C-O bond order lowest in ?
[Mn(CO)6]+
[Fe(CO)6]+
[Cr(CO)6]+
[V(CO)6]+
3http://books.google.co.in/books?id=MdDP4KnKnwcC&pg=PA120&lpg=PA120&dq=synergic+bonding&source=bl&ots=pQbfuiuDK-&sig=cCnutYyT4EkUOUQA2sNdDSa2Vy4&hl=en&ei=VEx4SpjILsSIkQWXqvS4Bg&sa=X&oi=book_result&ct=result&resnum=6#v=onepage&q=synergic%20bonding&f=false
1by the way,eure, the bond order would depend on the tendency of the d-orbitals of the following central
atoms to participate in bonding.more is the participating of unpaired d-electrons, more will be the bonding
electrons and more will be the bond order;thus i think, [Cr(CO)6)+] would have the highest
bond order and [V(CO)6]+ would have the lowest
1Mn+ = 3d54s1 in presence of CO effective configuration = 3d64s0.
Three lone pair for back bonding with vacant orbital of C in CO.
Fe0 = 3d64s2 in presence of CO effective configuration = 3d8 four lone pair for back bonding with CO.
Cr0 = 3d54s1
Effective configuration = 3d6.
Three lone pair for back bonding with CO.
V− = 3d44s2 effective configuration = 3d6 three lone pair for back bonding with CO.
Maximum back bonding is present in Fe(CO)5 there for CO bond order is lowest here.
1sorry for my previous misinterpretation.the solution goes this way:
Metallic carbonyls involve both sigma and pi bonding.sigma bond involves donation of the lone pair on
carbon of CO ligand into a hybridized vacant metal orbital. pi bond involves back donation of metal d e-
s into vacant antibonding pi* 2p orbital on CO.The larger the electrons in back-bonding,lesser
the bond-order of CO ligand.
thus,if ur question is right,then Fe and Mn involve a maximum of three lone pair for back bonding.
thus answer is both 'a' and 'b'.
the post of layak bhaiya is the solution of a similar question,that came in iit-2007.
