NEET – CHEMICAL BONDING AND MOLECULAR STRUCTURE – 2024

1. The angle between the overlapping of one sorbital and one p-orbital is
(a) °180
(b) °120
(c) °109 , 28′
(d) °120, °60
2. Equilateral shape has
(a) sp hybridisation
(b) sp2 hybridisation
(c) sp3 hybridisaiton
(d) sp3 hybridisation
3. Which of the following molecule does not have a linear arrangement of atoms?
(a) H2 S
(b) C2H2
(c) BeH2
(d) CO2
4. Which of the following does not apply to metallic bond?
(a) Overlapping valence orbitals
(b) Mobile valency electrons
(c) Delocalized electrons
(d) Highly directed bonds.
5. In which one of the following molecules the central atom said to adopt sp2 hybridization?
(a) BeF₂
(b) BF₃
(c) C₂H₂
(d) NH₃
6. H₂O has a non zero dipole moment while BeF2 has zero dipole moment because
(a) H2O molecule is linear while BeF2 is bent
(b) BeF2 molecule is linear while H2O is bent
(c) Fluorine has more electro negativity than oxygen
(d) Beryllium has more electro negativity than oxygen.

7. Among LiCl, BeCl₂ BCl₃ and CCl₄, the covalent bond character follows the order
(a) LiCl < BeCl₂ > BCl₃ > CCl₄
(b) BeCl₂ < BCl₃ < CCl₄ < LiCl
(c) LiCl < BeCl₂ < BCl₃ < CCl₄
(d) LiCl > BeCl₂ > BCl₃ > CCl₄
8. Which statement is NOT correct?

(a) A sigma bond is weaker than a S-bond.
(b) A sigma bond is stronger than a S-bond.
(c) A double bond is stronger than a single bond.
(d) A double bond is shorter than a single bond.
9. Which one shows maximum hydrogen bonding?
(a) H₂O
(b) H2 Se
(c) H2 S
(d) HF.
10. Linear combination of two hybridized orbitals belonging to two atoms and each having one electron leads to a
(a) Sigma bond
(b) Double bond
(c) Co-ordinate covalent bond
(d) Pi bond.
11. Which one of the following formulae does not correctly represent the bonding capacities of the two atoms involved?

12. In compound X, all the bond angles are exactly 28°109; X is
(a) Chloromethane
(b) Carbon tetrachloride
(c) Iodoform
(d) Chloroform.
13. Which of the following bonds will be most polar?
(a) N– Cl

(b) O– F
(c) N– F
(d) N– N
14. Which one of the following has the shortest carbon carbon bond length?
(a) Benzene
(b) Ethene
(c) Ethyne
(d) Ethane

16. Strongest hydrogen bond is shown by
(a) Water
(b) Ammonia
(c) Hydrogen fluoride
(d) Hydrogen sulphide.
17. Which of the following statements is not correct?
(a) Double bond is shorter than a single bond
(b) Sigma bond is weaker than a S(pi) bond
(c) Double bond is stronger than a single bond
(d) Covalent bond is stronger than hydrogen bond.
18. Among the following which compound will show the highest lattice energy?
(a) KF
(b) NaF
(c) CsF
(d) RbF
19. Which one of the following is the correct order of interactions?
(a) Covalent < hydrogen bonding < vander Waals < dipole-dipole
(b) vander Waals < hydrogen bonding < dipole < covalent
(c) vander Waals < dipole-dipole < hydrogen bonding < covalent
(d) Dipole-dipole < vander Waals < hydrogen bonding < covalent.
20. Strongest bond is in between
(a) CsF
(b) NaCl
(c) Both
(d) none of the above
21. Mark the incorrect statement in the following

(a) The bond order in the species O₂,  and  decreases as > O₂>
(b) The bond energy in a diatomic molecule always increases when an electron is lost
(c) Electrons in anti bonding M.O. contribute to repulsion between two atoms.
(d) With increase in bond order, bond length decreases and bond strength increases.
22. The weakest among the following types of bonds is
(a) ionic
(b) covalent
(c) metallic
(d) H–bond.

