Answer is following by appropriately matching the lists based on the information given in the paragraph. Ligands are broadly classified into two classes classical and non classical ligands depending on thier donor and acceptability. Bonding mechanism  which is non classical is called synergic bonding. 
In the following list-I contains some complex and list-II contains the characterestics of complex
List-IList-II(Compounds)(I) [Pt(NO2)2(en)2]2+(P) Stable according to EAN-Rule(II) [Cr(πC6H6)(NO)2](Q)  Both ligands act as ambidentate(III) [Ir(SCN)(SO4)(NH3)4](R) Bond order M - L bond > 1.0(IV)[V(C2H4)(CN)5](S)  Bond order of Ligands decrease(T) Coordination number of central metal is six 
Which of the following options has the correct combination considering List-I and List-II?
  1. I, Q
  2. III, T
  3. II, S
  4. IV, R


The correct option is C II, S
(a) In [Pt(NO2)2(en)2]2+
Complex both ligands are not ambidentate ligand. NO2 can act as ambidentate ligand, i.e. ONO (NitritoO) and NO2 (NitritoN) whereas en can't act as ambidentate ligand.
(b) Co-ordination number of complex (III) is more than 6 as SO4 co-ordinate through 2 O- sites.
(c) [V(C2H4)(CN)5] has zero electron(due to +5oxidation state). Hence, there is no metal ligand charge transfer. So, no partially double bond developed in M - L because of back bonding. Therefore bond order of M - L bond = 1.0.
(d) [Cr(πC6H6)(NO)2], chromium is in zero oxidation state according to MOT e will go to antibonding orbital of NO because of back bonding. Bond length increases and bond order decreases. 

So, the bond order between M+NO increase and MC6H6 decrease. 
Here, NO cordinates through only one bond and C6H6 cordinates through 3 bonds. So, 2[M NO bond order increases] and 3 MC||C decreases. 
Therefore overall bond order decreases. 

 Suggest corrections

Similar questions
View More

People also searched for
View More