InterviewSolution
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InterviewSolution
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Dedue the value of x in the following compounds. (a) `[Mo(CO)_(x)]` (b) `[Co_(2)(Co)_(x)]^(2-)` (c ) `Hx (Cr (CO)_(5)` (d) `HxCo(CO)_(4)` (e) `[Fe (C_(5)H_(5))_(2)]` (f) `[Cr (C_(6)H_(6))_(2)]` . |
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Answer» In `[Mo(CO)x]`, EAN of `Mo =54` Zo of `Mo=42` Oxidation state of Mo is zero No .of electrons donated by `x CO` group `=2x` EAN of Mo in `[Mo(CO)_(x)]` `42 - 0 + 2x =54` `:. x = (54 -42)/(2) =6` `x =6` Formula is `[Mo(CO)_(6)]` (b) `[Co_(2)(CO)x],` Atomic number `(Z) of Co =27` Oxidation state of `Co =0` In `[Co_(2) (CO)x]` , i.e one electrons is shared with other `Co` atom due to `Co -Co` bond which means contribution coming because of `Co -Co bond is 1` Far `EAN` of each `Co=36` `27 -0 + 1 + (2x)/(2) =36` `implies = 8` Formula is `Co_(2)(CO)_(8)` (c ) `HxCr(CO)_(5)` Atomic number `(Z)` of `Cr =24` Electrons gained by `5CO` group `=2 x 5 =10` EAN of `Cr = 36 =24 + 10 + x` `implies x =36 -24 -10 =2` Electrons gained by Co due to 4 CO group `=2 x 4 =8` EAN of `Co =36 =27 + 8+x` `x =36 - 27 -8 =1` Formula is `HCo(CO)_(4)` `EAN = 36` (g) `EAN =36` Refer to Section 7.14.1 (b) Point 2, Fig. 7.20 and 7.21 . |
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