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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.
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