Identify a complex having a monodentate ligan | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) monodentate ligand is a ligand that binds to the central metal atom through only one donor atom. $CN^{-}$ (cyanide) is a monodentate ligand. $C_2O

Vedclass · Accepted Answer

Tetracyanonickelate$(II)$ ion

Vedclass · Answer

(A) monodentate ligand is a ligand that binds to the central metal atom through only one donor atom. $CN^{-}$ (cyanide) is a monodentate ligand. $C_2O_4^{2-}$ (oxalate) is a bidentate ligand. $en$ (ethylenediamine) is a bidentate ligand. $SCN^{-}$ (thiocyanato) is an ambidentate ligand,which acts as a monodentate ligand. In the given options,Tetracyanonickelate$(II)$ ion,$[Ni(CN)_4]^{2-}$,contains the monodentate ligand $CN^{-}$. Therefore,option $A$ is correct.

List-$I$ (Complex)	List-$II$ (Primary valency and Secondary valency)
$(A)$ $[Co(en)_2 Cl_2] Cl$	$(I)$ $3, 6$
$(B)$ $[Pt(NH_3)_2 Cl(NO_2)]$	$(II)$ $2, 4$
$(C)$ $Hg[Co(SCN)_4]$	$(III)$ $3, 4$
$(D)$ $[Mg(EDTA)]^{2-}$	$(IV)$ $2, 6$

Identify a complex having a monodentate ligand from the following:

Similar Questions

The coordination numbers of $Co$ and $Al$ in $[Co(Cl)(en)_2]Cl$ and $K_3[Al(C_2O_4)_3]$,respectively,are ($en =$ ethane$-1,2-$diamine).

The sum of the coordination number and the oxidation number of the metal $M$ in the complex $[M(en)_2(C_2O_4)]Cl$ (where $en$ is ethylenediamine) is:

Which of the following is a $\pi$-acid ligand?

Which of the following ligands is not ambidentate?
$(P) = NO_3^-$,$(Q) = NO_2^-$,$(R) = CN^-$,$(S) = SCN^-$

Vedclass Test Series

Exam Paper Generator

Online Exam Module