NEETChemistryCoordination Compounds
Match List I with List II: List I (Complexes) List II (Hybridisation) (A) Ni ( CO ) 4 I sp 3 (B) Cu NH 3 4 2 + II dsp 2 (C) Fe NH 3 6 2 + III sp 3 d 2 (D) Fe H 2 O 6 2 + IV d 2 sp 3
Options
- AA – II, B – I, C – III, D – IV
- BA – I, B – II, C – III, D – IV
- CA – II, B – I, C – IV, D – III
- DA – I, B– II, C – IV, D – III
Correct answer
D. A – I, B– II, C – IV, D – III
Step-by-step solution
Ni is in zero oxidation state in [ Ni ( CO ) 4 ] . So, the electronic configuration of Ni is 3 d 8 4 s 2 . As CO is a strong ligand, it pushes all the electrons in the 3d orbital, therefore the hybridisation of [ N i ( C O ) 4 ] is sp 3 and it has tetrahedral geometry. It is diamagnetic due to the absence of unpaired electrons. Cu NH 3 4 2 + ion, oxidation state of Cu is +2 and its valence shell electronic configuration is 3 d 9 . So, there would be a rearrangement of electrons in Cu 2 + because of the NH 3 strong