24. Which of the following pairs will form the most stable ionic bond?
(a) Na and Cl
(b) Mg and F
(c) Li and F
(d) Na and F
25. Among the following orbital bonds, the angle is minimum between
(a) sp₃ bonds
(b) p_x and p_y orbitals
(c) H– O– H in water
(d) sp bonds.
26. Linus Pauling received the Nobel Prize for his work on
(a) atomic structure
(b) photosynthesis
(c) chemical bonds
(d) thermodynamics
27. The boiling point of p-nitrophenol is higher than that of o-nitrophenol because
(a) NO₂ group at p-position behave in a different way from that at o-position.
(b) intramolecular hydrogen bonding exists in pnitrophenol
(c) there is intermolecular hydrogen bonding in p-nitrophenol

(d) p-nitrophenol has a higher molecular weight than o-nitrophenol.
28. The distance between the two adjacent carbon atoms is largest in
(a) benzene
(b) ethene
(c) butane
(d) ethyne
29. Which of the following species is paramagnetic?
(a) O2^2-
(b) NO
(c) CO
(d) CN–
30. The BCl3 is a planar molecule whereas NCl3 is pyramidal because
(a) B-Cl bond is more polar than N-Cl bond
(b) N-Cl bond is more covalent than B-Cl bond
(c) nitrogen atom is smaller than boron atom
(d) BCl₃ has no lone pair but NCl₃ has a lone pair of electrons
31. The correct order of the O–O bond length in O₂, H₂O₂ and O₃ is
(a) O₂>O₃>H₂O₂
(b) O₃>H₂O₂>O₂
(c) O₂> H₂O₂> O₃
(d) H₂O₂> O₃> O₂
32. The ground state electronic configuration of valence shell electrons in nitrogen molecule (N₂) is written as KK s2s² , s^* 2s² , p2   ,  p2  2p2  bond order in nitrogen olecule is
(a) 0
(b) 1
(c) 2
(d) 3
33. The correct order of N- O bond lengths in NO ,  ,  and N₂O₄ is
(a) N₂O₄ >  >  > NO
(b) NO >   > N₂O₄ >
(c)  >  > N₂O₄  >NO
(d) NO > N₂O₄ >  >
34. Which of the following compounds has a -3 centre bond?
(a) Diborane
(b) Carbon dioxide
(c) Boron trifluroide
(d) Ammonia
35. The low density of ice compared to water is due to
(a) hydrogen-bonding interactions
(b) dipole-dipole interactions
(c) dipole-induced dipole interactions
(d) induced dipole-induced dipole interactions
36. The cylindrical shape of an alkyne is due to the fact that it has
(a) three sigma C –  C bonds
(b) two sigma C – C and one “ p “ C – C  bond
(c) three “ p “ C – C  bond
(d) one sigma C – C and two “ p “ C – C  bond
37. N₂ and O₂ are converted into monocations,  and respectively. Which of the following statements is wrong?
(a) In N₂, the N—Nbond weakens
(b) In O₂, the O—Obond order increases
(c) In O₂, paramagnetism decreases
(d)  becomes diamagnetic
38. The AsF5 molecule is trigonal bipyramidal. The hybrid orbitals used by the As atom for bonding are
(a)  ,  , s, P_x , P_y
(b)  , s, P_x  , P_z
(c) s, P_x  , P_y ,
(d)  , P_x , P_y , P_z
39. The number of anti-bonding electron pairs in  molecular ion on the basis of molecular orbital theory is, (Atomic number of O is )8
(a) 5
(b) 2
(c) 3
(d) 4
40. Among the following which one is not paramagnetic? [Atomic numbers: Be = 4, Ne = 10, As = 33, Cl = ]17
(a) Cl^–
(b) Be⁺
(c) Ne²⁺
(d) As+
41. In PO₄^3– ion, the formal charge on each oxygen atom and P—O bond order respectively are
(a) –0.75, 0.6
(b) – 0.75, 1.0
(c) – 0.75, 1.25
(d) –3, 1.25
42. Which of the following molecules is planar?
(a) SF₄

