## Chemical Bonding And Molecular Structure Multiple Choice Questions

**Question 1. The Sp ^{3}d^{2}-hybridization of the central atom of a molecule would be**

- Wouldsquareleadplanarto— geometry
- Tetrahedral geometry
- Trigonal bipyramidal geometry
- Octahedral geometry

**Answer:** 4. Octahedral geometry

One s, three p, and two d -orbitals mix up together to form six equivalent Sp^{3}d^{2}-hybrid orbitals. The molecules, in which these orbitals are involved, have octahedral geometry.

**Question 2. Which of the following is paramagnetic**

- N
^{2} - NO
- CO
- O
^{3}

**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,+1,-1
- +1,-1,-1
- +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 H ^{3}PO^{3} is**

- 0
- 1
- 2
- 3

**Answer:** 3. 2 Number of the —OH group in H_{3}PO_{3} is 2 and henceitis dibasic in nature.

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

- C
^{1}—C^{2}bondis a sp^{3}-sp^{3}tr – bond - C
^{2}— C^{3}bond is a sp^{3}-sp^{2}or- bond - C
^{1}—C^{2}bond is a sp^{3}-sp^{3}r-bond - C
^{1}—C^{2}bond is a sp^{2}-sp^{2}cr-bond

**Answer:** 3. C^{1}—C^{2} bond is a sp^{3}-sp^{2} r-bond

**Question 7. The paramagnetic behavior of B ^{2} Is due to the presence of—**

- 2 unpaired electrons π
_{b}MO - 2 unpaired electrons in π* MO
- 2 unpaired electrons in σ* MO
- 2 unpaired electrons in σ
_{b}MO

**Answer:** 1. The electronic configuration of B_{2} (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 POCl _{3} are-**

- sp,O
- sp
^{2}O - sp
^{3},O - dsp
^{2},1

**Answer:** 3. The Geometrical shape of the POC1_{3} molecule is tetrahedral. In POC1_{3} central P-atom is sp^{3} hybridised and has no lone pair.

**Question 9. CO is practically non-polar since—**

- The cr -electron drift from c to o is almost nullified by the n -electron drift from 0 to c
- The tr -electron drift from o to c is almost nullified by the 7r -electron drift from c to o
- The bond moment is low
- 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 NO _{2}, NO^{+}_{2}, and NO^{–}_{2}is—**

- \(\mathrm{NO}_2^{+}<\mathrm{NO}_2<\mathrm{NO}_2^{-}\)
- \(\mathrm{NO}_2<\mathrm{NO}_2^{-}<\mathrm{NO}_2^{+}\)
- \(\mathrm{NO}_2^{+}<\mathrm{NO}_2^{-}<\mathrm{NO}_2\)
- \(\mathrm{NO}_2<\mathrm{NO}_2^{+}<\mathrm{NO}_2^{-}\)

**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-**

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

**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**

- Six
- Two
- Eight
- 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 O _{2} and H_{2}O_{2}, the O — O bond lengths are 1.2lA and 1.48A respectively. In ozone, the average O —O bond length is**

- 1.28A
- 1.18A
- 1.44A
- 1.526A

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

**Question 14. In SOC1 _{2}, the Cl—S—Cl and Cl—S—O bond angles are—**

- 130°, 115°
- 106°, 96°
- 107°, 108°
- 96°, 106°

**Answer:** 4. 96°, 106°

IN SOC1_{2}, 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 XeF _{6} is experimentally determined to be a distorted octahedron. Its structure according to VSEPR theory is—**

- Octahedron
- Trigonal bipyramid
- Pentagonal bipyramid
- Tetragonal bipyramid

**Answer:** 3. Pentagonal bipyramid

In XeF_{6}, 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—**

- Is False
- Is True
- Cannot Be Evaluated Since the Data Is Not Sufficient
- 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 NF _{3} (102.3°) is smaller than NH_{3} (107.2°). This is because of—**

- Large size off compared to
- The large size compared to
- Opposite polarity of n in the two molecules
- Small size compared to a ton

**Answer:** 3. In NF_{3} molecules, the N—F bond pair is drawn

more towards the more electronegative F-atom. But in the NH_{3} 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 NF_{3}. As a consequence, the bond angle in NH_{3}(107.2°) is greater than that of NF_{3}(102.3°).

