Dedue the value of x in the following compounds.
(a) `[Mo(CO)_(x)]` (b) `[Co_(2)(Co)_(x)]^(2-)` .

Explanation

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 .