JEE Main201815 Apr 2018Morning ShiftChemistryCoordination CompoundsActual
The correct combination is :
Options
- A[ NiCl ₄ ]²⁻ - square-planar; [ Ni ( CN )₄ ]²⁻ - paramagnetic
- B[ Ni ( CN )₄ ]²⁻ -tetrahedral; [ Ni ( CO )₄ ] - paramagnetic
- C[ NiCl ₄ ]²⁻ - paramagnetic; [ Ni ( CO )₄ ] - tetrahedral
- D[ NiCl ₄ ]²⁻ - dimagnetic; [ Ni ( CO )₄ ] -square-planar
Correct answer
C. [ NiCl ₄ ]²⁻ - paramagnetic; [ Ni ( CO )₄ ] - tetrahedral
Step-by-step solution
[ Ni ( CN )₄ ]²⁻ is square planar, diamagnetic (0 unpaired electrons) with d s p^2 hybridisation. [ Ni ( CO )₄ ] - is tetrahedral,diamagnetic ( 0 unpaired electrons) with s p^3 hybridisation. [ NiCl ₄ ]²⁻ is tetrahedral, paramagnetic ( 2 unpaired electrons) with s p^3 hybridisation. Hence, the option (c) is the correct answer.