MCQs for Class 11 Chemical Bonding and Molecular Structure

Chemical Bonding And Molecular Structure Multiple Choice Questions

Question 1. The Sp³d²-hybridization of the central atom of a molecule would be

1. Wouldsquareleadplanarto— geometry
2. Tetrahedral geometry
3. Trigonal bipyramidal geometry
4. Octahedral geometry

Answer: 4. Octahedral geometry

One s, three p, and two d -orbitals mix up together to form six equivalent Sp³d²-hybrid orbitals. The molecules, in which these orbitals are involved, have octahedral geometry.

Question 2. Which of the following is paramagnetic

1. N²
2. NO
3. CO
4. O³

Answer: 2. No

From a consideration of electron distribution in molecular orbitals of NO, it is known that it has one unpaired electron. So, NO molecule is paramagnetic.

Question 3. In the electron-dot structure, calculate the formal charge from left to right nitrogen atom, \(\ddot{\mathrm{N}}=\mathrm{N}=\ddot{\mathrm{N}}-\)

1. -1,-1+1
2. -1,+1,-1
3. +1,-1,-1
4. +1,-1,+1

Answer: 2. -1,+1,-1

Formal charge = No. of valence electrons in the atom No. of unshared electrons \(-\frac{1}{2}\) No. of shared electrons.

• Formal charge on N-atom (1) =5-4- (4 ÷ 2) = -1
• Formal charge on N-atom (2) = 5- 0- (8 ÷ 2) = 1
• Formal charge onN-atom (3) = 5- 4- (4÷ 2) = -1

Question 4. Which of the following Compounds has the maximum volatility

Answer: 3. In O-hydroxy carboxylic acid, the —OH and —COOH groups are situated at two adjacent carbon atoms of the ring and are involved in intramolecular H-bond formation.

So, these molecules exist as discrete molecules and consequently, the compound has maximum volatility.

Question 5. The number of acid protons in H³PO³ is

1. 0
2. 1
3. 2
4. 3

Answer: 3. 2 Number of the —OH group in H₃PO₃ is 2 and henceitis dibasic in nature.

Question 6. In 2-butene, which of the following statements is true—

1. C¹ —C² bondis a sp³-sp³ tr – bond
2. C²— C³ bond is a sp³-sp² or- bond
3. C¹—C² bond is a sp³-sp³ r-bond
4. C¹—C² bond is a sp²-sp² cr-bond

Answer: 3. C¹—C² bond is a sp³-sp² r-bond

Question 7. The paramagnetic behavior of B² Is due to the presence of—

1. 2 unpaired electrons π_b MO
2. 2 unpaired electrons in π* MO
3. 2 unpaired electrons in σ* MO
4. 2 unpaired electrons in σ_b MO

Answer: 1. The electronic configuration of B₂ (10 electrons) is \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p x}\right)^1\left(\pi_{2 p y}\right)^1\)

Since the molecule contains two unpaired electrons in πbMO, it is paramagnetic.

Question 8. The state of hybridization of the central atom and the number of lone pairs over the central atom in POCl₃ are-

1. sp,O
2. sp²O
3. sp³,O
4. dsp²,1

Answer: 3. The Geometrical shape of the POC1₃ molecule is tetrahedral. In POC1₃ central P-atom is sp³ hybridised and has no lone pair.

Question 9. CO is practically non-polar since—

1. The cr -electron drift from c to o is almost nullified by the n -electron drift from 0 to c
2. The tr -electron drift from o to c is almost nullified by the 7r -electron drift from c to o
3. The bond moment is low
4. There is a triple bond between c and O

Answer: 1. The cr -electron drift from c to o is almost nullified by the n -electron drift from 0 to c

Oxygen donates an unshared pair of electrons to carbon and helps it to complete its octet by forming a dative n bond with it. As a result, a much stronger n -n-moment acts from oxygen to a carbon atom, and this moment is almost canceled by the cr -moment and the weak n -moment acting in the opposite direction. Hence, the polarity of the molecule is verylow.

Question 10. The increasing order of the O —N —0 bond angle in the species NO₂, NO⁺₂, and NO^–₂is—

1. \(\mathrm{NO}_2^{+}<\mathrm{NO}_2<\mathrm{NO}_2^{-}\)
2. \(\mathrm{NO}_2<\mathrm{NO}_2^{-}<\mathrm{NO}_2^{+}\)
3. \(\mathrm{NO}_2^{+}<\mathrm{NO}_2^{-}<\mathrm{NO}_2\)
4. \(\mathrm{NO}_2<\mathrm{NO}_2^{+}<\mathrm{NO}_2^{-}\)

Answer: None; the correct order is \(\mathrm{NO}_2<\mathrm{NO}_2^{-}<\mathrm{NO}_2^{+}\)

Question 11. The ground state electronic configuration of the CO molecule is-

1. \(1 \sigma^2 2 \sigma^2 1 \pi^2 3 \sigma^2\)
2. \(1 \sigma^2 2 \sigma^2 3 \sigma^2 1 \pi^2 2 \pi^2\)
3. \(1 \sigma^2 2 \sigma^2 1 \pi^2 3 \sigma^2 2 \pi^2\)
4. \(1 \sigma^2 1 \pi^4 2 \sigma^2 3 \sigma^2\)

Answer: 4. \(1 \sigma^2 1 \pi^4 2 \sigma^2 3 \sigma^2\)

The ground state outer electronic configuration of the CO molecule is \(1 \sigma^2 1 \pi^4 2 \sigma^2 3 \sigma^2\)

Question 12. In diborane, the number of electrons that account for bonding die bridges is

1. Six
2. Two
3. Eight
4. Four

Answer: 4. Four

In diborane, each bridging B——H——B bond is formed by two electrons. Hence, four electrons account for bonding the bridges.

Question 13. In O₂ and H₂O₂, the O — O bond lengths are 1.2lA and 1.48A respectively. In ozone, the average O —O bond length is

1. 1.28A
2. 1.18A
3. 1.44A
4. 1.526A

Answer: 1. Bond length is nearly average of O—O length in

Question 14. In SOC1₂, the Cl—S—Cl and Cl—S—O bond angles are—

1. 130°, 115°
2. 106°, 96°
3. 107°, 108°
4. 96°, 106°

Answer: 4. 96°, 106°

IN SOC1₂, the Cl —S —Cl bond angle is 96° and the Cl — S —O bond angle is 106°, since multiple bonds create more repulsions than single bonds.

