PSEB Class 11 Chemistry Chemical Bonding and Molecular Structure MCQs

Question 1. Among the following, which one is a wrong statement?

1. PH₅ and BiCl₅ do not exist.
2. pπ-dπ bonds are present in SO₂.
3. SeF₄ and CH₄ have the same shape.
4. \(\mathrm{I}_3^{+}\) has bent geometry.

Answer: 3. SeF₄ and CH₄ have the same shape.

Question 2. The outer orbitals of C in an ethene molecule can be considered to be hybridized to give three equivalent sp² orbitals. The total number of sigma (σ) and pi (π) bonds in ethene molecule is

1. 3 sigma (σ) and 2 pi (π) bonds
2. 4 sigma (σ) and 1 pi (π) bonds
3. 5 sigma (σ) and 1 pi (π) bonds
4. 1 sigma (σ) and 2 pi (π) bonds.

Answer: 3. 5 sigma (σ) and 1 pi (π) bonds

Question 3. In which of the following pairs both species have sp³ hybridization?

1. \(\mathrm{SiF}_4, \mathrm{BeH}_2\)
2. \(\mathrm{NF}_3, \mathrm{H}_2 \mathrm{O}\)
3. \(\mathrm{NF}_3, \mathrm{BF}_3\)
4. \(\mathrm{H}_2 \mathrm{~S}, \mathrm{BF}_3\)

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Answer: 2. \(\mathrm{NF}_3, \mathrm{H}_2 \mathrm{O}\)

NF₃ and H₂O are sp³-hybridised.

PSEB Class 11 Chemistry Chemical Bonding and Molecular Structure MCQs

Question 4. Which one of the following pairs is isostructural (i.e., having the same shape and hybridization)?

1. \(\left[\mathrm{BCl}_3\right]\) and \(\left.\mathrm{BrCl}_3\right]\)
2. \(\left[\mathrm{NH}_3\right]\) and \(\left.\mathrm{NO}_3^{-}\right]\)
3. \(\left[\mathrm{NF}_3\right]\) and \(\left.\mathrm{BF}_3\right]\)
4. \(\left[\mathrm{BF}_4^{-}\right]\) and \(\left.\mathrm{NH}_4^{+}\right]\)

Answer: 4. \(\left[\mathrm{BF}_4^{-}\right]\) and \(\left.\mathrm{NH}_4^{+}\right]\)

• \(\mathrm{BCl}_3 \Rightarrow s p^2\), trigonal planar
• \(\mathrm{BrCl}_3 \Rightarrow s p^3 d\), T-shaped
• \(\mathrm{NH}_3 \Rightarrow s p^3\) pyramidal
• \(\mathrm{NO}_3^{-} \Rightarrow s p^2\) trigonal planar
• \(\mathrm{NF}_3 \Rightarrow s p^3\), pyramidal
• \(\mathrm{BF}_3 \Rightarrow s p^2\), trigonal planar
• \(\mathrm{BF}_4^{-} \Rightarrow s p^3\), tetrahedral
• \(\mathrm{NH}_4^{+} \Rightarrow s p^3\), tetrahedral

Question 5. Which of the two ions from the list given below that have the geometry that is explained by the same hybridization of orbitals, \(\mathrm{NO}_2^{-}, \mathrm{NO}_3^{-}, \mathrm{NH}_2^{-}, \mathrm{NH}_4^{+},\), SCN^–?

1. \(\mathrm{NO}_2^{-}\) and \(\mathrm{NO}_3^{-}\)
2. \(\mathrm{NH}_4^{+}\) and \(\mathrm{NO}_3^{-}\)
3. \(\mathrm{SCN}^{-}\) and \(\mathrm{NH}_2^{-}\)
4. \(\mathrm{NO}_2^{-}\) and \(\mathrm{NH}_2^{-}\)

Answer: 1. \(\mathrm{NO}_2^{-}\) and \(\mathrm{NO}_3^{-}\)

⇒ \(\begin{array}{cc}
\text { Ions } & \text { Hybridisation } \\
\mathrm{NO}_2^{-} & s p^2 \\
\mathrm{NO}_3^{-} & s p^2 \\
\mathrm{NH}_2^{-} & s p^3 \\
\mathrm{NH}_4^{+} & s p^3 \\
\mathrm{SCN}^{-} & s p
\end{array}\)

PSEB Class 11 Chemistry Chemical Bonding and Molecular Structure MCQs

Question 6. In which of the following pairs of molecules/ions, do the central atoms have sp² hybridization?

