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Chapter 5: Problem 30

The shape of \(\mathrm{BF}_{3}\) molecule is

Short Answer

Expert verified
Answer: The shape of the BF₃ molecule is trigonal planar.

Step by step solution

01

Identify the central atom

In \(\mathrm{BF}_{3}\), boron (B) is the central atom with three fluorine (F) atoms bonded to it.
02

Determine the number of valence electrons

First, we need to find the total number of valence electrons in the molecule. Boron has 3 valence electrons while each fluorine atom has 7 valence electrons. So, the total number of valence electrons is: \(3 + 3 \times 7 = 3 + 21 = 24\).
03

Determine the number of bonding and lone electron pairs

In \(\mathrm{BF}_{3}\), boron forms three single bonds with three fluorine atoms, corresponding to 6 bonding electrons. This leaves 18 electrons, which can be arranged into 9 lone electron pairs—6 of these pairs will be attached to the three fluorine atoms, and the remaining 3 pairs will be assigned to the central boron atom. However, since boron is an exception and can have incomplete octets, it has no lone pairs, and all 9 lone pairs belong to the three fluorine atoms (3 lone pairs each).
04

Determine the electron pair geometry

Now, we need to determine the electron pair geometry around the central atom, which is the arrangement of all electron pairs (bonding and lone) around the central atom. Boron has three bonding electron pairs and no lone electron pairs, making the electron pair geometry trigonal planar.
05

Determine the molecular geometry

Since there are no lone electron pairs on the central atom (boron), the molecular geometry is the same as the electron pair geometry. Therefore, the shape of the \(\mathrm{BF}_{3}\) molecule is trigonal planar.

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Most popular questions from this chapter

Why are the participating atoms of a coordinate bond called the donor and the acceptor?

"The orbitals of two different shells of an atom combine and form one hybrid orbital." State whether the statement is true or false and justify your answer.

The number of \(\sigma\) bonds and \(\pi\) bonds in \(\mathrm{CO}_{2}\) molecule are \(\quad\) respectively.

Why is more energy required to break a sigma bond than a pi bond?

Why is \(\mathrm{BeCl}_{2}\) molecule linear, but \(\mathrm{BF}_{3}\) is trigonal planar?
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