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

- 1
- 2
- 3
- 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
- 2
- 3
- 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>2>3
- 2>13
- 3>2>1
- 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>2>3
- 2>1>3
- 3>2>1
- 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>2>3
- 2>1>3
- 3>2>1
- 1>3>2

**Answer:** 3. 3>2>1

**Question 23. Thenumberoflone pairs of electrons on the central atoms of H _{2}O, SnCl_{2} PCl_{3,} and XeF_{2} respectively, are**

- 2,1,1,3
- 2,2,1,3
- 3,1,1,2
- 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**

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

**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 CO _{2 }**

- HgCl
_{2} - SnCl
_{2} - C
_{2}H_{2} - NO
_{2}

**Answer:** 1. HgCl_{2}

**Question 26. The ground state magnetic property of B _{2} and C_{2} molecules will be—**

- B, paramagnetic and C
_{2}diamagnetic - B, diamagnetic C
_{2}paramagnetic - Botii are diamagnetic
- Both are paramagnetic

**Answer:** 1. B, paramagnetic and C_{2} diamagnetic

**MO electronic configuration of B _{2}**

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

**Mo electronic configuration of C _{2}**

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

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—**

- Square pyramidal
- Triangularbipyramidal
- Planar
- Pentagonal bipramidal

**Answer:** 3. Planar

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

∴** Central atom Xe:** sp^{3}d^{2} -hybridized

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

therefore XeF_{5} ionic planar.

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

- It has five filled anti-bonding molecular orbitals
- It is diamagnetic
- It has bond order one
- It is isoelectronic with neon

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

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

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

- H-O…shape
- S-H….O
- F-H….F
- 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**

- \(\mathrm{BF}_6^{3-}\)
- BH-4
- B(OH)-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—**

- Nitrogen cannot form an-inbound
- Single n-nbond is weaker than the single p-p bond
- N
_{2}O_{4}has two resonance structures - 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 NH_{3} to BiH_{3} due to the increase in size ofthe central atom.

**Question 32. The Square Of IF _{7} is**

- Square pyramid
- Trigonal bipyramid
- Octahedral
- Pentagonal bipyramid

**Answer:** 4. Pentagonal bipyramid

In IF_{7}, the hybridization of the central atom is sp^{3}d^{3} and its structure is a pentagonal bipyramid.

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

- sp, sp
^{2}, sp^{3} - Sp
^{2},sp,sp^{3} - Sp,sp
^{3},s^{2} - sp
^{2}, sp^{3}sp

**Answer:** 2. Sp^{2},sp,sp^{3}

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

- FeCl
_{2} - SnCl
_{2} - AlCl
_{3} - MgCl
_{2}

**Answer:** 3. AlCl_{3}

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, A1C1_{3} exhibits maximum covalency.

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

- Ferrous compounds are relatively more ionic than ferric compounds
- Ferrous compounds are less volatile than the corresponding ferric compounds
- Ferrous compounds are more easily hydrolyzed than the corresponding ferric compounds
- 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—**

- O-nitrophenol shows intramolecular-bonding
- O-nitrophenol shows intermolecular-bonding
- The melting point of o-nitrophenol is less than that of mand p-isomers
- 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**

- AsCl
_{3} - SbCl
_{3} - PCl
_{3} - NCL
_{3}

**Answer:** 2. SbCl_{3}

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: SbCl_{3} < ASC1_{3} < PC1_{3} < NCl_{3}

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

- PCI+6 and SiCl
_{4} - PF
_{5}and BrF_{5} - AlFg- and SF
_{6} - CO
_{6}– and NO-_{3}

**Answer:** 2. PF_{5} and BrF_{5}

In molecule PF5, \(H=\frac{1}{2}\) [5 + 5- 0 + 0] = 5. Hybridisation of central atom(P): sp_{3}d. Number of p lone pairs in the central atom (P): 0. **Shape of the molecule:** Trigonal bipyramidal.

For BrF_{5}, Hybridisation of central atom (Br): sp_{3}d_{2}. Number of loner pairs in the central atom (Br): 1. Shape ofthe molecule: square pyramidal.

**Question 39. The stability of the species Li _{2}, Li_{2}, and Li_{2} increases in the order of**

- Li
_{2}<Li+2<Li-_{2} - Li2
_{2}< Li^{+}_{2}< Li_{2} - Li
_{2}<Li^{–}_{2}<Li^{+}_{2} - Li
_{2}<Li_{2}<Li_{2}

**Answer:** 2. Li_{2}< Li-_{2} < Li_{2}

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

⇒ \(\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}\)

\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—**

- \(\mathrm{H}_2^{+}, \mathrm{He}_2^{2-}\)
- \(\mathrm{H}_2^{-}, \mathrm{He}_2^{2-}\)
- \(\mathrm{H}_2^{2+}, \mathrm{He}_2\)
- \(\mathrm{H}_2^{-}, \mathrm{He}_2^{2+}\)

