SbF4- Lewis structure

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SbF₄^– has one antimony atom and four fluorine atoms.

In SbF₄^– Lewis structure, there are four single bonds around the antimony atom, with four fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the antimony atom has one lone pair.

Also, there is a negative (-1) charge on the antimony atom.

Steps

Use these steps to correctly draw the SbF₄^– Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

#1 First draw a rough sketch

• First, determine the total number of valence electrons

In the periodic table, antimony lies in group 15, and fluorine lies in group 17.

Hence, antimony has five valence electrons and fluorine has seven valence electrons.

Since SbF₄^– has one antimony atom and four fluorine atoms, so…

Valence electrons of one antimony atom = 5 × 1 = 5
Valence electrons of four fluorine atoms = 7 × 4 = 28

Now the SbF₄^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 5 + 28 + 1 = 34

Learn how to find: Antimony valence electrons and Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 34 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 34 ÷ 2 = 17

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since antimony is less electronegative than fluorine, assume that the central atom is antimony.

Therefore, place antimony in the center and fluorines on either side.

• And finally, draw the rough sketch

#2 Mark lone pairs on the atoms

Here, we have a total of 17 electron pairs. And four Sb — F bonds are already marked. So we have to only mark the remaining thirteen electron pairs as lone pairs on the sketch.

Also remember that antimony is a period 5 element, so it can keep more than 8 electrons in its last shell. And fluorine is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines.

So for each fluorine, there are three lone pairs, and for antimony, there is one lone pair.

Mark the lone pairs on the sketch as follows:

#3 Calculate and mark formal charges on the atoms, if required

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For antimony atom, formal charge = 5 – 2 – ½ (8) = -1

For each fluorine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, the antimony atom has a charge, so mark it on the sketch as follows:

In the above structure, you can see that the central atom (antimony) forms an octet. And the outside atoms (fluorines) also form an octet. Hence, the octet rule is satisfied.

Therefore, this structure is the most stable Lewis structure of SbF₄^–.

And since the SbF₄^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: CH₃CH₂I Lewis structure

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External links

• https://homework.study.com/explanation/predict-the-shape-of-sbf4-on-the-basis-of-vsepr-theory.html
• https://www.chegg.com/homework-help/questions-and-answers/draw-lewis-structure-sbf4-q1589639
• https://brainly.com/question/23180559