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Coordination Compounds question

2022 · 28 Jun · Shift 1 · Q4
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Coordination Compounds question

2022 · 28 Jun · Shift 1 · Q4

JEE MainChemistryCoordination CompoundsMCQ+4 / −1
Given below are two statements : Statement I : [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2− is square planar and diamagnetic complex, with dsp2 hybridization for Ni but [Ni(CO)4][Ni(CO)_4][Ni(CO)4​] is tetrahedral, paramagnetic and with sp3-hybridication for Ni. Statement II : [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2− and [Ni(CO)4][Ni(CO)_4][Ni(CO)4​] both have same d-electron configuration have same geometry and are paramagnetic. In light the above statements, choose the correct answer from the options given below :
  1. A
    Both Statement I and Statement II are true.
  2. B
    Both Statement I and Statement II are false.
  3. C
    Statement I is correct but Statement II is false.
  4. D
    Statement I is false but Statement II is correct.
View written solutionFree

Correct answer: B

  1. Analyze Statement I

    We examine both complexes mentioned.

    (a) [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2−

    • Oxidation state of Ni: x+4(−1)=−2  ⟹  x=+2x + 4(-1) = -2 \implies x = +2x+4(−1)=−2⟹x=+2
    • So, Ni2+Ni^{2+}Ni2+ has configuration: Ni:[Ar]3d84s2⇒Ni2+:[Ar]3d8Ni: [Ar]3d^8 4s^2 \Rightarrow Ni^{2+}: [Ar]3d^8Ni:[Ar]3d84s2⇒Ni2+:[Ar]3d8
    • CN−CN^-CN− is a strong field ligand, so pairing occurs.
    • For d8d^8d8 with strong field in coordination number 4, the complex becomes square planar with dsp2dsp^2dsp2 hybridization.
    • All electrons are paired, so it is diamagnetic.

    Hence this part is correct.

    (b) [Ni(CO)4][Ni(CO)_4][Ni(CO)4​]

    • CO is a neutral ligand, so oxidation state of Ni is 000.
    • Configuration of Ni0Ni^0Ni0: [Ar]3d84s2[Ar]3d^8 4s^2[Ar]3d84s2
    • In [Ni(CO)4][Ni(CO)_4][Ni(CO)4​], nickel uses sp3sp^3sp3 hybridization and forms a tetrahedral complex.
    • All electrons become paired in the metal-ligand bonding situation, so [Ni(CO)4][Ni(CO)_4][Ni(CO)4​] is diamagnetic, not paramagnetic.

    Therefore, Statement I says "[Ni(CO)4][Ni(CO)_4][Ni(CO)4​] is tetrahedral, paramagnetic and with sp3sp^3sp3 hybridization".

    • Tetrahedral: true
    • sp3sp^3sp3: true
    • Paramagnetic: false

    So Statement I is false.

  2. Analyze Statement II

    Statement II says: [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2− and [Ni(CO)4][Ni(CO)_4][Ni(CO)4​] both have same ddd-electron configuration, same geometry and are paramagnetic.

    (a) [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2−

    • Oxidation state of Ni: x+4(−1)=−2  ⟹  x=+2x + 4(-1) = -2 \implies x = +2x+4(−1)=−2⟹x=+2
    • So, Ni2+Ni^{2+}Ni2+ is 3d83d^83d8.
    • Cl−Cl^-Cl− is a weak field ligand, so no pairing occurs.
    • Hence the complex is tetrahedral and paramagnetic.

    (b) [Ni(CO)4][Ni(CO)_4][Ni(CO)4​]

    • Ni oxidation state is 000.
    • Thus it is not d8d^8d8 in the same sense as Ni2+Ni^{2+}Ni2+; the metal is Ni0Ni^0Ni0 with valence configuration 3d84s23d^8 4s^23d84s2.
    • Geometry is tetrahedral.
    • But it is diamagnetic, not paramagnetic.

    So in Statement II:

    • same geometry: true
    • same ddd-electron configuration: commonly treated as not same oxidation-state electronic situation; in standard coordination chemistry comparison, [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2− has Ni2+(d8)Ni^{2+}(d^8)Ni2+(d8) while [Ni(CO)4][Ni(CO)_4][Ni(CO)4​] has Ni0Ni^0Ni0
    • both paramagnetic: false

    Therefore Statement II is false.

  3. Final conclusion

    • Statement I: False
    • Statement II: False

    Therefore, the correct option is: B\boxed{B}B​

  4. Comparison with stored answer

    Stored correct answer: B

    My derived answer matches the stored answer.

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