**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—**

- C
_{2} - N
_{2} - O
_{2} - S
_{2}

**Answer:** 1. C_{2}

**Question 42. The correct statement for the molecule, Csl _{3}**

**, is—**

- It contains Cs
^{+}, I^{–}and lattice I_{2}molecule - It is a covalent molecule
- It contains Cs
^{+}and I^{–}_{2}ions - It contains Cs
^{3+}and I^{–}ions

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

Cs cannot show a +3 oxidation state. So, Csl_{3} 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-**

- 3 and 4
- only 1
- 1 and 2
- 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—**

- BaSO
_{4} - SRSO
_{4} - CaSO
_{4} - BeSO
_{4}

**Answer:** 1. BaSO_{4}

In the NO_{4} 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 —**

- NO+
_{2} - NO-
_{2} - NO-
_{3} - NO
_{2}

**Answer:** 4. NO_{2}

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-**

- O
_{2} - B
_{2} - NO
- 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**

- 9
- 3
- 13
- 6

**Answer:** 1. 9

**Question 48. Which ofthe following compounds contain(s) no covalent bond(s)—KC1, PH _{3}, O_{2}, B_{2}H_{6}, H_{2}SO_{4}**

- KCL
- KCl, B
_{2}H_{6} - KClB
_{2}H_{6},PH_{3} - KCl, H
_{2}SO_{4}

**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—**

- H-
_{2} - H2-
_{2} - He2+
_{2} - 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 sp ^{2} hybridization—**

- NO-
_{2}And NH_{3} - BF
_{3}And NO-2 - NH-
_{2}And H_{2}O - BF
_{3}And NH-_{2}

**Answer:** 2. BF_{3} And NO-_{2}

\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-**

- CH
_{3}—CH_{2}—CH_{2}—CH_{3} - CH
_{3}—CH=CH—CH_{3} - CH
_{3}—C=C—CH_{3} - CH
_{2}=CH—CH_{2}—C=CH

**Answer:** 3. CH_{3}—C=C—CH_{3}

In the case of sp^{3}, sp^{2}, 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**

- C—H < C=C < C—O < C—C
- C—C < C=C < C—0 < C—H
- C—0<C—H<C—C<C=C
- 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^{–}**

- NO
^{–}_{2}and NO^{–}_{3} - NH
^{+}_{4}and NO^{–}_{3} - SCN
^{–}and NH^{–}_{2} - 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—**

- O
^{+}_{2} - O
^{–}_{2} - O
_{2}^{2-} - O
_{2}

**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**

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

**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 O _{2} to O_{2} ion, the electron adds on which one ofthe following orbitals—**

- π* -orbitals
- π – orbitals
- σ* orbital
- 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**

- \(\mathrm{O}_2^{2-}\)
- \(\mathrm{O}_2^{+}, \mathrm{NO}^{+}\)
- NO, CO
- N
_{2}, O_{2}

**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 O_{2} = 2.5, N0+ = 3, NO = 2.5 CO = 3, N_{2} = 3 and O_{2} = 2]

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

- H
_{2}O - BF
_{2} - NH-
_{2} - PCl
_{2}

**Answer:** 4. PCl_{3}

**Question 60. Bond order of 1.5 is shown by—**

- O+
_{2} - O-
_{2} - O2-
_{2} - O
_{2}

**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}\)

\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-**

- BC1
_{3}and BrCl_{3} - NH
_{3}and NO3- - NF
_{3}and BF_{3} - BF
_{4}and NH+_{4}

**Answer:** 4. BF_{2 }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—**

- SO
_{2} - NO
_{2} - CO
_{2} - H
_{2}O

**Answer:** 4. H_{2}O

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

- EU
^{2+} - Yb
^{2+} - Ce
^{2+} - Sm
^{2+}

**Answer:** 4. Sm^{2+}

⇒ \(\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—**

- SiF
_{4} - XeF
_{4} - BF
_{3} - SF
_{4}

**Answer:** 4. SF_{4}

SF_{4} has sp^{3}d -hybridization and see-saw shape with 4 bond pairs and 1 lone pair and resultant u=0.

