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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Jun 2007 16:39:32 IST
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Among the following metal carbonyls, the C-O bond order is lowest in (A) [Mn(CO)6]+ (B) [Fe(CO)5] (C) [Cr(CO)6] (D) [V(CO)6]-
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24 Jun 2007 16:47:36 IST
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plz help and give detail solution
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25 Jun 2007 21:00:52 IST
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can no one answer...! Question isn't so tough!
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25 Jun 2007 21:16:18 IST
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this was one of the controversial questions..( with almost evry 2nd institute providing a different answer )
implies such questions are meant to be left...
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............tseb eht ma i |
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25 Jun 2007 23:23:07 IST
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HERE ANSWER IS Fe(CO)5 by the help of electronic configuration of each ion and remember that CO is strong field ligend so responsible for low spin complex.1.Mn+ 3d54s1,3d64s0effective configuration where 3 lone pair for back bonding with vacent orbital of C in CO 2.Fe 0 3d64s2 in presence of CO effective configuration is 3d8 so 4 lone pair for back bonding with CO 3.Cr 03d54s1 effective configuration is 3d6 so 3 lone pair for back bonding with CO 4.V- =3d4s2 effective configuration is 3d6 3 lone pair for back bonding with CO so maximum back bonding is present in Fe(CO)5 so it has lowest bond order.
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Arun / Rashi - Authors Macromind MCQ of Chemistry from G.R Batla |
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4 Jul 2007 14:39:53 IST
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The answer is V(CO)6-. Because in anionic species back bonding is much more due to greater electron density.So bond order consequently decreases.For more refer to Atkins-Shriver(Inorganic chemistry).
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