
Which of the following orders are correct with respect to the strength of ${\text{X - X}}$ single \[{\text{\sigma }}\] covalent bonds?
(p) ${\text{Si - Si < Ge - Ge}}$ (q) ${\text{N - N < P - P}}$ (r) ${\text{Cl - Cl > F - F}}$ (s) ${\text{O - O < S - S}}$
(A) (p), (q) and (r) only
(B) (q), (r) and (s) only
(C) (q) and (s) only
(D) (p), (q), (r) and (s)
Answer
233.1k+ views
Hint: Higher the energy required to break the bond, higher will be the strength of the bond. This on the other hand, is related to bond order and bond length. Higher the bond order, shorter will be the bond length and shorter the bond length, more is the energy required to break the bond.
Complete step by step answer: Silicon $\left( {{\text{Si}}} \right)$ and germanium $\left( {{\text{Ge}}} \right)$ are both members of the carbon family, i.e., they are group 14 elements. Generally, the bond lengths tend to increase down a group and decrease along a period and higher the bond length, lesser will be the bond energy. The element – element bond energy of carbon is very large. So, carbon can form long straight or branched chain ${\text{C - C}}$ chains or rings of various sizes. But, down the group, the element - element bond energies decreases rapidly. So, the energies required to break the bonds decrease which means bond strength also decreases down the group. So ${\text{Si - Si}}$ bond strength is greater than ${\text{Ge - Ge}}$ bond strength. So (p) is a wrong order.
Due to the smaller size of nitrogen, the \[{\text{N - N}}\] bond length will be smaller than the \[{\text{P - P}}\] bond length. But the lone pair of electrons available on the two nitrogen atoms will repel each other more because of the small distance between them and thus lead to the unstability of the bond. So, the \[{\text{N - N}}\] bond strength will be weaker than the \[{\text{P - P}}\] bond strength. So, (q) is a correct order.
Among the halogens, fluorine is the smallest in size. It has a very high electronegativity and very small internuclear distance. So, the lone pairs of electrons on the fluorine atoms will repel each other and make the bond unstable and weak. So, the \[{\text{Cl - Cl}}\] bond is stronger than the \[{\text{F - F}}\] bond. So, (r) is also a correct order.
Due to the small size of the oxygen atom, the lone pairs of electrons on the oxygen atoms will repel the bond pair of the \[{\text{O - O}}\] bond to a greater extent than the lone pairs of electrons on the sulphur atoms of the \[{\text{S - S}}\] bond. So, the \[{\text{S - S}}\] bond is stronger than the \[{\text{O - O}}\] bond. So, (s) is also a correct order.
Thus, the correct option is B.
Note: The tendency of an atom to catenate depends upon the element – element bond strength. The elements with higher bond strength will have more tendency for catenation. Thus, carbon has more tendency for catenation than silicon and germanium, phosphorus has more tendency for catenation than nitrogen and sulphur has more tendency for catenation than oxygen.
Complete step by step answer: Silicon $\left( {{\text{Si}}} \right)$ and germanium $\left( {{\text{Ge}}} \right)$ are both members of the carbon family, i.e., they are group 14 elements. Generally, the bond lengths tend to increase down a group and decrease along a period and higher the bond length, lesser will be the bond energy. The element – element bond energy of carbon is very large. So, carbon can form long straight or branched chain ${\text{C - C}}$ chains or rings of various sizes. But, down the group, the element - element bond energies decreases rapidly. So, the energies required to break the bonds decrease which means bond strength also decreases down the group. So ${\text{Si - Si}}$ bond strength is greater than ${\text{Ge - Ge}}$ bond strength. So (p) is a wrong order.
Due to the smaller size of nitrogen, the \[{\text{N - N}}\] bond length will be smaller than the \[{\text{P - P}}\] bond length. But the lone pair of electrons available on the two nitrogen atoms will repel each other more because of the small distance between them and thus lead to the unstability of the bond. So, the \[{\text{N - N}}\] bond strength will be weaker than the \[{\text{P - P}}\] bond strength. So, (q) is a correct order.
Among the halogens, fluorine is the smallest in size. It has a very high electronegativity and very small internuclear distance. So, the lone pairs of electrons on the fluorine atoms will repel each other and make the bond unstable and weak. So, the \[{\text{Cl - Cl}}\] bond is stronger than the \[{\text{F - F}}\] bond. So, (r) is also a correct order.
Due to the small size of the oxygen atom, the lone pairs of electrons on the oxygen atoms will repel the bond pair of the \[{\text{O - O}}\] bond to a greater extent than the lone pairs of electrons on the sulphur atoms of the \[{\text{S - S}}\] bond. So, the \[{\text{S - S}}\] bond is stronger than the \[{\text{O - O}}\] bond. So, (s) is also a correct order.
Thus, the correct option is B.
Note: The tendency of an atom to catenate depends upon the element – element bond strength. The elements with higher bond strength will have more tendency for catenation. Thus, carbon has more tendency for catenation than silicon and germanium, phosphorus has more tendency for catenation than nitrogen and sulphur has more tendency for catenation than oxygen.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reactions (2025-26)

