Question

Calculate the crystal field stabilization energy valiue for the complex Cr(NH)J*. The crystal field splitting energy and pair

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Answer #1

Oxidation State for Chromium in the complex Cr(NH3)2 is (+2) and valance shell electronic configuration is d^{4}.

Now d-orbital can have two electronic configuration, such as high spin 13 (t2g) (eg) or low spin (t2g) (eg depending upon the splitting power of ligands. If the ligand is weak field ligand, the complex will be high spin complex. If the ligand is strong field ligand, the complex will be low spin complex.

Here comes a factor Pairing energy (P which is the energy penalty for putting two electrons in the same orbital, resulting from the electrostatic repulsion between electrons.

For d^{4} configuration first three electrons will go to the t_{2g} orbitals. If the crystal field splitting energy (Ao is lower than the pairing energy then the fourth electron will go to the e_{g} orbital, hence generate a high spin complex. If the crystal field splitting energy (Ao is higher than the pairing energy then the fourth electron will go to the t_{2g} orbital, hence we will end up with low spin complex.

4 1V 1 1 1 1 1 Low Spin 4, P High spin 4 P

In this complex Cr(NH3)2 , NH_{3} is acting as a strong field ligand as it is given in the question that the crystal field splitting energy (\bigtriangleup _{o} = 150 kJ mol^{-1}) is higher than the pairing energy  (P = 120 kJ mol^{-1}), hence the complex is a low spin complex.

+0.6 4 -40 Bary Centre 1111 0,4 4o degennate d-Or bi ta 1 Isotropic field Ligand field

Crystal Field Stabilization Energy (CFSE) = \bigtriangleup E_{ligand field} - \bigtriangleup E_{isotropic field}

\bigtriangleup E_{ligand field}=(- 0.4\bigtriangleup _{o}) * number of t2g electrons + (+ 0.6\bigtriangleup _{o}) * number of eg electrons + P * number of pair of paired electrons = (- 0.4\bigtriangleup _{o}) * 4 + (+ 0.6\bigtriangleup _{o}) * 0 + P * 1 = - 1.6 \bigtriangleup _{o} + P = (- 1.6 \times 150 + 120) kJ mol^{-1}

\bigtriangleup E_{isotropic field} =  P * number of pair of paired electrons = P * 0 = 0 kJ mol-1

CFSE = [(- 1.6 \times 150 + 120)- 0] kJ mol^{-1} = - 120 kJ mol^{-1} [ANSWER]

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