Question

Which of the molecules/species do not have perfect shape?
(A) $C{H_2}C{l_2}$
(B) $NF_4^ +$
(C) $SiC{l_4}$
(D) $BFC{l_2}$

Answer 1

Hint : The shape of a compound depends upon the hybridization of the central atom. Usually, VSEPR theory is used to determine the shape. The VSEPR theory is used to predict the shape of the molecules from the electron pairs that surround the central atoms of the molecule. Here, the central atom is the least electronegative element. VSEPR theory can be used to predict the shapes of the molecules of many compounds accurately.

Complete Step By Step Answer:
In the above question, only the compound $C{H_2}C{l_2}$ and $NF_4^ +$ do not have a perfect shape. While the compound $SiC{l_4}$ and $BFC{l_2}$ have a perfect shape. $C{H_2}C{l_2}$ has a tetrahedral shape but due to the polarity of the compound, the shape no longer remains perfect. The steric hindrance caused by two bulky chlorine atoms is very large which changes the shape. There is also a large electronegativity difference between carbon and chlorine.
The shape of $SiC{l_4}$ is tetrahedral while the shape of $BFC{l_2}$ is trigonal planar.
Hence, the correct option is A and B.

Note :
VSEPR theory is based on the repulsion that exists between electron pairs in the valence shell that causes the atoms to arrange themselves in a manner that minimizes this repulsion. Once the geometry of the molecule is understood, it becomes easier to understand its reactions. However, VSEPR theory also has some limitations. It fails to explain isoelectronic species. VSEPR theory does not shed any light on the compounds of transition metals. The structure of several such compounds cannot be correctly described by this theory.