Boron trichloride is a molecule that can be industrially produced by chlorinating boron oxide and carbon directly at 501°C.

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B2O3 + 3C + 3Cl2 ——–> 2BCl3 + 3CO

Corresponding trihalides of boron are produced during the reaction of boron with halogens.

Uses of BCl3 are immense. Boron trichloride is used in refining alloys of aluminum, zinc, magnesium, and copper.

The production of elemental boron needs BCl3 as a starting element.

The synthesis of organic compounds also involves this molecule. It is used as a reagent for cleaving C-O bonds in ethers.

Boron trichloride is also used in the manufacturing of electrical resistors, in the field of high energy fuels and rocket propellants.

With such wide use of boron trichloride, it is important to have knowledge about its properties. So in this article let’s dive deep into the world of boron trichloride.

How to Draw a Lewis Structure

The representation of electron distribution around the atoms, to predict the number and type of bonds that can be formed is basically known as lewis structure.

Now let’s see the proper steps to draw a lewis structure:-

1. Find the total number of valence electrons in a molecule:- Adding up the valence electrons of all the atoms in a molecule is the first step.

2. Choose a central atom and draw a skeletal structure:- Sketch a skeletal of the molecule with only single bonds. Now, the central atom is generally the least electronegative atom or atom with the most available sites.

3. Place the electrons around outside atoms:- After the skeleton formation with single bonds, the remaining electrons are placed around the atoms, to fulfill the octet of the atoms. Start filling from the electronegative atoms and proceed to electropositive atoms.

4. Check the octet rule for all the atoms:- Make sure all the atoms are fulfilling the octet. Otherwise fill-up the same by giving multiple bonds. Turn one lone pair of an electronegative atom into a bonding pair of the electron-deficient atom.

5. Check the formal charge of all the atoms:- See that all atoms have the lowest possible formal charge without violating the octet rule. The formula for the formal charge is

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Although we can draw the structures of many molecules with the lewis dot rule, there are still some exceptions to this rule.

Lewis structures are less useful for lanthanides and actinides.

BCl3 Lewis Structure

Let us apply the lewis dot rules and try to draw the structure of boron trichloride.

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First of all, we need to calculate the total valence electrons of this molecule,B = 3C l= 73Cl = 7*3=21So, total= 21+3= 24

Now, boron is less electronegative, which makes it the central atom. We need to draw a skeletal structure with single bonds only.

Out of 24 valence electrons, 6 electrons were used to make the single bonds in the skeleton. Thus, we are left with 18 electrons, which we need to distribute among the Cl atoms.

Each Cl atom will now have 6 electrons or in other words, 3 lone pairs on each Cl atom.

Lastly, we need to check if all the atoms are fulling the octet rule. We can see from the following image that the octet rule is fulfilled by all atoms.

Now, as per the octet rule, in order to complete its octet boron needs another 6 electrons in its outermost shell. So it shares 3 single bonds with chlorine in order to gain stability.

On the other hand, chlorine needs another one electron to complete its octet. Thus chlorine also shares a bond with boron’s electrons to fulfill the octet rule.

BCl3 Hybridization

Boron trichloride has Sp2 hybridization. By the use of one 2s orbital and two 2p orbitals in the excited state, it forms three, half-filled Sp2 hybrid orbitals.

Before assuming anything let us first learn a bit about Sp2 hybridization. So in this hybridization one S orbital and two P orbital mix together to form three equivalent orbitals. These orbitals lie on a plane and make 120° to each other.

Talking about boron trichloride, it also follows the same path and thus shows Sp2 hybridization.

Ground state configuration of boron- 1s2 2s2 2p1

In order to bond with three Cl atoms, the presence of three unpaired electrons is necessary. Thus one electron from 2s orbital is promoted to 2p sublevel with the involvement of energy.

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So, excited state configuration of boron- 1s2 2s1 2px1 2py1

Thus boron gets Sp2 hybridization with the use of one 2s orbital and two 2p orbitals in the excited state.

It then forms three