Question

List the order of increasing bond angle in the following molecule\[S{F_6}\],\[CC{l_{4}}\],\[BC{l_3}\],\[BeC{l_2}\]is:
A. \[S{F_6} < CC{l_4} < BC{l_3} < BeC{l_2}\]
B. \[BeC{l_2} < BC{l_3} < CC{l_4} < S{F_6}\]​
C. \[S{F_6} < CC{l_4} < BeC{l_2} < BC{l_3}\]​
D. \[BC{l_3} < BeC{l_2} < S{F_6} < CC{l_4}\]

Answer 1

Hint: Firstly, find the valence shell electron of an atom, to find the type of bond it will form. If we know the atomic number of the atom, the valence shell electron can be found out by using electronic configuration.

Complete step by step solution:
To find the bond angle, you should know about the hybridisation of the molecule to find its geometry. Hybridisation can be defined as the bonding of the hybrid orbitals having similar energy which are formed from the atomic orbitals of each atom with different energy. The requirement for finding the hybridisation, you should know about the valence shell electron of the atom for bonding.

In \[S{F_6}\]​, S-atom belongs to the group-16 having 6 electrons in the valence shell requires 2 electrons to fulfil octet and the F-atom requires only one electron to fulfil octet as it belongs to group 17. There are 6 F-atoms which are bonded with one S-atom because the S-atom has a vacant d-orbital to extend. The hybridization of the molecule is \[s{p^3}{d^2}\] and has octahedral geometry. The bond angle of the octahedral geometry is \[9{0^o}\].

In\[CC{l_{4}}\]​, C-atom belongs to the group-14 having 4 electron in the valence shell requires 4 electron to fulfil octet and the Cl-atom requires only one electron to fulfil octet as it belongs to group 17. There are 4 Cl-atoms which are bonded with one C-atom and the hybridization of the molecule is \[s{p^3}\] and have tetrahedral geometry. The bond angle of the octahedral geometry is \[10{9^o}\].

In \[BC{l_3}\]​, B-atom belongs to the group-13 having 3 electrons and forming the bond with 3 Cl-atoms by sharing one electron with each. There are 3 Cl-atoms which are bonded with one B-atom and the hybridization of the molecule is \[s{p^2}\] and have Trigonal planar geometry. The bond angle of the Trigonal planar geometry is \[12{0^o}\].

In \[BeC{l_2}\]​, Be-atom belongs to the group-2 having 2 electrons in the valence shell and forming the bond with 2 Cl-atoms by sharing one electron with each. There are 2 Cl-atoms which are bonded with one Be-atom and the hybridization of the molecule is sp and has linear geometry. The bond angle of the linear geometry is \[18{0^{o\;}}\].

The correct order for the increasing bond angle is \[S{F_6} < CC{l_4} < BC{l_3} < BeC{l_2}\].

Correct answer is option A.

Note: Geometry may be defined as the structure which includes all bond pairs, lone pairs whereas shape may be defined as the structure which only includes the bond pair.