(b) XeF₄
(c) NF₃
(d) SiF₄
43. The dipole moments of diatomic molecules AB and CD are 10.41 D and 10.27 D, respectively while  their bond distances are 2.82 and 2.67 , respectively. This indicate that
(a) bonding is %100 ionic in both the molecules
(b) AB has more ionic bond character than CD
(c) AB has lesser ionic bond character than CD
(d) bonding is nearly covalent in both the
molecules
44. Which one of the following arrangements represents the increasing bond orders of the given species?
(a) NO⁺ < NO< NO^–  <  
(b)  < NO^– <  NO < NO⁺
(c) NO ^– <   < NO< NO⁺
(d) NO < NO⁺ <  < NO^–
45. Which one of the following has the pyramidal shape?
(a)
(b) SO₃
(c) BF₃
(d) PF₃
46. Which one of the following molecules will form a linear polymeric structure due to hydrogen bonding?
(a) NH₃
(b) H₂O
(c) HCl
(d) HF
47. Among the following the electron deficient compound is:
(a) BCl₃
(b) CCl₄
(c) PCl₅
(d) BeCl₂
48. The relationship between the dissociation energy of N₂ and  is:
(a) Dissociation energy of > dissociation energy of N₂
(b) Dissociation energy of N₂ = dissociation energy of
(c) Dissociation energy of N₂ > dissociation energy of
(d) Dissociation energy of N₂ can either be lower or higher than the dissociation energy of

53. Main axis of a diatomic molecule is z, molecular orbital px and py overlap to form which of the following orbital?
(a)  – molecular orbital

(b)s – molecular orbital
(c)  – molecular orbital
(d) No bond will be formed.
54. In – H – Y, X and Y both the electronegative elements
(a) Electron density on X will increase and on H will decrease
(b) In both electron density will decrease
(c) In both electron density will increase
(d) Electron density will decrease on X and will increase on H

56. In NO3– ion number of bond pair and lone pair of electrons on nitrogen atom respectively are

(a) 2, 2
(b) 3, 1
(c) 1, 3
(d) 4, 0
57. Which of the following statements is not correct for sigma and pi-bonds formed between two carbon atoms?
(a) Sigma-bond determines the direction between carbon atoms but a pi-bond has no primary effect in this regard
(b) Sigma-bond is stronger than a pi-bond
(c) Bond energies of sigma- and pi-bonds are of the order of 264 kJ/mol and 347 kJ/mol, respectively
(d) Free rotation of atoms about a sigma-bond is allowed but not in case of a pi-bond
58. Among the following the pair in which the two species are not isostructural is
(a) SiF₄ and SF₄
(b)  and XeO₃
(c)   and
(d)  and SF₆
59. In  a regular octahedral molecule, MX₆ the number of X – M – X bonds at 180⁰ is (a) three
(b) two
(c) six
(d) four
60. H2Ois dipolar, whereas BeF2 is not. It is because
(a) the electronegativity of F is greater than that of O
(b) H2O involves hydrogen bonding whereas BeF2 is a discrete molecule
(c) H2O is linear and BeF2 is angular
(d) H2O is angular and BeF2 is linear

62. In BrF3 molecule, the lone pairs occupy equatorial positions to minimize
(a) lone pair – bond pair repulsion only
(b) bond pair – bond pair repulsion only
(c) lone pair – lone pair repulsion and lone pair bond pair repulsion
(d) lone pair – lone pair repulsion only
63. Which of the following is the electron deficient molecule?
(a) C₂H₆
(b) B₂H₆
(c) SiH₄
(d) PH_­
64. The correct sequence of increasing covalent character is represented by
(a) LiCl < NaCl < BeCl₂
(b) BeCl₂ < LiCl < NaCl
(c) NaCl < LiCl < BeCl₂
(d) BeCl₂ < NaCl < LiCl
65. Which of the following molecules has trigonal planar geometry?
(a) BF₃
(b) NH₃
(c) PCl₃
(d) IF₃
66. Which of the following would have a permanent dipole moment?
(a) SiF₄
(b) SF₄
(c) XeF₄
(d) BF₃
67. Which of the following is not a correct statement?
(a) The canonical structures have no real existence
(b) Every AB₅ molecule does in fact have square pyramidal structure
(c) Multiple bonds are always shorter than corresponding single bonds
(d) The electron-deficient molecules can act as Lewis acids
68. The number of unpaired electrons in a paramagnetic diatomic molecule of an element with atomic number 16 is
(a) 3
(b) 4
(c) 1
(d) 2
69. In which of the following molecules all the bonds are not equal?
(a) BF₃
(b) AlF₃
(c) NF₃
(d) ClF₃

