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ChemistryMediumMCQ2020 · 09 Jan Shift 2

Q32.Among the statements (a) −(d), the correct ones are: (a) Lithium has the highest hydration enthalpy among the alkali metals. (b) Lithium chloride is insoluble in pyridine. (c) Lithium cannot form ethynide upon its reaction with ethyne. (d) Both lithium and magnesium react slowly with H2O . (1) (a), (b) and (d) only (2) (a), (c) and (d) only (3) (b) and (c) only (4) (a) and (d) only

What This Question Tests

This question tests specific properties of lithium including its high hydration enthalpy, solubility of its halides, and its unique reactions with ethyne and water, which relate to its small size and diagonal relationship with magnesium.

Concepts Tested

Hydration enthalpy trendsSolubility of LiCl in organic solventsReactivity of Li with ethyneReactivity of Li and Mg with waterDiagonal relationship

📚 NCERT Sections This Tests

6.22 — What Happens When

Chemistry Class 12 · Chapter 6

76% match

6.22 What happens when (i) n-butyl chloride is treated with alcoholic KOH, (ii) bromobenzene is treated with Mg in the presence of dry ether, (iii) chlorobenzene is subjected to hydrolysis, (iv) ethyl chloride is treated with aqueous KOH, (v) methyl bromide is treated with sodium in the presence of dry ether, (vi) methyl chloride is treated with KCN? Answers to Some Intext Questions 6.1 6.2 (i) H2SO4 cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding acid, HI which is then oxidised by it to I2. 6.3 (i) ClCH2CH2CH2Cl (ii) ClCH2CHClCH3 (iii) Cl2CHCH2CH3 (iv) CH3CCl2CH3 191 Haloalkanes and Haloarenes Reprint 2025-26

5.29 — Amongst The Following Ions Which One Has The Highest Magnetic Moment Value?

Chemistry Class 11 · Chapter 5

75% match

5.29 Amongst the following ions which one has the highest magnetic moment value? (i) [Cr(H2O)6]3+ (ii) [Fe(H2O)6] 2+ (iii) [Zn(H2O)6]2+ 5.30 Amongst the following, the most stable complex is (i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6] 3+ (iii) [Fe(C2O4)3]3– (iv) [FeCl6] 3– 5.31 What will be the correct order for the wavelengths of absorption in the visible region for the following: [Ni(NO2)6] 4–, [Ni(NH3)6] 2+, [Ni(H2O)6] 2+ ? Answers to Some Intext Questions 5.1 (i) [Co(NH3)4(H2O)2]Cl3 (iv) [Pt(NH3)BrCl(NO2)]– (ii) K2[Ni(CN)4] (v) [PtCl2(en)2](NO3)2 (iii) [Cr(en)3]Cl3 (vi) Fe4[Fe(CN)6]3 5.2 (i) Hexaamminecobalt(III) chloride (ii) Pentaamminechloridocobalt(III) chloride (iii) Potassium hexacyanidoferrate(III) (iv) Potassium trioxalatoferrate(III) (v) Potassium tetrachloridopalladate(II) (vi) Diamminechlorido(methanamine)platinum(II) chloride 5.3 (i) Both geometrical (cis-, trans-) and optical isomers for cis can exist. (ii) Two optical isomers can exist. (iii) There are 10 possible isomers. (Hint: There are geometrical, ionisation and linkage isomers possible). (iv) Geometrical (cis-, trans-) isomers can exist. 5.4 The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents: [Co(NH3)5Br]SO4 + Ba2+ ® BaSO4 (s) [Co(NH3)5SO4]Br + Ba2+ ® No reaction [Co(NH3)5Br]SO4 + Ag+ ® No reaction [Co(NH3)5SO4]Br + Ag+ ® AgBr (s) 5.6 In Ni(CO)4, Ni is in zero oxidation state whereas in NiCl42–, it is in +2 oxidation state. In the presence of CO ligand, the unpaired d electrons of Ni pair up but Cl– being a weak ligand is unable to pair up the unpaired electrons. 5.7 In presence of CN–, (a strong ligand) the 3d electrons pair up leaving only one unpaired electron. The hybridisation is d 2sp 3 forming inner orbital complex. In the presence of H2O, (a weak ligand), 3d electrons do not pair up. The hybridisation is sp 3d 2 forming an outer orbital complex containing five unpaired electrons, it is strongly paramagnetic. 5.8 In the presence of NH3, the 3d electrons pair up leaving two d orbitals empty to be involved in d2sp3 hybridisation forming inner orbital complex in case of [Co(NH3)6]3+. In Ni(NH3)6 2+, Ni is in +2 oxidation state and has d 8 configuration, the hybridisation involved is sp 3d 2 forming outer orbital complex. 5.9 For square planar shape, the hybridisation is dsp 2. Hence the unpaired electrons in 5d orbital pair up to make one d orbital empty for dsp2 hybridisation. Thus there is no unpaired electron. Chemistry 140 Reprint 2025-26

1.23 — Suggest The Most Important Type Of Intermolecular Attractive Interaction In

Chemistry Class 11 · Chapter 1

74% match

1.23 Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane (ii) I2 and CCl4 (iii) NaClO4 and water (iv) methanol and acetone (v) acetonitrile (CH3CN) and acetone (C3H6O). 1.24 Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.