Q. No. 2 Bond formation is. Neither exothermic nor endothermic

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1 Q. No. 1 Which combination will give the strongest ionic bond? K + and Cl - K + and O 2- Ca 2+ and Cl - Ca 2+ and O 2- In CaO, the polarizability of O 2- is very less therefore it has the maximum ionic character and strongest ionic bond. Q. No. 2 Bond formation is Always exothermic Always endothermic Neither exothermic nor endothermic Sometimes exothermic and sometimes endothermic During bond formation potential energy decreases and energy is liberated out. Q. No. 3 Element X is strongly electropositive and element Y is strongly electronegative. Both are univalent. The compound formed would be X + Y - X - Y + X - Y X Y Element X is a metal and element Y is a non-metal compound formed will be X + Y - since they are univalent. Q. No. 4 The bond between two identical non-metal atoms has a pair of electrons Unequally shared between the two Transferred fully from one atom to another With identical spins Equally shared between them Two identical non-metal atoms are forming covalent bonds in which electron pair is equally shared. Q. No. 5 An ionic compound A + B - is most likely to be formed when: The ionization energy of A high and electron affinity of B is low The ionization energy of A is low and electron affinity of B is high Both, the ionization energy of A and electrons affinity of B are high Both, the ionization energy of A and electron affinity of B are low Ionic compound is most likely to be formed when metal atom can loose electron very easily and when non-metal atom can gain electron very easily. Q. No. 6 The lattice energy of KF, KCl, KBr and KI follow the order: KF> KCl> KBr> KI KI> KBr> KCl> KF KF> KCl> KI> KBr KI> KBr> KF> KCl In KI the polarizability of I - is maximum.

2 According to the Fajan s Rule its covalent character is maximum and lattice energy is the least sequence is KF> KCl> KBr> KI. Q. No. 7 In which of the following species the bonds are non-directional? NCl 3 RbCl BeCl 2 BCl 3 Smaller is the size of cation and bigger is the size of anion then covalent character in the ionic compounds is maximum and vice-versa. In BeCl 2 size of cation is the least while in RbCl size of cation is maximum. Q. No. 8 The compound with the highest degree of covalency is: NaCl MgCl 2 AgCl CsCl AgCl has the highest degree of covalency because Ag + has pseudo noble gas configuration and due to very weak shielding effect of d-subshell the nuclear charge become very strong. Q. No. 9 The fluorine molecules is formed by: p-p orbitals (sideways overlap) p-p orbitals (end-to-end overlap) sp-sp orbitals s-s orbitals Fluorine molecule possess one σ bond formed by end to end overlap between two p- orbitals. Q. No. 10 Covalency favoured in the following case: Smaller cation Larger anion Large charge on cation and anions All the above All the above Q. No. 11 The molecular size of ICl and Br 2 is approximately same, but b.p. if ICl is about 40 0 C higher than that of Br 2. It is because: ICl bond is stronger than Br -Br bond IE of iodine < IE of bromine ICl is polar while Br 2 is nonpolar I has larger size than Br ICl is a polar molecule while Br 2 is non-polar. Due to stronger dipole-dipole attraction b.pt of ICl is higher than Br 2. Q. No. 12 Which of the following molecules does not have coordinate bonds? PH + 4

3 NO 2 O 3 CO 2-3 Q. No. 13 The compound which contains both ionic and covalent bonds is CH 4 H 2 KCN KCl Q. No. 14 The total number of electrons that taken part in forming the bonds in N 2 is N N Q. No. 15 Which of the following compound is covalent? H 2 CaO KCl Na 2 S H-H Q. No. 16 Which pair of elements can form multiple bond with itself and oxygen? F, N N, Cl N, P N, C Q. No The correct order of increasing C-O bond strength of CO, CO 3, CO2 is: 2- CO < CO < CO < 2-3 < < 2-3 < 2 CO CO CO CO CO CO CO< CO2 < CO 2-3