70. The electro negativity difference between N and F is greater than that between N and H yet the dipole moment of NH31.5( D) is larger than that of NF30.2( D) . This is because
(a) in NH3 the atomic dipole and bond dipole are in the same direction whereas in NF₃ these are in opposite directions
(b) in NH₃ as well as NF₃ the atomic dipole and bond dipole are in opposite directions
(c) in NH₃ the atomic dipole and bond dipole are in the opposite directions whereas in NF₃ these are in the same direction
(d) in NH₃ as well as in NF₃ the atomic dipole and bond dipole are in the same direction

74. The correct order of C–O bond length among CO,   , CO₂ is
(a) CO< < CO₂
(b) < CO₂ < CO
(c) CO< CO2 <
(d) CO₂ < < CO
75. The angular shape of ozone molecule (O₃) consists of:
(a) 1 sigma and 2 pi bonds
(b) 2 sigma and 2 pi bonds
(c) 1 sigma and 1 pi bonds
(d) 2 sigma and 1 pi bonds

 
78. What is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OH to a gas?
(a) Dipole-dipole interaction
(b) Covalent bonds
(c) London dispersion force
(d) Hydrogen bonding
79. In which of the following molecules / ions BF₃ ,  ,   and H₂O , the central atom is sp² hybridized ?
(a)  and H₂O
(b)   and H₂O
(c) BF₃ and
(d)   and   
80. In which of the following pairs of molecules/ ions, the central atoms have sp2 hybridization?
(a)  and NH₃
(b) BF₃ and
(c)   and HO₂
(d) BF₃ and

82. Some of the properties of the two species, and  H₃O⁺ are described below. Which one of them is correct?
(a) Similar in hybridization for the central atom with different structures.
(b) Dissimilar in hybridization for the central atom with different structures.
(c) isostructural with same hybridization for the central atom.
(d) Isostructural with different hybridization for the central atom.

Solutions  :

1.ANS: (A)

The overlap between s- and p –orbital’s occurs along inter nuclear axis axis and hence the angle is 180⁰
2.Ans: (b) Equilateral or triangular planar shape involves sp² hybridization
3.Ans: (a) For linear arrangement of atoms the hybridisation should be sp(linear shape, 180⁰ angle) Only H₂S has sp^3- hybridization and hence has angular shape while C₂H₂ , BeH₂ and CO₂ all involve sp- hybridization and ehcne have linear arrangement of atoms.
4.Ans: (d) In metallic bonds each ions is surrounded by equal no of oppositely charged ions hence have electrostatic attraction on all sides and hence do nto have directional characteristics.
5.Ans: (b) BF₃ involves sp² – hybridization
6.Ans: (b) BeF₂ is linear and hence has zero dipole moment while H₂O being a bent molecule , has a finite or non zero dipole moment.

7. Ans: (c) As move in period form Li® Be® B ® C , the electro negativity (EN) increases and hence the EN difference between the element and Cl decreases and accordingly the covalent character increases.
8.Ans: (a) A s bond is stronger than a p bond. Sigma s bonds are formed by head on overlap of unhybridised s-s , p- p , or s-p orbitals and hybridised orbitals (sp , sp² , sp³ , sp³d and sp³d²) hence s bonds are strong bonds. Where as Pi(p) bonds are formed by side ways overlap of unhybridised p –  and d oribtals hence p bonds are weak bonds.

9.Ans: (d) With the increase of electro negativity and decrease in size of the atom to which hydrogen is covalently linked , the strength of hydrogen bond increases. As F is most electronegative thus HF shows maximum strength of hydrogen bond.

10.Ans: (a) Linear combination of two hybridized orbital’s leads to the formation of sigma bond.
11. Ans: (d)

12.Ans: (d) CCl₄ has sp³ hybridisation, tetrahedral geometry and  all bond angles of 109⁰ 28^’

13.Ans: (c) Polarity of the bond depends upon the electro negativity differences of the two atoms forming the bond. Greater the electro negativity differences, more is the polarity of the bond.

