CBSE 12th Chemistry Board Exam 2020: Important Questions & Answers from Chapter 9 - Coordination Compounds
Check the important questions and answers from chapter Coordination Compounds for the upcoming CBSE Class 12th Chemistry Examination 2020.
The students who are appearing for the CBSE Class 12th Chemistry Examination 2020 can check the important questions and answers form the chapter Coordination Compounds. These chapter-wise important questions are based on the NCERT textbook, previous year papers and sample papers.
Key Points to be mentioned while writing the answers to the below mentioned important questions:
Question 1- Name these coordination compounds according to the IUPAC system of nomenclature:
(i) [Co(NH3)4 (H2O)Cl]Cl2
(iii) [CoCl2(en)2] +
Answer: (i) Tetraammineaquachlorocobalt (III) chloride
(ii) Dichlorobis(ethane-1, 2-diamine)chromium (III) chloride
(v) Mercury tetrathiocyanatocobaltate(III)
Answer: Shape: Square Planar; Hybridisation: dsp2; Magnetic behaviour: Diamagnetics
Question 3- Explain:
(i) [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3- is weakly paramagnetic.
(ii) [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
Answer: (i) Fe is in +3 oxidation state in both the cases. In the presence of CN–, the 3d electrons pair up leaving only one unpaired electron. d2sp3 hybridisation forms an inner orbital complex and is weakly paramagnetic. In the presence of H2O, 3d electrons do not pair up as it is a weak ligand. The hybridisation involved here is sp3d2 which forms an outer orbital complex. Thus, [Fe(H2O)6]3+ is strongly paramagnetic as it contains five unpaired electrons.
(ii) In [Co(NH3)6]3+, the d-electrons of Co3+ gets paired leaving behind two empty d-orbitals and undergo d2sp3 hybridization. Therefore, [Co(NH3)6]3+ is an inner orbital complex, whereas in [Ni(NH3)6]2+ the d-electrons of Ni2+ do not pair up and use outer 4d subshell. Therefore, [Ni(NH3)6]2+ is an outer orbital complex.
Question 4- Write the formulas for the following:
(i) Potassium trioxalatochromate(III)
(ii) Pentaamminecarbonatocobalt(III) chloride
(iii) Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride
(iv) Pentaamminecarbonatocobalt(III) chloride
Answer: (i) K3[Cr(C2O4)3]
Question 5- For [Fe(en)2Cl2] Cl, state
(i) the oxidation number of iron
(ii) the hybrid orbitals and the shape of the complex
(iii) the magnetic behaviour of the complex
(iv) the number of geometrical isomers
(v) whether there is an optical isomer also
(vi) name of the complex. (At. no. of Fe = 26).
Answer: (i) the oxidation number of iron is +3.
(ii) the hybridization will be d2sp3 and the shape of the complex will be octahedral.
(iii) the complex shows paramagnetic behaviour as it has one unpaired electron in the d-orbital.
(iv) the geometrical isomers of the complex are cis and trans.
(v) only cis-isomer shows optical isomerism
(vi) name of the complex: Dichlorobis(ethylenediamine)iron (III) chloride
Question 6- Define:
(i) Crystal field splitting
(ii) Linkage isomerism
(iii) Ambidentate ligand
(iv) Denticity of a ligand
Answer: (i) Crystal field splitting: In an octahedral coordination entity with six ligands surrounding the metal atom/ion, there will be repulsion between the electrons in metal d orbitals and the electrons (or negative charges) of the ligands. Such a repulsion is more when the metal d orbital is directed towards the ligand than when it is away from the ligand. Thus, the d x2 −y2 and d z2 orbitals which point towards the axes along the direction of the ligand will experience more repulsion and will be raised in energy; and the dxy, dyz and dxz orbitals which are directed between the axes will be lowered in energy relative to the average energy in the spherical crystal field. Thus, the degeneracy of the d orbitals has been removed due to ligand electron-metal electron repulsions in the octahedral complex to yield three orbitals of lower energy, t2g set and two orbitals of higher energy, eg set. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted by Δo (the subscript o is for octahedral).
(ii) Linkage isomerism: It arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand, NCS–, which may bind through the nitrogen to give M–NCS or through sulphur to give M–SCN. Jørgensen discovered such behaviour in the complex [Co(NH3)5(NO2)]Cl2, which is obtained as the red form, in which the nitrite ligand is bound through oxygen (–ONO), and as the yellow form, in which the nitrite ligand is bound through nitrogen (–NO2).
(iii) Ambidentate ligand: Ligand which can ligate through two different atoms is called ambidentate ligand. Examples of such ligands are the NO2– and SCN– ions. NO2– ion can coordinate either through nitrogen or through oxygen to a central metal atom/ion. Similarly, SCN– ion can coordinate through the sulphur or nitrogen atom.
(iv) Denticity of a ligand: When a di- polydentate ligand uses its two or more donor atoms to bind a single metal ion, it is said to be a chelate ligand. The number of such ligating groups is called the denticity of the ligand. Such complexes, called chelate complexes tend to be more stable than similar complexes containing unidentate ligands.
Question 7- Give Reasons:
(i) Low spin octahedral complexes of nickel are unknown.
(ii) The π-complexes are known for transition elements only.
(iii) For many metals, CO is a stronger ligand than NTL.
Answer: (i)The electronic configuration of nickel: [Ar] 3d84s2. This means that it can only form 2 types of complexes. Square planar (dsp2) in the presence of strong ligand and tetrahedral (sp3) in the presence of weak ligand. There are 4 empty orbitals in Ni while octahedral complexes require 6 empty orbitals. Therefore, low spin octahedral complexes of nickel are unknown.
(ii) Due to the presence of empty d-orbitals in transition metals, they can accept electron pairs from ligands containing π electrons and forms ic-bonding complexes.
(iii) Due to the greater magnitude of Δ0, CO produces strong fields which cause more splitting of d-orbitals. Also, it is able to form a π-bond due to back bonding.
Question 8- State:
(i) Names of two complexes which are used in treatments.
(ii) Double salt is different from a complex.
Answer: (i) (a) Pt(NH3)2Cl2] for curing cancer.
(b) EDTA for curing lead poisoning.
(ii) Double salt is different from a complex as it dissociates itself completely in its constituent ions (in aqueous solution) while complex doesn’t dissociate itself into constituent ions.
Question 9- Mention the isomerism shown by the following:
(ii) [Co(NH3)6] [Cr(CN)6]
Answer: (i) Ionisation Isomerism
(ii) Coordination Isomerism
Question 10- (a) Draw the structures of geometrical isomers:
(ii) Pt(NH3) 2Cl2)
(b) Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)4 2+ ion, given that β4 for this complex is 2.1 × 1013.
Answer: (a) (i)
(b) 1/ β4 = 4.7 × 10–14
The above-mentioned list of important questions and answers is based on the latest syllabus prescribed by the CBSE. The students appearing for the Class 12th CBSE Chemistry Examination 2020 are advised to go through the below-mentioned links to prepare for the upcoming examination: