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ICSE Class IX Board Exam 2021 : Chemistry

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Hemkanta Maiti
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(www.tiwariacademy.com) (Chapter 9)(Coordination compounds) XII Intext Questions 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) 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 32 Free web support in Education (www.tiwariacademy.com) (Chapter 9)(Coordination compounds) XII 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 Question 9.3: Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers: i. K[Cr(H2O)2(C2O4)2 ii. [Co(en)3]Cl3 iii. [Co(NH3)5(NO2)](NO3)2 iv. [Pt(NH3)(H2O)Cl2] Answer i. Both geometrical (cis-, trans-) isomers for can exist. Also, optical isomers for cis-isomer exist. Trans-isomer is optically inactive. On the other hand, cis-isomer is optically active. (ii) Two optical isomers for exist. 33 Free web support in Education (www.tiwariacademy.com) (Chapter 9)(Coordination compounds) XII 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. 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. 34 Free web support in Education (www.tiwariacademy.com) (Chapter 9)(Coordination compounds) XII 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 [NiCl 4]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. 35 Free web support in Education (www.tiwariacademy.com) (Chapter 9)(Coordination compounds) XII 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, [NiCl 4]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. 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, 36 Free web support in Education (www.tiwariacademy.com) (Chapter 9)(Coordination compounds) XII 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. 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 37 Free web support in Education (www.tiwariacademy.com) (Chapter 9)(Coordination compounds) XII NH3 being a strong field ligand causes the pairing. Therefore, Ni can undergo d2sp3 If NH3 causes the pairing, then only one 3d orbital is empty. Thus, it cannot undergo hybridization. d2sp3 hybridization. Therefore, it undergoes sp3d2 hybridization. Hence, it is an inner orbital complex. Hence, it forms an outer orbital complex. 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 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 38 Free web support in Education (www.tiwariacademy.com) (Chapter 9)(Coordination compounds) XII 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. Cyanide is The crystal field is octahedral. Water is a a strong field ligand. Therefore, the weak field ligand. Therefore, the arrangement of the electrons in arrangement of the electrons in is is t2g3eg2. T2g5eg0. Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron. Question 9.11: Calculate the overall complex dissociation equilibrium constant for the Cu(NH 3)42+ ion, given that 4 for this complex is 2.1 1013. Answer 4 = 2.1 1013 The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, 4. 39 Free web support in Education

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