Match List I with List II: List I (Complex/ion) List II (Shape/geometry) A. [Pt(Cl2)(NH3)2] (I) Octahedral B. [Co(NH3)6]Cl3 (II) Trigonal bipyramidal C. [NiCl4]2- (III) Square planar D. [Fe(CO)5] (IV) Tetrahedral Choose the correct answer from the options given below:

A. [Pt(Cl2)(NH3)2] dsp2 Square planar (III). B. [Co(NH3)6]Cl3 d2sp3 Octahedral (I). C. [NiCl4]2- sp3 Tetrahedral (IV). D. [Fe(CO)5] dsp3 Trigonal bipyramidal (II). Matches: A-III, B-I, C-IV, D-II. Platinum(II) complexes are square planar; Cobalt(III) hexammine is octahedral. Class 12 Chemistry: Coordination Compounds

Match List I with List II: List I (Complex/ion) List II (Shape/geometry) A. [Pt(Cl2)(NH3)2] (I) Octahedral B. [Co(NH3)6]Cl3 (II) Trigonal bipyramidal C. [NiCl4]2- (III) Square planar D. [Fe(CO)5] (IV)

A. [Pt(Cl2)(NH3)2] dsp2 Square planar (III). B. [Co(NH3)6]Cl3 d2sp3 Octahedral (I). C. [NiCl4]2- sp3 Tetrahedral (IV). D. [Fe(CO)5] dsp3 Trigonal bipyramidal (II). Matches: A-III, B-I, C-IV, D-II. Platinum(II) complexes are square planar; Cobalt(III) hexammine is octahedral. Class 12 Chemistry: Coordination Compounds

Match List I with List II: List I (Complex/ion) List II (Shape/geometry) A. [Pt(Cl2)(NH3)2] (I) Octahedral B. [Co(NH3)6]Cl3 (II) Trigonal bipyramidal C. [NiCl4]2- (III) Square planar D. [Fe(CO)5] (IV) Tetrahedral Choose the correct answer from the options given below:

A. [Pt(Cl2)(NH3)2] dsp2 Square planar (III). B. [Co(NH3)6]Cl3 d2sp3 Octahedral (I). C. [NiCl4]2- sp3 Tetrahedral (IV). D. [Fe(CO)5] dsp3 Trigonal bipyramidal (II). Matches: A-III, B-I, C-IV, D-II. Platinum(II) complexes are square planar; Cobalt(III) hexammine is octahedral. Class 12 Chemistry: Coordination Compounds

Match List I with List II: List I (Complex/ion) List II (Shape/geometry) A. [Pt(Cl2)(NH3)2] (I) Octahedral B. [Co(NH3)6]Cl3 (II) Trigonal bipyramidal C. [NiCl4]2- (III) Square planar D. [Fe(CO)5] (IV)

A. [Pt(Cl2)(NH3)2] dsp2 Square planar (III). B. [Co(NH3)6]Cl3 d2sp3 Octahedral (I). C. [NiCl4]2- sp3 Tetrahedral (IV). D. [Fe(CO)5] dsp3 Trigonal bipyramidal (II). Matches: A-III, B-I, C-IV, D-II. Platinum(II) complexes are square planar; Cobalt(III) hexammine is octahedral. Class 12 Chemistry: Coordination Compounds

Match List I with List II: List I (Complex/ion) List II (Shape/geometry) A. [Pt(Cl2)(NH3)2] (I) Octahedral B…

Year	2024	Total
Questions	3	3

Year

2024

Total

Questions

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

List I (Complex/ion)

List II (Shape/geometry)

[Pt(Cl₂)(NH₃)₂]

(I) Octahedral

[Co(NH₃)₆]Cl₃

(II) Trigonal bipyramidal

[NiCl₄]²⁻

(III) Square planar

[Fe(CO)₅]

(IV) Tetrahedral

Solution & Explanation

Related Formula

dsp² arrow Square Planar, d²sp³ arrow Octahedral, sp³ arrow Tetrahedral, dsp³ arrow Trigonal Bipyramidal

Core Logic

A. $[Pt(Cl₂)(NH₃)₂] dsp²$ Square planar (III). B. $[Co(NH₃)₆]Cl₃ d²sp³$ Octahedral (I). C. $[NiCl₄]²⁻ sp³$ $[NiCl_4]^{2-} \implies sp^3 \implies$ Tetrahedral (IV). D. $[Fe(CO)₅] dsp³$ $[Fe(CO)_5] \implies dsp^3 \implies$ Trigonal bipyramidal (II).

Step 1: Conclusion

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

Pattern Recognition

Platinum(II) complexes are square planar; Cobalt(III) hexammine is octahedral.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

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Q84 neet_2024_05_may_morning Types of Ligands - Ambidentate Ligands

Which one of the following is an ambidentate ligand?

A. (1) Ethylenediaminetetraacetate ion
B. (2) Oxalate
C. (3) Ethane-1,2-diamine
D. (4) Thiocyanate

Solution

Related Formula

Ambidentate ligand: Contains 2 different donor atoms, but coordinates via only one at a time.

Core Logic

Thiocyanate ion ( $SCN^-$ $SCN^-$) can bind through either nitrogen ( $NCS^-$ $NCS^-$) or sulfur ( $SCN^-$ $SCN^-$). Hence, it is an ambidentate ligand.

Step 1: Conclusion

Ambidentate ligand is Thiocyanate.

Pattern Recognition

Classic ambidentate ligands: $NO₂^-, SCN^-, CN^-$ $NO_2^-, SCN^-, CN^-$.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q90 neet_2024_05_may_morning Isomerism in Coordination Complexes

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. (1) A-III, B-I, C-II, D-IV
B. (2) A-I, B-III, C-II, D-IV
C. (3) A-III, B-I, C-IV, D-II
D. (4) A-II, B-IV, C-III, D-I

Solution

Related Formula

Square Planar MA₂B₂ arrow Geometrical, [M(en)₃] arrow Optical

Core Logic

A. $[Pt(NH₃)₂Cl₂]$ $[Pt(NH_3)_2Cl_2]$ shows cis-trans geometrical isomerism (III). B. $[Co(en)₃]³⁺$ $[Co(en)_3]^{3+}$ chiral tris-bidentate complex exhibits optical isomerism (I). C. $[Co(NH₃)₅NO₂]Cl₂$ $[Co(NH_3)_5NO_2]Cl_2$ contains ambidentate $NO₂^-$ $NO_2^-$ ligand showing linkage isomerism (IV). D. $[Cr(H₂O)₆]Cl₃$ $[Cr(H_2O)_6]Cl_3$ exhibits hydrate/solvate isomerism (II).

Step 1: Conclusion

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

Pattern Recognition

Cisplatin $arrow$ $\rightarrow$ Geometrical; $[M(en)₃] arrow$ $[M(en)_3] \rightarrow$ Optical; $NO₂^- arrow$ $NO_2^- \rightarrow$ Linkage; $H₂O arrow$ $H_2O \rightarrow$ Solvate.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Solution & Explanation

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

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Solution

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

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Solution & Explanation

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Step 1: Conclusion

Pattern Recognition

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