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Coordination Compounds question

2023 · 25 Jan · Shift 1 · Q21
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Coordination Compounds question

2023 · 25 Jan · Shift 1 · Q21

JEE MainChemistryCoordination CompoundsNumerical+4 / −1
The number of paramagnetic species from the following is ‾\underline{\hspace{2cm}}​. [Ni(CN)4]2−,[Ni(CO)4],[NiCl4]2−[Fe(CN)6]4−,[Cu(NH3)4]2+[Fe(CN)6]3− and [Fe(H2O)6]2+\mathrm{{[Ni{(CN)_4}]^{2 - }},[Ni{(CO)_4}],{[NiC{l_4}]^{2 - }}}\mathrm{{[Fe{(CN)_6}]^{4 - }},{[Cu{(N{H_3})_4}]^{2 + }}}\mathrm{{[Fe{(CN)_6}]^{3 - }}\,and\,{[Fe{({H_2}O)_6}]^{2 + }}}[Ni(CN)4​]2−,[Ni(CO)4​],[NiCl4​]2−[Fe(CN)6​]4−,[Cu(NH3​)4​]2+[Fe(CN)6​]3−and[Fe(H2​O)6​]2+
Numerical answer
View written solutionFree

Correct answer: 4

  1. Find oxidation state and ddd-electron count for each complex

We check whether each species has unpaired electrons.

  1. [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2−
  • Let oxidation state of Ni be xxx.
  • x+4(−1)=−2⇒x=+2x+4(-1)=-2 \Rightarrow x=+2x+4(−1)=−2⇒x=+2
  • So Ni is Ni2+Ni^{2+}Ni2+, i.e. 3d83d^83d8.
  • CN−CN^-CN− is a strong field ligand, so this complex is square planar for d8d^8d8 Ni(II).
  • Square planar d8d^8d8 has all electrons paired.

Therefore, [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2− is diamagnetic.

  1. [Ni(CO)4][Ni(CO)_4][Ni(CO)4​]
  • CO is neutral, so Ni oxidation state is 000.
  • Ni0:3d84s2Ni^0 : 3d^8 4s^2Ni0:3d84s2, effectively d10d^{10}d10 in the complex.
  • Tetrahedral d10d^{10}d10 has no unpaired electrons.

Therefore, [Ni(CO)4][Ni(CO)_4][Ni(CO)4​] is diamagnetic.

  1. [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2−
  • x+4(−1)=−2⇒x=+2x+4(-1)=-2 \Rightarrow x=+2x+4(−1)=−2⇒x=+2
  • So Ni is Ni2+Ni^{2+}Ni2+, i.e. d8d^8d8.
  • Cl−Cl^-Cl− is a weak field ligand, giving tetrahedral geometry.
  • Tetrahedral d8d^8d8 is high spin with 2 unpaired electrons.

Therefore, [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2− is paramagnetic.

  1. [Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4−
  • x+6(−1)=−4⇒x=+2x+6(-1)=-4 \Rightarrow x=+2x+6(−1)=−4⇒x=+2
  • So Fe is Fe2+Fe^{2+}Fe2+, i.e. d6d^6d6.
  • CN−CN^-CN− is strong field, so octahedral low spin complex.
  • Configuration: t2g6eg0t_{2g}^6 e_g^0t2g6​eg0​
  • No unpaired electrons.

Therefore, [Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4− is diamagnetic.

  1. [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+
  • NH3NH_3NH3​ is neutral, so Cu oxidation state is +2+2+2.
  • Cu2+Cu^{2+}Cu2+ is d9d^9d9.
  • A d9d^9d9 complex has 1 unpaired electron.

Therefore, [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+ is paramagnetic.

  1. [Fe(CN)6]3−[Fe(CN)_6]^{3-}[Fe(CN)6​]3−
  • x+6(−1)=−3⇒x=+3x+6(-1)=-3 \Rightarrow x=+3x+6(−1)=−3⇒x=+3
  • So Fe is Fe3+Fe^{3+}Fe3+, i.e. d5d^5d5.
  • CN−CN^-CN− is strong field, so octahedral low spin.
  • Configuration: t2g5eg0t_{2g}^5 e_g^0t2g5​eg0​
  • This has 1 unpaired electron.

Therefore, [Fe(CN)6]3−[Fe(CN)_6]^{3-}[Fe(CN)6​]3− is paramagnetic.

  1. [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}[Fe(H2​O)6​]2+
  • H2OH_2OH2​O is neutral, so Fe oxidation state is +2+2+2.
  • Fe2+Fe^{2+}Fe2+ is d6d^6d6.
  • H2OH_2OH2​O is a weak field ligand, so octahedral high spin.
  • Configuration: t2g4eg2t_{2g}^4 e_g^2t2g4​eg2​
  • This has 4 unpaired electrons.

Therefore, [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}[Fe(H2​O)6​]2+ is paramagnetic.

  1. Count paramagnetic species

Paramagnetic complexes are:

  • [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2−
  • [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}[Cu(NH3​)4​]2+
  • [Fe(CN)6]3−[Fe(CN)_6]^{3-}[Fe(CN)6​]3−
  • [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}[Fe(H2​O)6​]2+

Total number of paramagnetic species =4=4=4.

4\boxed{4}4​

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