Question 15. The structure of XeF₆ is experimentally determined to be a distorted octahedron. Its structure according to VSEPR theory is—

1. Octahedron
2. Trigonal bipyramid
3. Pentagonal bipyramid
4. Tetragonal bipyramid

Answer: 3. Pentagonal bipyramid

In XeF₆, Xe is surrounded by 6 bond pairs and one lone pair. So, according to VSPER theory, the geometry (geometry of electron pairs) is pentagonal bipyramid.

Question 16. In the case of heteronuclear diatomics of the type AB, where A is more electronegative than B, bonding MO resembles the character of A more than that of B. The statement—

1. Is False
2. Is True
3. Cannot Be Evaluated Since the Data Is Not Sufficient
4. Is True Only FOr Certaqin Systems

Answer: 2. Cannot Be Evaluated Since Data Is Not Sufficient

As A is more electronegative, there is less energy difference between the atomic orbital of A and bonding M.O. Hence, bonding M.O. resembles A more closely.

Question 17. The bond angle in NF₃ (102.3°) is smaller than NH₃ (107.2°). This is because of—

1. Large size off compared to
2. The large size compared to
3. Opposite polarity of n in the two molecules
4. Small size compared to a ton

Answer: 3. In NF₃ molecules, the N—F bond pair is drawn

more towards the more electronegative F-atom. But in the NH₃ molecule, the N—H bond pair is drawn more towards the more electronegative N-atom. Therefore, the extent of bp-bp repulsion in NH3 is more than that in NF₃. As a consequence, the bond angle in NH₃(107.2°) is greater than that of NF₃(102.3°).

Question 18. The compound that will have a permanent dipole moment among the following is

1. 1
2. 2
3. 3
4. 4

Answer: 1. Permanent dipole moment means a non-zero value of dipole moment. So, only compound (1) has a permanent dipole moment.

Question 19. Among the following structures, the one which is not a resonating structure of others is—

1. 1
2. 2
3. 3
4. 4

Answer: 

Structure (4) is not a resonance structure because it involves shifting a pair of electrons as well as an H-atom

Question 20. The correct order of decreasing length ofthe bond as indicated by the arrow in the following structures is—

1. 1>2>3
2. 2>13
3. 3>2>1
4. 1>3>2

Answer: 3. In general, C=C and C—C bond lengths are respectively 1.33A and 1.54A

In structure I, n -electrons are delocalized over Ci — C2 and C2— C3 bonds. In structure II a pair of n electrons are delocalised Over C5-C6, C6-C7C6-C8

Question 21. The correct order of decreasing H—C—H angle in the following molecules is

1. 1>2>3
2. 2>1>3
3. 3>2>1
4. 1>3>2

Answer: Overlapping is maximum when orbitals overlap “endon” i.e., via a -bonding, n -bonds overlap laterally. The overlap in cyclopropane is neither end-on nor lateral but in between. So, it is intermediate between cr -and n bonding.

So, in order of decreasing H —C —H angle: 2 >1 > 3

Question 22. The correct order of decreasing length ofthe bond as indicated by the arrow in the following structures is—

1. 1>2>3
2. 2>1>3
3. 3>2>1
4. 1>3>2

Answer: 3. 3>2>1

Question 23. Thenumberoflone pairs of electrons on the central atoms of H₂O, SnCl₂ PCl_3, and XeF₂ respectively, are

1. 2,1,1,3
2. 2,2,1,3
3. 3,1,1,2
4. 2,1,2,3

Answer: 4. Number of lone pairs of electrons present in the hybrid orbital, L = H- X- D [Where H: no. of orbitals involved in the hybridization, X: no. of monovalent atoms surrounding the central atom, D: no. of bivalent atoms attached to the central atom.

Question 24. The number of car and n -bonds present between the two carbon atoms of calcium carbide are respectively

1. 1σ, 1π- bond
2. 1σ, 2π- bond
3. 2σ, 1π- bond
4. \(1 \sigma, 1 \frac{1}{2} \pi \text {-bond }\)

Answer: 2. 1σ, 2π- bond

In calcium carbide, two carbon atoms are bonded by a triple bond. Thus between two carbon atoms, 1 cr, and 2/r bonds are present.

Question 25. Which of the following molecules have a shape like CO₂ 

1. HgCl₂
2. SnCl₂
3. C₂H₂
4. NO₂

Answer: 1. HgCl₂

Question 26. The ground state magnetic property of B₂ and C₂ molecules will be—

1. B, paramagnetic and C₂ diamagnetic
2. B, diamagnetic C₂ paramagnetic
3. Botii are diamagnetic
4. Both are paramagnetic

Answer: 1. B, paramagnetic and C₂ diamagnetic

MO electronic configuration of B₂

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^1\left(\pi_{2 p_y}\right)^1\)

Mo electronic configuration of C₂

⇒ \(\operatorname{KK}\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\)

In BO, there are unpaired electrons present which is paramagnetic. But in C9 there is no impaired electron and henceitis diamagnetic.

Question 27. The shape of XeFg- is—

1. Square pyramidal
2. Triangularbipyramidal
3. Planar
4. Pentagonal bipramidal

Answer: 3. Planar

⇒ \(\text { Here, } \mathrm{H}=\frac{1}{2}(8+5-0+1)=7\)

∴ Central atom Xe: sp³d² -hybridized

No. of lone pair of electronic Xe, L = (7-5-0) = 2

therefore XeF₅ ionic planar.

Question 28. Which statements are correct for the peroxide ion—

1. It has five filled anti-bonding molecular orbitals
2. It is diamagnetic
3. It has bond order one
4. It is isoelectronic with neon

Answer: 1. MO electronic configuration of 0|~ (peroxide ion):

\(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\) \(\left(\pi_{2 p x}\right)^2\left(\pi_{2 p y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2\)

Bond Order \(=\frac{N_b-N_a}{2}=\frac{10-8}{2}=1\)

Question 29. Which ofthe following has the strongest H-bond

1. H-O…shape
2. S-H….O
3. F-H….F
4. F-H….O

Answer: 3. F-H….F

Since fluorine is the most electronegative element F —H bond is more polar which forms the strongest H-bonding [F—H-—F] among the given compounds.

