
Which of the following molecule does not show tetrahedral shape
(a) CCl4
(b) SiCl4
(c) SF4
(d) CF4
Answer
161.1k+ views
Hint: An ammonia (NH3) molecule has sp3 hybridization, and as expected its structure must be tetrahedral. But due to the presence of lone pairs of electrons, it has a pyramidal structure. Similarly, water molecules also have sp3 hybridization but in the presence of lone it has V-shaped structure.
Complete step by step solution:We all know that the geometry of a molecule can easily be determined by the application of the hybridization concept.
Whereas, for the determination of the shape of any given molecule the concept of VSEPR theory is used.
Therefore, to know the correct shape of the given molecule first we will predict the hybridization. After that we will look at the presence of lone pairs of electrons.
The hybridization of a molecule can be calculated by counting the total number of sigma bonds and lone pairs of electrons.
In CCl4 there are 4 sigma bonds present around the carbon atom. The value of 4 sigma bonds will be equal to sp3 hybridization. Because all the four valence electrons of a carbon atom are engaged in bonding with four chlorine atoms, therefore CCl4, does not have any lone pair of electrons i.e., it has tetrahedral geometry as well as tetrahedral shape.
Like CCl4 , the SiCl4 and CF4 also contain 4 sigma bonds without any lone pair of electrons. Therefore, they also have tetrahedral geometry as well as tetrahedral shape.
Unlike CCl4 ,SiCl4 and CF4 , In SF4 molecule the sulphur has six valence electrons, in which four are involved in the formation of sigma bonds with fluorine atoms and the two-electrons remain as an unshared pair of electrons. Therefore, the total sum of sigma bond and lone pairs in SF4 will be 5. The value of 5 sigma bonds will be equal to sp3d hybridization and the geometry will be trigonal bi-pyramidal. But due to the presence of a lone pair its shape will be see-saw.
Therefore from the above explanation we can say option (c) will be the correct option:
Note: Hybridization is the process of mixing atomic orbitals to produce new hybrid orbitals. These hybrid orbitals have lower energy than unhybridized orbitals. Based on the mixing of orbitals hybridization can be sp, sp2, sp3, sp3d and sp3d2.
Complete step by step solution:We all know that the geometry of a molecule can easily be determined by the application of the hybridization concept.
Whereas, for the determination of the shape of any given molecule the concept of VSEPR theory is used.
Therefore, to know the correct shape of the given molecule first we will predict the hybridization. After that we will look at the presence of lone pairs of electrons.
The hybridization of a molecule can be calculated by counting the total number of sigma bonds and lone pairs of electrons.
In CCl4 there are 4 sigma bonds present around the carbon atom. The value of 4 sigma bonds will be equal to sp3 hybridization. Because all the four valence electrons of a carbon atom are engaged in bonding with four chlorine atoms, therefore CCl4, does not have any lone pair of electrons i.e., it has tetrahedral geometry as well as tetrahedral shape.
Like CCl4 , the SiCl4 and CF4 also contain 4 sigma bonds without any lone pair of electrons. Therefore, they also have tetrahedral geometry as well as tetrahedral shape.
Unlike CCl4 ,SiCl4 and CF4 , In SF4 molecule the sulphur has six valence electrons, in which four are involved in the formation of sigma bonds with fluorine atoms and the two-electrons remain as an unshared pair of electrons. Therefore, the total sum of sigma bond and lone pairs in SF4 will be 5. The value of 5 sigma bonds will be equal to sp3d hybridization and the geometry will be trigonal bi-pyramidal. But due to the presence of a lone pair its shape will be see-saw.
Therefore from the above explanation we can say option (c) will be the correct option:
Note: Hybridization is the process of mixing atomic orbitals to produce new hybrid orbitals. These hybrid orbitals have lower energy than unhybridized orbitals. Based on the mixing of orbitals hybridization can be sp, sp2, sp3, sp3d and sp3d2.
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