Maharashtra Board Class 11 Chemistry Chapter 5 Chemical Bonding Solutions

Chemical Bonding Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 5 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 5 Exercise Solutions

1. Select and write the most appropriate alternatives from the given choices.

 

Question A. Which molecule is linear?
(a) SO3
(b) CO2
(c) H2S
(d) Cl2O
Answer: (b) CO2
In simple words: CO2 has a linear shape because the central carbon atom forms two double bonds with oxygen atoms, and there are no lone pairs, resulting in a 180° bond angle.

🎯 Exam Tip: Understanding VSEPR theory is key to predicting molecular geometries like linearity; common linear molecules often have sp hybridization.

 

Question B. When the following bond types are listed in decreasing order of strength (strongest first). Which is the correct order ?
(a) covalent > hydrogen > van der waals
(b) covalent > vander waal's > hydrogen
(c) hydrogen > covalent > vander waal's
(d) vander waal's > hydrogen > covalent.
Answer: (a) covalent > hydrogen > van der waals
In simple words: Covalent bonds are the strongest as they involve electron sharing, followed by hydrogen bonds which are specific intermolecular attractions, and then weak van der Waals forces.

🎯 Exam Tip: Rank bond strengths by considering the nature of the interaction-intramolecular covalent bonds are strongest, followed by intermolecular hydrogen bonds, and then general intermolecular van der Waals forces.

 

Question C. Valence Shell Electron Pair repulsion (VSEPR) theory is used to predict which of the following:
(a) Energy levels in an atom
(b) the shapes of molecules and ions.
(c) the electrone getivities of elements.
(d) the type of bonding in compounds.
Answer: (b) the shapes of molecules and ions.
In simple words: VSEPR theory helps predict the three-dimensional arrangement of atoms in a molecule or ion by minimizing repulsion between electron pairs in the valence shell.

🎯 Exam Tip: VSEPR theory is primarily used for predicting molecular geometry, which is crucial for understanding a molecule's physical and chemical properties.

 

Question D. Which of the following is true for CO2?

	C=O bond	CO2 molecule
A	polar	non-polar
B	non-polar	polar
C	polar	polar
D	non-polar	non-polar

Answer: (A) polar, non-polar

	C=O bond	CO2 molecule
A	polar	non-polar

In simple words: The C=O bonds are polar due to electronegativity difference, but because the CO2 molecule is linear, these bond dipoles cancel each other out, making the overall molecule non-polar.

🎯 Exam Tip: Remember that molecular polarity depends on both bond polarity and molecular geometry; even if individual bonds are polar, a symmetrical molecule can be non-polar.

 

Question E. Which O2 molecule is paramagnetic. It is explained on the basis of :
(a) Hybridisation
(b) VBT
(c) MOT
(d) VSEPR
Answer: (c) MOT
In simple words: Molecular Orbital Theory (MOT) correctly predicts the presence of two unpaired electrons in the oxygen molecule, which explains its paramagnetic nature, unlike other theories.

🎯 Exam Tip: Paramagnetism of O2 is a classic example where MOT provides a more accurate description of bonding and electronic configuration than simpler theories like VBT.

 

Question F. The angle between two covalent bonds is minimum in:
(a) CH4
(b) C2H2
(c) NH3
(d) H2O
Answer: (d) H2O
In simple words: Water (H2O) has the smallest bond angle among these because its two lone pairs exert greater repulsion on the two bond pairs compared to the single lone pair in NH3 or no lone pairs in CH4 and C2H2.

🎯 Exam Tip: Lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion; this order helps explain deviations from ideal bond angles predicted by VSEPR theory.

 

2. Draw

 

Question A. Lewis dot diagrams for the folowing
(a) Hydrogen (H2)
(b) Water (H2O)
(c) Carbon dioxide (CO2)
(d) Methane (CH4)
(e) Lithium Fluoride (LiF)
Answer:
(i) H2: `\( H : H \)` or `\( H - H \)`
(ii) Water (H2O): `\( H : \ddot{O} : H \)` or `\( H - \ddot{O} - H \)`
(iii) Carbon dioxide (CO2): `\( \ddot{O} :: C :: \ddot{O} \)` or `\( O = C = O \)`
(iv) Methane (CH4): `\[ \begin{matrix} & H \\ & | \\ H : & C & : H \\ & | \\ & H \end{matrix} \]` or `\[ \begin{matrix} & H \\ & | \\ H - & C & - H \\ & | \\ & H \end{matrix} \]`
(v) Lithium Fluoride (LiF): `\( Li^+ : \ddot{F} : ^- \)` or `\( Li - \ddot{F} : \)`
[Note: H atom in H2 and Li atom in LiF attain the configuration of helium (a duplet of electrons).]
In simple words: Lewis dot diagrams show valence electrons as dots to illustrate bonding and non-bonding pairs, helping to understand how atoms achieve stable electron configurations, usually an octet or a duplet for hydrogen.