Question 30. B cannot form which ofthe following anions

1. \(\mathrm{BF}_6^{3-}\)
2. BH-4
3. B(OH)-4
4. BO-2

Answer: 1. \(\mathrm{BF}_6^{3-}\)

The stability of hydrides of group-15 decreases from NH3 to BiH3 due to an increase in the size of the central atom.

Question 31. Which of the following statements is wrong—

1. Nitrogen cannot form an-inbound
2. Single n-nbond is weaker than the single p-p bond
3. N₂O₄ has two resonance structures
4. The stability of hydrides increases from nh3 to bih3 due to the increase in size ofthe central atom.

Answer: 4. The stability of hydrides increases from NH₃ to BiH₃ due to the increase in size ofthe central atom.

Question 32. The Square Of IF₇ is

1. Square pyramid
2. Trigonal bipyramid
3. Octahedral
4. Pentagonal bipyramid

Answer: 4. Pentagonal bipyramid

In IF₇, the hybridization of the central atom is sp³d³ and its structure is a pentagonal bipyramid.

Question 33. The hybridization orbitals of N-atoinin NO-3, NO+2, and NH+4 are respectively—

1. sp, sp², sp³
2. Sp²,sp,sp³
3. Sp,sp³,s²
4. sp², sp³ sp

Answer: 2. Sp²,sp,sp³

Question 34. Among die following, die maximum covalent character is shown by

1. FeCl₂
2. SnCl₂
3. AlCl₃
4. MgCl₂

Answer: 3. AlCl₃

The ionic potential (0) of the cations increases with the increase in cationic charge and decrease in cationic radii. Thus, the resulting compound possesses a more covalent
character. So, A1C1₃ exhibits maximum covalency.

Question 35. Iron exhibits +2 and +3 oxidation states. Which of the following statements about incorrect—

1. Ferrous compounds are relatively more ionic than ferric compounds
2. Ferrous compounds are less volatile than the corresponding ferric compounds
3. Ferrous compounds are more easily hydrolyzed than the corresponding ferric compounds
4. Ferrous oxide is more basic than ferric oxide

Answer: 3. Ferrous compounds are more easily hydrolyzed than the corresponding ferric compounds

The tendency of hydration increases with a decrease in the size of a cation. Ferrous ions are larger than ferric ions. Consequently, the ferric ion will be more easily hydrolyzed than the ferrous ion.

Question 36. Ortho-nitrophenol is less soluble in water than p – and m nitrophenols because—

1. O-nitrophenol shows intramolecular-bonding
2. O-nitrophenol shows intermolecular-bonding
3. The melting point of o-nitrophenol is less than that of mand p-isomers
4. O-nitrophenol is more volatile than those of m-and prisoners

Answer: 1. o-nitrophenol shows intramolecular H-bonding

In o-nitrophenol, —OH & —N02 groups are situated at two adjacent carbon atoms of the ring and involved in intramolecular bonding. So, o-nitrophenol is less soluble in water than p- and m- m-nitrophenol.

Question 37. The molecule having the smallest bond angle is

1. AsCl₃
2. SbCl₃
3. PCl₃
4. NCL₃

Answer: 2. SbCl₃

The bond angle of a molecule increases with the increases in electronegativity or with the decrease in size of the central atom. Hence, the correct order of bond angle is: SbCl₃ < ASC1₃ < PC1₃ < NCl₃

Question 38. In which of the following pairs the two species are not isostructural

1. PCI+6 and SiCl₄
2. PF₅ and BrF₅
3. AlFg- and SF₆
4. CO₆– and NO-₃

Answer: 2. PF₅ and BrF₅

In molecule PF5, \(H=\frac{1}{2}\) [5 + 5- 0 + 0] = 5. Hybridisation of central atom(P): sp₃d. Number of p lone pairs in the central atom (P): 0. Shape of the molecule: Trigonal bipyramidal.

For BrF₅, Hybridisation of central atom (Br): sp₃d₂. Number of loner pairs in the central atom (Br): 1. Shape ofthe molecule: square pyramidal.

Question 39. The stability of the species Li₂, Li₂, and Li₂ increases in the order of

1. Li₂<Li+2<Li-₂
2. Li2₂< Li⁺₂ < Li₂
3. Li₂<Li^–₂<Li⁺₂
4. Li₂<Li₂<Li₂

Answer: 2. Li₂< Li-₂ < Li₂

⇒ \(\mathrm{Li}_2:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\)

⇒ \(\begin{aligned}
& \mathrm{Li}_2^{+}:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^1 \\
& \mathrm{Li}_2^{-}:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^1
\end{aligned}\)

\(\begin{array}{|c|c|c|c|}
\hline \mathbf{B . 0} & \mathbf{L i}_{\mathbf{2}} & \mathbf{L i}_{\mathbf{2}}^{+} & \mathbf{L i}_{\mathbf{2}}^{-} \\
\hline \frac{N_b-N_a}{2} & \frac{2-0}{2}=1 & \frac{1-0}{2}=0.5 & \frac{2-1}{2}=0.5 \\
\hline
\end{array}\)

Question 40. In which ofthe following pairs of molecules/ions, both the species are not likely to exist—

1. \(\mathrm{H}_2^{+}, \mathrm{He}_2^{2-}\)
2. \(\mathrm{H}_2^{-}, \mathrm{He}_2^{2-}\)
3. \(\mathrm{H}_2^{2+}, \mathrm{He}_2\)
4. \(\mathrm{H}_2^{-}, \mathrm{He}_2^{2+}\)

Answer: 3. \(\mathrm{H}_2^{2+}, \mathrm{He}_2\)

⇒ \(\mathrm{H}_2^{2+}:\left(\sigma_{1 s}\right)^0\)

⇒ \(\mathrm{He}_2:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\)

\(\begin{array}{|c|c|c|}
\hline \text { B.o. } & \mathbf{H}_2^{2+} & \mathbf{H e}_{\mathbf{2}} \\
\hline \frac{N_b-N_a}{2} & 0 & \frac{2-2}{2}=0 \\
\hline
\end{array}\)

Question 41. Which one of the following molecules is expected to exhibit diamagnetic behavior—

1. C₂
2. N₂
3. O₂
4. S₂

Answer: 1. C₂

Question 42. The correct statement for the molecule, Csl₃, is—

1. It contains Cs⁺, I^– and lattice I₂ molecule
2. It is a covalent molecule
3. It contains Cs⁺ and I^–₂ ions
4. It contains Cs³⁺ and I^– ions

Answer: 3. It contains Cs+ and I-3 ions

Cs cannot show a +3 oxidation state. So, Csl₃ is formulated as Cs+ and I3 ions. It is a typical ionic compound.