4 O = C = O CO 2 has double bond. exhibits resource and bond order = 1.33 CO has triple bond. CO 2- < CO < CO CO Q. No. 18 Resonance structures can be written for: O 3 NH 3 CH 4 H 2 O Q. No. 19 Correct statement regarding molecules SF 4, CF 4 and XeF 4 are: 2, 0 and 1 lone pairs of central atom respectively 1, 0 and 1 lone pairs of central atom respectively 0, 0 and 2 lone pairs of central atom respectively 1, 0 and 2 lone pairs of central atom respectively Q. No. 20 The pairs of species with similar shape is: PCl 3, NH 3 CF 4, SF 4 PbCl 2, CO 2 PF 5, IF 5 Both PCl 3 and NH 3 has 3 bond pairs and one lone pair and they have pyramidal shape. Q. No. 21 The geometrical arrangement of orbitals and shape of I - 3 are respectively: Trigonal Bipyramidal geometry, linear shape Hexagonal geometry, T-shape Triangular planar geometry, triangular shape Tetrahedral geometry, pyramidal shape Geometrical arrangement of orbitals is trigonal Bipyramidal and shape is linear.

5 Q. No. 22 The shape of the noble gas compound XeF 4 is: Square planar Distorted tetrahedral Tetrahedral Octahedral Shape of XeF 4 is square planar. Q. No. 23 The molecule exhibiting maximum number of non-bonding electron pairs (1.p.) around the central atom is: XeOF 4 XeO 2 F 2 XeF3 - XeO 3 Q. No. 24 Which is the following pairs of species have identical shapes? + - NO2 and NO2 PCl 5 and BrF 5 XeF4 andicl - 4 TeCl 4 and XeO 4 Both XeF4 andicl - 4 have 2 lone pair and 4 bond pairs and are having square planar shapes. Q. No. 25 Which is not correctly matched? XeO 3 - Trigonal bipyramidal ClF 3 - bent T-shape XeOF 4 - Square pyramidal XeF 2 - Linear shape XeO 3 has 1 lone pair and 3 bond pair. Shape is trigonal pyramidal. Q. No. 26 The compound MX 4 is tetrahedral. The number of XMX angles in the compound is: Three Four Five Six

6 Q. No. 27 Which of the following are isoelectronic and isostructural? , ClO 3, SO3 NO, CO - 2- NO 3, CO3 SO 3,NO ClO 3, CO CO 3, ClO3 Both have 32 electrons and have trigonal shape. Q. No. 28 In which of the following pairs, both the species have the same hybridization? - (I) SF 4, XeF 4 (II) I, XeF + (III) ICl 4, SiCl4 I, II II, III II, IV I, II, III (IV) ClO,PO Both I3 -, XeF 2 have 2 bond pair and 3 lone pair. hybridization is sp 3 d. Q. No. 29 Which of the following molecule is theoretically not possible? SF 4 OF 2 OF 4 O 2 F 2 OF 4 is not possible because O will be having more than 8 electrons. Q. No. 30 The strength of bonds formed by 2s-2s, 2p-2p and 2p-2s overlap has the order: s- s> p-p> p- s s- s> p- s> p-p p-p> p- s> s- s p-p> s-s> p- s Extent of overlapping decreases in the order p-p> p- s> s- s. Q. No. 31 C 2 H 2 is isostructural with: H 2 O 2

7 NO 2 SnCl 2 CO 2 H -C C -H It is having linear shape since both the C are sp hybridised without lone pair. O = C = O It is also having linear shape. Q. No. 32 When iodine is dissolved in aqueous potassium iodide, the shape of the species formed is : Linear Angular Triangular See-saw I + KI K + [ I ] 2 3 I - 3 is having linear shape. - Q. No. 33 Which one of the following is the correct set with respect to molecule, hybridization and shape? BeCl 2, sp 2, linear BeCl 2, sp 2, triangular planar BCl 3, sp 2, triangular planar BCl 3, sp 3, tetrahedral B is having 3 bond pair without lone pair. Its shape is trigonal planar and sp 2 hybridised. Q. No. 34 The hybridization of the central atom in ICl + 2 is: dsp 2 sp sp 2 sp 3 Cl - I - Cl I is having 2 bond pair and 2 lone pair therefore it is sp 3 hybridised. Q. No. 35 The state of hybridization of the central atom is not the same as in the others: B in BF 3 O in H 3 O + N in NH 3 P in PCl 3 B in BF 3 is sp 2 hybridised while other central atoms are sp 3 hybridised. Q. No. 36 Which of the following statements us incorrect for PCl 5?

