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ChemistryMediumMCQ2006 · Unknown

Q82.Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of 2- fluoroethanol is (1) Eclipse, Gauche, Anti (2) Gauche, Eclipse, Anti (3) Eclipse, Anti, Gauche (4) Anti, Gauche, Eclipse JEE Main 2006 JEE Main Previous Year Paper

What This Question Tests

This question assesses the understanding of conformational stability, specifically for 2-fluoroethanol, highlighting the exceptional stabilization of the Gauche conformation due to intramolecular hydrogen bonding.

Concepts Tested

Conformations (Eclipse, Anti, Gauche)Torsional strainSteric strainIntramolecular hydrogen bonding

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8.6Isomerism Isomers And This Phenomenon Is Termed

Chemistry Class 12 · Chapter 8

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8.6 ISOMERISM isomers and this phenomenon is termed as position isomerism. For example, the The phenomenon of existence of two or more molecular formula C3H8O represents twocompounds possessing the same molecular alcohols: formula but different properties is known as isomerism. Such compounds are called OH as isomers. The following flow chart shows different types of isomerism. CH3CH2CH2OH CH3−CH-CH3 Propan-1-ol Propan-2-ol 8.6.1 Structural Isomerism Compounds having the same molecular (iii) Functional group isomerism: Two or formula but different structures (manners more compounds having the same molecular in which atoms are linked) are classified as formula but different functional groups structural isomers. Some typical examples are called functional isomers and this of different types of structural isomerism are phenomenon is termed as functional group given below: isomerism. For example, the molecular (i) Chain isomerism: When two or more formula C3H6O represents an aldehyde and compounds have similar molecular formula but a ketone: Isomerism Structural isomerism Stereoisomerism Chain Position Functional Metamerism Geometrical Optical isomerism isomerism group isomerism isomerism isomerism Reprint 2025-26 organic chemistry – some basic principles and techniques 271 in understanding the reactivity of organic O H compounds and in planning strategy for their   synthesis. CH3−C-CH3 CH3−CH2—C= O In the following sections, we shall learn Propanone Propanal some of the principles that explain how these (iv) Metamerism: It arises due to different reactions take place. alkyl chains on either side of the functional 8.7.1 Fission of a Covalent Bond group in the molecule. For example, C4H10O represents methoxypropane (CH3OC3H7) and A covalent bond can get cleaved either by : (i) ethoxyethane (C2H5OC2H5). heterolytic cleavage, or by (ii) homolytic cleavage.8.6.2 Stereoisomerism In heterolytic cleavage, the bond breaks The compounds that have the same in such a fashion that the shared pair of constitution and sequence of covalent bonds electrons remains with one of the fragments. but differ in relative positions of their atoms After heterolysis, one atom has a sextet or groups in space are called stereoisomers. electronic structure and a positive charge and This special type of isomerism is called as the other, a valence octet with at least one lone stereoisomerism and can be classified as pair and a negative charge. Thus, heterolytic geometrical and optical isomerism. + cleavage of bromomethane will give C H3 and

1.23Suggest The Most Important Type Of Intermolecular Attractive Interaction In

Chemistry Class 11 · Chapter 1

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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.

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

Chemistry Class 11 · Chapter 5

72% 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