N – Cl              O –F        N – F                N-N

3.0- 3.0           3.5-4.0   3.0-4.0         3.0-3.0

As the electro negativity difference between N and f is maximum hence this bond is most polar.

14.Ans: (c) The bond length decreases in the order sp³ > sp² > sp.

Because of the triple bond, the carbon – carbon bond distance in ehtyne is shortest.

15.Ans: (b)CO₂ has sp – hybridization and is linear. SO₂ and  are planar (sp²) while  is tetrahendral (sp³)

16.Ans: (c) H- F shows strongest H bonds
17.Ans: (b) Sigma bond is stronger than p – bond. The electrons in the p bond are loosely held. The bond is easily broken and is more reactive than s – bond. Energy released during sigma

bond formation is always  more than p bond because of greater extent of overlapping.
18.ns: (b) For compounds containing ions of same charge, lattice energy increases as the size the ions decreases. Thus NaF has highest lattice energy. The size of cation is in the order Na⁺ < K⁺  , Rb⁺< Cs⁺

19. Ans: (b) The strength of the interactions follows the order vander Waal’s  < hydrogen – bonding  < dipole – dipole < covalent.

20.Ans: (a) According to Fajan rules, ionic character increases with increase in size of the cation and decrease in size of the anion. Thus, CsF has higher ionic character than NaCl and hence bond in CsF is stronger than NaCl

21.Ans: (b) The removal of an electron from a diatomic molecule may increase the bond order as in the conversion O₂(2) ® (2.5) or decreases the bond order as in the conversion, N₂(3.0)® (2.5) As a result , the bond energy may increase or decrease, thus. Statement (b) is incorrect
22.Ans: (d)H bond is the weakest
23.Ans: (b)BH_3­ has sp² hybridization and hence does not have tetrahedral structure while others have tetrahedral structure.
24.Ans: (b) the stability of the ionic bond depends uon the lactic energy which is expected to be more between Mg and F due to +2 charge on Mg atom.
25.Ans: (b) The angle between the bonds formed by p_x and P_v orbital’s is the minimum i.e. 90⁰
26.Ans: (c) Chemical bonds.
27.Ans: (c) The b.p of p –nitrphenol is higher than that of o –nitrophenol because in p-nitrophenol there is intermolecular H bonding but in o – nitrophenol it is intramolecular H bonding

28.Ans: (c) The C – C bond distance decreases as the multiplicity of the bond increases. Thus , bond distance decreases in the order. Butane (1.54  ) > benzene (1.39  ) > ethane (1.34 ) > ethyne (1.20 ) . Thus in butane , C – C bond distance is the largest,

29.Ans: (b) Paramagnetism is caused by the presence of atoms, ions or molecules with unpaired electrons. In NO the presence of unpaired electron is clear. Therefore it is paramagnetic.

30.Ans: (d) As there is no lone pair on boron in BCl₃ therefore no repulsion takes palace. But there isa lone pair on nitrogen in NCl₃ . Therefore repulsion takes place. Thus BCl₃ is planar molecule but  NCl₃ is a pyramidal molecules.
31.Ans: (d) The bond length of O – O is O₂ ios 1.21  , in H₂O₂ it is 1.48  and in O₃ It is 1.28 .

Correct order of bond length is H₂O₂ > O₃ > O₂

The bond represented by dots form the 3 – centred electron pair bond. The ides of three centred electron pair bond B –H –B briges is necessitated because diborane does not have sufficient electrons to form normal equivalent  bonds. It has only 12 electrons instead of14 required to give simple ethane like structure of diborane.

35.Ans: (a)We know that due to polar nature, water molecules are held together by intermolecular hydrogen bonds. The structure of ice is open with large number of vacant space, therefore the density of ice is less than water.

36.Ans: (d) In alkynes the hybridisation is sp i.e. each carbon atom undergoes sp hybridisation to form two sp – hybrid orbitals. The two 2p – orbitals remain hybridised. Hybrid orbitals form one sigma and two unsubsidised orbitals from p bonds.

Hence two p bond and one sigma bond between C – C lead to cylindrical shape.

