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

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Coordination Compounds

The coordination number of Th in

K_{4} [Th (C_{2} O_{4})_{4} (OH_{2})_{2}]

is:

(C_{2} O_{4}^{2 -} = O

xalato

)

(A)14
(B)6
(C)8
(D)10✓ Correct

Answer: (D)

Solution

K X_{4} [Th (C X_{2} O X_{4}) X_{4} (OH X_{2}) X_{2}]

each oxalate is bidentate, binding through two oxygen atoms, so the four oxalates occupy

4 \times 2 = 8

sites, while the two water molecules are monodentate and add

2

more. The coordination number of thorium is therefore

8 + 2 = 10

, which its large size readily accommodates.

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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