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Question: In which of the following species the bond angle around the central atom is equal to \({120^{\text{o...

In which of the following species the bond angle around the central atom is equal to 120o{120^{\text{o}}}
A.BF3B{F_3}
B.BCl3BC{l_3}
C.BF4B{F_4}
D.SO3S{O_3}

Explanation

Solution

To answer this question you must recall the VSEPR theory. The Valence shell electron pair repulsion theory proposes that the hybridized orbitals in an atom arrange themselves in such a way so as to minimize the repulsion between them, hence determining the geometry of a molecule on the basis of its hybridization. The bonds will be at 120o{120^{\text{o}}} if the hybridization is sp2s{p^2}

Complete answer:
First we consider BF3B{F_3} molecules.
Boron has atomic number 5 and has 3 valence electrons. One electron from its 2s2s orbital is excited to its 2p2porbital. The boron atom undergoes hybridization of sp2s{p^2} and forms three hybrid orbitals. The geometry of the molecule is trigonal planar and the bond angle around the central atom is equal to 120o{120^{\text{o}}}
First we consider BCl3BC{l_3} molecules.
Boron has atomic number 5 and has 3 valence electrons. One electron from its 2s2s orbital is excited to its 2p2p orbital. The boron atom undergoes hybridization of sp2s{p^2} and forms three hybrid orbitals. The geometry of the molecule is trigonal planar and the bond angle around the central atom is equal to 120o{120^{\text{o}}}
Now we consider BF4B{F_4} molecules.
Boron is an electron deficient atom and has only three electrons present. It cannot form more than 4 covalent bonds. The fourth bond is formed by fluorine and it is a coordinate bond. There are 4 bonds in the molecule and the shape is tetrahedral.
Now we consider SO3S{O_3} molecule.
Sulphur has six valence electrons. It forms two bonds with each oxygen atom, one σ\sigma and one π\pi bond. Thus only three electrons are hybridized. The hybridization of the molecule is sp2s{p^2} and shape of the molecule is trigonal planar and the bond angle around the central atom is equal to 120o{120^{\text{o}}}

Thus, the correct answer is A, B and D.

Note:
The concept of mixing of atomic orbitals in order to form new hybrid orbitals that possess different shapes and energies as compared to the original parent atomic orbitals is known as hybridisation. Hybrid orbitals are suitable to form chemical bonds of equal energies. Also hybridization of orbitals leads to the formation of more stable compounds because hybrid orbitals have lower energy than the unhybrid orbitals.