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Coordination Compounds question

2023 · 8 Apr · Shift 2 · Q1
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Coordination Compounds question

2023 · 8 Apr · Shift 2 · Q1

JEE MainChemistryCoordination CompoundsMCQ+4 / −1

Match List I with List II

LIST I
Coordination Complex		 LIST II
Number of unpaired electrons
A. [Cr(CN)6]3−\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}[Cr(CN)6​]3− I. 0
B. [Fe(H2O)6]2+\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}[Fe(H2​O)6​]2+ II. 3
C. [Co(NH3)6]3+\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}[Co(NH3​)6​]3+ III. 2
D. [Ni(NH3)6]2+\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}[Ni(NH3​)6​]2+ IV. 4

Choose the correct answer from the options given below:

  1. A
    A-IV, B-III, C-II, D-I
  2. B
    A-II, B-I, C-IV, D-III
  3. C
    A-II, B-IV, C-I, D-III
  4. D
    A-III, B-IV, C-I, D-II
View written solutionFree

Correct answer: C

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

  • A. [Cr(CN)6]3−\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}[Cr(CN)6​]3−

    Let oxidation state of Cr be xxx. x+6(−1)=−3⇒x=+3x+6(-1)=-3 \Rightarrow x=+3x+6(−1)=−3⇒x=+3 So, Cr3+\mathrm{Cr}^{3+}Cr3+ has configuration: Cr:[Ar]3d54s1⇒Cr3+:3d3\mathrm{Cr}: [\mathrm{Ar}]3d^54s^1 \Rightarrow \mathrm{Cr}^{3+}: 3d^3Cr:[Ar]3d54s1⇒Cr3+:3d3 For d3d^3d3, number of unpaired electrons is always 333.

    So, A →\to→ II.

  • B. [Fe(H2O)6]2+\left[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6\right]^{2+}[Fe(H2​O)6​]2+

    Water is neutral, so Fe is +2+2+2. Fe2+:3d6\mathrm{Fe}^{2+}: 3d^6Fe2+:3d6 H2O\mathrm{H_2O}H2​O is a weak field ligand, so this is high-spin octahedral. High-spin d6d^6d6 configuration has 444 unpaired electrons.

    So, B →\to→ IV.

  • C. [Co(NH3)6]3+\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}[Co(NH3​)6​]3+

    NH3\mathrm{NH_3}NH3​ is neutral, so Co is +3+3+3. Co3+:3d6\mathrm{Co}^{3+}: 3d^6Co3+:3d6 In Co3+\mathrm{Co}^{3+}Co3+ complexes, NH3\mathrm{NH_3}NH3​ causes sufficiently large splitting; this is low-spin octahedral. Low-spin d6d^6d6 configuration is: t2g6eg0t_{2g}^6 e_g^0t2g6​eg0​ Hence, unpaired electrons =0=0=0.

    So, C →\to→ I.

  • D. [Ni(NH3)6]2+\left[\mathrm{Ni}(\mathrm{NH}_3)_6\right]^{2+}[Ni(NH3​)6​]2+

    Ni2+:3d8\mathrm{Ni}^{2+}: 3d^8Ni2+:3d8 In octahedral field, d8d^8d8 configuration is: t2g6eg2t_{2g}^6 e_g^2t2g6​eg2​ Hence, number of unpaired electrons =2=2=2.

    So, D →\to→ III.

  1. Final matching

A→II,B→IV,C→I,D→IIIA \to II, \quad B \to IV, \quad C \to I, \quad D \to IIIA→II,B→IV,C→I,D→III

  1. Compare with options

This corresponds to Option C.

  1. Comparison with stored correct answer

Stored correct answer = C

So, the derived answer agrees with the stored answer.

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