The total number of non-bonding electrons in O is 6.

The total 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 atom has 7 valence electrons, and oxygen atom has 6 valence electrons Each of the three blue-colored oxygen atoms forms one double bond with bromine atom, and two lone pairs are present on each of them. The red-colored oxygen atom forms one single bond with the bromine atom, and six lone pairs are present on it. Therefore the Lewis structure of ${\mathbf{BrO}}_{\mathbf{4}}^ -$ is attached in the image.

Learn more:

1. Molecular shape around the central atoms in the amino acid glycine:

2. How many molecules will be present on completion of 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.

Answer from: Quest

the answer is a for sure mark brainliest i had got this right on usatestprep

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.