Answer
Verified
396.6k+ views
Hint: 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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE