question_answer

The number of isomers for the complexes X and Y are, respectively, \[X={{[Pt(N{{H}_{3}})(N{{H}_{2}}OH)(N{{O}_{2}})(py)]}^{\oplus }}\] \[Y=\left[ Cr \right[\left( N{{H}_{3}} \right){{\left( OH \right)}_{2}}C{{l}_{3}}]\]

A) 2 and 3

B) 3 and 3        

C) 3 and 4              

D) 4 and 3

Correct Answer: B

Solution :

   [b] For X:- \[{{[Mabcd]}^{n\pm }}:\]: For complexes with general formula Mabcd, three geometrical isomers [a], [b] and [c] are possible.

	A	B	C
\[a-b\]	cis	trans	cis
\[c-d\]	cis	trans	cis
\[a-d\]	cis	cis	trans
\[a-c\]	trans	cis	cis
\[b-d\]	trans	cis	cis

For example \[{{[P{{t}^{+2}}(N{{H}_{3}})\,(N{{H}_{2}}OH)\,(N{{O}_{2}})\,(py)]}^{\oplus }}\,\,N{{O}_{2}}^{\Theta }\] For Y:- \[CN=6\] Geometry is octahedral. Three isomers are possible. I.  II. III.