A) \[{{[Mn{{(CO)}_{6}}]}^{+}}\]
B) \[[Fe{{(CO)}_{5}}]\]
C) \[[Cr{{(CO)}_{6}}]\]
D) \[{{[V\,{{(CO)}_{6}}]}^{-}}\]
Correct Answer: B
Solution :
| Greater the extent of \[d\pi -p\pi \] back bonding, smaller will be the bond order of \[CO\] bond in metal carbonyls. More are the number of valence shell electrons (d - electrons), maximum is the chance of \[p\pi -d\pi \] back bonding and thus lowest will be the bond order of \[CO\] bond. |
| The number of d-etectrons of metal in given complexes are as follows: |
| [a] \[{{[Mn{{(CO)}_{6}}]}^{+}}\] |
| \[Mn\]is present as \[M{{n}^{+}},\] thus the electronic configuration is \[3{{d}^{5}}4{{s}^{1}}.\] |
| [b] \[[Fe{{(CO)}_{5}}]\] |
| Fe is present in zero oxidation state, thus the electronic configuration will be \[3{{d}^{6}}4{{s}^{0}}.\] |
| [c] \[[Cr{{(CO)}_{6}}]\] |
| \[Cr\]is present in zero oxidation state, thus the electronic configuration is \[3{{d}^{5}}4{{s}^{1}}.\] |
| [d] \[{{[V{{(CO)}_{6}}]}^{-}}\] |
| V is present as \[{{V}^{+}}\] in given complex. |
| Thus, the electronic configuration is \[3{{d}^{3}}4{{s}^{1}}.\] |
| Thus, \[[Fe{{(CO)}_{5}}]\]has lowest bond order of \[CO\]bond. |
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