Based on VBT, $[F e C l_{6}]^{3 -}$ has $C l^{-}$ as weak ligands. $F e$ is $s p^{3} d^{2}$ hybridized (outer orbital complex) having five unpaired electrons. $[F e (C N)_{6}]^{3 -}$ has $C N^{-}$ as strong ligand. $F e$ is $d^{2} s p^{3}$ hybridized (inner orbital complex) having only one unpaired electron. Hence, $[F e C l_{6}]^{3 -}$ is more paramagnetic than $[F e (C N)_{8}]^{3 -}$ .
If true enter 1, else enter 0.

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Solution

Based on VBT, $[F e C l_{6}]^{3 -}$ has $C l^{-}$ as weak ligands. Therefore, $F e$ is $s p^{3} d^{2}$ hybridized (outer orbital complex) having five unpaired electrons.
$[F e (C N)_{6}]^{3 -}$ has $C N^{-}$ as strong ligands. Therefore, $F e$ is $d^{2} s p^{3}$ hybridized (inner orbital complex) having only one unpaired electron. Hence, $[F e C l_{6}]^{3 -}$ is more paramagnetic than $[F e (C N)_{8}]^{3 -}$ .
Hence, the correct answer is 1.

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