According to Crystal Field Theory..all three are octahedral complexes a) [Fe(CN)6]4- CN- is strong Field ligand(have -acceptor orbitals:Here *) .That means all electrons o(CFSE) > P(pairing energy) ..Therefore the fourth electron occupies t2g orbital with configuration t2g4eg0. The energy you are asking is LFSE ( ligand field stabilization energy) and not Crystal Field SE which is o For electron occupying t2g orbital complex is stabilized by factor of 0.40o For electron occupying eg orbital complex is destabilized by factor of 0.60o Fe is in Fe2+ state Outer Electronic configuration of Fe: 3d64s2 Outer Electronic configuration of Fe2+ : 3d6 All 6 electrons occupy t2g,LFSE = 6 x 0.40o = 2.4o c)[Co(NH3)6]3+ ..There is a reason for doing third part first(will make you better understand) NH3 is intermediate field ligand(has few or no -acceptor/donor orbitals) .It can act as both weak field and strong field ligand. For first row Transition metal complexes .. For M2+ ..NH3 is weak field and for M3+, it is strong field ligand. For 2nd(Ru3+) and higher row elements o increases..so,NH3 is Strong field ligand . Here it is Co3+...so NH3 acts as strong field ligand Electronic config of Co3+ = 3d6 All 6 electrons occupy t2g,LFSE = 6 x 0.40o = 2.4o b) [Ru(NH3)6]3+ For 2nd(Ru3+) and higher row elements o increases..so,NH3 is Strong field ligand . Outer electronic config of Ru3+ : 4d5 All 5 electrong in t2g,LFSE = 5 x 0.40o = 2o Solution According to Crystal Field Theory..all three are octahedral complexes a) [Fe(CN)6]4- CN- is strong Field ligand(have -acceptor orbitals:Here *) .That means all electrons o(CFSE) > P(pairing energy) ..Therefore the fourth electron occupies t2g orbital with configuration t2g4eg0. The energy you are asking is LFSE ( ligand field stabilization energy) and not Crystal Field SE which is o For electron occupying t2g orbital complex is stabilized by factor of 0.40o For electron occupying eg orbital complex is destabilized by factor of 0.60o Fe is in Fe2+ state Outer Electronic configuration of Fe: 3d64s2 Outer Electronic configuration of Fe2+ : 3d6 All 6 electrons occupy t2g,LFSE = 6 x 0.40o = 2.4o c)[Co(NH3)6]3+ ..There is a reason for doing third part first(will make you better understand) NH3 is intermediate field ligand(has few or no -acceptor/donor orbitals) .It can act as both weak field and strong field ligand. For first row Transition metal complexes .. For M2+ ..NH3 is weak field and for M3+, it is strong field ligand. For 2nd(Ru3+) and higher row elements o increases..so,NH3 is Strong field ligand . Here it is Co3+...so NH3 acts as strong field ligand Electronic config of Co3+ = 3d6 All 6 electrons occupy t2g,LFSE = 6 x 0.40o = 2.4o b) [Ru(NH3)6]3+ For 2nd(Ru3+) and higher row elements o increases..so,NH3 is Strong field ligand . Outer electronic config of Ru3+ : 4d5 All 5 electrong in t2g,LFSE = 5 x 0.40o = 2o.

1. According to Crystal Field Theory..all three are octahedral complexes
a) [Fe(CN)6]4-
CN- is strong Field ligand(have -acceptor orbitals:Here *) .That means all electrons o(CFSE) >
P(pairing energy) ..Therefore the fourth electron occupies t2g orbital with configuration t2g4eg0.
The energy you are asking is LFSE ( ligand field stabilization energy) and not Crystal Field SE
which is o
For electron occupying t2g orbital complex is stabilized by factor of 0.40o
For electron occupying eg orbital complex is destabilized by factor of 0.60o
Fe is in Fe2+ state
Outer Electronic configuration of Fe: 3d64s2
Outer Electronic configuration of Fe2+ : 3d6
All 6 electrons occupy t2g,LFSE = 6 x 0.40o = 2.4o
c)[Co(NH3)6]3+ ..There is a reason for doing third part first(will make you better understand)
NH3 is intermediate field ligand(has few or no -acceptor/donor orbitals) .It can act as both weak
field and strong field ligand. For first row Transition metal complexes .. For M2+ ..NH3 is weak
field and for M3+, it is strong field ligand.
For 2nd(Ru3+) and higher row elements o increases..so,NH3 is Strong field ligand .
Here it is Co3+...so NH3 acts as strong field ligand
Electronic config of Co3+ = 3d6
All 6 electrons occupy t2g,LFSE = 6 x 0.40o = 2.4o
b) [Ru(NH3)6]3+
For 2nd(Ru3+) and higher row elements o increases..so,NH3 is Strong field ligand .
Outer electronic config of Ru3+ : 4d5
All 5 electrong in t2g,LFSE = 5 x 0.40o = 2o
Solution
According to Crystal Field Theory..all three are octahedral complexes
a) [Fe(CN)6]4-
CN- is strong Field ligand(have -acceptor orbitals:Here *) .That means all electrons o(CFSE) >
P(pairing energy) ..Therefore the fourth electron occupies t2g orbital with configuration t2g4eg0.
The energy you are asking is LFSE ( ligand field stabilization energy) and not Crystal Field SE

2. which is o
For electron occupying t2g orbital complex is stabilized by factor of 0.40o
For electron occupying eg orbital complex is destabilized by factor of 0.60o
Fe is in Fe2+ state
Outer Electronic configuration of Fe: 3d64s2
Outer Electronic configuration of Fe2+ : 3d6
All 6 electrons occupy t2g,LFSE = 6 x 0.40o = 2.4o
c)[Co(NH3)6]3+ ..There is a reason for doing third part first(will make you better understand)
NH3 is intermediate field ligand(has few or no -acceptor/donor orbitals) .It can act as both weak
field and strong field ligand. For first row Transition metal complexes .. For M2+ ..NH3 is weak
field and for M3+, it is strong field ligand.
For 2nd(Ru3+) and higher row elements o increases..so,NH3 is Strong field ligand .
Here it is Co3+...so NH3 acts as strong field ligand
Electronic config of Co3+ = 3d6
All 6 electrons occupy t2g,LFSE = 6 x 0.40o = 2.4o
b) [Ru(NH3)6]3+
For 2nd(Ru3+) and higher row elements o increases..so,NH3 is Strong field ligand .
Outer electronic config of Ru3+ : 4d5
All 5 electrong in t2g,LFSE = 5 x 0.40o = 2o