8 Its three P - Cl bond lengths are equal It involves sp 3 d hybridization It has an regular geometry Its shape is trigonal Bipyramidal PCl 5 has trigonal Bipyramidal geometry. Q. No The number of sp - s sigma bonds in benzene are : None of these In benzene every C-atom is sp 2 hybridised which is forming 6σ bonds with s-orbital of H-atom. Q. No. 38 During the complete combustion of methane CH 4, what change in hybridization does the carbon atom undergo? sp 3 to sp sp 3 to sp 2 sp 2 to sp sp 2 to sp 3 Q. No. 39 Select pair of compounds in which both have different hybridization but have same molecular geometry? BF 3, BrF 3 - ICl 2,BeCl2 BCl 3, PCl 3 PCl 3, NCl 3 ICl - 2 has 2 bond pair and 3 lone pair therefore shape is linear and BeCl 2 has 2 bond pair without lone pair. Therefore shape is linear. Q. No. 40 Which of the following statements is incorrect for sigma and pi-bonds formed between two carbon atoms? Sigma-bond is stronger than a pi-bond Bond energies of sigma and pi-bonds are of the order of 264 kj/mol Free rotation of surroundings atoms about a sigma-bond is allowed but not in case of a pi-bond Sigma-bond determines the direction between carbon atoms but a pi-bond has no primary effect in this regard Sigma bonds are stronger than π bonds. Therefore bond energy of σ bonds must be higher than π bonds. Q. No. 41 Assuming the bond direction to the z-axis, which of the overlapping of atomic orbitals of two atom (A) and (B) will result in bonding?

9 (I) s-orbital of A and p x orbital of B (II) s-orbital of A and p z orbital of B (III) p y -orbital of A and p z orbital of B (IV) s-orbital of both (A) and (B) I and IV I and II III and IV II and IV Since the bond direction is the z-axis, therefore overlapping of s-orbital of A and p z orbital of B will form σ bond along z-axis. Since s-orbital is directionally independent. Therefore s-orbital of both (A) and (B) will also result the bond. Q. No. 42 The hybridization of atomic orbitals of central atom Xe in XeO 4, XeO 2 F 2 and XeOF 4 respectively. sp 3, sp 3 d 2, sp 3 d 2 sp 3 d, sp 3 d, sp 3 d 2 sp 3, sp 3 d 2, sp 3 d sp 3, sp 3 d, sp 3 d 2 Q. No. 43 For exhibiting tetravalency, carbon atoms have to be excited. Now, which of the following statements is true? Excitation occurs before bonding Bonding occurs before excitation Both bonding and excitation occur simultaneously Two bonds are formed first, then excitation occurs followed by formation of another two bonds. Both bonding and excitation occur simultaneously Q. No. 44 Which of the following statements is true about bonding and excitation? Energy required for excitation of carbon atoms (96 Kcal/mol) is less than energy released in bonding Energy required for excitation is more than energy released in bonding Energy required for excitation is equal to the energy released in bonding None of these After the formation of bond the molecule become more stable, therefore released is much higher than energy required for excitation of C-atom. Q. No. 45 Which of the following statements is true about hybridization? Only those atomic orbitals can be hybridized which do not differ much in shape. Only those atomic orbitals can be hybridized which do not differ much in energy. Only those atomic orbitals can be hybridized which do not differ much in size. Only those atomic orbitals can be hybridized which do not differ much in overlap integrals.

10 Only those atomic orbitals participates in the hybridization process which have some energy nearly some energy. Q. No. 46 Which of the following statement is true about hybridization? Hybridization generates new set of atomic orbitals identical in shape but not in size and energy. Hybridization generates new set of atomic orbitals identical in size but not in shape and energy. Hybridization generates new set of atomic orbitals identical energy but not in shape and size. Hybridization generates new set of atomic orbitals identical in shape, size and energy. Hybridization generates new set of atomic orbitals which are having same shape, size and energy. Q. No. 47 Which of the following statements is true about hybridization? Hybrid orbitals frequently undergo linear overlaps making sigma bonds. Hybrid orbitals frequently undergo lateral overlaps making π -bonds. In other words, there are several compounds in which π -bonds are formed using hybrid orbitals. Hybrid orbitals are molecular orbitals. A hybrid orbital bigger in size makes shorter bond. Hybrid orbitals always undergo linear overlaps and form sigma bonds. Q. No. 48 In 2sp hybridization, 2s-orbital can be mixed with Only 2p x Only 2p y Only 2p z Any one of 2p x, 2p y and 2p z Since the s-orbital is directionally independent therefore it can be mixed with any one of 2p x, 2p y and 2p z Q. No. 49 In 2sp 2 orbital, character of 2p z orbital will be Always 33.33% Always 0% Always 66.66% Either 33.33% or 0% In 2sp 2 orbital character of 2p z orbital may be 33.33% or 0% - because sometime s and p x and p y orbitals are participating in the mixing process. Q. No. 51 Energy content of a molecule will be less if Bond energy is more Bond energy is less Magnitude of overlap is very less None of these If bond energy is more then more amount of energy is liberated out in the formation of bond in molecule and therefore energy content of molecule become lesser. Q. No. 52 Which of the following orders is correct for electronegativity?