1. \(\mathrm{NO}_2^{-}\) and \(\mathrm{NH}_3\)
2. \(\mathrm{BF}_3\) and \(\mathrm{NO}_2^{-}\)
3. \(\mathrm{NH}_2\) and \(\mathrm{H}_2 \mathrm{O}\)
4. \(\mathrm{BF}_3\) and \(\mathrm{NH}_2^{-}\)

Answer: 2. \(\mathrm{BF}_3\) and \(\mathrm{NO}_2^{-}\)

The hybridisation of calculated as

H = \(\frac{1}{2}\left[\begin{array}{r}
\left(\begin{array}{l}
\text { No. of electrons } \\
\text { in valence shell } \\
\text { of atom }
\end{array}\right)+\left(\begin{array}{l}
\text { No. of monovalent } \\
\text { atoms around } \\
\text { central atom }
\end{array}\right) \\
-\left(\begin{array}{l}
\text { Charge on } \\
\text { cation }
\end{array}\right)+\left(\begin{array}{l}
\text { Charge on } \\
\text { anion }
\end{array}\right)
\end{array}\right]\)

∴ For \(\mathrm{BF}_3, H=\frac{1}{2}[(3)+(3)-(0)+(0)]\)

= \(3 \Rightarrow s p^2\) hybridisation

For \(\mathrm{NO}_2^{-}, H=\frac{1}{2}[(5)+(0)-(0)+(1)]\)

= \(3 \Rightarrow s p^2\) hybridisation.

Question 7. In which one of the following species the central atom has the type of hybridization which is not the same as that present in the other three?

1. \(\mathrm{SF}_4\)
2. \(\mathrm{I}_3\)
3. \(\mathrm{SbCl}_5^{2-}\)
4. \(\mathrm{PCl}_5\)

Answer: 3. \(\mathrm{SbCl}_5^{2-}\)

Hybridization of the central atom can be calculated as

H = \(\frac{1}{2}\left[\begin{array}{c}
\left(\begin{array}{l}
\text { No. of valence } \\
\text { electrons in the } \\
\text { central atom }
\end{array}\right)+\left(\begin{array}{l}
\text { No. of monovalent } \\
\text { atoms around } \\
\text { central atom }
\end{array}\right) \\
-\left(\begin{array}{l}
\text { Charge on } \\
\text { cation }
\end{array}\right)+\left(\begin{array}{l}
\text { Charge on } \\
\text { anion }
\end{array}\right)
\end{array}\right]\)

Applying this formula we find that all the given species except [SbCl₅]^2- have central atoms with sp³d (corresponding to H = 5) hybridization. In [SbCl₅]^2-, Sb is sp³d² hybridized.

Question 8. In which of the following molecules the central atom does not have sp³ hybridization?

1. \(\mathrm{CH}_4\)
2. \(\mathrm{SF}_4\)
3. \(\mathrm{BF}_4\)
4. \(\mathrm{NH}_4^{+}\)

Answer: 2. \(\mathrm{SF}_4\)

For neutral molecules,

No. of electron pairs = No. of atoms bonded to it + 1/2[Gp. number of the central atoms – Valency of the central atom]

∴ For \(\mathrm{CH}_4\), number of \(e^{-}\) pairs = \(4+\frac{1}{2}[4-4]=4\left(s p^3\right.\) hybridisation)

For \(\mathrm{SF}_4\), number of \(e^{-}\) pairs \(=4+\frac{1}{2}[6-4]=5\left(s p^3 d\right.\) hybridisation)

For ions,

No. of electron pairs = No. of atoms bonded to it +1/2 [Gp. no. of central atom – Valency of central atom ± No. of electrons equals to the units of charge]

∴ For \(\mathrm{BF}_4^{-}\), number of \(e^{-}\) pairs = \(4+\frac{1}{2}[3-4+1]\)

= \(4(s p^3\) hybridisation)

∴ For \(\mathrm{NH}_4^{+}\), number of \(e^{-}\) pairs = \(4+\frac{1}{2}[5-4-1]\)

= \(4\left(s p^3 \text { hybridisation }\right)\)

Question 9. Some of the properties of the two species, NO₃^– and H₃O⁺ are described below. Which one of them is correct?

1. Dissimilar in hybridization for the central atom with different structures.
2. Isostructural with the same hybridization for the central atom.
3. Isostructural with different hybridization for the central atom.
4. Similar in hybridization for the central gap atom with different structures.

Answer: 1. Dissimilar in hybridization for the central atom with different structures.

No. of electron pairs at the central atom = No. of atoms bonded to it + 1/2[Group number of the central atom – Valency of the central atom ± No. of electrons equals the units of charge]

No. of electron pairs at the central atom in \(\mathrm{NO}_3^{-}\) = 3 + 1/2(5-6+1) = 3 (sp² hybridization).

No. of electron pairs at the central atom in \(\mathrm{H}_3 \mathrm{O}^{+}=3+\frac{1}{2}[6-3-1]=4\) (sp³ hybridization)

Question 10. In which of the following molecules/ions BF₃, NO^–₂, NH^–_2, and H₂O, the central atom is sp² hybridized?

