Verified Solution • 2025–26 Session moderate

QJEEM24Ch9da3: JEE Main 2024 Chemistry - Class 12 JEEM24Ch9da3 Solution: Select the option with correct property -

Select the option with correct property -

Multiple Choice Options

  • A [Ni(CO)4]\left[\mathrm{Ni}(\mathrm{CO})_4\right] and [NiCl4]2−\left[\mathrm{NiCl}_4\right]^{2-} both Paramagnetic
  • B [Ni(CO)4]\left[\mathrm{Ni}(\mathrm{CO})_4\right] and [NiCl4]2−\left[\mathrm{NiCl}_4\right]^{2-} both Diamagnetic
  • C [NiCl4]2−\left[\mathrm{NiCl}_4\right]^{2-} Diamagnetic, [Ni(CO)4]\left[\mathrm{Ni}(\mathrm{CO})_4\right] Paramagnetic
  • D [Ni(CO)4]\left[\mathrm{Ni}(\mathrm{CO})_4\right] Diamagnetic, [NiCl4]2−\left[\mathrm{NiCl}_4\right]^{2-} Paramagnetic

Step-by-Step Verified Solution

[Ni(CO)4]→\left[\mathrm{Ni}(\mathrm{CO})_4\right] \rightarrow diamagnetic, sp3\mathrm{sp}^3 hybridisation, number of unpaired electrons =0=0

[NiCl4]2−,→\left[\mathrm{NiCl}_4\right]^{2-}, \rightarrow paramagnetic, sp3\mathrm{sp}^3 hybridisation, number of unpaired electrons =2=2

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