Question

The modern formula of the above complexes based on the number of free $C{l}^{-}$ ions per formula unit will be:

	Formula	Colour	$C{l}^{-}$ ions in solution per formula unit
(A)	$CrC{l}_3.6H_{2}O$	violet	3
(B)	$CrC{l}_3.6H_{2}O$	light green	2
(C)	$CrC{l}_3^{\prime }6H_{2}O$	dark green	1

• A. (a)$\left[Cr\right(H_{2}O{)}_6]C{l}_3$; (b)$\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2\cdot H_{2}O$; (c)$\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl\cdot 2H_{2}O$
• B. (a)$\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2\cdot H_{2}O$; (b)$\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl\cdot 2H_{2}O$;(c)$\left[Cr\right(H_{2}O{)}_6]C{l}_3$;
• C. (a)$\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2\cdot H_{2}O$;(b)$\left[Cr\right(H_{2}O{)}_6]C{l}_3$; (c)$\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl\cdot 2H_{2}O$
• D. None of the Above

Answer 1

The correct option is A (a)$\left[Cr\right(H_{2}O{)}_6]C{l}_3$; (b)$\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2\cdot H_{2}O$; (c)$\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl\cdot 2H_{2}O$

(a)$\left[Cr\right(H_{2}O{)}_6]C{l}_3$; three chloride ions
(b)$\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2\cdot H_{2}O$; 2 chloride ions
(c)$\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl\cdot 2H_{2}O$ one chloride ions