1. \(\mathrm{NH}_2^{-}\) and \(\mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{NO}_2^{-}\) and \(\mathrm{H}_2 \mathrm{O}\)
3. \(\mathrm{BF}_3\) and \(\mathrm{NO}_2^{-}\)
4. \(\mathrm{NO}_2^{-}\) and \(\mathrm{NH}_2^{-}\)

Answer: 3. \(\mathrm{BF}_3\) and \(\mathrm{NO}_2^{-}\)

∴ \(\mathrm{BF}_3 \rightarrow s p^2, \mathrm{NO}_2^{-} \rightarrow s p^2, \mathrm{NH}_2^{-} \rightarrow s p^3, \mathrm{H}_2 \mathrm{O} \rightarrow s p^3\)

Question 11. Among the following, the pair in which the two species are not isostructural is

1. \(\mathrm{SiF}_4\) and \(\mathrm{SF}_4\)
2. \(\mathrm{IO}_3^{-}\) and \(\mathrm{XeO}_3\)
3. \(\mathrm{BH}_4^{-}\) and \(\mathrm{NH}_4^{+}\)
4. \(\mathrm{PF}_6^{-}\) and \(\mathrm{SF}_6\)

Answer: 1. \(\mathrm{SiF}_4\) and \(\mathrm{SF}_4\)

SiF₄ has a symmetrical tetrahedral shape which is due to sp³ hybridisation of the centrai silicon atom. SF₄ has distorted tetrahedral or see-saw geometry which arises due to sp³d hybridisation of a central sulfur atom and due to the presence of one lone pair of electrons in one of the equatorial hybrid orbitals.

Question 12. Which of the following two are isostructural?

1. \(\mathrm{XeF}_2, \mathrm{IF}_2^{-}\)
2. \(\mathrm{NH}_3, \mathrm{BF}_3\)
3. \(\mathrm{CO}_3^{2-}, \mathrm{SO}_3^{2-}\)
4. \(\mathrm{PCl}_5, \mathrm{ICl}_5\)

Answer: 1. \(\mathrm{XeF}_2, \mathrm{IF}_2^{-}\)

Compounds having the same shape with the same hybridization are known as isostructural.

∴ \(\mathrm{XeF}_2, \mathrm{IF}_2^{-}\) + both are sp³d hybridised linear molecules

Question 13. The bond length between hybridized carbon atoms and other carbon atoms is minimal in

1. Propene
2. Propyne
3. Propane
4. Butane.

Answer: 2. Propyne

The C-C bond length = 1.54 Å, C = C bond length = 1.34 Å, and C: C bond length = 1.20 Å.

Since propyne has a triple bond, therefore it has a minimum bond length.

Question 14. Which of the following has sp²-hybridisation?

1. \(\mathrm{BeCl}_2\)
2. \(\mathrm{C}_2 \mathrm{H}_2\)
3. \(\mathrm{C}_2 \mathrm{H}_6\)
4. \(\mathrm{C}_2 \mathrm{H}_4\)

Answer: 4. \(\mathrm{C}_2 \mathrm{H}_4\)

BeCl₂ and C₂H₂ have sp-hybridisation and C₂H₆ has sp³-hybridisation. C₂H₄ has sp² hybridisation.

Chemical Bonding and Molecular Structure MCQs PSEB Class 11

Question 15. When the hybridization state of the carbon atom changes from sp³ to sp² and finally to sp, the angle between the hybridized orbitals

1. Decreases gradually
2. Decreases considerably
3. Is not affected
4. Increases progressively.

Answer: 4. Increases progressively.

The angle increases progressively, sp³ (109° 28′), sp² (120°), sp (180°).

Question 16. Which one of the following has the shortest carbon-carbon bond length?

1. Benzene
2. Ethene
3. Ethyne
4. Ethane

Answer: 3. Ethyne

There is a triple bond in the ethyne molecule (H – C ≡ C – H) and due to this triple bond, the carbon-carbon bond distance is the shortest in ethyne.

Question 17. Which structure is linear?

1. \(\mathrm{SO}_2\)
2. \(\mathrm{CO}_2\)
3. \(\mathrm{CO}_3^{2-}\)
4. \(\mathrm{SO}_4^{2-}\)

Answer: 2. \(\mathrm{CO}_2\)

The CO₂ molecule is sp-hybridized and thus, it is linear while \(\mathrm{CO}_3^{2-}\) is planar (sp² -hybridized), SO₂ is an angular molecule with sp² hybridization and \(\mathrm{SO}_4^{2-}\) is tetrahedral (sp³-hybridised).

Question 18. A sp³ hybrid orbital contains

1. 1/4 s-character
2. 1/2 s-character
3. 1/3 s-character
4. 2/3 s-character.

Answer: 1. 1/4 s-character

sp³ orbital has 1/4(25%) s-character.

Question 19. The complex ion [Co(NH₃)₆]³⁺ is formed by sp³d² hybridization. Hence the ion should possess

1. Octahedral geometry
2. Tetrahedral geometry
3. Square planar geometry
4. Tetragonal geometry.

Answer: 1. Octahedral geometry

According to VSEPR theory, a molecule with 6 bond pairs must be octahedral.

Question 20. Which of the following molecules does not have a linear arrangement of atoms?

1. \(\mathrm{H}_2 \mathrm{~S}\)
2. \(\mathrm{C}_2 \mathrm{H}_2\)
3. \(\mathrm{BeH}_2\)
4. \(\mathrm{CO}_2\)

Answer: 1. \(\mathrm{H}_2 \mathrm{~S}\)

For the linear arrangement of atoms, the hybridization is sp (bond angle = 180°).

