
Find out the similarities between ${ I }_{ 2 }{ Cl }_{ 6 }$ and ${ Al }_{ 2 }{ Cl }_{ 6 }$.
A. both have ${ 3C-4e }^{ - }$ bond.
B. both have ${ sp }^{ 3 }$ hybridization for the central atom.
C. both are non-polar.
D. all are correct.
Answer
486.6k+ views
Hint: Electronic configuration of I = $\left[ Kr \right] { 4d }^{ 10 }{ 5s }^{ 2 }{ 5p }^{ 5 }$
Electronic configuration of Al = $\left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 1 }$
Electronic configuration of Cl = $\left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 5 }$
${ I }_{ 2 }{ Cl }_{ 6 }$ and ${ Al }_{ 2 }{ Cl }_{ 6 }$ are present in dimer form to complete their octet and to increase its stability.
Complete answer:
A. ${ I }_{ 2 }{ Cl }_{ 6 }$
Hybridization of each I atom = ${ sp }^{ 3 }{ d }^{ 2 }$
It has a planar structure.
In this, there are two bonds of ${ 3C-4e }^{ - }$.
In this, there are four bonds of ${ 2C-4e }^{ - }$.
B. ${ Al }_{ 2 }{ Cl }_{ 6 }$:
Hybridization of each Al atom = ${ sp }^{ 3 }$
It has a non-polar structure.
In this, there are two bonds of ${ 3C-4e }^{ - }$.
In this, there are four bonds of ${ 2C-4e }^{ - }$.
As we see ${ I }_{ 2 }{ Cl }_{ 6 }$ and ${ Al }_{ 2 }{ Cl }_{ 6 }$ have ${ sp }^{ 3 }{ d }^{ 2 }$ and ${ sp }^{ 3 }$ hybridization respectively. But both of these contain bonds of 3C .
Hence, the correct option is A.
Additional Information:
In science , hybridisation (or hybridization ) is the idea of mixing nuclear orbitals to shape new half and half orbitals reasonable for the qualitative description of nuclear bonding properties. Hybridized orbitals are helpful in the clarification of the state of sub-atomic orbitals for particles . It is a fundamental piece of valence bond theory. Sometimes, trained along with the valence shell electron-pair shell (VSEPR) hypothesis , valence bond and hybridization are in certainty not identified with the VSEPR model.
Note: The possibility to make a mistake is that you may choose option C. As the bonding in ${ I }_{ 2 }{ Cl }_{ 6 }$ is similar to ${ B }_{ 2 }{ H }_{ 6 }$, so you may think it is a nonpolar molecule but when you see the difference in their electronegativities then you will find it is a polar covalent bond.
Electronic configuration of Al = $\left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 1 }$
Electronic configuration of Cl = $\left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 5 }$
${ I }_{ 2 }{ Cl }_{ 6 }$ and ${ Al }_{ 2 }{ Cl }_{ 6 }$ are present in dimer form to complete their octet and to increase its stability.
Complete answer:
A. ${ I }_{ 2 }{ Cl }_{ 6 }$
Hybridization of each I atom = ${ sp }^{ 3 }{ d }^{ 2 }$
It has a planar structure.
In this, there are two bonds of ${ 3C-4e }^{ - }$.
In this, there are four bonds of ${ 2C-4e }^{ - }$.
B. ${ Al }_{ 2 }{ Cl }_{ 6 }$:
Hybridization of each Al atom = ${ sp }^{ 3 }$
It has a non-polar structure.
In this, there are two bonds of ${ 3C-4e }^{ - }$.
In this, there are four bonds of ${ 2C-4e }^{ - }$.

As we see ${ I }_{ 2 }{ Cl }_{ 6 }$ and ${ Al }_{ 2 }{ Cl }_{ 6 }$ have ${ sp }^{ 3 }{ d }^{ 2 }$ and ${ sp }^{ 3 }$ hybridization respectively. But both of these contain bonds of 3C .
Hence, the correct option is A.
Additional Information:
In science , hybridisation (or hybridization ) is the idea of mixing nuclear orbitals to shape new half and half orbitals reasonable for the qualitative description of nuclear bonding properties. Hybridized orbitals are helpful in the clarification of the state of sub-atomic orbitals for particles . It is a fundamental piece of valence bond theory. Sometimes, trained along with the valence shell electron-pair shell (VSEPR) hypothesis , valence bond and hybridization are in certainty not identified with the VSEPR model.
Note: The possibility to make a mistake is that you may choose option C. As the bonding in ${ I }_{ 2 }{ Cl }_{ 6 }$ is similar to ${ B }_{ 2 }{ H }_{ 6 }$, so you may think it is a nonpolar molecule but when you see the difference in their electronegativities then you will find it is a polar covalent bond.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
