JEE Main202228 Jul 2022Evening ShiftChemistryCoordination CompoundsActual
Match List-I with List-II List-I (Complex) List-II (Hybridization) A Ni CO 4 I sp 3 B Ni CN 4 2 - II sp 3 d 2 C Co CN 6 3 - III d 2 sp 3 D CoF 6 3 - IV dsp 2 Choose the correct answer from the options given below
Options
- AA - IV , B - I , C - III , D - II
- BA - I , B - IV , C - III , D - II
- CA - I , B - IV , C - II , D - III
- DA - IV , B - I , C - II , D - III
Correct answer
B. A - I , B - IV , C - III , D - II
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 3 d orbital, therefore the hybridisation of Ni ( CO ) 4 is sp 3 and it has tetrahedral geometry. It is diamagnetic due to the absence of unpaired electrons. In [ Ni ( CN ) 4 ] 2 - , there is Ni 2 + ion for which the electronic configuration in the valence shell is 3 d 8 4 s 0 . In presence of strong field CN - ions, all the electrons are paired up. The empt