Calculate the number of unpaired electrons and LFSE of [ Cr( NH3)6 ]+3 ion .

 Calculate the number of unpaired electrons and LFSE of [ Cr( NH₃)₆ ]⁺³ ion .

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

The valence shell electronic configuration  of Cr ⁺³ ion  : [Ar] 3d³ . 

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

The arrangement of the complex ion is octahedral . The octahedral  structure of  the above complex ion are shown below,

Ligand NH₃ which is a strong field Ligand . Under the influence of the octahedral crystal field, the five degenerate d-orbital’s  of Cr ⁺³ ions  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 three electrons  in the 3d³ degenerate orbital’s , so according to crystal field theory the possible arrangement  is t_2g³ e_g⁰ .

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

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

Since NH₃ is a strong field Ligand , so it is low spin complex  and from the above splitting diagram ,it is clear that the complex ion have three unpaired electron .

Due to presence of three unpaired electron , the [ Cr( NH₃)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,

3 x ( -4 Dq_o ) + 0 x 6 Dq_o  

= - 12 Dq_o +0  

=  -  12 Dq_o

_{Summary :}

_{What is ligand field stabilization energy ?}

Calculate the number of unpaired electrons and LFSE of [ Cr( NH₃)₆ ]⁺³ ion .

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