Only H₂S has sp³-hybridisation and hence it has an angular shape while C₂H₂, BeH_2, and CO₂ all involve sp-hybridization and hence, have a linear arrangement of atoms.

Question 21. In which one of the following molecules the central atom can be said to adopt sp² hybridization?

1. \(\mathrm{BeF}_2\)
2. \(\mathrm{BF}_3\)
3. \(\mathrm{C}_2 \mathrm{H}_2\)
4. \(\mathrm{NH}_3\)

Answer: 2. \(\mathrm{BF}_3\)

BF₃ involves sp²-hybridisation.

Question 22. Equilateral shape has

1. sp hybridization
2. sp² hybridisation
3. sp³ hybridisation
4. dsp³ hybridisation

Answer: 2. sp² hybridization

Equilateral or triangular planar shape involves sp² hybridization, for example, BCl₃.

Question 23. The correct order of energies of molecular orbitals of N₂ molecules is

1. \(\sigma 1 s<\sigma^* 1 s< \sigma 2 s<\sigma^* 2 s<\sigma 2 p_z<\sigma^* 2 p_z\) \(<\left(\pi 2 p_x=\pi 2 p_y\right) <\left(\pi^* 2 p_x=\pi^* 2 p_y\right)\)
2. \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x\right.\left.=\pi 2 p_y\right)\) \(<\left(\pi^* 2 p_x=\pi^* 2 p_y\right)<\sigma 2 p_z<\sigma^* 2 p_z\)
3. \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x=\pi 2 p_y\right)\) \(<\sigma 2 p_z<\left(\pi^* 2 p_x=\pi^* 2 p_y\right)<\sigma^* 2 p_z\)
4. \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\sigma 2 p_z<\left(\pi 2 p_x=\pi 2 p_y\right)\) \(<\left(\pi^* 2 p_x=\pi^* 2 p_y\right)<\sigma^* 2 p_z\)

Answer: 3. \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x=\pi 2 p_y\right)\) \(<\sigma 2 p_z<\left(\pi^* 2 p_x=\pi^* 2 p_y\right)<\sigma^* 2 p_z\)

For molecules Li₂, Be₂, Br₂, C_2, and N₂ the order of energies of various molecular orbitals is \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x=\pi 2 p_y\right)<\sigma 2 p_z<\left(\pi^* 2 p_x\right. \left.=\pi^* 2 p_y\right) <\sigma^* 2 p_z\)

Question 24. Consider the following species: CN⁺, CN^–, NO and CN. Which one of these will have the highest bond order?

1. NO
2. CN^–
3. CN⁺
4. CN

Answer: 2. CN^–

NO(15): \((\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^* 2 s\right)^2,\left(\sigma 2 p_z\right)^2, \left(\pi 2 p_x\right)^2=\left(\pi 2 p_y\right)^2,\left(\pi^* 2 p_x\right)^1=\left(\pi^* 2 p_y\right)^0\)

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

∴ \(\mathrm{CN}^{-}(14):(\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^* 2 s\right)^2,\left(\pi 2 p_x\right)^2=\left(\pi 2 p_y\right)^2,\left(\sigma 2 p_z\right)^2\)

B.O. = \(\frac{10-4}{2}=3\)

CN(13): \((\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^* 2 s\right)^2,\left(\pi 2 p_x\right)^2=\left(\pi 2 p_y\right)^2,\left(\sigma 2 p_z\right)^1\)

B.O. = \(\frac{9-4}{2}=2.5\)

∴ \(\mathrm{CN}^{+}(12):(\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^* 2 s\right)^2,\left(\pi 2 p_x\right)^2=\left(\pi 2 p_y\right)^2\)

B.O. \(=\frac{8-4}{2}=2\)

Hence, CN^– has the highest bond order.

Chemical Bonding and Molecular Structure MCQs PSEB Class 11

Question 25. Which one of the following pairs of species have the same bond order?

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

Answer: 2. \(\mathrm{CN}^{-}, \mathrm{CO}\)

Molecular orbital electronic configurations and bond order values are \(\mathrm{O}_2(16): \sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2,\pi^* 2 p_x^1=\pi^* 2 p_y^1\)

B.O. = \(\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(10-6)=2\)

∴ \(\mathrm{NO}^{+}(14): \sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2\)

B.O. = \(\frac{1}{2}(10-4)=3\)

∴ \(\mathrm{CN}^{-}(14): \sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2\)

B.O. = \(\frac{1}{2}(10-4)=3\)

CO(14): \(1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2\)

B.O. = \(\frac{1}{2}(10-4)=3\)

∴ \(N_2\) (14): \(\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2\)

B.O. = \(\frac{1}{2}(10-4)=3\)

∴ \(\mathrm{O}_2^{-}(17)=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2,\pi^* 2 p_x^2=\pi^* 2 p_y^1\)

B.O. = \(\frac{1}{2}(10-7)=1.5\)

NO(15): \(\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2, \pi^* 2 p_x^1\)

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

Question 26. Which of the following is paramagnetic?