37. Ans: (b) We know that in O₂ bond, the order is 2 and in  bond, the order is 1.5. therefore the wrong statement is (b)

38.Ans: (c) The electric configuration of As is

As = 1s² , 2s² , 2p⁶ , 3s¹ 3P_x  3P_y¹ 3P_z¹ 3d¹

Thus the hybridisation involved in the AsF₅  molecule is trigonal bipyramidal. So, the hybrid orbitals used area s., P_x , P_y , P_z , d_z².

44.Ans: (b)1.In molecular geometry, bond length or bond distance is defined as the average distance between nuclei of two bonded atoms in a molecule. It is a transferable property of a bond between atoms of fixed types, relatively independent of the rest of the molecule.2.The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. 3.Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms.

45.Ans: (d) PF₃ has pyramidal shape phosphorus exists in sp³ hybridiation state hence it exists in tetrahedral shape. But due to presence of lone pair its shape is pyramidal.F – H – F – H – F – H – F

46.Ans: (d) HF form linear polymer structure due to hydrogen bonding.

47.Ans: (a) Boron in BCl₃ has 6 electrons in outer most shell. Hence BCl₃ is a electron pair deficient

48.Ans: (c) Dissociation energy of any molecules depends upon bond order. Bond order in N₂ molecule is 3 wile bond order in N⁺₂ is 2.5 Further we know that more the Bond order, more is the stability and more is the B.D. E

49.Ans: (a) We know that the electrostatic force that bonds the oppositely charged ions which are formed by transfer of electron from one atom to another is called ionic bond. We also know that cation and anion are oppositely charged particles therefore they form ionic bond in crystal.
50.Ans: (c) In P – O bond, p bond is formed by the side wise overlapping of d- orbital of P and p –orbital of oxygen. Hence it is formed by pp and dp overlapping

51.Ans: (c)

We know that bond angles of NH₃ = 107⁰ .,

SCl₂ molecule has a bond angle of 105⁰  and has sp₃ hybridization.

 has 4 bond pairs and no lone pairs and is sp³ hybridized with the geometry and shape both being tetrahedral with a bond angle of 109.5⁰  and so it has the highest bond angle among the four.52.
Ans: (c) In XeF₂ and . Both XeF₂ and  are sp³ d hybridized and have planar shape.

53.Ans: (a) For p overlap the lobes of the atomic orbitals are perpendicular to the line joining the nuclei.

54.(a)
In X – H — Y,  X and Y both are electronegative element (i.e. attracts the electron pair) then electron density on x will increase and on H will decrease

55.Ans: (b)

N,B, and C are the elements of second period. These elements cannot form p_π−d_π bonds as they do not possess d− orbitals in their valence shell. On the other hand, S, an element of third period has, 3d orbitals and, hence, can form p_π−d_π bonds with oxygen. In this bonding, the d− orbitals of S overlap with the p – orbitals of the O atom.

56.Ans: (d)

p(pi) bond, so  it must be having  higher bond energy  than p (pi) bond.

58.Ans: (a)
→IO₃​⁻ has 4 pairs (3 bond pair + 1 lone pair )
→XeO₃​ has 4 pairs [3 bond pair + 1 lone pair] so they are isostructural
→PF⁻ has 6 pairs [6 bond pair] & SF₆​ has 6 pairs [6 bond pairs]so they are isostructural.
→BH₄⁻​ has 4 pairs [4 bond pair] & NH₄⁺​ has 6 pairs [4 bond pairs]so they are isostructural.
→CO₃²⁻​ has 3 pairs [3 bond pair ] &NO₂⁻​ has 3 pair [2 bond pair + 1 lone pair] so they are isostructural.
→SIF₄​ has 4 pairs [4 bond pair ] &SF₄​ has 5 pair [4 bond pair + 1 lone pair] so they are not isostructural.

60.Ans: (d) In a linear symmetrical molecule like BeF₂ , the bond angle between three atoms is 180⁰ , hence the polarity due to one bond is cancelled by the equal polarity due to other bond, while it si not so in angular molecules, like H₂O

61. Ans: (a)Only those d orbital whose lobes are directed along X, Y and Z directions hybridsis with s and p orbitals. In other three d orbitals namely d_xy , d_yz  and d_xz , the lobes are at an angle of 45⁰ from both axis, hence the extent of their overlap with s and p orbitals is much lesser than  and  orbitals.

