February 24, 2024

Here, you can find comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 9 on Coordination Compounds. These solutions include easy-to-follow, step-by-step explanations. Widely appreciated by class 12 Science students, these Chemistry Coordination Compounds Solutions are invaluable for efficiently completing homework assignments and preparing for exams. All the questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 9 are available here at no cost. Page No 244: Question 9.1: Write the formulas for the following coordination compounds: (i) Tetraamminediaquacobalt(III) chloride (ii) Potassium tetracyanonickelate(II) (iii) Tris(ethane−1,2−diamine) chromium(III) chloride (iv) Amminebromidochloridonitrito-N-platinate(II) (v) Dichloridobis(ethane−1,2−diamine)platinum(IV) nitrate (vi) Iron(III) hexacyanoferrate(II) ANSWER: (i)  (ii)  (iii)  (vi)  (v)  (vi)  Page No 244: Question 9.2: Write the IUPAC names of the following coordination compounds: (i) [Co(NH3)6]Cl3 (ii) [Co(NH3)5Cl]Cl2 (iii) K3[Fe(CN)6] (iv) K3[Fe(C2O4)3] (v) K2[PdCl4] (vi) [Pt(NH3)2Cl(NH2CH3)]Cl ANSWER: (i) Hexaamminecobalt(III) chloride (ii) Pentaamminechloridocobalt(III) chloride (iii) Potassium hexacyanoferrate(III) (iv) Potassium trioxalatoferrate(III) (v) Potassium tetrachloridopalladate(II) (vi) Diamminechlorido(methylamine)platinum(II) chloride Page No 247: Question 9.3: Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers: ANSWER: Trans-isomer is optically inactive. On the other hand, cis-isomer is optically active. (ii) Two optical isomers for  exist. Two optical isomers are possible for this structure. (iii)  A pair of optical isomers: It can also show linkage isomerism. and It can also show ionization isomerism. (iv) Geometrical (cis-, trans-) isomers of can exist. Page No 247: Question 9.4: Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl are ionization isomers. ANSWER: When ionization isomers are dissolved in water, they ionize to give different ions. These ions then react differently with different reagents to give different products. Page No 254: Question 9.5: Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic. ANSWER: Ni is in the +2 oxidation state i.e., in d8 configuration. There are 4 CN− ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN− ion is a strong field ligand, it causes the pairing of unpaired 3d electrons. It now undergoes dsp2 hybridization. Since all electrons are paired, it is diamagnetic. In case of [NiCl4]2−, Cl− ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired 3d electrons. Therefore, it undergoes sp3 hybridization. Since there are 2 unpaired electrons in this case, it is paramagnetic in nature. Page No 254: Question 9.6: [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? ANSWER: Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl4]2− is paramagnetic. In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2. But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are present in this case, [Ni(CO)4] is diamagnetic. Page No 254: Question 9.7: [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain. ANSWER: In both and , Fe exists in the +3 oxidation state i.e., in d5 configuration. Since CN− is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in the d-orbital. Therefore, On the other hand, H2O is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5. Therefore, Thus, it is evident that is strongly paramagnetic, while is weakly paramagnetic. Page No 254: Question 9.8: Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex. ANSWER: Oxidation state of cobalt = +3 Oxidation state of Ni = +2 Electronic configuration of cobalt = d6 Electronic configuration of nickel = d8 NH3 being a strong field ligand causes the pairing. Therefore, Cobalt can undergo d2sp3 hybridization.Hence, it is an inner orbital complex. If NH3 causes the pairing, then only one 3d orbital is empty. Thus, it cannot undergo d2sp3hybridization. Therefore, it undergoes sp3d2 hybridization.Hence, it forms an outer orbital complex. Page No 254: Question 9.9: Predict the number of unpaired electrons in the square planar [Pt(CN)4]2− ion. ANSWER: In this complex, Pt is in the +2 state. It forms a square planar structure. This means that it undergoes dsp2 hybridization. Now, the electronic configuration of Pd(+2) is 5d8. CN− being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in Page No 254: Question 9.10: The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory. ANSWER: Mn is in the +2 oxidation state. Mn is in the +2 oxidation state. The electronic configuration is d5. The electronic configuration is d5. The crystal field is octahedral. Water is a weak field ligand. Therefore, the arrangement of the electrons in is t2g3eg2. The crystal field is octahedral. Cyanide is a strong field ligand. Therefore, the arrangement of the electrons in  isT2g5eg0.     Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron. Page No 256: Question 9.11: Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)42+ ion, given that β4 for this complex is 2.1 × 1013. ANSWER: β4 = 2.1 × 1013 The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, β4. Page No 258: Question 9.1: Explain the bonding in coordination compounds in terms of Werner’s postulates. ANSWER: Werner’s postulates explain the bonding in coordination compounds as follows: (i) A metal exhibits two types of valencies namely, primary and secondary valencies. Primary valencies are satisfied by negative ions while secondary valencies are satisfied by both negative and neutral ions. (In modern terminology, the primary valency corresponds to the oxidation number of the metal ion, whereas the secondary valency refers to the coordination number of the metal ion. (ii) A metal ion has a definite number of secondary valencies around the central atom. Also, these valencies project in a specific direction in the space assigned to the definite geometry of the coordination compound. (iii) Primary valencies are usually ionizable, while secondary valencies are non-ionizable. Page No 258: Question 9.2: FeSO4 solution mixed with …

NCERT Solutions for Class 12 Science Chemistry Chapter 9 – Coordination Compounds Read More »