11 3 2 sp C> sp C> spc 2 3 spc> sp C> sp C 2 3 sp C> spc> sp C 3 2 sp C> spc> sp C In sp-hybridised C, s character is 50% therefore size of orbital is very small and electron load is strongly attracted by the nucleus and therefore it is highly electronegative. As the s-character decreases, electronegativity decreases. Q. No. 53 CH 2 = C = CH2 In this molecule (allene) All three C-atoms are sp 2 hybridised Both terminal C-atoms are sp 2 hybridised while central C-atom is sp-hybridised Both terminal C-atoms are sp-hybridised while central C-atom is sp 2 -hybridised None of these The C in this compound has 3σ bonds while other C have 4σ bonds. Q. No. 54 The nodal plane in the π - bond of ethane is located in: The molecular plane A plane parallel to the molecule plane A plane perpendicular to the molecular plane which bisects the carbon-carbon σ - bond at right angle A plane perpendicular to the molecular plane which contains the carbon-carbon bond Q. No. 55 Which of the following leads to bonding? Bonding occurs when two orbitals having φ = ( + ) lobes are end to end with each other. Q. No. 56 Which of the following is paramagnetic? O 2-2 NO CO

12 CN - NO has unpaired electron therefore it is paramagnetic. Q. No. 57 The ground state electronic configuration of valence shell electrons in nitrogen molecule (N 2 ) is written ( σ 2s) ( σ 2s) ( π 2p) ( σ 2p) nitrogen molecule is: Bond order = [ Nb - Na ] 2 1 = 8-2 = 3 2 Q. No. 58 The ion that is isoelectronic with CO is: O 2- N 2 + O + CN - CO and CN - are isoelectronic. Q. No. 59 Which one of the following is not paramagnetic? NO N 2 + CO O 2 In CO all the electrons are paired up Hence, the bond order in Q. No. 60 When N 2 is formed from N 2 + bond-order and when O 2 is formed from O 2 + bondorder : Increases Decreases Increases, decreases Decreases, increases For N 2 +, bonding electrons are 9 and antibonding electrons are Bond order = = While for N 2 bond order = 3 therefore bond order increases For O 2 +, bond order = 2.5 while for O 2 bond order = 2 Q. No. 61 Which of the following have identical bond order? (I) CN - (II) O 2 (III) NO +

13 (IV) CN - I, III II, IV I, II, III I, IV Both CN - and NO + have bond order = 3. Q. No. 62 Among unpaired electron is present in: KO 2 only NO 2 + and BaO 2 KO 2 and AlO2 - BaO 2 only O2 - has one unpaired electron. Q. No Bond order of O, O, O < 2-2 < 2 < < 2 - < 2 < < 2 < 2 - < < 2 + < 2 - < 2-2 O O O O O O O O O O O O O O O O and O2-2 is in order: Bond order of O 2 = 2, O 2 = 2.5, O 2 = 1.5, O 2 = 1 O < O < O < O Passage Text According to MOT, two atomic orbitals overlap resulting in the formation of molecular orbitals. Number of atomic orbitals overlapping together is equal to the molecular orbital formed. The two atomic orbital thus formed by LCAO (linear combination of atomic orbital) in the same phase or in the different phase are known as bonding and antibonding molecular orbitals respectively. The energy of bonding molecular orbital is lower than that of the pure atomic orbital by an amount. This known as the stabilization energy. The energy of antibonding molecular orbital is increased by (destabilization energy). Q. No. 64 Which among the following pairs contain both paramagnetic species O 2-2 and N- 2 O2 - and N 2 O 2 and N 2 O 2 and N2 - O 2 possess 2 unpaired electron whereas N2 - both are paramagnetic species. possess one unpaired electron therefore Q. No. 65 Which of the following statement(s) is true Higher the bond order lesser the bond length Higher the bond order greater the bond length Higher the bond order lesser the bond energy Higher the bond order lesser the number of bonds Higher the bond order then more it s the number of multiple bonds present and lesser

