The coordination number of Th in $\mathrm{K}_{4}\left[\mathrm{Th}\left(\mathrm{C}_{2}…

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 $)$
  1. 14
  2. 6
  3. 8
  4. 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 $=2 \times 4+2=10$

Asked in: JEE Main 2019 (11 Jan Shift 2)

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