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

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

∴ \(\mathrm{O}_2^{-}(17)\) superoxide has one unpaired electron. \(\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y^1\)

Question 27. The pair of species that has the same bond order in the following is

1. \(\mathrm{CO}, \mathrm{NO}^{+}\)
2. \(\mathrm{NO}^{-}, \mathrm{CN}^{-}\)
3. \(\mathrm{O}_2, \mathrm{~N}_2\)
4. \(\mathrm{O}_2, \mathrm{~B}_2\)

Answer: 1. \(\mathrm{CO}, \mathrm{NO}^{+}\)

CO = 6 + 8 = 14electrons

NO⁺ = 7 + B – 1 = 14 electrons

Electronic configuration of \(\mathrm{NO}^{+}:\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2\)

Electronic configuration of \(\mathrm{CO}:\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2 \sigma 2 p_z^2\)

So, both have bond order = \(\frac{10-4}{2}=3\)

PSEB Class 11 Chemistry Chapter 4 MCQs with Answers

Question 28. In which of the following ionization processes does the bond energy increase and the magnetic behavior change from paramagnetic to diamagnetic?

1. \(\mathrm{O}_2 \rightarrow \mathrm{O}_2^{+}\)
2. \(\mathrm{C}_2 \rightarrow \mathrm{C}_2^{+}\)
3. \(\mathrm{NO} \rightarrow \mathrm{NO}^{+}\)
4. \(\mathrm{N}_2 \rightarrow \mathrm{N}_2^{+}\)

Answer: 3. \(\mathrm{NO} \rightarrow \mathrm{NO}^{+}\)

Molecular orbital configuration of \(\mathrm{O}_2(16): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^1 \pi^* 2 p_y{ }^1\)

⇒ Paramagnetic-Bond order =\(\frac{10-6}{2}=2\)

∴ \(\mathrm{O}_2^{+}(15): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^1\)

⇒ Paramagnetic- Bond order = \(\frac{10-5}{2}=2.5\)

∴ \(\mathrm{C}_2(12): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2\)

⇒ Diamagnetic ; Bond order \(=\frac{8-4}{2}=2\)

∴ \(\mathrm{C}_2^{+}(11): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y{ }^1\)

⇒ Paramagnetic; Bond order = \(\frac{7-4}{2}=1.5\)

∴ \(\mathrm{NO}(15): \sigma 1 s^2 \sigma^{\star} 1 s^2 \sigma 2 s^2 \sigma^{\star} 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^1\)

⇒ Paramagnetic;Bond order \(=\frac{10-5}{2}=2.5\)

∴ \(\mathrm{NO}^{+}(14): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2\)

⇒ Diamagnetic; Bond order = \(\frac{10-4}{2}=3\)

∴ \(\mathrm{~N}_2(14): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2 \sigma 2 p_z^2 \)

⇒ Diamagnetic; Bond order = \(\frac{10-4}{2}=3\)

∴ \(\mathrm{~N}_2^{+}(13): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2 \sigma 2 p_z^1\)

⇒ Paramagnetic Bond order = \(\frac{9-4}{2.5}=2.5\)

Thus from NO → NO⁺, bond order increases i.e., bond energy increases, and magnetic behaviour changes from paramagnetic to diamagnetic

Question 29. The pair of species with the same bond order is

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

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

⇒ \(\begin{array}{ll}
\mathrm{O}_2^{2-} \rightarrow 1.0 & \mathrm{~B}_2 \rightarrow 1.0 \\
\mathrm{O}_2^{+} \rightarrow 2.5 & \mathrm{NO}^{+} \rightarrow 3.0 \\
\mathrm{NO} \rightarrow 2.5 & \mathrm{CO} \rightarrow 3.0 \\
\mathrm{~N}_2 \rightarrow 3.0 & \mathrm{O}_2 \rightarrow 2.0
\end{array}\)

PSEB Class 11 Chemistry Chapter 4 MCQs with Answers

Question 30. During the change of O₂ to O^–₂ ion, the electron adds on which one of the following orbitals?

1. π* orbital
2. π orbital
3. σ* orbital
4. σ orbital

Answer: 1. π* orbital

Electronic configuration of O₂(16) \(\sigma(1 s)^2 \sigma^*(1 s)^2 \sigma(2 s)^2 \sigma^*(2 s)^2 \sigma\left(2 p_z\right)^2 \pi\left(2 p_x\right)^2=\pi\left(2 p_y\right)^2 \pi^*\left(2 p_x\right)^1=\pi^*\left(2 p_y\right)^1\)

Question 31. Which of the following is isoelectronic?

1. \(\mathrm{CO}_2, \mathrm{NO}_2\)
2. \(\mathrm{NO}_2^{-}, \mathrm{CO}_2\)
3. \(\mathrm{CN}^{-}, \mathrm{CO}\)
4. \(\mathrm{SO}_2, \mathrm{CO}_2\)

Answer: 3. \(\mathrm{CN}^{-}, \mathrm{CO}\)

In CO, the number of electrons = 6 + 8 = 14 Molecular orbital electronic configuration of

CO: \((\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\left(\sigma 2 p_z\right)^2\)

CN^– have also gets (6 + 7 + 1) 14 electrons and the configuration is similar to that of CO.

