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Sunday, January 19, 2020

Which of the Complex of the following pairs has the highest value of CFSE?

(1) [Co(CN)6]3-  and [Co(NH3)6]3+
(2) [Co(NH3)6]3+ and [CoF6]3-
(3) [Co(H2O)6]3+  and [Rh(H2O)6]3+
(4) [Co(H2O)6]2+ and [Co(H2O)6]3+

SOLUTION:
(1)  CN is the stronger ligand than NH3 therefore CFSE of [Co(CN)6]3-  will be more than  [Co(NH3)6]3+
(2) NH3 is stronger ligand than F therefore CFSE of [Co(NH3)6]3+ will be more than [CoF6]3- .
(3) Co belong to 3d series whereas The Rh belong to 4d series. More the value of n more is CFSE therefore CFSE of  [Rh(H2O)6]3+  is more than [Co(H2O)6]3+ .
(4) Oxidation number of Co in [Co(H2O)6]3+ is more than the Oxidation number of [Co(H2O)6]2+  therefore, CFSE of [Co(H2O)6]3+ is more than  [Co(H2O)6]2+ .



Related Questions:

Why all the tetrahedral Complexes are high spin Complexes?

Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?

Why [Mn(H2O)6]+2 is colourless although in which Mn+2 ion had five unpaired electrons ?

Why [FeF6]3– is colourless whereas [CoF6]3– is coloured

Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 is colour although both have +2 oxidation state and 3d8 configuration ?

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