Question

The crystal field stabilisation energy of $ [Co{(N{H_3})_6}]C{l_3} $ is
a. $ - 7.2{\Delta _o} $
b. $ - 0.4{\Delta _o} $
c. $ - 2.4{\Delta _o} $
d. $ - 3.6{\Delta _o} $

Answer 1

Hint :Stabilization of the Crystal Field The energy of the electron configuration in the ligand field is subtracted from the energy of the electronic configuration in the isotropic field to get energy. The CSFE will be determined by a number of criteria, including: Geometry is the study of geometry (which changes the d-orbital splitting patterns) The number of d-electrons in a molecule.

Complete Step By Step Answer:
The breaking of degeneracies of electron orbital states, commonly d or f orbitals, due to a static electric field created by a surrounding charge distribution is described by crystal field theory (CFT) (anion neighbors). Various spectroscopies of transition metal coordination complexes, particularly optical spectra, have been described using this idea (colors). Some magnetic characteristics, hues, hydration enthalpies, and spinel structures of transition metal complexes are satisfactorily explained by CFT.
It is given by the formula $ \Delta E = \left( {\dfrac{3}{5}{\Delta _{{\text{oct}}}}} \right) + \left( { - \dfrac{2}{5}{\Delta _{{\text{oct }}}}} \right) $
The cobalt ion in $ \left[ {{\text{Co}}{{\left( {{\text{N}}{{\text{H}}_3}} \right)}_6}} \right]{\text{C}}{{\text{l}}_3} $ undergoes $ {\text{d}}{{\text{s}}^2}{{\text{p}}^3} $ hybridization to form inner orbital or low spin or spin paired complex.
6 electrons are present in $ {t_{2g}} $ level and 0 electrons are present in $ {e_g} $ level.
The crystal field stabilization energy of $ \left[ {{\text{Co}}{{\left( {{\text{N}}{{\text{H}}_3}} \right)}_6}} \right]{\text{C}}{{\text{l}}_3} $ is
 $ \Delta E = 0\left( {\dfrac{3}{5}{\Delta _{{\text{oct}}}}} \right) + 6\left( { - \dfrac{2}{5}{\Delta _{{\text{oct }}}}} \right) $
CFSE = $ 6 \times ( - 0.4\Delta 0) + 0 \times \left( {0.6{\Delta _0}} \right) = - 2.4{\Delta _0} $
Hence option c is correct
The following elements have an impact on this splitting:
the metal ion's properties
the oxidation state of the metal In comparison to the spherical field, a greater oxidation state causes more splitting.
the ligands' arrangement around the metal ion
the metal's coordination number (i.e. tetrahedral, octahedral...)
the ligands that surround the metal ion's nature The bigger the difference between the high and low energy d groups, the stronger the action of the ligands.

Note :
The attraction between the positively charged metal cation and the negative charge on the non-bonding electrons of the ligand causes the interaction between a transition metal and ligands, according to crystal field theory. The idea is built upon the energy shifts that occur when the five degenerate d-orbitals are surrounded by an array of point charges made up of ligands. The electrons from the ligand will be closer to some of the d-orbitals and further away from others as the ligand approaches the metal ion, resulting in a loss of degeneracy.