JEE Main 2023 Co – ordination Compounds with Solution

Question :

The set which does not have ambidentate ligand(s) is

Options :

(a) C₂O₄^2-, ethylene diamine, H₂O

(b) EDTA^4-, NCS^–, C₂O₄^2-

(c) NO₂^–, C₂O₄^2-, EDTA^4-

(d) C₂O₄^2-, NO₂^–, NCS^–

JEE Main 2023 Co – ordination Compounds with Solution

Answer :

Correct answer : Option (a)

Ambidentate ligand :

• The ligand which have donor atoms but in forming complexes only one donor atom is attached to the metal atom at a given time. Such ligands are called ambidentate ligands.

Examples :

• M $\leftarrow$ NO₂ [ Nitrito – N ] and M $\leftarrow$ ONO [ Nitrito – O ]
• M $\leftarrow$ CN [ Cyano – C ] and M $\leftarrow$ NC [ Isocyano ]

In above question

C₂O₄^2- :

• C₂O₄^2- is bidentate which have two donor atoms and have the ability to link with central metal at two positions

en ( ethylene diamine ) :

• en ( ethylene diamine ) is bidentate which have two donor atoms of Nitrogen and have the ability to link with central metal atom

H₂O :

• H₂O is Mono dentate which have one donor atom that donates only one electron pair to central metal atom

hence , in Option (a) No ambidentate ligand is present

Concept Involved :

• Ambidentate Ligands (Coordination Chemistry)

Examples for Practice :

Find the central atom of ( i ) oxidation number ( ii) Co ordination number (iii) Effective atomic number

1. K₄[ Fe(CN)₆ ]

i) Oxidation state of Central metal atom :

4 [ +1] + x + 6( – 1) = 0

x – 2 = 0

x = 2

ii) Co ordination number of central metal atom :

Co – ordination number : Number of ligands x denticity = 6×1 = 6

iii) Effective atomic number [ EAN ] :

= Atomic number of Central metal atom – oxidation number + 2 x co ordination number

= 26 – 2 + 2 x 6

= 24 + 12 = 36 [ stable ]

2 .K₃ [ CO ( C₂O₄)₃ ]

i) Oxidation state of central metal atom

3 [ +1] + x + 3( -2) = 0

3 + x – 6 = 0

x = +3

ii) Co – ordination number of central metal atom :

Co – ordination number : Number of ligands x denticity = 3 x 2

iii) Effective atomic number [ EAN ] of K₃ [ CO ( C₂O₄)₃ ] :

= Atomic number of Central metal atom – oxidation number + 2 x co ordination number

= 27 – 3 + 2 x 6 = 36 [ stable ]

3. [ Fe ( CN₆]^-3

i) Oxidation state of Central metal atom :

x + 6 ( – 1) = – 3

x – 6 = -3

x = +3

ii) Co – ordination number of central metal atom :

Co – ordination number : Number of ligands x denticity = 6 x 1

iii) Effective atomic number [ EAN ] of [ Fe ( CN₆]^-3

= Atomic number of Central metal atom – oxidation number + 2 x co ordination number

= 26 – 3 + 2 x 6

= 23 + 12 = 35 [ stable ]

4. [ CO (NH₃)₆]Cl₃

i) Oxidation state of Central metal atom :

= x + 6(0) + 3(-1) = 0

x = +3

ii) Co – ordination number of central metal atom :

Co – ordination number : Number of ligands x denticity = 6 x 1 = 6

iii) Effective atomic number [ EAN ] of [ CO (NH₃)₆]Cl₃

= Atomic number of Central metal atom – oxidation number + 2 x co ordination number

27 – 3 + 2 x 6 = 24 + 12 = 36

5. The oxidation number of Co in [ Co(en)₃]₂(SO₄)₃ is

Options :

(a) +2

(b) +4

(c) +3

(d) +6

Answer :

oxidation number of Co in [ Co(en)₃]₂(SO₄)₃

2x + 3(0) + 3( -2) = 0

2x = 6

x = +3

Correct Answer : Option (c)

6. The oxidation state of Cr in [ Cr(NH₃)₄Cl₂]⁺

Options :

(a) +1

(b) 0

(c) +3

(d) +2

Answer :

The oxidation state of Cr in [ Cr(NH₃)₄Cl₂]⁺

x + 4(0) + 2(-1) = +1

x = +3

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