🎯 Exam Tip: When drawing Lewis structures, always start by counting total valence electrons, then arrange atoms, form single bonds, and distribute remaining electrons as lone pairs to satisfy the octet rule (or duplet for hydrogen).

 

Question B. Diagram for bonding in ethene with sp² Hybridisation.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ईथेन (\(C_2H_4\)) अणु में sp² संकरण को दर्शाता है। इसमें प्रत्येक कार्बन परमाणु तीन sp² संकरित ऑर्बिटलों और एक असंकरित p ऑर्बिटल के साथ दिखाया गया है। sp² ऑर्बिटल सिग्मा बंधन बनाते हैं, जबकि p ऑर्बिटल पाई बंधन बनाते हैं, जिससे कार्बन-कार्बन डबल बॉन्ड का निर्माण होता है।
In simple words: The ethene molecule uses sp² hybridization where each carbon forms three sigma bonds and one pi bond, leading to a planar geometry around each carbon and a carbon-carbon double bond.

🎯 Exam Tip: For sp² hybridization, remember that it involves one s and two p orbitals, forming three sp² hybrid orbitals at 120° angles, with one unhybridized p orbital remaining to form a pi bond.

 

Question C. Lewis electron dot structures of
(a) HF
(b) C2H6
(c) C2H4
(d) CF3Cl
(e) SO2
Answer:
(i) HF: `\( H : \ddot{F} : \)` or `\( H - \ddot{F} : \)`
(ii) C2H6: `\[ \begin{matrix} H & H \\ | & | \\ H : C : C : H \\ | & | \\ H & H \end{matrix} \]` or `\[ \begin{matrix} H & H \\ | & | \\ H - C - C - H \\ | & | \\ H & H \end{matrix} \]`
(iii) C2H4: `\[ \begin{matrix} H \\ \diagdown \\ C = C \\ \diagup \\ H \end{matrix} \]` (with dots representing shared electrons for clarity, as `\[ \begin{matrix} H & H \\ \text{.} & \text{.} \\ \text{.} H - & C & :: C & - H \text{.} \\ \text{.} & \text{.} \\ H & H \end{matrix} \]` or `\[ \begin{matrix} H \\ \diagdown \\ C :: C \\ \diagup \\ H \end{matrix} \]` where `::` denotes double bond with electron pairs)
(iv) CF3Cl: `\[ \begin{matrix} & : \ddot{Cl} : \\ & | \\ : \ddot{F} - & C & - \ddot{F} : \\ & | \\ & : \ddot{F} : \end{matrix} \]`
(v) SO2: `\( \ddot{O} = \ddot{S} = \ddot{O} \)` (The image shows `::S:: or Ö=S=O`. This implies resonance structures. The `Ö=S=O` with dots represents a valid Lewis structure. Let's ensure formal charge is considered if necessary, but for drawing, `\( \ddot{O} = \ddot{S} = \ddot{O} \)` is a common simplified representation for the double bond resonance hybrid. A more precise one is `\( \ddot{O} - \ddot{S} = \ddot{O} \)` and its resonance form.)
In simple words: Lewis structures display the bonding and non-bonding valence electrons of atoms in a molecule, showing how they achieve stability, typically by fulfilling the octet rule through sharing or transferring electrons.

🎯 Exam Tip: Always verify that all atoms in your Lewis structures (except hydrogen) satisfy the octet rule, and that the total number of valence electrons used matches the calculated sum for the molecule.

 

Question D. Draw orbital diagrams of
(a) Fluorine molecule
(b) Hydrogen fluoride molecule
Answer:
(a) Fluorine molecule (F2):
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फ्लोरीन अणु (\(F_2\)) में p-p सिग्मा ओवरलैप को दर्शाता है। इसमें दो फ्लोरीन परमाणुओं के 2pz ऑर्बिटल एक-दूसरे के साथ अक्षीय रूप से ओवरलैप करते हुए दिखाए गए हैं, जिससे एक मजबूत सिग्मा बंधन बनता है। यह इलेक्ट्रॉन घनत्व को अंतरनाभिकीय अक्ष पर केंद्रित करता है।
(b) Hydrogen fluoride molecule (HF):
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र हाइड्रोजन फ्लोराइड अणु (HF) में s-p सिग्मा ओवरलैप को दर्शाता है। इसमें हाइड्रोजन परमाणु का 1s ऑर्बिटल और फ्लोरीन परमाणु का 2pz ऑर्बिटल एक-दूसरे के साथ अक्षीय रूप से ओवरलैप करते हुए दिखाए गए हैं, जिससे एक सिग्मा बंधन बनता है। यह इलेक्ट्रॉन घनत्व को अंतरनाभिकीय अक्ष पर केंद्रित करता है।
In simple words: Orbital diagrams show how atomic orbitals overlap to form molecular bonds, illustrating the spatial arrangement of electron density between atoms for molecules like F2 (p-p overlap) and HF (s-p overlap).

