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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.
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