Q1 / 90 · JEE Main 2019 (11 Jan Shift 2) · Chemistry

The coordination number of Th in $\mathrm{K}_{4}\left[\mathrm{Th}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]$ is: $\left(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}=\right.$ oxalato $)$

  • A
    14
  • B
    6
  • C
    8
  • Correct answer
    D
    10

Solution

$\mathrm{K}_{4}\left[\mathrm{Th}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]$ $\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ (oxalato) : bidentate ligand $\mathrm{H}_{2} \mathrm{O}$ (aqua): monodentate $\therefore$ Co-ordination no. of Th $=2 \times 4+2=10$

Hence, the correct answer is option: 'D'.

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