BH4- Lewis structure

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BH₄^– (borohydride) has one boron atom and four hydrogen atoms.

In BH₄^– Lewis structure, there are four single bonds around the boron atom, with four hydrogen atoms attached to it, and none of the atoms has a lone pair.

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

Steps

Here’s how you can easily draw the BH₄^– Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, boron lies in group 13, and hydrogen lies in group 1.

Hence, boron has three valence electrons and hydrogen has one valence electron.

Since BH₄^– has one boron atom and four hydrogen atoms, so…

Valence electrons of one boron atom = 3 × 1 = 3
Valence electrons of four hydrogen atoms = 1 × 4 = 4

Now the BH₄^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 3 + 4 + 1 = 8

Learn how to find: Boron valence electrons and Hydrogen valence electrons

• Second, find the total electron pairs

We have a total of 8 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 = 8 ÷ 2 = 4

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Hence, here we have to assume that the central atom is boron.

Therefore, place boron in the center and hydrogens on either side.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have a total of 4 electron pairs. And four B — H bonds are already marked. So we do not have to mark any electron pair as a lone pair on the sketch.

#3 If needed, mention formal charges on the atoms

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

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

For boron atom, formal charge = 3 – 0 – ½ (8) = -1

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

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

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

Therefore, this structure is the most stable Lewis structure of BH₄^–.

And since the BH₄^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: SOF₄ Lewis structure

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External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/BH4–Lewis-structure.html
• https://homework.study.com/explanation/what-are-the-formal-charges-on-each-of-the-atoms-in-the-bh4-ion-hint-draw-the-lewis-dot-structure-of-the-ion.html
• https://www.numerade.com/ask/question/identify-the-correct-structure-for-the-following-ion-bh4-54593/
• https://www.chegg.com/homework-help/questions-and-answers/bh4-draw-lewis-structure-number-valence-electrons-formal-charge-q35322862