In BrF₃,both bond pairs as well as lone pairs of electrons are present. Due to thee presence of lone pairs of electrons (lp) in the valence shell, the bond angle is contracted and the molecule takes the T –shape. This is due to greater repulsion between two lone pairs or between a lone pair and a bond pair than between the two bond pairs.

63.Ans: (b)

The compound, of which central atom is octetless known as electron deficient compound. Hence B₂H₆ is electron deficient compound.

64.Ans: (c) As difference of electro negativity increases  % ionic character increases and covalent character decreases i.e. negativity difference decreases i.e. negativity difference decreases covalent character increases.

Further  greater the charge on the cation more will be its covalent character. Be has maximum (+2) charge.

65.(a)  is sp² hybridised. So, it is trigonal planner. NH₃ , PCl₃ has sp³ hybridisation hence has trigonal bipyramidal shape, IF₃ , has sp³d hybridization and has linear shape.   

66.Ans: (b)

SF₄ has permanent dipole moment.

SF₄ has sp³d hybridization and see saw shape ( irregular geometry)

70.Ans: (a) In NH₃ the atomic dipole and bond dipole are in the same direction whereas in NF₃ these are in opposite direction so in the former case they are added up whereas in the later case net result is reduction of dipole moment. It has been shown in the following figure.

the structure  is sp³. But in SCl₄ , sp³dhybridisationis sp³ . but in SCl₄ , sp³d hybridisation present so shape is different which is see saw.

74.Ans: (c)
The higher the bond order, the shorter the C–O bond length will be. CO has a bond order of 3 (this is incorrect; it has a bond-order of at least 2.5). The carbonate anion has an average bond order of 1.33. Carbon dioxide has a bond order of 2. Hence, the order of increasing C–O bond lengths is: CO<CO₂< .

75.Ans: (d)The shape of ozone molecule is

Thus smaller is bond angle. The correct order of bond angle is NO₂     NO_­2     NO₂ I.e.(b) is correct

Ans: (c)

(a) NH⁺₄  :   sp³ hybridisation

(b) CH₄ :  sp³ hybridisation

(c) SF₄ :sp³ d hybridisation

(d)  : sp³ hybridisation

Correct choice (c)

84.Ans: (b)

ons)

= ​×(6−3)

=1.5.

88.Ans: (a)

89.Ans: (a)​&B2​ Bond order is one

90.Ans: (a)Molecular orbital configuration of O₂​(total  e− = 16)The electronic confiugration of oxygen is given as :

NO of electrons in CO= 6 +8 = 14

No of electrons in NO⁺ = 7+8 -1 = 14

CO and NO⁺ are isoelectronic species.

Isoelectronic species have identical and order/

98.Ans: (d)
In NO→NO⁺, an electron is removed from antibonding molecular orbital so bond order increases and nature changes from paramagnetic to diamagnetic.  99.
99.Ans: (d)

104.(d) According to V.S.E.P.R.T lone-lone pair repulsion is maximum because lone pair electron held by nuclei of one atom there for occupy more space.

Repulsion ⇒⇒ lone pair-lone pair > lone pair-bond pair > bond pair-bond pair

119 (b) For compounds containing ions of same charge, lattice energy increases as the size of ions decreases. Thus NaF has highest lattice energy.

            61=7

Therefore, the correct order of energy would be 5f > 6p > 5p > 4d.

Hence, correct option is 1

126 (3) On going down the group thermal stability order for H₂E decreases because H – E bond energy decreases.

Order for stability :

H₂Po < H₂Te < H₂Se < H₂S < H₂O   

127(3) C₂ is the diatomic molecular species which has only π bonds according to molecular orbital theory.

128(3) H₂ is noble gas element and they are monoatomic.

129(4)BF₃ , BeF₂ , CO₂ & 1, 4- dichloro benzene all are symmetrical structure. The   species with polar bonds have zero dipole moment if resultant of all dipoles is cancelled due to symmetrical structure.

130 (4)