Question 43. For which of the following molecules is significant μ≠ 0-

1. 3 and 4
2. only 1
3. 1 and 2
4. only 3

Answer: 1. 3 and 4

Question 44. Which of the following alkaline earth metal sulfates has its hydration enthalpy greater than its lattice enthalpy—

1. BaSO₄
2. SRSO₄
3. CaSO₄
4. BeSO₄

Answer: 1. BaSO₄

In the NO₄ ion, the N atom is sp -hybridized. O=N=0

Question 45. In which of the following molecule Orion the hybridization state of the-atom is sp —

1. NO+₂
2. NO-₂
3. NO-₃
4. NO₂

Answer: 4. NO₂

There is no unpaired electron in the MO electronic configuration of the CO molecule. Thus, it is not paramagnetic.

Question 46. Which one of the following is not paramagnetic-

1. O₂
2. B₂
3. NO
4. CO

Answer: 4. CO

KC1 is an electrovalent compound, it exists as K+ and Cl- ions

Question 47. The total number of one pair of electrons I-3 ion is

1. 9
2. 3
3. 13
4. 6

Answer: 1. 9

Question 48. Which ofthe following compounds contain(s) no covalent bond(s)—KC1, PH₃, O₂, B₂H₆, H₂SO₄

1. KCL
2. KCl, B₂H₆
3. KClB₂H₆,PH₃
4. KCl, H₂SO₄

Answer: 1. KCL

KC1 is an electrovalent compound, it exists as K+ and Cl- ions

Question 49. According to molecular orbital theory, which of the following will not be a viable molecule—

1. H-₂
2. H2-₂
3. He2+₂
4. He+₂

Answer: 2. MO electronic configuration of H2-2 \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}\right)^2\)

Therefore Bond Order \(=\frac{2-2}{2}=0\)

Question 50. In which of the following pairs of molecules/ions, do the central atoms have sp² hybridization—

1. NO-₂ And NH₃
2. BF₃ And NO-2
3. NH-₂ And H₂O
4. BF₃ And NH-₂

Answer: 2. BF₃ And NO-₂

\(\begin{array}{|c|c|c|}
\hline \begin{array}{c}
\text { Molecule / } \\
\text { ion }
\end{array} & \text { Value of } \mathbf{H} & \begin{array}{c}
\text { Type of } \\
\text { hybridisation }
\end{array} \\
\hline \mathrm{BF}_3 & \mathrm{H}=\frac{1}{2}(3+3-0+0) & s p^2 \\
& =3 & \\
\hline \mathrm{NO}_2^{-} & \mathrm{H}=\frac{1}{2}(5+0-0+1) & s p^2 \\
\hline
\end{array}\)

Question 51. Considering the state of hybridization ofC-atoms, find out the molecule among the following which is linear-

1. CH₃—CH₂—CH₂—CH₃
2. CH₃—CH=CH—CH₃
3. CH₃—C=C—CH₃
4. CH₂=CH—CH₂—C=CH

Answer: 3. CH₃—C=C—CH₃

In the case of sp³, sp², and sp hybridized carbons, the bond angle is 109°28′; 120° and 180° respectively. So, only image- is linear (excluding H-atoms)

Question 52. Which ofthe following structures is the most preferred and
hence of the lowest energy for SO3-

Answer: 4. Has a maximum number of covalent bonds and hence is of the lowest energy.

Question 53. The correct order of increasing bond length of C—H C—O, C—C, and C=C is

1. C—H < C=C < C—O < C—C
2. C—C < C=C < C—0 < C—H
3. C—0<C—H<C—C<C=C
4. C—H<C—0<C—C<C=C

Answer: 4. C—H<C—0<C—C<C=C

⇒ \(\begin{aligned}
& \mathrm{C}-\mathrm{H}<\mathrm{C}=\mathrm{C}<\mathrm{C}-\mathrm{O}<\mathrm{C}-\mathrm{C} \\
& 107 \mathrm{pm} \quad 134 \mathrm{pm} \quad 141 \mathrm{pm} \quad 154 \mathrm{pm}
\end{aligned}\)

Question 54. Which of the two ions from the list given below have the geometry that is explained by the same hybridization of orbitals, NO^–₂, NO^–₃, NH^–₂, NH^–₄, SCN^–

1. NO^–₂ and NO^–₃
2. NH⁺₄ and NO^–₃
3. SCN^– and NH^–₂
4. NO^–₂ and NH^–₂

Answer: 1. Increasing the order of bond length is

⇒ \(\begin{array}{|c|c|c|c|c|c|}
\hline \text { Ions } & \mathrm{NO}_3^{-} & \mathbf{N O}_2^{-} & \mathbf{N H}_2^{-} & \mathbf{N H}_4^{+} & \mathbf{S C N}^{-} \\
\hline \text { Hybridisation } & s p^2 & s p^2 & s p^3 & s p^3 & s p \\
\hline
\end{array}\)

Question 55. Which has the minimum bond length—

1. O⁺₂
2. O^–₂
3. O₂^2-
4. O₂

Answer: 1. O⁺₂

⇒ \(\mathrm{O}_2^*: \mathrm{KK}\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_1}\right)^2\left(\pi_{2 p}\right)^2\left(\pi_{2 p}^*\right)^1\)

⇒ \(\begin{aligned}
& \mathrm{O}_2^{-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^1 \\
& \mathrm{O}_2^{2-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x^*}^*\right)^2\left(\pi_{2 p_y}^*\right)^2
\end{aligned}\)

⇒ \(\mathrm{O}_2: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y}^*\right)^1\)

Question 56. Four diatomic species are listed below. Identify the correct order in which the bond order is increasing

1. \(\mathrm{NO}<\mathrm{O}_2^{-}<\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}\)
2. \(\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}\)
3. \(\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}\)
4. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

Answer: 4. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

NO(7=8=15)

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\)

Question 57. During the change of O₂ to O₂ ion, the electron adds on which one ofthe following orbitals—

1. π* -orbitals
2. π – orbitals
3. σ* orbital
4. σ- orbital

Answer: 1. π* -orbitals

⇒ \(\mathrm{O}_2: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y}^*\right)^1\)

⇒ \(\mathrm{O}_2^{-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p y}^*\right)^1\)

Thus, the electron goes into the π*-orbital.

