Please refer to Coordination Compounds Class 12 Chemistry Important Questions with answers below. These solved questions for Chapter 9 Coordination Compounds in NCERT Book for Class 12 Chemistry have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these solved problems properly as these will help them to get better marks in your class tests and examinations. You will also be able to understand how to write answers properly. Revise these questions and answers regularly. We have provided Notes for Class 12 Chemistry for all chapters in your textbooks.

Important Questions Class 12 Chemistry Chapter 9 Coordination Compounds

All Coordination Compounds Class 12 Chemistry Important Questions provided below have been prepared by expert teachers of Standard 12 Chemistry. Please learn them and let us know if you have any questions.

Very Short Answer Questions :

Question. Write the state of hybridization, shape and IUPAC name of the complex [Co(NH₃)₆]³⁺.
(Atomic no. of Co = 27) 
Answer : Oxidation of cobalt in[Co(NH₃)₆]³⁺ is +3.

Question. Write the name, the structure and the magnetic behaviour of each one of the following complexes :
(i) [Pt(NH₃)Cl(NO₂)]
(ii) [Co(NH₃)₄Cl₂]Cl
(iii) Ni(CO)₄
(At. nos. Co = 27, Ni = 28, Pt = 78)
Answer : (i) [Pt(NH₃)₂Cl(NO₂)] :
Diamminechloridonitrito-N-platinum(II) It is square planar and diamagnetic.
(ii) [Co(NH₃)₄Cl₂]Cl : Tetraamminedichloridocobalt (III) chloride
It is octahedral and diamagnetic.
(iii) Ni(CO)₄ : Tetracarbonylnickel(0)
It is tetrahedral and diamagnetic

Question. What will be the correct order for the wave lengths of absorption in the visible region for the following :
[Ni(NO₂)₆]^4–, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer : All the complex ions are derived from Ni²⁺ ion with different ligands. From the position of the ligands in the spectrochemical series the order of field strength is
           H₂O < NH₃ < NO₂^–
That means Δ_o for NO₂^– is maximum and so it would absorb the radiation of shorter wavelength having high energy. So, the order of absorption of the correct wavelengths is
[Ni(H₂O)₆]²⁺    >    [Ni(NH₃)₆]²⁺+    >    [Ni(NO₂)₆]^4–
(λ about 700 nm)     (about 500 nm)         (< 500 nm)

Question. Draw the structures of :
(i) Ni(CO)₄
(ii) Fe(CO)₅
Answer :

Question. Name the following coordination entities and describe their structures.
(i) [Fe(CN)₆]^4–
(ii) [Cr(NH₃)₄Cl₂]⁺
(iii) [Ni(CN)₄]^2– 
Answer : (i) [Fe(CN)₆]^4– : Hexacyanidoferrate(II) ion Hybridisation – d²sp³
Structure : Inner orbital octahedral complex

Question. Describe for any two of the following complex ions, the type of hybridization, shape and magnetic property :
(i) [Fe(H₂O)₆]²⁺
[At. Nos. Fe = 26] 
Answer : 

Question. Write the hybridization and shape of the following complexes :
(i) [CoF₆]^3–
(ii) [Ni(CN)₄]^2–
(Atomic number : Co = 27, Ni = 28)
Answer : (i) Oxidation state of Co ion in [CoF₆]^3– is +3.

The complex ion has square planar geometry and is diamagnetic in nature.

Question. Write the IUPAC name and indicate the shape of the complex ion [Co(en)₂Cl(ONO)]⁺.
[At. no. Co = 27] 
Answer : IUPAC name : Chloridobis(ethylenediamine) nitrito-O-cobalt(III) ion. Shape is octahedral.

Question. Which of the following is more stable complex and why?
[Co(NH₃)₆]³⁺ and [Co(en)₃]³⁺ 
Answer : [Co(en)₃]³⁺  is more stable complex than [Co(NH₃)₆]³⁺ due to chelate eect as it forms rings.

