| \[{{\left[ Pd\left( F \right)\left( Cl \right)\left( Br \right)\left( I \right) \right]}^{2}}\] has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilisation energy [CFSE] of \[{{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{n\text{ }\text{ }6}}\], respectively, are |
| [Note: Ignore the pairing energy] |
A) 0 BM and -2.4 \[{{\Delta }_{0}}\]
B) 5.92 BM and 0
C) 1.73 BM and -2.0 \[{{\Delta }_{0}}\]
D) 2.84 BM and -1.6 \[{{\Delta }_{0}}\]
Correct Answer: C
Solution :
[c] Complex \[{{\left[ Pd\left( F \right)\left( Cl \right)\left( Br \right)\left( I \right) \right]}^{2}}\](square planar geometry)-has 3 geometrical isomers. \[\therefore {{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{3\text{ }\text{ }6}}\text{ }=\text{ }{{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{3}}\] O.S. of Fe = +3 \[\therefore \] No. of unpaired \[{{e}^{-}}s\] in \[F{{e}^{+3}}\text{ }=\text{ }1\] \[F{{e}^{+3}}=3\,{{d}^{5}}\] \[\therefore \] E.C. according to CFT = \[{{t}_{2{{g}^{5}}}}{{e}_{{{g}^{0}}}}\] \[\therefore \] Spin only magnetic moment \[=\sqrt{1\left( 1+2 \right)}=\sqrt{3}=1.73BM\] \[\therefore \] CFSE\[=-4{{\Delta }_{0}}\times 5=-2{{\Delta }_{0}}\]
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