Question 58. The pair of species with the same bond order is

1. \(\mathrm{O}_2^{2-}\)
2. \(\mathrm{O}_2^{+}, \mathrm{NO}^{+}\)
3. NO, CO
4. N₂, O₂

Answer: 1. \(\mathrm{O}_2^{2-}\)

⇒ \(\mathrm{O}_2^{2-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2\)

⇒ \(\mathrm{B}_2: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^1\left(\pi_{2 p_y}\right)^1\)

\(\begin{array}{|c|c|c|}
\hline \text { Bond order } & \mathbf{O}_2^{2-} & \mathbf{B}_2 \\
\hline \frac{N_b-N_a}{2} & \frac{8-6}{2}=1 & \frac{4-2}{2}=1 \\
\hline
\end{array}\)

Note: B.O. of O₂ = 2.5, N0+ = 3, NO = 2.5 CO = 3, N₂ = 3 and O₂ = 2]

Question 59. Which of the following species contains three bond pairs and one lone pair around the central atom—

1. H₂O
2. BF₂
3. NH-₂
4. PCl₂

Answer: 4. PCl₃

Question 60. Bond order of 1.5 is shown by—

1. O+₂
2. O-₂
3. O2-₂
4. O₂

Answer: 2. O-₂

⇒ \(\mathrm{O}_2^{+}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\)

⇒ \(\begin{aligned}
& \mathrm{O}_2^{+}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^1 \\
& \mathrm{O}_2^{2-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x^*}^*\right)^2\left(\pi_{2 p_y}^*\right)^2 \\
& \mathrm{O}_2: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y^*}\right)^1
\end{aligned}\)

\(\begin{array}{|c|c|c|c|c|}
\hline \begin{array}{c}
\text { Bond } \\
\text { order }
\end{array} & \mathbf{O}_2^{+} & \mathbf{O}_2^{-} & \mathbf{0}_2^{2-} & \mathbf{O}_2 \\
\hline \frac{N_b-N_a}{2} & \frac{8-3}{2}=2.5 & \frac{8-5}{2}=1.5 & \frac{8-6}{2}=1 & \frac{8-4}{2}=2 \\
\hline
\end{array}\)

Question 61. The following pairs are isostructural-

1. BC1₃ and BrCl₃
2. NH₃ and NO3-
3. NF₃ and BF₃
4. BF₄ and NH+₄

Answer: 4. BF₂ and NH+₄

If some bond pairs and lone pairs are the same for the given pairs, they are isostructural.

Question 62. Which contains no n -bond—

1. SO₂
2. NO₂
3. CO₂
4. H₂O

Answer: 4. H₂O

Question 63. Which ofthe following lanthanoid ions is diamagnetic (Atnos. Ce=58, Sm=62,Eu=63, Yb70)-

1. EU²⁺
2. Yb²⁺
3. Ce²⁺
4. Sm²⁺

Answer: 4. Sm²⁺

⇒ \(\begin{aligned}
& ; \mathrm{Sm}^{2+}(\mathrm{Z}=62):[\mathrm{Xe}] 4 f^6 \\
& \mathrm{Yb}^{2+}(\mathrm{Z}=70):[\mathrm{Xe}] 4 f^{14} \\
& \mathrm{Ce}^{2+}(\mathrm{Z}=58):[\mathrm{Xe}] 4 f^1 5 d^1 \\
& \mathrm{Eu}^{2+}(\mathrm{Z}=63):[\mathrm{Xe}] 4 f^7
\end{aligned}\)

Question 64. Which of the following is paramagnetic—

CN^–

NO⁺

CO

O^–₂

Answer: 4. O^–₂

MO electronic configuration of O-2(17):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi 2_{p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^1\)

Owing to the presence of one unpaired electron, it is paramagnetic in nature.

Question 65. Which of the following is a polar molecule—

1. SiF₄
2. XeF₄
3. BF₃
4. SF₄

Answer: 4. SF₄

SF₄ has sp³d -hybridization and see-saw shape with 4 bond pairs and 1 lone pair and resultant u=0.

Question 66. XeF₂ is isostructural with-

1. SbCl₃
2. BaCL₂
3. TeF₂
4. ICI-₂

Answer: 4. ICI-₂

Question 67. Which species has a plane triangular shape-

1. N₃
2. NO-₃
3. NO₂
4. CO₂

Answer: 2. NO-₃

Question 68. Which has the maximum dipole moment—

1. CO₂
2. CH₄
3. NH₃
4. NF₃

Answer: 3. NH₃

In NH₃, H is less electronegative than N and hence dipole moment of each N —H bond is towards N and creates a high net dipole moment.

Question 69. The formation ofthe oxide ion O^2-(g), form oxygen atom requires first an exothermic and then an endothermic step as shown below

⇒ \(\begin{aligned}
& \mathrm{O}(\mathrm{g})+e^{-} \rightarrow \mathrm{O}^{-}(\mathrm{g}) ; \Delta H_f^0=-141 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\
& \mathrm{O}^{-}(\mathrm{g})+e^{-} \rightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta H_f^0=+780 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Thus, the process of formation of O^2- in the gas phase is unfavorable even though O^2- is isoelectronic with neon. It is because—

1. Electron repulsion outweighs the stability gained by achieving noble gas configuration
2. O-ion has a comparatively smaller size than o-atom
3. Oxygen is more electronegative
4. The addition of electronic o results in large-size ion

Answer: 1. Electron repulsion outweighs the stability gained by achieving noble gas configuration

Electron repulsion predominates over the stability gained by achieving noble gas configuration. Hence, the formation of O^2- in the gas phase is unfavorable.

