Question

Which of the following molecules has a planar structure?
a.) ${O_2}S{F_2}$
b.) $OS{F_2}$
c.) $Xe{F_4}$
d.) $ClO_4^ -$

Answer 1

The hybridisation of any molecule can tell about its structure. The molecules with $s{p^3}$ hybridisation are square planar and with $s{p^2}$ hybridisation are trigonal planar in nature.
We can find hybridisation of any molecule by the formula -
H = $\dfrac{1}{2}$[V + M - C + A]
Where H is the number of orbitals involved in hybridisation
V is the valence electrons in an atom
M is the number of monovalent atoms attached to central metal atom
C is the charge of cation and
A is the charge of anion.

Complete answer:
We know that planar structure is the one in which metal atoms and all its ligands are in one plane. Let us first see the structures of all these molecules given and then we can easily find out our answer.
The easiest way to find the geometry or shape of a molecule is to see its hybridisation.
We can find hybridisation of any molecule by the formula -
H = $\dfrac{1}{2}$[V + M - C + A]
Where H is the number of orbitals involved in hybridisation
V is the valence electrons in an atom
M is the number of monovalent atoms attached to central metal atom
C is the charge of cation and
A is the charge of anion.
So, the first molecule is ${O_2}S{F_2}$.
It has a central metal atom Sulphur which has valence electrons = 6.
This molecule has only two monovalent atoms linked which are fluorine. No other charge is present.
Thus, the hybridisation is -
H = $\dfrac{1}{2}$[6 + 2]
H = 4
This means $s{p^3}$ hybridised and so tetrahedral in shape.
The second molecule we have is $OS{F_2}$.
It has a central metal atom Os which has 6 valence electrons. Further, the monovalent atoms attached are two.
So, H = $\dfrac{1}{2}$[6 + 2]
H = 4
Again, here the hybridisation is $s{p^3}$ and shape is tetrahedral.
Now, the third molecule is $Xe{F_4}$ with Xe as the central metal atom. The Xe has 8 valence electrons and four monovalent atoms.
Thus, H = $\dfrac{1}{2}$[8 + 4]
H = 6
So, the hybridisation is $s{p^3}{d^2}$ and it has two lone pairs. Thus, the shape is square planar.
So, the fourth molecule is $ClO_4^ -$.
It has Cl as the central atom which has 7 valence electrons. No monovalent atom is attached. But it has one negative charge which will be added.
Thus, H = $\dfrac{1}{2}$[7 + 1]
H = 4
So, the hybridisation is $s{p^3}$ and shapes tetrahedral.
So, the correct answer is “Option C”.

Note: Oxygen atoms are attached in some molecules to the central atom. But these can not be considered because oxygen is not monovalent. It is divalent in nature. The atoms which are only monovalent are to be considered.