Question. Describe the type of hybridization, shape and magnetic property of [Co(NH₃)₄Cl₂]⁺.
[Given : At. no. of Co = 27] 
Answer :

Question. State a reason for the following situation :
CO is a stronger complexing reagent than NH₃.
Answer : In CO both lone pair of electrons and vacant π* orbitals are present. Hence, it acts as electron pair σ donor as well as p acceptor by back bonding. Hence, M—CO bond is stronger.
M  ^π⇔_σ  CO
NH₃ is electron pair donor only. Accumulation of negative charge on the metal ion takes place, hence M—NH₃ bond is weaker.
M ← NH₃

Question. Write the state of hybridization, the shape and the magnetic behaviour of the following complex entities :
(i) [Cr(NH₃)₄Cl₂]Cl
(ii) [Co(en)₃]Cl₃
(iii) K₂[Ni(CN)₄] 
Answer : 

Short Answer Questions :

Question. Giving a suitable example, explain the following :
Crystal field splitting
Answer : The splitting of the degenerate d-orbitals into three orbitals of lower energy, t_2g set and two orbitals of higher energy eg set due to the interaction of ligand in an octahedral crystal field is known as crystal field splitting in an octahedral field.

Question. What is meant by crystal field splitting energy?
On the basis of crystal field theory, write the electronic configuration of d⁴ in terms of t_2g and eg in an octahedral field when
(i) Δ_o > P
(ii) Δ_o < P
Answer : (i) The difference of energy between two states of splitted d-orbitals is called crystal field splitting energy. It is denoted by Δ or 10 Dq.
For octahedral Δ_o, for tetrahedral it is Δ_t and for square planar Δ_sp.

Question. Write the name, the state of hybridization, the shape and the magnetic behaviour of the following complexes :
[CoCl₄]^2–, [Ni(CN)₄]^2–, [Cr(H₂O)₂(C₂O₄)₂]^–
(At. no. : Co = 27, Ni = 28, Cr = 24) 
Answer : [CoCl₄]^2– Tetrachloridocobaltate(II) ion
[Ni(CN)₄]^2– Tetracyanonickelate(II) ion
[Cr(H₂O)₂(C₂O₄)₂]^– Diaquadioxalatochromate(III) ion

Question. Out of NH₃ and CO, which ligand forms a more stable complex with a transition metal and why? 
Answer : In CO both lone pair of electrons and vacant π* orbitals are present. Hence, it acts as electron pair σ donor as well as p acceptor by back bonding. Hence, M—CO bond is stronger.
M  ^π⇔_σ  CO
NH₃ is electron pair donor only. Accumulation of negative charge on the metal ion takes place, hence M—NH₃ bond is weaker.
M ← NH₃

Question. Explain the following term giving a suitable example :
Crystal field splitting in an octahedral field.
Answer : The splitting of the degenerate d-orbitals into three orbitals of lower energy, t_2g set and two orbitals of higher energy eg set due to the interaction of ligand in an octahedral crystal field is known as crystal field splitting in an octahedral field.

Question. Write the IUPAC name, deduce the geometry and magnetic behaviour of the complex K₄[Mn(CN)₆].
[Atomic no. of Mn = 25] 
Answer : Mn atom (Z = 25)

IUPAC name : Potassium hexacyanomanganate (II)
Geometry : Octahedral
No. of unpaired electrons, n = 1
Magnetic behaviour : paramagnetic.

Question. Why is CO a stronger ligand than Cl–?
Answer : Because CO has vacant molecular orbitals with which it can form π-bond with metal through back donation.