Question 70. In which of the following pairs, both the species are not isostructural—

1. SiCl₄, Pcl+₄
2. Diamond, siC
3. NH₃,PH₃
4. XeF₄, XeO₄

Answer: 4. XeF₄, XeO₄

Diamond and silicon carbide (SiC), are both isostructural because their central atom is sp₃ hybridized and both have tetrahedral arrangements.

Both NH₃ and PH₃ have pyramidal geometry.

XeF₄ has sp³d² hybridisation while XeO₄ has sp³ hybridisation.

Hence, XeF₄ and XeO₄ are notisostructural.

Question 71. Decreasing order of stability is-

1. \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)
2. \(\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}>\mathrm{O}_2>\mathrm{O}_2^{+}\)
3. \(\mathrm{O}_2>\mathrm{O}_2^{+}>\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}\)
4. \(\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}>\mathrm{O}_2^{+}>\mathrm{O}_2\)

Answer: 1. \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

Order of stability ∞ bond order.

therefore The order of stability ofthe given species,

⇒ \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

Bond order: 2.5 2 1.5 1

Question 72. Which one of the following compounds §hows the presence of intramolecular hydrogen bond-

1. HCN
2. Cellulose
3. Conc.Acetic Acid
4. H₂O₂

Answer: 2. Cellulose

Only cellulose can perform intramolecular bonding whereas the other compounds can perform intermolecular-bonding only.

Question 73. Which of the following pairs of ions are isoelectronic and isostructural-

1. \(\mathrm{ClO}_3^{-}, \mathrm{SO}_3^{2-}\)
2. \(\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\)
3. \(\mathrm{ClO}_3^{-}, \mathrm{CO}_3^{2-}\)
4. \(\mathrm{SO}_3^{2-}, \mathrm{CO}_3^{2-}\)

Answer: 2. \(\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\)

Both CO²–₃ and NO₃ have a total of 32 electrons and both are triangular planar shape

Question 74. Among the following, which one is a wrong statement—

1. Pn-dn bonds are present in SO₂
2. SeF₄ and CH₄ have the same shape
3. 1+₃ has bent geometry
4. PH₃ and BiCl₅ do not exist

Answer: 2. SeF₄ and CH₄ have same shape

The shape of CH₄(sp₃) is regular tetrahedron. However, in SeF₄, the Se atom is sp3d-hybridized -hybridised and the presence of two lone pairs makes the molecule see-saw shaped.

Question 75. The hybridizations of atomic orbitals of nitrogen in NO+₂ NO-₃ and NH+₄ respectively are-

1. Sp², sp³ and sp
2. sp, sp² and sp³
3. sp², sp and sp³
4. sp, sp³ and sp²

Answer: 2. sp, sp² and sp³

⇒ \(\mathrm{NO}_2^{\oplus}(s p) \text {-linear; } \mathrm{NO}_3^{-}\left(s p^2\right) \rightarrow \text { trigonal planar }\mathrm{NH}_4^{\oplus}\left(s p^3\right) \text {-tetrahedral }\)

Question 76. Consider the molecules CH, NH₃, and H₂0. Which of the given statements is false—

1. The h—c—h bond angle in ch4, the h—n—h bond angle in nh3 & the h—o —h bond angle in h,0 are all greater than 90°
2. The h—o —h bond angle in h20 is larger than the h—c—h bond angle in ch₄
3. The h—o —h bond anglein h20 is smaller than the h —n—h bond anglein nh₃
4. The h—c—h bond angle in ch4 is larger than the h—n—h bond anglein nh

Answer: 2. The h—o —h bond angle in h20 is larger than the h—c—h bond angle in ch4

Question 77. Predict the correct order among the following—

1. Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
2. Lone pair-lone pair > bond pair-bond pair > lone pairlonepair
3. Bond pair-bond pair > lone pair-bond pair > lone pairlonepair
4. Lone pair-bond pair > bond pair-bond pair > lone pairlonepair

Answer: 1. Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair

According to VSEPR theory, Ip — Ip repulsion > Ip- bp repulsion > bp-bp repulsion.

Question 78. The charge-forming electron pair in the carbonation CH₃C=C exists in

1. sp -orbital
2. 2p-orbital
3. sp³ -orbital
4. sp² -orbital

Answer: 1. sp -orbital

CH₃CHC-, triply bonded carbon atoms are sp hybridized. Thus forming an electron pair is in sp orbital

Question 79. Match the compound given in column 1 with the hybridization and shape given in column 2 and mark the correct option:

Code: 1 2 3 4

1. 4-1- 2- 3
2. 1- 3- 4- 2
3. 1- 2- 4-3
4. 4- 3- 1 – 2

Answer: 2. 1- 3- 4- 2

Question 80. Which of the following pairs of species have the bond order

1. O₂, NO⁺
2. CN^–, CO
3. N₂,O₂^–
4. CO, NO

Answer: 2. Total no. of electrons present in CN = 14

• Total no. electrons present in CO = 14
• MO electronic configuration of CO

⇒ \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 p_z}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\)

MO electronic Configuration of CN-

⇒ \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\sigma_{2 p_z}\right)^2\)

⇒ \(\text { Bond order of } \mathrm{CO}=\frac{1}{2}(8-2)=3\)

⇒ \(\text { Bond order of } \mathrm{CN}^{-}=\frac{1}{2}(8-2)=3\)

Question 81. The species, having angles of 120° is—

1. ClF₃
2. NCL₃
3. BCl₃
4. PH₃

Answer: 3. BCl₃

Question 82. Match the interhalogen compounds of column I with the geometry in column 2 and assign the correct code:

Code: (1) (2) (3) (4)

1. 3-1- 4- 2
2. 5- 4- 3- 2
3. 4- 3 – 2- 1
4. 3- 4 -1 -2

Answer: 1. 3-1- 4- 2

• XX ⇒ linear (sp³d)
• XX3 ⇒ T-shape (sp³d)
• XXg ⇒ square pyramidal (sp³d²)
• XXy ⇒ pentagonal bipyramidal (sp³d³)

