BH4- Lewis structure

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.

Alternative method: Lewis structure of BH₄^–

Rough sketch

• 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

Lone pair

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.

Formal charge

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:

Final structure

The final structure of BH₄^– comprises a central boron atom linked to four hydrogen atoms through single covalent bonds. In this arrangement, the boron atom satisfies the octet rule by forming four total bonds, utilizing the additional electron to complete its valence shell. Within this layout, each hydrogen atom achieves a stable duet. This configuration represents the most stable state for the ion because it results in a formal charge of -1 on the central boron atom, while each hydrogen atom maintains a formal charge of zero. Thus, this specific electronic distribution serves as the definitive and most accurate Lewis representation of the borohydride ion.

To properly represent this as a polyatomic ion, the entire Lewis structure is enclosed within square brackets. The overall charge of 1- is then written as a superscript outside the brackets at the top right, indicating that the structure possesses one additional electron beyond the valence count of the neutral atoms.

Next: SOF₄ Lewis structure

External video

• BH4- Lewis Structure: How to Draw the Lewis Structure for the BH4 – – YouTube • Wayne Breslyn

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