Answer
Verified
36.6k+ views
Hint: Crystal field stabilization energy: Crystal field stabilization energy is the energy of electronic configuration in the ligand field - Electronic configuration in the isotropic field.
Complete answer step-by-step:
> The splitting pattern in each complex compound is octahedral. If a strong field ligand is present, then back pairing will take place as in the case of B and D, and splitting energy can never be zero, since no electron can go into ${ e }_{ g }$ orbitals.
But if a strong field is not present, then back pairing will not take place as in the case of A and C.
> The splitting pattern in ${ K }_{ 3 }\left[ Fe{ F }_{ 6 } \right]$ is;
Atomic number of Fe = ${ 26 }$
Electronic configuration = $\left[ Ar \right] { 3d }^{ 6 }{ 4s }^{ 2 }$
For, ${ Fe }^{ +3 } = \left[ Ar \right] { 3d }^{ 5 } or { t }_{ 2g }^{ 3 }{ e }_{ g }^{ 2 }$
The crystal field splitting energy for octahedral complexes are given by;
CFSE = ${ -0.4\times }{ t }_{ 2g }{ +0.6\times }{ e }_{ g }$
By putting the values, we get
CFSE = ${ -0.4\times 3 }{ +0.6\times 2 }$
CFSE = 0
Now, for ${ K }_{ 2 }\left[ Mn{ F }_{ 6 } \right]$
Atomic number of Mn = ${ 25 }$
Electronic configuration = $\left[ Ar \right] { 3d }^{ 5 }{ 4s }^{ 2 }$
For, ${ Mn }^{ +3 } = \left[ Ar \right] { 3d }^{ 4 }or { t }_{ 2g }^{ 3 }{ e }_{ g }^{ 1 }$
The crystal field splitting energy for octahedral complexes are given by;
CFSE = ${ -0.4\times }{ t }_{ 2g }{ +0.6\times }{ e }_{ g }$
By putting the values, we get
CFSE = ${ -0.4\times 3 }{ +0.6\times 1 }$
CFSE = ${ -0.6 } \Delta _{ 0 }$
Hence, we can say that the value for crystal field stabilization energy is zero for ${ K }_{ 3 }\left[ Fe{ F }_{ 6 } \right]$.
The correct option is C.
Note: The possibility to make a mistake is that you may use the crystal field splitting energy formula for tetrahedral, not octahedral. But all the compounds are octahedral so do not confuse between them.
Complete answer step-by-step:
> The splitting pattern in each complex compound is octahedral. If a strong field ligand is present, then back pairing will take place as in the case of B and D, and splitting energy can never be zero, since no electron can go into ${ e }_{ g }$ orbitals.
But if a strong field is not present, then back pairing will not take place as in the case of A and C.
> The splitting pattern in ${ K }_{ 3 }\left[ Fe{ F }_{ 6 } \right]$ is;
Atomic number of Fe = ${ 26 }$
Electronic configuration = $\left[ Ar \right] { 3d }^{ 6 }{ 4s }^{ 2 }$
For, ${ Fe }^{ +3 } = \left[ Ar \right] { 3d }^{ 5 } or { t }_{ 2g }^{ 3 }{ e }_{ g }^{ 2 }$
The crystal field splitting energy for octahedral complexes are given by;
CFSE = ${ -0.4\times }{ t }_{ 2g }{ +0.6\times }{ e }_{ g }$
By putting the values, we get
CFSE = ${ -0.4\times 3 }{ +0.6\times 2 }$
CFSE = 0
Now, for ${ K }_{ 2 }\left[ Mn{ F }_{ 6 } \right]$
Atomic number of Mn = ${ 25 }$
Electronic configuration = $\left[ Ar \right] { 3d }^{ 5 }{ 4s }^{ 2 }$
For, ${ Mn }^{ +3 } = \left[ Ar \right] { 3d }^{ 4 }or { t }_{ 2g }^{ 3 }{ e }_{ g }^{ 1 }$
The crystal field splitting energy for octahedral complexes are given by;
CFSE = ${ -0.4\times }{ t }_{ 2g }{ +0.6\times }{ e }_{ g }$
By putting the values, we get
CFSE = ${ -0.4\times 3 }{ +0.6\times 1 }$
CFSE = ${ -0.6 } \Delta _{ 0 }$
Hence, we can say that the value for crystal field stabilization energy is zero for ${ K }_{ 3 }\left[ Fe{ F }_{ 6 } \right]$.
The correct option is C.
Note: The possibility to make a mistake is that you may use the crystal field splitting energy formula for tetrahedral, not octahedral. But all the compounds are octahedral so do not confuse between them.
Recently Updated Pages
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
What is the difference between solvation and hydra class 11 chemistry JEE_Main
IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main
Sodium chloride is purified by passing hydrogen chloride class 11 chemistry JEE_Main
Hydrogen readily combines with nonmetals and thus it class 12 chemistry JEE_Main
Other Pages
An electric bulb has a power of 500W Express it in class 11 physics JEE_Main
The cell in the circuit shown in the figure is ideal class 12 physics JEE_Main
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main
For an electromagnet the core should have A High retentivity class 12 physics JEE_Main
Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main
Normality of 03 M phosphorus acid H3PO3 is A 05 B 06 class 11 chemistry JEE_Main