[Pd(F)(Cl)(Br)(I)]2– has n number of
geometrical isomers. Then, the spin-only
magnetic moment and crystal field stabilisation
energy [CFSE] of [Fe(CN)6]n–6, respectively,
are:
[Note : Ignore the pairing energy]
Solution
Complex [Pd(F)(Cl)(Br)(I)]<sup>2–</sup> (square planar
geometry)-has 3 geometrical isomers.
<br><br>$\therefore$ n = 3
<br><br>[Fe(CN)<sub>6</sub>]<sup>n–6</sup> becomes [Fe(CN)<sub>6</sub>]<sup>3-</sup>
<br><br>Here Oxidation state of Fe = +3
<br><br>$\therefore$ Fe<sup>+3</sup> = [Ar]3d<sup>5</sup>4s<sup>0</sup>
<br><br>CN<sup>-</sup> is strong field ligand so it pairing of electrons happens.
<br><br>E.C. according to CFT = ($t_{2g}^5e{g^0}$)
<br><br>$\therefore$ No. of unpaired e<sup>-</sup> = 1
<br><br>Then Magnetic moment = $\sqrt {n\left( {n + 2} \right)}$ = $\sqrt 3$ = 1.73 B.M
<br><br>CFSE = [(–0.4 × 5) + (0.6 × 0)] $\Delta$<sub>0</sub>
= –2.0 $\Delta$<sub>0</sub>
About this question
Subject: Chemistry · Chapter: Coordination Compounds · Topic: Ligands and Coordination Number
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