Which one of the following is an ambidentate ligand?

Year	2024	Total
Questions	3	3

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2024

Total

Questions

Solution & Explanation

Core Logic

An ambidentate ligand is a unidentate ligand that has more than one potential donor atom but binds to the central metal atom through only one donor atom at a time. (1) Ethylenediaminetetraacetate ion ( $EDTA⁴⁻$ $EDTA^{4-}$) is a hexadentate ligand. (2) Oxalate ( $C₂O₄²⁻$ $C_2O_4^{2-}$) is a didentate ligand. (3) Ethane-1,2-diamine ( $en$ $en$) is a didentate ligand. (4) Thiocyanate ( $SCN^-$ $SCN^-$) is a unidentate ligand that can bind through either the Sulphur atom ( $M-SCN$ $M-SCN$, thiocyanato) or the Nitrogen atom ( $M-NCS$ $M-NCS$, isothiocyanato). Therefore, it is an ambidentate ligand.

Step 1: Final Conclusion

Thiocyanate is an ambidentate ligand.

Pattern Recognition

Common ambidentate ligands are $NO₂^-$ $NO_2^-$, $SCN^-$ $SCN^-$, and $CN^-$ $CN^-$. They exhibit linkage isomerism in coordination complexes.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

More Coordination Compounds Previous-Year Questions

Q82 neet_2026_03_may_morning Geometry of Complexes

Match List I with List II :

List I (Complex/ion)	List II (Shape/geometry)
A. $[Pt(Cl₂)(NH₃)₂]$ $[Pt(Cl_2)(NH_3)_2]$	(I) Octahedral
B. $[Co(NH₃)₆]Cl₃$ $[Co(NH_3)_6]Cl_3$	(II) Trigonal bipyramidal
C. $[NiCl₄]²⁻$ $[NiCl_4]^{2-}$	(III) Square planar
D. $[Fe(CO)₅]$ $[Fe(CO)_5]$	(IV) Tetrahedral

Choose the correct answer from the options given below :

A. A-I, B-III, C-IV, D-II
B. A-III, B-IV, C-I, D-II
C. A-IV, B-I, C-III, D-II
D. A-III, B-I, C-IV, D-II

Solution

Core Logic

(A) $[Pt(Cl₂)(NH₃)₂]$ $[Pt(Cl_2)(NH_3)_2]$: Pt(II) is a $5d⁸$ $5d^8$ system. $4d⁸$ $4d^8$ and $5d⁸$ $5d^8$ complexes are almost always square planar with $dsp²$ $dsp^2$ hybridization regardless of ligand strength. (Matches III) (B) $[Co(NH₃)₆]Cl₃$ $[Co(NH_3)_6]Cl_3$: Co(III) with Coordination Number 6. Has $d²sp³$ $d^2sp^3$ hybridization, making it octahedral. (Matches I) (C) $[NiCl₄]²⁻$ $[NiCl_4]^{2-}$: Ni(II) is a $3d⁸$ $3d^8$ system. With weak field ligand $Cl^-$ $Cl^-$, it undergoes $sp³$ $sp^3$ hybridization, forming a tetrahedral geometry. (Matches IV) (D) $[Fe(CO)₅]$ $[Fe(CO)_5]$: Fe(0) is a $3d⁸ 4s²$ $3d^8 4s^2$ system. Under the influence of strong field ligand CO, electrons pair up to form $3d¹⁰$ $3d^{10}$, leaving $4s$ $4s$ and $4p$ $4p$ empty for $dsp³$ $dsp^3$ hybridization. Shape is trigonal bipyramidal. (Matches II)

Step 1: Final Match

A $arrow$ $\rightarrow$ III B $arrow$ $\rightarrow$ I C $arrow$ $\rightarrow$ IV D $arrow$ $\rightarrow$ II

Pattern Recognition

Pt(II) and Pd(II) complexes with CN=4 are universally square planar. $Ni²⁺$ $Ni^{2+}$ with weak ligands ( $Cl^-$ $Cl^-$) is tetrahedral, but with strong ligands ( $CN^-$ $CN^-$) is square planar. Fe with 5 CO ligands is the classic $dsp³$ $dsp^3$ trigonal bipyramidal example.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q90 neet_2026_03_may_morning Isomerism in Coordination Compounds

Match List I with List II :

List-I (Complex)	List-II (Type of isomerism)
A. $[Pt(NH₃)₂Cl₂]$ $[Pt(NH_3)_2Cl_2]$	I. Optical
B. $[Co(en)₃]³⁺$ $[Co(en)_3]^{3+}$	II. Solvate
C. $[Co(NH₃)₅NO₂]Cl₂$ $[Co(NH_3)_5NO_2]Cl_2$	III. Geometrical
D. $[Cr(H₂O)₆]Cl₃$ $[Cr(H_2O)_6]Cl_3$	IV. Linkage

Choose the correct answer from the options given below :

A. A-III, B-I, C-II, D-IV
B. A-I, B-III, C-II, D-IV
C. A-III, B-I, C-IV, D-II
D. A-II, B-IV, C-III, D-I

Solution

Core Logic

(A) $[Pt(NH₃)₂Cl₂]$ $[Pt(NH_3)_2Cl_2]$: Square planar complex of $MA₂B₂$ $MA_2B_2$ type. Forms strictly cis and trans isomers, thus exhibiting Geometrical isomerism. (Matches III) (B) $[Co(en)₃]³⁺$ $[Co(en)_3]^{3+}$: Octahedral complex with three bidentate ligands ( $M(AA)₃$ $M(AA)_3$ type). It lacks a plane of symmetry, so it exhibits Optical isomerism (forms non-superimposable mirror images). (Matches I) (C) $[Co(NH₃)₅NO₂]Cl₂$ $[Co(NH_3)_5NO_2]Cl_2$: Contains $NO₂^-$ $NO_2^-$, an ambidentate ligand that can bind through N (nitro) or O (nitrito). This exhibits Linkage isomerism. (Matches IV) (D) $[Cr(H₂O)₆]Cl₃$ $[Cr(H_2O)_6]Cl_3$: Can exchange water molecules with counter ions (e.g., forming $[Cr(H₂O)₅Cl]Cl₂ \cdot H₂O$ $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$). This exhibits Hydration or Solvate isomerism. (Matches II)

Step 1: Final Match

A $arrow$ $\rightarrow$ III B $arrow$ $\rightarrow$ I C $arrow$ $\rightarrow$ IV D $arrow$ $\rightarrow$ II

Pattern Recognition

Ambidentate ligands ( $NO₂, SCN$ $NO_2, SCN$) = Linkage isomerism. Exchangeable water molecules = Solvate isomerism. $M(AA)₃$ $M(AA)_3$ bidentate chelates = Optical isomerism. $MA₂B₂$ $MA_2B_2$ square planar = Geometrical isomerism.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

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Pattern Recognition

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Which one of the following is an ambidentate ligand?

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Coordination Compounds Previous-Year Questions

Solution

Core Logic

Step 1: Final Match

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Match

Pattern Recognition

Chapter Mix

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