**Question 66. XeF _{2} is isostructural with-**

- SbCl
_{3} - BaCL
_{2} - TeF
_{2} - ICI-
_{2}

**Answer:** 4. ICI-_{2}

**Question 67. Which species has a plane triangular shape-**

- N
_{3} - NO-
_{3} - NO
_{2} - CO
_{2}

**Answer:** 2. NO-_{3}

**Question 68. Which has the maximum dipole moment—**

- CO
_{2} - CH
_{4} - NH
_{3} - NF
_{3}

**Answer:** 3. NH_{3}

In NH_{3}, 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—**

- Electron repulsion outweighs the stability gained by achieving noble gas configuration
- O-ion has a comparatively smaller size than o-atom
- Oxygen is more electronegative
- 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—**

- SiCl
_{4}, Pcl+_{4} - Diamond, siC
- NH
_{3},PH_{3} - XeF
_{4}, XeO_{4}

**Answer:** 4. XeF_{4}, XeO_{4}

Diamond and silicon carbide (SiC), are both isostructural because their central atom is sp_{3} hybridized and both have tetrahedral arrangements.

Both NH_{3} and PH_{3} have pyramidal geometry.

XeF_{4} has sp^{3}d^{2} hybridisation while XeO_{4} has sp^{3} hybridisation.

Hence, XeF_{4} and XeO_{4} are notisostructural.

**Question 71. Decreasing order of stability is-**

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

**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-**

- HCN
- Cellulose
- Conc.Acetic Acid
- H
_{2}O_{2}

**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-**

- \(\mathrm{ClO}_3^{-}, \mathrm{SO}_3^{2-}\)
- \(\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\)
- \(\mathrm{ClO}_3^{-}, \mathrm{CO}_3^{2-}\)
- \(\mathrm{SO}_3^{2-}, \mathrm{CO}_3^{2-}\)

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

Both CO^{2}–_{3} and NO_{3} have a total of 32 electrons and both are triangular planar shape

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

- Pn-dn bonds are present in SO
_{2} - SeF
_{4}and CH_{4}have the same shape - 1+
_{3}has bent geometry - PH
_{3}and BiCl_{5}do not exist

**Answer:** 2. SeF_{4} and CH_{4} have same shape

The shape of CH_{4}(sp_{3}) is regular tetrahedron. However, in SeF_{4}, 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-**

- Sp
^{2}, sp^{3 }and sp - sp, sp
^{2}and sp^{3} - sp
^{2}, sp and sp^{3} - sp, sp
^{3}and sp^{2}

**Answer:** 2. sp, sp^{2} and sp^{3}

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

**Question 76. Consider the molecules CH, NH _{3}, and H_{2}0. Which of the given statements is false—**

- 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°
- The h—o —h bond angle in h20 is larger than the h—c—h bond angle in ch
_{4} - The h—o —h bond anglein h20 is smaller than the h —n—h bond anglein nh
_{3} - 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—**

- Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
- Lone pair-lone pair > bond pair-bond pair > lone pairlonepair
- Bond pair-bond pair > lone pair-bond pair > lone pairlonepair
- 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 CH _{3}C=C exists in**

- sp -orbital
- 2p-orbital
- sp
^{3}-orbital - sp
^{2}-orbital

**Answer:** 1. sp -orbital

CH_{3}CHC-, 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**

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

**Answer:** 2. 1- 3- 4- 2

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

- O
_{2}, NO^{+} - CN
^{–}, CO - N
_{2},O_{2}^{–} - 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—**

- ClF
_{3} - NCL
_{3} - BCl
_{3} - PH
_{3}

**Answer:** 3. BCl_{3}

**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)**

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

**Answer:** 1. 3-1- 4- 2

- XX ⇒ linear (sp
^{3}d) - XX3 ⇒ T-shape (sp
^{3}d) - XXg ⇒ square pyramidal (sp
^{3}d^{2}) - XXy ⇒ pentagonal bipyramidal (sp
^{3}d^{3})

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

- CN
- NO
- CN
^{+} - 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-**

- Three
- One
- Four
- Two

**Answer:** 4. Two

**Question 85. The decreasing order of bond angle is—**

- BeCl
_{2}> NO_{2}> SO_{2} - BeCl
_{2}> SO_{2}> NO_{2} - SO
_{2}> BeCl_{2}> NO_{2} - SO
_{2}>NO_{2}>BeCl_{2}

**Answer:** 1. BeCl_{2} > NO_{2} > SO_{2}

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—**

- NH
_{3} - NF
_{3} - SO
_{2} - BF
_{3}

**Answer:** 4. BF_{3}

BF_{3} has zero dipole moment.