CN^– and CO are isoelectronic species.

Question 32. Which species does not exhibit paramagnetism?

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

Answer: 3. \(\mathrm{CO}\)

In ‘CO’ (14 electrons), there is no unpaired electron in its molecular orbital. Therefore, this does not exhibit paramagnetism.

Question 33. The number of anti-bonding electron pairs in O₂^2- molecular ion on the basis of the molecular orbital theory is (Atomic number of O is 8.)

1. 3
2. 2
3. 5
4. 4

Answer: 4. 4

∴ \(\mathrm{O}_2^{2-}(18):(\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^* 2 s\right)^2,\left(\sigma 2 p_z\right)^2,\left(\pi 2 p_x\right)^2, \left(\pi 2 p_y\right)^2,\left(\pi^{\star} 2 p_x\right)^2,\left(\pi^{\star} 2 p_y\right)^2\)

Thus the number of electrons in O₂^2- ion is 8(4 pairs).

Question 34. Which of the following species is paramagnetic?

1. \(\mathrm{CO}\)
2. \(\mathrm{CN}^{-}\)
3. \(\mathrm{O}_2^{2-}\)
4. \(\mathrm{NO}\)

Answer: 4. \(\mathrm{NO}\)

As per their molecular orbital electronic configurations \(\mathrm{CO}, \mathrm{CN}^{-}\) and \(\mathrm{O}_2^{2-}\) are diamagnetic and NO is paramagnetic.

Question 35. Which of the following is an incorrect statement?

1. The bond orders of \(\mathrm{O}_2^{+}, \mathrm{O}_2, \mathrm{O}_2^{-}\) and \(\mathrm{O}_2^{2-}\) are 2.5, 2, 1.5 and 1, respectively.
2. C₂ molecule has four electrons in its two degenerate Tt molecular orbitals
3. \(\mathrm{H}_2^{+}\) ion has one electron.
4. \(\mathrm{O}_2^{+}\) ion is diamagnetic.

Answer: 4. \(\mathrm{O}_2^{+}\) ion is diamagnetic.

∴ \(\mathrm{O}_2^{+}: \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^1\)

Due to the presence of one unpaired electron, \(\mathrm{O}_2^{+}\) is paramagnetic in nature.

PSEB Class 11 Chemistry Chapter 4 MCQs

Question 36. Identify a molecule that does not exist.

1. \(\mathrm{He}_2\)
2. \(\mathrm{Li}_2\)
3. \(\mathrm{C}_2\)
4. \(\mathrm{O}_2\)

Answer: 1. \(\mathrm{He}_2\)

He₂ does not exist as it has zero bond order \(\mathrm{He}_2: \sigma 1 s^2, \sigma^* 1 s^2\)

Bond order = \(\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(2-2)=0\)

Question 37. Which of the following diatomic molecular species has only π bonds according to Molecular Orbital Theory?

1. Be₂
2. O₂
3. N₂
4. C₂

Answer: 4. C₂

∴ \(\mathrm{Be}_2(8): K K \sigma(2 s)^2 \sigma^*(2 s)^2\)

• \(\mathrm{O}_2(16): K K \sigma(2 s)^2 \sigma^*(2 s)^2 \sigma\left(2 p_z\right)^2 \pi\left(2 p_x\right)^2 \pi\left(2 p_y\right)^2 \pi^*\left(2 p_x\right)^1 \pi^*\left(2 p_y\right)^1\)
• \(\mathrm{~N}_2(14): K K \sigma(2 s)^2 \sigma^*(2 s)^2 \pi\left(2 p_x\right)^2 \pi\left(2 p_y\right)^2 \sigma\left(2 p_z\right)^2\)
• \(\mathrm{C}_2(12): K K \sigma(2 s)^2 \sigma^*(2 s)^2 \pi\left(2 p_x\right)^2 \pi\left(2 p_y\right)^2\)

Therefore, C₂ contains 2 π bonds as, it has 4 electrons in two pi-molecular orbitals.

Question 38. Which of the following is paramagnetic?