Question 83. Consider the following species: CN+, CN-, NO, and CN Which one of these will have the highest bond order—

1. CN
2. NO
3. CN⁺
4. CN^–

Answer: 4. CN^–

• \(\text { B.O. of } \mathrm{NO}=\frac{10-5}{2}=2.5\)
• \(\text { B.O. of } \mathrm{CN}^{-}=\frac{10-4}{2}=3 \text {; }\)
• \(\text { B.O. of } \mathrm{CN}=\frac{9-4}{2}=2.5 \text {; }\)
• \(\text { B.O. of } \mathrm{CN}^{+}=\frac{8-4}{2}=2\)

Question 84. In the structure of C1F3, the number of lone pairs of electrons on central atom ‘Cl’ is-

1. Three
2. One
3. Four
4. Two

Answer: 4. Two

Question 85. The decreasing order of bond angle is—

1. BeCl₂ > NO₂ > SO₂
2. BeCl₂ > SO₂ > NO₂
3. SO₂ > BeCl₂ > NO₂
4. SO₂>NO₂>BeCl₂

Answer: 1. BeCl₂ > NO₂ > SO₂

Compound \(\begin{aligned}
& \mathrm{BeCl}_2>\mathrm{NO}_2>\mathrm{SO}_2 \\
& 180^{\circ} \quad 132^{\circ} \quad 119.5^{\circ} \\
&
\end{aligned}\)

Question 86. The dipole moment is minium in—

1. NH₃
2. NF₃
3. SO₂
4. BF₃

Answer: 4. BF₃

BF₃ has zero dipole moment.

Question 87. In BF₃, the B —F bond length is 1.30 A when BF3 is allowed to be treated with mMe₃ N, it forms an adduct, Me₃ N→BF₃, and the bond length of —F in the adduct is-

1. Greater than 1.30A
2. Smaller than 1.30A
3. Equal to 1.30A
4. None of these

Answer: 1. Greater than 1.30A

In BF₃, there is backbonding in between fluorine and boron due to the presence of -orbital in boron.

back bonding imparts double-bond characteristics

As BF₃ forms adduct, the back bonding Is no longer present and thus double bond characteristic disappears. Hence, the bond becomes a bit longer than earlier (1.30A).

Question 88. The total number of antibonding electrons present in O₂ will be

1. 6
2. 8
3. 4
4. 2

Answer: 1. MO electronic configuration of O₂

\(\begin{aligned}
& \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2 \\
& \left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y}^*\right)^1
\end{aligned}\)

Hence, the correct B.O. is O+2 → O-2 → 2-2

Question 89. Which ofthe following represents the correct bond order

1. \(\mathrm{O}_2^{+}<\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)
2. \(\mathrm{O}_2^{+}<\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)
3. \(\mathrm{O}_2^{2-}>\mathrm{O}_2^{+}>\mathrm{O}_2^{-}\)
4. \(\mathrm{O}_2^{+}>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

Answer: 4. \(\mathrm{O}_2^{+}>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

Question 90. In the O₃ molecule, the formal charge on the central O-atom is—

1. 0
2. -1
3. -2
4. +1

Answer: 4. +1

Lewis gave the structure of the O₃ molecule as

Using the relation, Formal charge = [Total no, of valence electrons in the free atom] – [Total no. of non-bonding (lone pair) electrons]-\(\frac{1}{2}\)[Total no. of bonding (shared) electrons]

The formal charge on central O -atom i.e., no. 1 = +1

Question 91. Four diatomic species are listed below in different sequences. Which of these represents the correct order of their increasing order

1. \(\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}<\mathrm{NO}<\mathrm{O}_2^{-}\)
2. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)
3. \(\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}\)
4. \(\mathrm{NO}<\mathrm{C}_2^{2-}<\mathrm{O}_2^{-}<\mathrm{He}_2^{+}\)

Answer: 2. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

According to molecular orbital theory, the energy level ofthe given molecules are

⇒ \(\mathrm{C}_2^{2-}-K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\sigma_{2 p_z}\right)^2\)

⇒ \(\begin{aligned}
& \text { B.O. }=\frac{1}{2}[10-4]=3 \\
& \mathrm{He}_2^{+}-\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^1
\end{aligned}\)

⇒ \(\begin{aligned}
& \text { B.O. }=\frac{1}{2}[2-1]=\frac{1}{2}=0.5 \\
& \mathrm{O}_2^{-}-K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2 \\
& \quad\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^1
\end{aligned}\)

⇒ \(\begin{aligned}
& \text { B.O. }=\frac{1}{2}[10-7]=1.5 \\
& \text { NO : }-K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2 \\
& \qquad\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y}^*\right)^0
\end{aligned}\)

⇒ \(\text { B.O. }=\frac{1}{2}[8-3]=2.5\)

So, the correct order of their increasing bond order is He+2 < O-₂ < NO < C²-₂

Question 92. Which of the following molecules has more than one lone pair

1. SO₂
2. XeF₂
3. SiF₄
4. CH₄

Answer: 2. XeF₂

Question 93. The ASF₅ molecule is trigonal bipyramidal. The hybrid orbitals used by the As atoms for bonding are—

1. \(d_{x^2-y^2}, d_{z^2}, s, p_x, p_y\)
2. \(d_{x y}, s, p_x, p_y, p_z\)
3. \(d_{x^2-y^2}, s, p_x, p_y\)
4. \(s, p_x, p_y, p_z, d_z^2\)

Answer: 4. \(s, p_x, p_y, p_z, d_z^2\)

AsF₅ has sp³d hybridisation. In sp²d hybridization, the dz² orbital is used along with the ‘s’ and three ‘p’ orbitals to form three equatorial bonds and two equally strong axial bonds for a trigonal bipyramid

Question 94. H₂O is polar, whereas BeF₂ is not because—

1. Electronegativity ofF is greater than that of O
2. H₂O involves H-bonding, whereas, BeF₂ is a discrete molecule
3. H₂O is angular and BeF₂ is linear
4. H₂O is linear and BeF₂ is angular