Question. For the complex [Fe(en)₂Cl₂]Cl, identify the following :
(i) Oxidation number of iron
(ii) Hybrid orbitals and shape of the complex
(iii) Magnetic behaviour of the complex
(iv) Number of its geometrical isomers
(v) Whether there may be optical isomer also.
(vi) Name of the complex. 
Answer : (i) [Fe(en)₂Cl₂]Cl
x + 0 × 2 + (–1) × 2 + (–1) × 1 = 0      ∴      x = 3
Oxidation number of iron = 3
(ii) d²sp³ hybridisation and octahedral shape.
(iii) Paramagnetic due to presence of one unpaired electron.
(iv) 2, cis and trans isomers.
(v) cis-[Fe(en)₂Cl₂] has optical isomer.
(vi) dichloridobis(ethane-1,2-diamine)iron(III) chloride

Question. Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved :
(i) [CoF₄]^2–
(ii) [Cr(H₂O)₂(C₂O₄)₂]^–
(iii) [Ni(CO)₄]
(Atomic number : Co = 27, Cr = 24, Ni = 28)
Answer : 

Question. State a reason for each of the following situations :
(i) Co²⁺ is easily oxidised to Co³⁺ in presence of a strong ligand.
Answer : (i) In presence of strong field ligand Co(II) has electronic configuration t_2g⁶ e_g¹

It can easily lose one electron present in e_g orbital to give stable t_2g⁶ configuration. This is why Co²⁺ is easily oxidised to Co³⁺ in the presence of strong field ligand.

Question. Compare the following complexes with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units :
(i) [Ni(CN)₄]^2–
(ii) [NiCl₄]^2–
(iii) [CoF₆]^3–
[At. nos. : Ni = 28; Co = 27] 
Answer : 

Question. How is the stability of a co-ordination compound in solution decided? How is the dissociation constant of a complex defined? 
Answer : The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. The magnitude of the equilibrium constant for the association, quantitatively expresses the stability.
The instability constant or dissociation constant of coordination compound is defined as the reciprocal of the formation constant.

Question. Using valence bond theory, explain the geometry and magnetic behaviour of [Co(NH₃)₆]³⁺.
(At. no. of Co = 27) 
Answer : Oxidation of cobalt in[Co(NH₃)₆]³⁺ is +3.

Question. Explain the following :
(i) Low spin octahedral complexes of nickel are not known.
(ii) The p-complexes are known for transition elements only.
(iii) CO is a stronger ligand than NH₃ for many metals.
Answer : (i) Nickel forms octahedral complexes mainly in +2 oxidation state which has 3d⁸ configuration. In presence of strong field ligand also it has two unpaired electrons in eg orbital.

Hence, it does not form low spin octahedral complexes.
(ii) The transition metals/ions have empty d orbitals into which the electron pairs can be donated by ligands containing π electrons.
For example : CH₂=CH₂ and C₆H₆, C₅H₅^–.
(iii) Co is stronger ligand than NH₃ because CO has vacant molecular orbitals with which it can form π-bond with metal through back donation.

Question. What do you understand by stepwise stability constant and overall stability constant of a coordination compound? How are these two constants related? 
Answer : Stability constant of each step of complex formation reaction is called stepwise stability constant . It is denoted by K. Stability constant of overall complex formation reaction is called overall  stability constant. It is denoted by β.
The stepwise and overall stability constant are therefore related as follows:
β₄ = K₁ × K₂ × K₃ × K₄ or more generally,
β_n = K₁ × K₂ × K₃ × K₄ … K_n

Question. For the complex [NiCl₄]^2–, write
(i) the IUPAC name
(ii) the hybridization type
(iii) the shape of the complex.
(Atomic no. of Ni = 28) 
Answer : (i) Tetrachloridonickelate(II) ion Ni atom (Z = 28)

The complex ion has tetrahedral geometry and is paramagnetic due to the presence of unpaired electrons

Question. Why is [NiCl₄]^2– paramagnetic but [Ni(CO)₄] is diamagnetic? (At. no. : Cr = 24, Co = 27, Ni = 28)
Answer : [NiCl₄]^2– contains Ni²⁺ ion with 3d⁸ configuration.

The complex has all paired electrons hence, it is diamagnetic.