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Question

Most complex compounds have either a coordination number of 4 or 6 and, in both cases, they show stereomerism. They may show geometrical isomerism or optical isomerism or both.

List-I contains complex compounds and List-II contains the number of stereomers they can form and the type of stereomers.
List-I List-II(I) K[PdNH3BrClF](P) Geometrical isomerism(II) [Co(NH3)3BrClF](Q) Optical isomerism(III) [Cr(NH3)3Br2Cl](R) 3(IV) [FeBr2(en)2]NO3(S) 5(T) 6(U) 2
Which of the following options has the correct combination considering compounds in List-I and number of their stereoisomers and type in List-II?

A
(II), (P)(U) and (III), (Q)(U)
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B
(II), (P)(Q)(T) and (IV), (P)(U)
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C
(II), (P)(Q)(S) and (III), (P)(R)
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D
(II), (P)(Q)(U) and (III), (P)(U)
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Solution

The correct option is C (II), (P)(Q)(S) and (III), (P)(R)
Here GI stands for geometrical isomers and SI stands for stereoisomers.

(I) is square planar so, it can show only GI.
Square planar complex of type [Mabcd] exhibit 3 GI.

(II) has 4 GI in which one is chiral so, a total of 5 SI.

(III) has 3 GI and all are optically inactive. So, a total of 3 SI.

(IV) has two GI, in which cis is Optically active.
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