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Chemistry · Coordination Compounds

JEE Main 2026 — 23 January, Evening Shift — Question 65

Total number of unpaired electrons present in the central metal atoms/ions of [Ni(CO)4],[NiCl4]2−,[PtCl2(NH3)2],[Ni(CN4)]2−\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],\left[\mathrm{NiCl}_{4}\right]^{2-},\left[\mathrm{PtCl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right],\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-} and [Pt(CN4)]2−\left[\mathrm{Pt}\left(\mathrm{CN}_{4}\right)\right]^{2-} is ____\_\_\_\_ .

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

In   [Ni(CO)4],Ni0:3d84s2\text{In\; } [\mathrm{Ni(CO)_4}], \quad \mathrm{Ni^0} : 3d^{8}4s^{2} Hybridisation   state:sp3\text{Hybridisation\; state} : sp^{3} Unpaired   electrons=0\text{Unpaired\; electrons} = 0 In   [NiCl4]2−,Ni2+:3d8\text{In\; } [\mathrm{NiCl_4}]^{2-}, \quad \mathrm{Ni^{2+}} : 3d^{8} Hybridisation   state:sp3\text{Hybridisation\; state} : sp^{3} Unpaired   electrons=2\text{Unpaired\; electrons} = 2 In   [PtCl4]2−,Pt2+:5d8\text{In\; } [\mathrm{PtCl_4}]^{2-}, \quad \mathrm{Pt^{2+}} : 5d^{8} Hybridisation   state:dsp2\text{Hybridisation\; state} : dsp^{2} Unpaired    electrons=0\text{Unpaired \; electrons} = 0 In   [Ni(CN)4]2−,Ni2+:3d8\text{In\; } [\mathrm{Ni(CN)_4}]^{2-}, \quad \mathrm{Ni^{2+}} : 3d^{8} Hybridisation   state:dsp2\text{Hybridisation\; state} : dsp^{2} Unpaired    electrons=0\text{Unpaired \; electrons} = 0 In   [Pt(NH3)2Cl2],Pt2+:5d8\text{In\; } [\mathrm{Pt(NH_3)_2Cl_2}], \quad \mathrm{Pt^{2+}} : 5d^{8} Hybridisation   state:dsp2\text{Hybridisation\; state} : dsp^{2} Unpaired   electrons=0\text{Unpaired \;electrons} = 0

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Exam
JEE Main 2026
Paper
23 January, Evening Shift
Subject
Chemistry
Chapter
Coordination Compounds
Topic
Theories of Bonding in Coordination Compounds

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