Answer
Verified
396.3k+ views
Hint: Take note of the oxidation state of the $Xe$ atom and then consider the electronic configuration. From this, calculate the number of lone pairs and bond pairs. This will give you an idea about the hybridization as well as the geometry of the molecule.
Complete solution:
First, let us consider the electronic configuration of $Xe$ in its ground state. The atomic number of xenon is 54. Its electronic configuration is $[Kr]4{{d}^{10}}5{{s}^{2}}5{{p}^{6}}$. Usually, it is inert, but it can form bonds with different atoms by promoting its electrons to the $5d$ orbital and hybridization.
Both the ions that need to be considered here have a +1 charge, since fluorine cannot have lost one of its electrons, let us assume that it is xenon that has lost one. So, the configuration will be $[Kr]4{{d}^{10}}5{{s}^{2}}5{{p}^{5}}$. While drawing the electronic configuration, let us ignore the $4d$ orbital and only consider the $5s,\text{ }5p,\text{ and 5d}$ orbitals for the time being.
So, the configuration will be:
- For $Xe{{F}_{3}}^{+}$
To form bonds with 3 fluorine atoms, xenon will require 2 more free electrons, so it will promote one of the electrons from the $5p$ orbital to the $5d$ orbital and carry out the hybridization to form 5 $s{{p}^{3}}d$ orbitals. The hybridized configuration will be:
Now the three electrons will form bonds with 3 fluorine atoms and the geometry of the molecule will become trigonal bipyramidal. There are 2 lone pairs and 3 bond pairs present; the lone pairs always occupy the equatorial positions in the trigonal bipyramidal geometry. Thus, the shape of the molecule will be:
Here, we can see the bent T-shape formed by the atoms.
- For $Xe{{F}_{5}}^{+}$
The formation of this ion will be similar to that of the $Xe{{F}_{3}}^{+}$ ion. Since, xenon has to form bonds with 5 fluorine atoms, it will need 5 free electrons, so it will promote 2 electrons from the $5p$ orbital to the $5d$ orbital. And carry out the hybridization to form 6 $s{{p}^{3}}{{d}^{2}}$ orbitals. The hybridized configuration will be:
Now, the unpaired electrons will pair with the 5 fluorine atoms and the molecule will become octahedral. There are 5 bond pairs and 1 lone pair present. The geometry will be:
Here, we can see the square pyramid formed by the atoms.
The geometries of the $Xe{{F}_{3}}^{+}$ ion and $Xe{{F}_{5}}^{+}$ ion respectively is bent T-shaped and square pyramidal respectively.
Hence, the correct answer is ‘B. bent T-shaped, square pyramidal’
Note:
Always consider the charge that is present on the ions, if the charge is ignored then it will not be possible to draw the hybridization diagram. When xenon has to bond with an odd number of atoms, it always has to have a charge. Otherwise, one unpaired electron will remain and make the molecule unstable.
Complete solution:
First, let us consider the electronic configuration of $Xe$ in its ground state. The atomic number of xenon is 54. Its electronic configuration is $[Kr]4{{d}^{10}}5{{s}^{2}}5{{p}^{6}}$. Usually, it is inert, but it can form bonds with different atoms by promoting its electrons to the $5d$ orbital and hybridization.
Both the ions that need to be considered here have a +1 charge, since fluorine cannot have lost one of its electrons, let us assume that it is xenon that has lost one. So, the configuration will be $[Kr]4{{d}^{10}}5{{s}^{2}}5{{p}^{5}}$. While drawing the electronic configuration, let us ignore the $4d$ orbital and only consider the $5s,\text{ }5p,\text{ and 5d}$ orbitals for the time being.
So, the configuration will be:
- For $Xe{{F}_{3}}^{+}$
To form bonds with 3 fluorine atoms, xenon will require 2 more free electrons, so it will promote one of the electrons from the $5p$ orbital to the $5d$ orbital and carry out the hybridization to form 5 $s{{p}^{3}}d$ orbitals. The hybridized configuration will be:
Now the three electrons will form bonds with 3 fluorine atoms and the geometry of the molecule will become trigonal bipyramidal. There are 2 lone pairs and 3 bond pairs present; the lone pairs always occupy the equatorial positions in the trigonal bipyramidal geometry. Thus, the shape of the molecule will be:
Here, we can see the bent T-shape formed by the atoms.
- For $Xe{{F}_{5}}^{+}$
The formation of this ion will be similar to that of the $Xe{{F}_{3}}^{+}$ ion. Since, xenon has to form bonds with 5 fluorine atoms, it will need 5 free electrons, so it will promote 2 electrons from the $5p$ orbital to the $5d$ orbital. And carry out the hybridization to form 6 $s{{p}^{3}}{{d}^{2}}$ orbitals. The hybridized configuration will be:
Now, the unpaired electrons will pair with the 5 fluorine atoms and the molecule will become octahedral. There are 5 bond pairs and 1 lone pair present. The geometry will be:
Here, we can see the square pyramid formed by the atoms.
The geometries of the $Xe{{F}_{3}}^{+}$ ion and $Xe{{F}_{5}}^{+}$ ion respectively is bent T-shaped and square pyramidal respectively.
Hence, the correct answer is ‘B. bent T-shaped, square pyramidal’
Note:
Always consider the charge that is present on the ions, if the charge is ignored then it will not be possible to draw the hybridization diagram. When xenon has to bond with an odd number of atoms, it always has to have a charge. Otherwise, one unpaired electron will remain and make the molecule unstable.
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