**Question 87. In BF _{3}, the B —F bond length is 1.30 A when BF3 is allowed to be treated with mMe_{3} N, it forms an adduct, Me_{3} N→BF_{3}, and the bond length of —F in the adduct is-**

- Greater than 1.30A
- Smaller than 1.30A
- Equal to 1.30A
- None of these

**Answer:** 1. Greater than 1.30A

In BF_{3}, there is backbonding in between fluorine and boron due to the presence of -orbital in boron.

back bonding imparts double-bond characteristics

As BF_{3} 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 O _{2} will be**

- 6
- 8
- 4
- 2

**Answer:** 1. MO electronic configuration of O_{2}

& \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**

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

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

**Question 90. In the O _{3} molecule, the formal charge on the central O-atom is—**

- 0
- -1
- -2
- +1

**Answer:** 4. +1

Lewis gave the structure of the O_{3} 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**

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

**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**

- SO
_{2} - XeF
_{2} - SiF
_{4} - CH
_{4}

**Answer:** 2. XeF_{2}

**Question 93. The ASF _{5} molecule is trigonal bipyramidal. The hybrid orbitals used by the As atoms for bonding are—**

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

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

AsF_{5} has sp^{3}d hybridisation. In sp^{2}d hybridization, the dz^{2} 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. H _{2}O is polar, whereas BeF_{2} is not because—**

- Electronegativity ofF is greater than that of O
- H
_{2}O involves H-bonding, whereas, BeF_{2}is a discrete molecule - H
_{2}O is angular and BeF_{2}is linear - H
_{2}O is linear and BeF_{2}is angular

**Answer:** 3. H_{2}O is angular and BeF_{2} is linear

Because of the linear shape, dipole moments cancel each other in BeF_{2} (F—Be—F) and thus, it is non-polar, whereas H_{2}O is V-shaped and hence, it is polar

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

- C1F
_{3}and l-3 - BrF
_{3}and NH_{3} - CH
_{4}and NH+_{4} - XeO
_{3}and NH_{3}

**Answer:** 1. C1F_{3} and l-_{3}

- C1F
_{3}(sp_{3}d, T-shape); I_{3}(sp_{3}d, linear) - BrF
_{3}(sp_{3}d, T-shape); NH_{3}(sp_{3}, pyramidal) - CH
_{4}(sp_{3}, Tetrahedral); NH_{3}(sp_{3}, Tetrahedral) - XeO
_{3}(sp_{3}, Pyramidal); NH_{3}(sp_{3}, Pyramidal)

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

- \(\mathrm{NO}_3^{-} \text {and } \mathrm{CO}_3^{2-}\)
- SO
_{2}and NH2- - XeF
_{2}and CO_{2} - H
_{2}O and NH_{3}

**Choose The Correct Option**

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

**Answer:** 3. 2 and 3

- NO-
_{3}(sp^{2}, trigonal planar); - CO2-
_{3}(sp^{2}, trigonal planar); - Same hybridization and the same shape.
- SO
_{2}(sp^{2}, bent); NH2 (sp^{2}, bent)

**Different hybridization but the same shapes.**

- XeF
_{2}(sp^{2}d, linear); CO_{2}(sp, linear) - Different hybridization but the same shapes.
- H
_{2}O (sp^{3}angular); NH_{3}(sp^{3}, pyramidal)

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

- SO
_{2} - H
_{2}S<SO_{2} - SO
_{2}<H_{2}S - SBH
_{3}<NO+2

**Choose the correct one**

- 1 and 4
- 2, 1 and 4
- 1 and 3
- 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-**

- I-3,BeF
_{2} - O
_{3},SO_{2} - BF
_{2},ICI_{3} - BrF-
_{4},XeF_{4}

**Answer:** 3. BF_{3}, ICI_{3}

- BF
_{3}—Trigonal planar - IC1
_{3}—T-shape

**Question 99. Which ofthe following order is correct—**

- A1C1
_{3}< MgCl_{2}< NaCl: polarising power - CO > CO
_{2}> HCO-_{2}> CO2-3: bond length - BeCl
_{2}< NF_{3}< NH_{3}:dipole moment - H
_{2}S > NH_{3}> SiH_{4}> BF_{3}: bond angle

**Answer:** 3. A1C1_{3} < MgCl_{2} < 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} > HCO_{2}> CO_{2} > CO

**The correct order of bond angle:**

BF_{3} > SiH_{4} > NH_{3} > H_{2}S

120° 109°28/ 107° 94°

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

- N
_{2}H_{2} - N
_{2}H_{4} - NH
_{3} - NH-
_{2}

**Answer:** 1. N_{2}H_{2}

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

- N
_{2}and Cn- - o+
_{2}and no - F-
_{2}and O+2 - B
_{2}–_{2}and CN+

**Answer:** 1. N_{2 }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 O _{2}(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 O² _{2}-(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):**

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