1. N₂
2. H₂
3. Li₂
4. O₂

Answer: 4. O₂

∴ \(\mathrm{N}_2(14): K K \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2=\pi 2 p_y^2 \sigma 2 p_z^2\); Diamagnetic

• \(\mathrm{H}_2(2): \sigma l s^2\); Diamagnetic
• \(\mathrm{Li}_2(6): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2\); Diamagnetic
• \(\mathrm{O}_2(16): \sigma 1 s^2 \sigma^{\star} 1 s^2 \sigma 2 s^2 \sigma^{\star} 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2\)
• \(\pi^* 2 p_x^1=\pi^* 2 p_y^1 ;\) Paramagnetic

Question 39. Decreasing order of stability of \(\mathrm{O}_2, \mathrm{O}_2^{-}, \mathrm{O}_2^{+}\) and \(\mathrm{O}_2^{2-}\)

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

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

∴ \(\mathrm{O}_2(16): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^1=\pi^* 2 p_y^1\)

Bond order \(=\frac{1}{2}(8-4)=2\)

∴ \(\mathrm{O}_2^{2-}(18): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y^2\)

Bond order = \(\frac{1}{2}(8-6)=1\)

∴ \(\mathrm{O}_2^{-}(17): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y{ }^1\)

Bond order \(=\frac{1}{2}(8-5)=1.5\)

∴ \(\mathrm{O}_2^{+}(15): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^1\)

Bond order \(=\frac{1}{2}(8-3)=2.5\)

As, bond order ∝ stability

The decreasing order of stability is \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

PSEB Class 11 Chemistry Chapter 4 MCQs

Question 40. The correct bond order in the following species is

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

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

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

B.O.: 1.5 = 2.5 3.0

Question 41. Bond order of 1.5 is shown by

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

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

Configuration of \(\mathrm{O}_2(16)\): \(\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^1 \pi^* 2 p_y^1\)

Bond order \(=\frac{\begin{array}{l}\text { No. of } e^{-} \text {in } \\ \text { bonding M.O. }-\end{array} \begin{array}{c}\text { No. of } e^{–} \text {in } \\ \text { antibonding M.O. }\end{array}}{2}\)

Bond order of \(\mathrm{O}_2^{+}=\frac{10-5}{2}=2.5\)

Bond order of \(\mathrm{O}_2^{-}=\frac{10-7}{2}=1.5\)

Bond order of \(\mathrm{O}_2^{2-}=\frac{10-8}{2}=1.0\)

Bond order of \(\mathrm{O}_2=\frac{10-6}{2}=2\)

Question 42. Which of the following has the minimum bond length?

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

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

Electronic configuration \(\mathrm{O}_2: \mathrm{KK}(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_y\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\left(\pi^* 2 p_x\right)^1\left(\pi^* 2 p_y\right)^1\)

Bond order = \(\frac{1}{2}(8-4)=2\)

⇒ \(\mathrm{O}_2^{+}\): Bond order = \(\frac{1}{2}(8-3)=2 \frac{1}{2}\)

⇒\(\mathrm{O}_2^{-}\): Bond order = \(\frac{1}{2}(8-5)=1 \frac{1}{2}\)

⇒ \(\mathrm{O}_2^{2-}: \text { Bond order }=\frac{1}{2}(8-6)=1\)

As bond order increases, bond length decreases

Question 43. The pairs of species of oxygen and their magnetic behavior are noted below. Which of the following presents the correct description?

1. \(\mathrm{O}_2^{-}, \mathrm{O}_2^{2-}\) – Both diamagnetic
2. \(\mathrm{O}^{+}, \mathrm{O}_2^{2-}\) – Both paramagnetic
3. \(\mathrm{O}_2^{+}, \mathrm{O}_2\) – Both paramagnetic
4. \(\mathrm{O}, \mathrm{O}_2^{2-}\) – Both paramagnetic

Answer: 3. \(\mathrm{O}_2^{+}, \mathrm{O}_2\) – Both paramagnetic

O⁺₂ and O₂ are paramagnetic in nature as they contain one and two unpaired electrons respectively.

Question 44. Which one of the following species does not exist under normal conditions?

1. \(\mathrm{Be}_2^{+}\)
2. \(\mathrm{Be}_2\)
3. \(\mathrm{B}_2\)
4. \(\mathrm{Li}_2\)

Answer: 2. \(\mathrm{Be}_2\)

Be₂ does not exist.

Be₂ has an electronic configuration of: \(\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2\)

∴ Bond order = \(\frac{4-4}{2}=0\) Thus, \(\mathrm{Be}_2\) does not exist.

Question 45. According to MO theory which of the lists ranks the nitrogen species in terms of increasing bond order?

1. \(\mathrm{N}_2^{2-}<\mathrm{N}_2^{-}<\mathrm{N}_2\)
2. \(\mathrm{N}_2<\mathrm{N}_2^{2-}<\mathrm{N}_2^{-}\)
3. \(\mathrm{N}_2^{-}<\mathrm{N}_2^{2-}<\mathrm{N}_2\)
4. \(\mathrm{N}_2^{-}<\mathrm{N}_2<\mathrm{N}_2^{2-}\)

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

According to MOT, the molecular orbital electronic configuration of

∴ \(\mathrm{N}_2(14):(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\left(\sigma 2 p_z\right)^2\)

B.O = \(\frac{10-4}{2}=3\)

∴ \(\mathrm{~N}_2^{-}(15):(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\)

∴ \(\left(\sigma 2 p_z\right)^2\left(\pi^* 2 p_x\right)^1\)

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

∴ \(\mathrm{~N}_2^{2-}(16):(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\)

B.O. = \(\frac{10-6}{2}=2\)

Hence, bond order increases as: \(\mathrm{N}_2^{2-}<\mathrm{N}_2^{-}<\mathrm{N}_2\)

Question 46. Right order of dissociation energy N₂ and \(\mathrm{N}_2^{+}\) is

1. \(\mathrm{N}_2>\mathrm{N}_2^{+}\)
2. \(\mathrm{N}_2=\mathrm{N}_2^{+}\)
3. \(\mathrm{N}_2^{+}>\mathrm{N}_2\)
4. None.

