Question

Amongst the following, \[{\text{S}}{{\text{F}}_{\text{6}}}\] ​, \[{\text{B}}{{\text{F}}_{{\text{6}}}}^{{\text{3 - }}}\] , \[{\text{S}}{{\text{F}}_{\text{4}}}\] ​, \[{\text{AlF}}_6^{3 - }\] , \[{\text{P}}{{\text{H}}_{\text{5}}}\] , \[{\text{NC}}{{\text{l}}_{\text{5}}}\], \[{\text{SC}}{{\text{l}}_{\text{6}}}\] ​, the total number of species which does/do not exist is:
A)5
B)4
C)6
D)3

Answer 1

Hint: To answer this question, you should recall the concept of molecular orbital theory. The central atom cannot accommodate more peripheral atoms after completing its octet. Those with more peripheral atoms do not exist.

Complete Step by step solution:
We know that a molecule cannot accommodate more peripheral bonding atoms after completing its octet. Let's analyse each of the options systematically.
Amongst hexahalides of sulfur, only \[{\text{S}}{{\text{F}}_{\text{6}}}\] is exceptionally stable for a steric reason. In \[{\text{SC}}{{\text{l}}_{\text{6}}}\]​, smaller S cannot accommodate six larger \[C{l^ - }\] ions, hence \[{\text{SC}}{{\text{l}}_{\text{6}}}\] does not exist.
1. \[{\text{P}}{{\text{H}}_{\text{5}}}\]does not exist as there is a large difference in the energies of s, p and d orbitals of P and hence it does not undergo \[{\text{s}}{{\text{p}}^{\text{3}}}{\text{d}}\] hybridisation.
2. \[{\text{NC}}{{\text{l}}_{\text{5}}}\] does not exist as vacant d-orbital are absent in N atom.
3. \[{\text{B}}{{\text{F}}_{{\text{6}}}}^{{\text{3 - }}}\] does not exist due to the absence of vacant d-orbital in B and hence, it cannot exceed its covalency beyond four.
4. Sulfur \[{\text{S}}{{\text{F}}_{\text{4}}}\] has an oxidation number +4. This oxidation number is possible for sulfur because adding two electrons fills its 3p orbitals.
5. \[{\text{AlF}}_6^{3 - }\] exists due to the presence of vacant p orbitals in Al atom.
Thus, out of the given compounds that do not exist are \[{\text{SC}}{{\text{l}}_{\text{6}}}\] , \[{\text{B}}{{\text{F}}_{{\text{6}}}}^{{\text{3 - }}}\] ,\[{\text{P}}{{\text{H}}_{\text{5}}}\] , \[{\text{NC}}{{\text{l}}_{\text{5}}}\].
Therefore, we can conclude that the correct answer to this question is option B.

Note: You can also use molecular orbital theory to find molecules that do not exist. Place the available bonding electrons in bonding and anti-bonding orbitals. Those molecules in which the bond order is zero, do not exist.