Question Oxidation State for Chromium in the complex is (+2) and valance shell electronic configuration is .

Now d-orbital can have two electronic configuration, such as high spin or low spin 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 which is the energy penalty for putting two electrons in the same orbital, resulting from the electrostatic repulsion between electrons.

For configuration first three electrons will go to the orbitals. If the crystal field splitting energy is lower than the pairing energy then the fourth electron will go to the orbital, hence generate a high spin complex. If the crystal field splitting energy is higher than the pairing energy then the fourth electron will go to the orbital, hence we will end up with low spin complex. In this complex , is acting as a strong field ligand as it is given in the question that the crystal field splitting energy is higher than the pairing energy , hence the complex is a low spin complex. Crystal Field Stabilization Energy (CFSE) =   * number of t2g electrons + * number of eg electrons + P * number of pair of paired electrons = * 4 + * 0 + P * 1 = =  P * number of pair of paired electrons = P * 0 = 0 kJ mol-1

CFSE = = [ANSWER]

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