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Coordination Compounds question

2023 · 31 Jan · Shift 1 · Q13
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Coordination Compounds question

2023 · 31 Jan · Shift 1 · Q13

JEE MainChemistryCoordination CompoundsMCQ+4 / −1
Cobalt chloride when dissolved in water forms pink colored complex X‾\underline{\mathrm{X}}X​ which has octahedral geometry. This solution on treating with conc HCl\mathrm{HCl}HCl forms deep blue complex, Y‾\underline{\mathrm{Y}}Y​ which has a Z‾\underline{\mathrm{Z}}Z​ geometry. X,Y\mathrm{X}, \mathrm{Y}X,Y and Z\mathrm{Z}Z, respectively, are
  1. A
    X=[Co(H2O)6]2+,Y=[CoCl4]2−,Z=\mathrm{X}=\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}, \mathrm{Y}=\left[\mathrm{CoCl}_{4}\right]^{2-}, \mathrm{Z}=X=[Co(H2​O)6​]2+,Y=[CoCl4​]2−,Z= Tetrahedral
  2. B
    X=[Co(H2O)6]2+,Y=[CoCl6]3−,Z=\mathrm{X}=\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}, \mathrm{Y}=\left[\mathrm{CoCl}_{6}\right]^{3-}, \mathrm{Z}=X=[Co(H2​O)6​]2+,Y=[CoCl6​]3−,Z= Octahedral
  3. C
    X=[Co(H2O)4Cl2]+,Y=[CoCl4]2−,Z=\mathrm{X}=\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]^{+}, \mathrm{Y}=\left[\mathrm{CoCl}_{4}\right]^{2-}, \mathrm{Z}=X=[Co(H2​O)4​Cl2​]+,Y=[CoCl4​]2−,Z= Tetrahedral
  4. D
    X=[Co(H2O)6]3+,Y=[CoCl6]3−,Z=\mathrm{X}=\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}, \mathrm{Y}=\left[\mathrm{CoCl}_{6}\right]^{3-}, \mathrm{Z}=X=[Co(H2​O)6​]3+,Y=[CoCl6​]3−,Z= Octahedral
View written solutionFree

Correct answer: A

  1. Identify the pink octahedral complex formed when cobalt chloride dissolves in water

When CoCl2\mathrm{CoCl_2}CoCl2​ dissolves in water, the cobalt(II) ion gets hydrated to form the hexaaqua complex:

[Co(H2O)6]2+\mathrm{[Co(H_2O)_6]^{2+}}[Co(H2​O)6​]2+

This complex is pink in color and has octahedral geometry.

So,

X=[Co(H2O)6]2+X = \mathrm{[Co(H_2O)_6]^{2+}}X=[Co(H2​O)6​]2+
  1. Effect of concentrated HCl\mathrm{HCl}HCl

On adding concentrated HCl\mathrm{HCl}HCl, the chloride ion concentration increases greatly. Due to ligand substitution, water molecules are replaced by chloride ions:

[Co(H2O)6]2++4Cl−⇌[CoCl4]2−+6H2O\mathrm{[Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O}[Co(H2​O)6​]2++4Cl−⇌[CoCl4​]2−+6H2​O

The complex formed is:

Y=[CoCl4]2−Y = \mathrm{[CoCl_4]^{2-}}Y=[CoCl4​]2−

This complex is deep blue in color.

  1. Geometry of [CoCl4]2−\mathrm{[CoCl_4]^{2-}}[CoCl4​]2−

The complex [CoCl4]2−\mathrm{[CoCl_4]^{2-}}[CoCl4​]2− is a four-coordinate cobalt(II) complex with weak-field ligand Cl−\mathrm{Cl^-}Cl−, and it has tetrahedral geometry.

So,

Z=TetrahedralZ = \text{Tetrahedral}Z=Tetrahedral
  1. Match with options
  • Option A: X=[Co(H2O)6]2+,  Y=[CoCl4]2−,  Z=X=\mathrm{[Co(H_2O)_6]^{2+}},\; Y=\mathrm{[CoCl_4]^{2-}},\; Z=X=[Co(H2​O)6​]2+,Y=[CoCl4​]2−,Z= Tetrahedral ✔️
  • Option B: Wrong blue complex
  • Option C: Wrong initial pink octahedral complex
  • Option D: Wrong oxidation state

Therefore, the correct option is A.

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