Question

Identify the pair in which the geometry of the species is T-shaped and square pyramidal, respectively.
(A) $Cl{{F}_{3\,}}\,and\,IO_{4}^{-}$
(B) $ICl_{2}^{-}\,and\,IC{{l}_{5}}$
(C) $IO_{3}^{-}\,and\,I{{O}_{2}}F_{2}^{-}$
(D) $XeO{{F}_{2}}\,and\,XeO{{F}_{4}}$

Answer 1

Hint: The number of bond pairs and lone pairs in the molecule giving the total electron pair number is used to determine the geometry given from the VSEPR theory.

Complete step by step solution:
The geometry of the molecule can be determined by counting the number of bond pairs and lone pairs present in it, which account for the VSEPR electron pair that is further related to the possible geometry.
For example, taking $XeO{{F}_{2}}\,and\,XeO{{F}_{4}}$ molecule pairs.
- Firstly, the total valence electrons of the molecule are counted. We have, Xe with eight, O with six and the F with seven valence electrons.
Then, the total valence electrons in$XeO{{F}_{2}}$ molecule is $8+6+(2\times 7)=\,28$ and in $XeO{{F}_{4}}$ molecule is $8+6+(4\times 7)=\,42$.
- The number of valence electrons on the bonded atoms is obtained by multiplying the number of atoms attached to the central atom by 8.
In $XeO{{F}_{2}}$ molecule there is one oxygen and two fluorine atoms bonded to the xenon atom. That is, three atoms are bonded. So, the bonded electrons are $3\times 8=\,24$and in $XeO{{F}_{4}}$ molecule, it is $5\times 8=\,40$.
- Now, the lone pair can be obtained for the molecule, which is accounted for by the remaining electrons in the total valence electrons that did not bind in the bonded valence electrons.
In $XeO{{F}_{2}}$ molecule, the non-bonded electrons will be $28-24=4$electrons (2 lone pairs) and in $XeO{{F}_{4}}$ molecule, it is $42-40=2$(one lone pair).
- Therefore, we get the number of bond pairs and lone pairs in $XeO{{F}_{2}}$ molecule to be 3 and 2. The VSEPR electron pair number = 5 with 3 bond pairs and 2 lone pairs, has a T-shaped geometry.
And the $XeO{{F}_{4}}$ molecule having VSEPR electron pair number = 6 with 5 bond pairs and one lone pair, has a square- pyramidal geometry.

Thus, the molecule pair with geometry of the species is T-shaped and square pyramidal, respectively is option (D)- $XeO{{F}_{2}}\,and\,XeO{{F}_{4}}$.

Note: Similarly, in the remaining molecules from the VSEPR electron pair number the geometry is found to be as follows:
$Cl{{F}_{3\,}}\,and\,IO_{4}^{-}$ have trigonal bipyramidal and see-saw geometry respectively.
$ICl_{2}^{-}\,and\,IC{{l}_{5}}$ have linear and square pyramidal respectively.
$IO_{3}^{-}\,and\,I{{O}_{2}}F_{2}^{-}$ have trigonal pyramidal and see-saw respectively.
It should also be noted that the charge on the molecule is also added to the total valence electrons such that the negative charge is added and the positive charge is subtracted from it.