Co Ordination Compounds Question 108
Question: 50 mL of 0.2 M solution of a compound with empirical formula $ CoCl_3.4NH_3 $ on treatment with excess of $ AgNO_3(aq) $ yields 1.435g of $ AgCl $ Ammonia is not removed by treatment with concentrated $ H_2SO_4 $ . The formula of the compound is:
Options:
A) $ [ Co{{( NH_3 )}_4} ]Cl_3 $
B) $ [ Co( NH_3 )4Cl_2 ]Cl $
C) $ [ Co{{( NH_3 )}_4} ]Cl_3 $
D) $ [ CoCl_3( NH_3 ) ]{{( NH_3 )}_3} $
Show Answer
Answer:
Correct Answer: B
Solution:
Mole of $ AgCl=\frac{1.435}{143.5}=0.01$
;$50mL,of,0.2 $ M complex $ \equiv $ 0.01 mole
$ [ Co{{( NH_3 )}_4}Cl_2 ]Cl\to {{[ CoNH_3 )}_4}Cl_2{{]}^{+}}+C{r^{-}} $
BETA
