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Coordination Compounds question

2023 · 31 Jan · Shift 1 · Q13
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  5. /2023 · 31 Jan · Shift 1 · Q13

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
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