Calculate the number of unpaired electrons and LFSE of [ Fe ( H2O)6 ]+3 ion .

Calculate the number of unpaired electrons and LFSE of [ Fe ( H₂O)₆ ]⁺³ ion .

The IUPAC  name of the above complex ion is hexaaquo iron(III) ion. The oxidation state of iron in this complex ion is +3 .

Coordination number( C.N ) of iron is 6 .

The valence shell electronic configuration  of Fe ⁺³ ion  : [Ar] 3d⁵ .  

The arrangement of the complex ion is octahedral . The octahedral structure of this complex are as follows ,

Ligand H₂O, which is a weak  field Ligand . Under the influence of the octahedral crystal field, the five degenerate d-orbital’s  of Fe ⁺³ ion  are splitted into two sets of energetically different orbital’s. 

These two sets are energetically lower t_2g orbital and energetically higher e_g orbital . 

As there are five  electrons  in the 3d⁵ degenerate orbital’s , so according to crystal field theory the possible arrangement  is t_2g³ e_g² .

Here, the crystal field splitting energy ( 10 Dq_o ) is lower than pairing energy (P).

That is , 10 Dq_o  <  P

So, no pairing of electrons will be occured  in any sets of orbital’s . 

The crystal field splitting of degenerate d-orbital’s of  Fe ⁺³ ion  under the influence of Ligand weak field Ligand  H₂O are as follows,

Since H₂O is a weak field Ligand , so it should be high  spin complex  and from the above splitting diagram ,it is clear that the complex ion have five unpaired electron .

Due to presence of five  unpaired electron , the [ Fe( H₂O)₆ ]⁺³ complex ion exhibit paramagnetic properties .

Calculation of Ligand field stabilization energy .

The Ligand field stabilization energy ( LFSE ) of the above complex ion is,

      3x( -4 Dq_o ) + 2 x 6 Dq_o  
      = - 12 Dq_o + 12 Dq_o 
      =  0  Dq_o .

Summary : 

Calculate the number of unpaired electrons and LFSE of [ Fe ( H₂O)₆ ]⁺³ ion .

What is ligand field stabilization energy ?

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Calculate the spin only magnetic momentum  µ of  K3 [ Mn(CN)6]  compound  .