NEETChemistryCoordination Compounds
Match List-I with List-II List-I Coordination Complex List-II Number of unpaired electrons A. Cr ( CN ) 6 3 - I. 0 B. Fe H 2 O 6 2 + II. 3 C. Co NH 3 6 3 + III. 2 D. Ni NH 3 6 2 + IV. 4 Choose the correct answer from the options given below:
Options
- AA-II, B-IV, C-I, D-III
- BA-III, B-IV, C-I, D-II
- CA-II, B-I, C-IV, D-III
- DA-IV, B-III, C-II, D-I
Correct answer
A. A-II, B-IV, C-I, D-III
Step-by-step solution
Cr ( CN ) 6 3 - ion, oxidation state of Cr is +3 and its valence shell electronic configuration is 3 d 3 . There are 3 unpaired electrons in 3 d orbital. (A). Cr ( CN ) 6 3 - No. of unpaired electrons = 3 Fe H 2 O 6 2 + ion, oxidation state of Fe is +2 and its valence shell electronic configuration is 3 d 6 . There are 4 unpaired electrons in 3d orbital. So, you can say the hybridisation here would be sp 3 d 2 . (B). Fe H 2 O 6 2 + No. of unpaired electrons = 4 Co NH 3 6 3 + in this oxidation state of central metal