JEE MAIN - Chemistry (2019 - 11th January Evening Slot - No. 7)
The coordination number of Th in K4[Th(C2O4)4(OH2)2] is :
(C2O$${_4^{2 - }}$$ = Oxalato)
(C2O$${_4^{2 - }}$$ = Oxalato)
14
10
8
6
Explanation
Oxalato (C2O42–) is a bidentate and H2O is unidentate
ligand.
4C2O42– creates 8 covalent bonds.
2H2O creates 2 covalent bonds.
$$ \therefore $$ Around Th 10 coordinate covalent bonds will be present.
4C2O42– creates 8 covalent bonds.
2H2O creates 2 covalent bonds.
$$ \therefore $$ Around Th 10 coordinate covalent bonds will be present.
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