14 is the bond length. Q. No. 67 Which of the following combination of orbitals is correct? There no nodal plane present in σ s and σ p bonding molecular orbital. Q. No. 68 Which of the following statements is not correct regarding bonding molecular orbitals? Bonding molecular orbitals possess less energy than the atomic orbitals from which they are formed Bonding molecular orbitals have low electron density between the two nuclei Electron in bonding molecular contributes to the attraction between atoms They are formed when the lobes of the combining atomic orbitals have the same sign Bonding molecular orbitals have high electron density between the two nuclei. Q. No. 69 The BCl 3 is a planar molecule whereas NCl 3 is pyramidal because: BCl bond is more polar than N - Cl bond N - Cl bond is more covalent that B - Cl bond Nitrogen atom is smaller than boron atom BCl 3 has no lone pair electrons but NCl 3 has a lone pair of electrons Q. No. 70 The correct order of the O-O bond length in O 2, H 2 O 2 and O 3 is : O2 > O3 > H2O2 O3 > H2O2 > O2 O2 > H2O2 > O3 H2O2 > O3 > O2 H 2 O 2 has a single bond whereas in O 3 due to resonance O-O bond length will be more than double bond nut less than single bond while in O 2 bond length is least due to double bonds. H O > O > O Q. No. 71 The structure and hybridization of Si(CH 3 ) 4 is :

15 Bent, sp Trigonal, sp 2 Octahedral, sp 3 d Tetrahedral, sp 3 Q. No. 72 In which of the following bond angle is maximum: NH 3 NH + 4 PCl 3 SCl 2 In NH + 4 hybridization of N is sp3 without lone pair therefore bond angle is maximum while others have lone pair so bond angle is contracted. Q. No. 73 BrF 3 molecule, the lone pairs occupy equatorial position to minimize. Lone pair bond pair repulsion only Bond pair-bond pair repulsion only Lone pair-lone pair repulsion and lone pair-bond pair repulsion Lone pair-lone pair repulsion only When the lone pairs lies at the equatorial position then total number of repulsion gets minimized. Q. No. 74 H 2 O is dipolar, whereas BeF 2 is not. It is because: The electronegativity of F is greater than that of O H 2 O involves hydrogen bonding whereas BeF 2 is a discrete molecule H 2 O is linear and BeF 2 is angular H 2 O is angular and BeF 2 is linear H 2 O is dipoles because the shape of H 2 O is distorted i.e. angular due to 2 lone pair and hybridization is sp 3. BeF 2 is non polar because its shape is non-distorted. Q. No. 75 The ratio of π and σ bonds in benzene is: 1 : 2 1 : 4 1 : 5 1 : 6 In benzene there are 3π bonds and 12σ bonds therefore ration is 1 : 4.

16 Q. No. 76 In graphite each carbon atom is : sp 2 hybridised sp 3 hybridised sp hybridised Not hybridised sp 2 hybridised Q. No. 77 Both BF 2 and NF 3 are covalent but BF 3 molecule is non-polar while NF 3 is polar because: Atomic size of boron is smaller than nitrogen BF 3 is planar but NF 3 is pyramidal Boron is a metal while nitrogen is gas BF bond has no dipole moment while NF bond has dipole BF 3 molecule is non-polar because its shape is non-distorted (planar) while NF 3 molecule is polar because its shape is distorted (pyramidal). Q. No. 78 The highest dipole moment is of: CF 4 CH 3 OH CO 2 CH 3 F Electronegativity of F is maximum. Q. No. 79 Which one of the following is the most polar bond? N-F N-N N-Cl O-F In N-F bond the difference of electronegativity is maximum. Q. No The hybridization of atomic number orbitals of N in NO 2,NO3 4 are respectively: sp, sp 2, sp 3 sp, sp 3, sp 2 sp 2, sp, sp 3 sp 2, sp 3, sp Q. No. 81 Which of the following molecules does not have a dipole moment? IBr CHCl 3 CH 2 Cl 2 BF 3

