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problem from jee 2007
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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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