Answer: 1. \(\mathrm{N}_2>\mathrm{N}_2^{+}\)

∴ \(\mathrm{N}_2(14):(\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^{\star} 2 s\right)^2\) \(\left(\pi 2 p_x\right)^2,\left(\pi 2 p_y\right)^2,\left(\sigma 2 p_z\right)^2\)

In \(N_2\), bond order \(=\frac{N_b-N_a}{2}=\frac{10-4}{2}=3\)

In \(\mathrm{N}_2^{+}\), bond order \(=\frac{9-4}{2}=2 \cdot 5\)

As the bond order in \(\mathrm{N}_2\) is more than \(\mathrm{N}_2^{+}\) so the dissociation energy of \(\mathrm{N}_2\) is higher than \(\mathrm{N}_2^{+}\)

Question 47. N₂ and O₂ are converted into monocations, \(\mathrm{N}_2^{+}\) and \(\mathrm{O}_2^{+}\) respectively. Which is wrong?

1. In \(\mathrm{O}_2^{+}\), paramagnetism decreases.
2. \(\mathrm{N}_2^{+}\) becomes diamagnetic.
3. In \(\mathrm{N}_2^{+}\), the N-N bond weakens.
4. In \(\mathrm{O}_2^{+}\), the O-O bond order increases.

Answer: 2. \(\mathrm{N}_2^{+}\) becomes diamagnetic.

Diamagnetism is caused due to the absence of unpaired electrons. But in \(\mathrm{N}_2^{+}\) there is an unpaired electron So, it is paramagnetic.

Question 48. N₂ and O₂ are converted into monoanionic N^–₂ and O^–₂ respectively, which of the following statements is wrong?

1. In \(\mathrm{O}_2^{-}\) bond length increases.
2. \(\mathrm{N}_2^{-}\) becomes diamagnetic.
3. In \(\mathrm{N}_2^{-}\), N-N bond weakens.
4. In \(\mathrm{O}_2^{-}\), the O-O bond order decreases.

Answer: 2. \(\mathrm{N}_2^{-}\) becomes diamagnetic.

∴ \(\mathrm{N}_2^{-}\) becomes paramagnetic due to one unpaired electron in \(\pi^* 2 p_x\), orbital.

Question 49. The ground state electronic configuration of valence shell electrons in nitrogen molecule (N₂) is written as \(K K, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2\). Hence the bond order in nitrogen molecules is

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

Answer: 2. 3

Number of electrons in bonding orbitals, \(\mathrm{N}_b\) = 10 and number of electrons in antibonding orbitals, \(\mathrm{N}_a\) = 4

Therefore bond order = 1/2(\(\mathrm{N}_b\) – \(\mathrm{N}_a\)) = 1/2(10 – 4) = 3

Question 50. Which of the following molecules has the highest bond order?

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

Answer: 3. \(\mathrm{O}_2^{+}\)

The bond order of \(\mathrm{O}_2^{+}=2.5, \mathrm{O}_2^{2-}=1\), \(\mathrm{O}_2^{-}\) = 1 .5 and that of O₂ = 2

Question 51. Which one of the following compounds shows the presence of an intramolecular hydrogen bond?

1. H₂O₂
2. HCN
3. Cellulose
4. Concentrated acetic acid

Answer: 3. Cellulose

H₂O₂, HCN and cons. CH₃COOH forms intermolecular hydrogen bonding while cellulose has intramolecular hydrogen bonding.

Question 51. In X – H….Y, X, and Y both are electronegative elements. Then

1. Electron density on X will increase and on H will decrease
2. In both electron density will increase
3. In both electron density will decrease
4. On X electron density will decrease and on H increases.

Answer: 1. Electron density on X will increase and on H will decrease

∴ \({ }^{\delta-} X-\mathrm{H}^{\delta+} \ldots . . . Y\), the electrons of the covalent bon<l are shifted towards the more electronegative atom’ This partially positively charged H-atom forms hydrogen bond with the other more electronegative atom.

Question 52. Strongest hydrogen bond is shown by

1. Water
2. Ammonia
3. Hydrogen fluoride
4. Hydrogen sulfide.

Answer: 3. Hydrogen fluoride

H – F shows the strongest H-bonds because fluorine is the most electronegative.

Question 53. Which one shows maximum hydrogen bonding?

1. \(\mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{H}_2 \mathrm{Se}\)
3. \(\mathrm{H}_2 \mathrm{~S}\)
4. \(\mathrm{HF}\)

Answer: 1. \(\mathrm{H}_2 \mathrm{O}\)

H₂O shelves maximum H-bonding because each H₂O molecule is linked to four H₂O molecules through H-bonds.