🎯 Exam Tip: Orbital overlap diagrams visually represent how atomic orbitals combine to form covalent bonds; remember that sigma bonds result from head-on overlap and pi bonds from lateral overlap.

 

3. Answer the following questions

 

Question A. Distinguish between sigma and pi bond.
Answer:

σ (sigma) bond	π (pi) bond
1. It is formed when atomic orbitals overlap along internuclear axis.	1. It is formed when atomic orbitals overlap side-ways (laterally).
2. Electron density is high along the axis of the molecule (i.e., internuclear axis).	2. Electron density is zero along the axis of the molecule (i.e., internuclear axis).
3. In the formation of sigma bond, the extent of overlap is greater, hence, more energy is released.	3. In the formation of pi bond, the extent of overlap is less, hence, less energy is released.
4. It is a strong bond.	4. It is a weak bond.
5. Formation of sigma bonds involves s-s, s-p, p-p overlap and overlap between hybrid orbitals.	5. Formation of pi bonds involves p-p or d-d overlap. The overlap between hybrid orbitals is not involved.

In simple words: Sigma bonds are stronger and formed by direct, head-on overlap of orbitals along the internuclear axis, leading to high electron density between nuclei; pi bonds are weaker, formed by sideways overlap of p-orbitals, with electron density above and below the internuclear axis.

🎯 Exam Tip: Remember that single bonds are always sigma bonds, while double bonds consist of one sigma and one pi bond, and triple bonds consist of one sigma and two pi bonds, reflecting their different strengths and electron distributions.

 

Question B. Display electron distribution around the oxygen atom in water molecule and state shape of the molecule, also write H-O-H bond angle.
Answer:
Electron distribution around oxygen atom in water molecule:
Shape of water molecule: Angular or V shaped H-O-H bond angle = 104°35′
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र जल के अणु (\(H_2O\)) में ऑक्सीजन परमाणु के चारों ओर इलेक्ट्रॉन वितरण और अणु की कोणीय संरचना को दर्शाता है। ऑक्सीजन परमाणु पर दो एकाकी इलेक्ट्रॉन युग्म और दो बॉन्ड युग्म होते हैं। इसमें ऑक्सीजन का sp³ संकरित ऑर्बिटल, हाइड्रोजन के 1s ऑर्बिटल के साथ सिग्मा बॉन्ड बनाता है, जिसके परिणामस्वरूप H-O-H बॉन्ड कोण 104°35′ होता है।
In simple words: In a water molecule, the central oxygen atom has two lone pairs and two bond pairs (with hydrogen), leading to sp³ hybridization and an angular (V-shaped) geometry with a bond angle of 104°35′, due to lone pair-bond pair repulsions.

🎯 Exam Tip: The presence of lone pairs significantly influences molecular geometry and bond angles by causing greater repulsion than bond pairs, reducing the ideal bond angles predicted by hybridization alone.

 

Question C. State octet rule. Explain its inadequecies with respect to
(a) Incomplete octet
(b) Expanded octet
Answer:
Statement: During the formation of chemical bond, atom loses, gains or shares electrons so that its outermost orbit (valence shell) contains eight electrons. Therefore, the atom attains the nearest inert gas electronic configuration.
(a) Molecules with incomplete octet: e.g. BF3, BeCl2, LiCl
In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.
Li in LiCl has only two electrons, Be in BeCl2 has four electrons while B in BF3 has six electrons in the valence shell.
(b) Molecules with expanded octet: Some molecules like SF6, PCl5, H2SO4 have more than eight electrons around the central atom.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र SF6, PCl5 और H2SO4 जैसे अणुओं के लुईस संरचनाओं को दर्शाता है जहाँ केंद्रीय परमाणु पर आठ से अधिक इलेक्ट्रॉन होते हैं, जिससे अष्टक नियम का विस्तार होता है। SF6 में सल्फर के चारों ओर 12 इलेक्ट्रॉन, PCl5 में फास्फोरस के चारों ओर 10 इलेक्ट्रॉन और H2SO4 में सल्फर के चारों ओर 12 इलेक्ट्रॉन दिखाए गए हैं।
SF6; 12 electrons around sulphur
PCl5; 10 electrons around phosphorus
H2SO4; 12 electrons around sulphur
In simple words: The octet rule states that atoms form bonds to achieve eight valence electrons, but it has limitations, including molecules with an incomplete octet (like BF3, BeCl2) that are stable with fewer than eight electrons, and molecules with an expanded octet (like SF6, PCl5) where the central atom holds more than eight electrons.

