Correct Answer - B

`Ag_(2)CrO_(4)hArr2Ag+CrO_(4)^(2-), K_(sp)=(2s)^(2)xxs=4s^(3), s=((K_(sp))/(4))^(1//3)=((1.1xx10^(-12))/(4))^(1//3)`

`=0.65xx10^(-4)`

`AgCl hArr Ag^(+)+Cl^(-),K_(sp)=sxxs=s^(2)`,

`s=sqrt(K_(sp))=sqrt(1.8xx10^(-10))=1.34xx10^(-5)`

`AgBr hArr Ag^(+)+Br^(-), K_(sp)=sxxs=s^(2)`,

`s=sqrt(K_(sp))=sqrt(5.0xx10^(-13))=0.71xx10^(-6)`

`AgI hArr Ag^(+)+I, K_(sp)=sxxs=s^(2)`,

`s=sqrt(K_(sp))=sqrt(8.3xx10^(-17))=0.9xx10^(-8)`

As solubility of `Ag_(2)CrO_(4)` is highest, it will be precipitated last of all.