+ form four complexes with four different ligands which are [Cr(Cl)6]3–, [Cr()6]3+, [Cr()6]3+ and [Cr(CN)6]3–. The order of CFSE (0) in these complexes is in the order :

Question

$Cr_3$+ form four complexes with four different ligands which are [Cr(Cl)6]3–, [Cr($H_2O$)6]3+, [Cr($NH_3$)6]3+ and [Cr(CN)6]3–. The order of CFSE (0) in these complexes is in the order :

Prashant Kotian · Accepted Answer

Correct answer: A. Because central atom metal are same but Ligands are different So, CFSE Strength of ligands - - 3 2 CN > NH > H O > Cl 100. For an octahedral complex, which of the following d- electron configuration will give maximum CFSE? (a) High spin d6 – 0.4 0 (b) Low spin d4 – 1.6 0 (c) Low spin d5 – 2.0 0 (d) High spin d7 – 0.8 0 CFSE for octahedral complexes CFSE = 2 0 [0.4( ) 0.6( ') ] g n t n eg nP So, value is maximum for low spin 5 d –2.0 0 101. The value of CFSE () for complexes given below follow the order: (I) [Co(NH3)6]3+ (II) [Rh(NH3)6]3+ (III) [Ir(NH3)6]3+ (a) I < II < III

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