1.

Match each coordination compound in List-I with an appropriate pair of charateristics from List-II and select the correct answer using the code given below the lists : (en = H_(2)NCH_(2)CH_(2)NH_(2), atomic nos. : Ti = 22, Cr = 24, Co = 27, Pt = 78) {:(,"List-I",,"List-II"),(("P"),[Cr(NH_(3))_(4)Cl_(2)]Cl,1.,"Paramagnetic and exhibits ionisation isomerism"),(("Q"),[Ti(H_(2)O)_(5)Cl](NO_(3))_(2),2.,"Diamagnetic and exhibits cis-trans isomerism"),(("R"),[Pt(en)(NH_(3))(Cl)]NO_(3),3.,"Paramagnetic and exhibits cis-trans isomerism"),(("S"),[Co(NH_(3))_(4)(NO_(3))_(2)]NO_(3),4.,"Diamagnetic and exhibits ionisation isomerism"):}

Answer»

`{:("P","Q","R","S"),(4,2,3,1):}`
`{:("P","Q","R","S"),(3,1,4,2):}`
`{:("P","Q","R","S"),(2,1,3,4):}`
`{:("P","Q","R","S"),(1,3,4,2):}`

Solution :P : `Cr^(3+)` has `3d^(3)` configuration with 3 unpaired electrons. Hence,it shows PARAMAGNETIC behaviour. Complex of the type `MA_(4)B_(2)`shows cis-trans isomerism.
Q : `Ti^(3+)` has `3d^(1)` configuration with 1 unpaired electron. Hence, it is paramagnetic. Complex gives `Cl^(-)` and `NO_(3)^(-)`ions in solution. Hence, it shows ionisation isomerism.
R : `Pt^(2+)` has `3d^(8)` configuration but ligands are strong field ligands. Hence, all electrons pair up. Hence, it is diamagnetic and FORMS square planar complex. It also shows ionisation isomerism.
S : `Co^(3+)` has `3d^(6)` configuration. But ligands present are strong enough to CAUSE electron pairing. Hence, it will be diamagnetic. As it is a complex of the type `MA_(4)B_(2)`, it exhibits cis-trans isomerism.


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