Match each coordination compound in List-I with an appropriate pair of characteristics from List - II and select the correct answer using the code given below the lists .
[en = `H_(2)NCH_(2)CH_(2)NH_(2)` , At 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"), ((S) [Co(NH_(3))_(4) (NO_(3))_(2)]NO_(3) ,, 4."Diamagnetic and exhibits ionisation isomerism"):}`

Options

A. `{:("P","Q","R","S"),(4,2,3,1):}`
B. `{:("P","Q","R","S"),(3,1,4,2):}`
C. `{:("P","Q","R","S"),(2,1,3,4):}`
D. `{:("P","Q","R","S"),(1,3,4,2):}`

Answer

B

Explanation

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.