17 IBr Q. No. 82 The high density of water compared to ice is due to: Hydrogen bonding interactions Dipole-dipole interactions Dipole-induced dipole interactions Induced dipole induced dipole interactions Each water molecule is tetrahedrally surrounded by 4 water molecules. Q. No. 84 The boiling point of alcohol is higher than ether due to: Hydrogen bonding Large size of alcohol Presence of -OH group High molecular weight In alcohol H is directly attached to O therefore it exhibit hydrogen bonding due to which its boiling point is higher than ether. Q. No. 85 The correct order of boiling point is: I> II> III III> II> I II> I> III III> I> II Due to intramolecular H-bonding it cannot from the polymeric chain, therefore its boiling point is the least. Cannot form the intramolecular H-bond therefore it forms the intermolecular H-bond and boiling point is maximum. Also forms the intermolecular H-bond but due to lesser molecular cut its b.pt is less than

18 Q. No. 86 The boiling points at atmospheric pressure of HF, H 2 S, NH 3 can be arranged in the following order: HF> NH3 > H2S HF> H2S> NH3 HF< H2S< NH3 HF< NH3 < H2S H-F is exhibits the strongest intermolecular H-bonding therefore its boiling point is maximum. NH 3 also exhibit intermolecular H-bonding but weaken than HF. H 2 S exhibits the weak dipole-dipole attraction. Q. No. 87 Which statement is correct? m.p. of H 2 O, NH 3, HF are maximum in their respective group due to intermolecular H- bonding b.p. of CH 4 out of CH 4, SiH 4, GeH 4 and SnH 4 is least due to weak intermolecular force of attraction Formic acid forms dimer by H-bonding All are correct H 2 O, NH 3, HF exhibit the intermolecular H-bonding therefore m.pt is maximum. B.pt of CH 4 is least because its molecular wt. is the least. Formic acid forms dimer by H- bonding as H is directly attached to O. Q. No. 88 Which of the following is not true about H 2 O molecule? The molecular has µ = 0 The molecular can act as a base Shows abnormally high boiling point in comparison to the hydrides of other elements of oxygen group The molecule has a bent shape. H 2 O is having distorted shape (bent shape) therefore its dipole moment is not zero. It can act as a lewis base and shows abnormally high boiling point due to intermolecular H-bonding. Q. No. 89 Which of the following does not contain any co-ordinate bond? H 3 O + BF4 - HF - 2 NH + 4 S+ S- - F H - F It does not contain any coordinate bond. Q. No. 90 The boiling points of methanol, water and dimethyl ether are respectively 65 0 C, C and C. Which of the following best explains these wide variations in b.p.? The molecular mass increases from water (18) to methanol (32) to diethyl ether (74) The extent of H-bonding decreases from water to methanol while it is absent in ether The extent of intramolecular H-bonding decreases from ether to methanol to water The density of water is 1.00 g ml -1, methanol g ml -1 and that of diethyl ether is g ml -1

19 Water is having the strongest intermolecular H-bond as it has no alkyl radicals present in its. Then comes methanol while dimethyl ether has no H-bonds is it. Q. No. 91 In ice, the length of H-bonds: Is less than that of covalent bonds Is greater than that of covalent bonds Is same as that of covalent bonds Can be less grater or same as that of covalent bonds Covalent bonds are shorter than H-bonds. Q. No. 92 The correct order of strength of H-bond is: H...F> H... O> H...N H...N> H... O> H...F H... O> H...N> H...F H...F> H...N> H...O Strength of H-bonds depend upon electronegativity of atom attached to H and the sequences is F> O> N. Q. No. 93 o-nitrophenol can be easily steam distilled whereas p-nitrophenol cannot be. This is because of : Strong intermolecular hydrogen bonding in o-nitrophenol Strong intramolecular hydrogen bonding in o-nitrophenol Strong intramolecular hydrogen bonding in p-nitrophenol Dipole moment of p-nitrophenol is larger than that of o-nitrophenol o-nitrophenol can be easily steam distilled because its boiling point is less as it exhibits intramolecular H-bonding and cannot form chain like structure. Q. No. 94 What is not true about ice? It has open cage like structure It has less density than water Each O atom is surrounded by 4 H atoms Each O atom has four H-bonds around it When ice melts to form liquid water than some of the hydrogen bonds are broken up and H 2 O molecules which are free they occupy the interstitial vacant spaces resulting in the contraction in volume.

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