`Ba^(2+)` ions, `CN^(-) & Co^(2+)` ions form a water soluble complex with `Ba^(2+)` ions as free cations for a `0.01M` solution of this complex, osmot
`Ba^(2+)` ions, `CN^(-) & Co^(2+)` ions form a water soluble complex with `Ba^(2+)` ions as free cations for a `0.01M` solution of this complex, osmotic pressure `=0.984` atm `&` degree of dissociation `=75%`. Then find ccordination number of `Co^(2+)` ion in this complex (`T=300K, R=0.082L "mol"^(-1) K^(-1)`)
1 Answers
Say `C.N.=x`
`0.984=i CRT`
`0.984=i0.01xx0.082xx300=ixx0.246`
`i=4=1+(n-1)alpha`
`rArr n=5`
Charge on co-ordination sphere`=` charge on Coblat ion `-` charge on `x` cyanide ions `=-(x-2)`
i.e. co-ordination sphere is `[Co(CN)_(x)]^(-(x-2))`
Change on Barium ion is `+2`
THus, formula of the comples will be `Ba_((x-2))[Co(CN)_(x)]_(2)` by charge balance.
`x-2+2=5`
`x=5`
`therefore CN=5`
`therefore` Formula is `Ba_(3)[Co(CN)_(5))_(2)`.