Answer: 3. H₂O is angular and BeF₂ is linear

Because of the linear shape, dipole moments cancel each other in BeF₂ (F—Be—F) and thus, it is non-polar, whereas H₂O is V-shaped and hence, it is polar

Question 95. Which of the following have the same hybridization but are not isostructural

1. C1F₃ and l-3
2. BrF₃ and NH₃
3. CH₄ and NH+₄
4. XeO₃ and NH₃

Answer: 1. C1F₃ and l-₃

1. C1F₃ (sp₃d, T-shape); I₃ (sp₃d, linear)
2. BrF₃ (sp₃d, T-shape); NH₃ (sp₃, pyramidal)
3. CH₄ (sp₃, Tetrahedral); NH₃ (sp₃, Tetrahedral)
4. XeO₃ (sp₃, Pyramidal); NH₃ (sp₃, Pyramidal)

Question 96. Which of the following pairs have different hybridization and the same shape—

1. \(\mathrm{NO}_3^{-} \text {and } \mathrm{CO}_3^{2-}\)
2. SO₂ and NH2-
3. XeF₂ and CO₂
4. H₂O and NH₃

Choose The Correct Option

1. 1 and 4
2. 2 and 4
3. 2 and 3
4. None of these

Answer: 3. 2 and 3

• NO-₃ (sp², trigonal planar);
• CO2-₃ (sp², trigonal planar);
• Same hybridization and the same shape.
• SO₂ (sp², bent); NH2 (sp², bent)

Different hybridization but the same shapes.

• XeF₂ (sp²d, linear); CO₂ (sp, linear)
• Different hybridization but the same shapes.
• H₂O (sp³angular); NH₃ (sp³, pyramidal)

Question 97. Which of the following is correct regarding bond angles—

1. SO₂
2. H₂S<SO₂
3. SO₂<H₂S
4. SBH₃<NO+2

Choose the correct one

1. 1 and 4
2. 2, 1 and 4
3. 1 and 3
4. None of these

Answer: 1. 1 and 4

⇒ \(\begin{array}{ccccr}
\mathrm{SbH}_3 & \mathrm{H}_2 \mathrm{O} & \mathrm{H}_2 \mathrm{~S} & \mathrm{SO}_2 & \mathrm{NO}_2^{+} \\
91.3^{\circ} & 104.5^{\circ} & 109.5^{\circ} & 120^{\circ} & 180^{\circ}
\end{array}\)

Question 98. Which pair of molecules does not have an identical structure-

1. I-3,BeF₂
2. O₃,SO₂
3. BF₂,ICI₃
4. BrF-₄,XeF₄

Answer: 3. BF₃, ICI₃

1. BF₃ —Trigonal planar
2. IC1₃ —T-shape

Question 99. Which ofthe following order is correct—

1. A1C1₃ < MgCl₂ < NaCl: polarising power
2. CO > CO₂ > HCO-₂ > CO2-3: bond length
3. BeCl₂ < NF₃ < NH₃ :dipole moment
4. H₂S > NH₃ > SiH₄ > BF₃: bond angle

Answer: 3. A1C1₃ < MgCl₂ < NaCl: polarising power

The polarising power of cations increases with the increasing charge.

⇒ \(\stackrel{(+1)}{\mathrm{NaCl}}<\stackrel{(+2)}{\mathrm{MgCl}_2}<\stackrel{(+3)}{\mathrm{AlCl}_2}\)

⇒ \(\text { Bond order } \propto \frac{1}{\text { Bond length }}\)

\(\text { Bond order }=\frac{\text { Bond order of each } \mathrm{C}-\mathrm{O} \text { bond }}{\text { Total no. of resonating structures }}\)

⇒ \(\begin{aligned}
& \mathrm{CO} \rightarrow \mathrm{C} \equiv \mathrm{O} \Rightarrow \frac{2+1}{1}=3.0 \\
& \mathrm{CO}_2 \rightarrow \mathrm{O}=\mathrm{C}=\mathrm{O} \Rightarrow \frac{2+2}{2}=2.0
\end{aligned}\)

Hence, the decreasing order of bond length is, CO2-₃ > HCO₂> CO₂ > CO

The correct order of bond angle:

BF₃ > SiH₄ > NH₃ > H₂S
120° 109°28/ 107° 94°

Question 100. Which of the following has the maximum % of s- s-character

1. N₂H₂
2. N₂H₄
3. NH₃
4. NH-₂

Answer: 1. N₂H₂

Question 101. Which of the following pairs does not have the same bond order-

1. N₂ and Cn-
2. o+₂ and no
3. F-₂ and O+2
4. B₂–₂ and CN+

Answer: 1. N₂ and Cn-

M.O. electronic configuration of N2(14):

⇒ \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_z}\right)^2\)

⇒ \(\text { B.O. }=\frac{1}{2}\left(\mathrm{~N}_b-\mathrm{N}_a\right)=\frac{1}{2}(10-4)=3.0\)

M.O. electronic configuration of CN-(14):

⇒ \(\begin{aligned}
& \quad\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_z}\right)^2 \\
& \text { B.O. }=\frac{1}{2}(10-4)=3.0
\end{aligned}\)

M.O. electronic configuration of O₂(15):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\)

⇒ \(\text { B.O. }=\frac{1}{2}(10-5)=2.5\) \text { B.O. }=\frac{1}{2}(10-5)=2.5

M.O. electronic configuration of NO(15):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\)

⇒ \(\text { B.O. }=\frac{1}{2}(10-5)=2.5\)

M.O. electronic configuration of F-2( 19):

⇒ \(\begin{aligned}
& K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2 \\
& \left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2\left(\sigma_{2 p_z}\right)^1 \\
&
\end{aligned}\)

⇒ \(\text { B.O. }=\frac{1}{2}(10-8)=1.0\)

M.O. electronic configuration of O²₂-(18):

⇒ \(\begin{aligned}
& K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2 \\
& \text { B.O. }=\frac{1}{2}(10-8)=1.0
\end{aligned}\)

M.O. electronic configuration of B²-₂(12):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\)

⇒ \(\text { B.O. }=\frac{1}{2}(8-4)=2.0\)

M.O. electronic configuration of CN+(12):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2 ; \text { B.O. }=\frac{1}{2}(8-4)=2.0\).