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 , hence shows paramagnetic behaviour . complex gives `Cl^(-)` and `NO_(3)^(-)` ions in solution hence , shows ionisation isomerism .
R : `Pt^(2+)` has `3d^(8)` configuration but ligands are strong field ligands hence it forms square planar complex . Thus , all electrons are paired and it also exhibits ionisation isomerism .
S : `co^(3+)` has `3d^(6)` configuration . But , ligands present are strong enough to cause electron pairing , hence , it shows diamagnetic behaviour and exhibits cis-trans isomerism as it is `Ma_(4) b_2` type complex .