🎯 Exam Tip: While the octet rule is a useful guideline, be aware of its exceptions, especially for elements in period 3 and beyond, which can form expanded octets due to the availability of d-orbitals.

 

Question D. Explain in brief with one example:
(a) Ionic bond
(b) covalent bond
(c) co-ordinate bond
Answer:
(a) Formation of calcium chloride (CaCl2):
(i) The electronic configurations of calcium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or (2, 8, 8, 2)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
(ii) Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
(iii) Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
(iv) During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
(v) Calcium atom changes into \(Ca^{2+}\) ion while the two chlorine atoms change into two \(Cl^-\) ions. These ions are held together by strong electrostatic force of attraction.
(vi) The formation of ionic bond(s) between Ca and Cl can be shown as follows:
`\( : \ddot{Cl} \cdot + Ca + \cdot \ddot{Cl} : \implies : \ddot{Cl} : ^- + Ca^{2+} + : \ddot{Cl} : ^- \)`
`\( 2,8,7 \quad 2,8,8,2 \quad 2,8,7 \quad \implies \quad 2,8,8 \quad 2,8,8 \quad 2,8,8 \)`
`\( Cl^- + Ca^{2+} + Cl^- \implies CaCl_2 \)` or `\( Ca^{2+}(Cl^-)_2 \)`
(b) Formation of Cl2 molecule:
(i) The electronic configuration of Cl atom is [Ne] 3s² 3p⁵.
(ii) It needs one more electron to complete its valence shell.
(iii) When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.
(iv) The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl2 molecule is formed.
`\( : \ddot{Cl} \cdot + \cdot \ddot{Cl} : \implies : \ddot{Cl} : \ddot{Cl} : \)` OR `\( Cl - Cl \)`
(c) co-ordinate bond:
(i) A coordinate bond is a type of covalent bond where both of the electrons that form the bond originate from the same atom
(ii) An atom with a lone pair of electrons (non-bonding pair of electrons) is capable of forming a coordinate bond.
(iii) For example, reaction of ammonia with boron trifluoride: Before the reaction, nitrogen (N) in ammonia has eight valence electrons, including a lone pair of electrons. Boron (B) in boron trifluoride has only six valence electrons, so it is two electrons short of an octet. The two unpaired electrons form a bond between nitrogen and boron, resulting in complete octets for both atoms. A coordinate bond is represented by an arrow. The direction of the arrow indicates that the electrons are moving from nitrogen to boron. Thus, ammonia forms a coordinate bond with boron trifluoride.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र अमोनिया (\(NH_3\)) और बोरोन ट्राइफ्लोराइड (\(BF_3\)) के बीच एक उपसहसंयोजक बंधन के निर्माण को दर्शाता है। अमोनिया का नाइट्रोजन परमाणु एक एकाकी इलेक्ट्रॉन युग्म दान करता है, जिसे बोरोन परमाणु ग्रहण करता है, जिससे \(H_3N \to BF_3\) का निर्माण होता है। यह तीर की दिशा से स्पष्ट होता है कि इलेक्ट्रॉन नाइट्रोजन से बोरोन की ओर जा रहे हैं।
(iv) Once formed, a coordinate covalent bond is the same as any other covalent bond.
In simple words: An ionic bond forms through the complete transfer of electrons, a covalent bond involves the sharing of electrons, and a coordinate bond is a type of covalent bond where one atom donates both shared electrons.

🎯 Exam Tip: When explaining bond types, always provide a clear definition along with a representative example and illustrate electron movement (transfer or sharing) where applicable.

 

Question E. Give reasons for need of Hybridisation.
Answer:
The concept of hybridization was introduced because the valence bond theory failed to explain the following points:
(i) Valencies of certain elements:
The maximum number of covalent bonds which an atom can form equals the number of unpaired electrons present in its valence shell. However, valence bond theory failed to explain how beryllium, boron and carbon forms two, three and four covalent bonds respectively.
(a) Beryllium: The electronic configuration of beryllium is 1s² 2s². The expected valency is zero (as there is no unpaired electron) but the observed valency is 2 as in BeCl2.
(b) Boron: The electronic configuration of boron is 1s² 2s² 2p¹. The valency is expected to be 1 but it is 3 as in BF3.
(c) Carbon: The electronic configuration of carbon is 1s² 2s² 2px¹ 2py¹. The valency is expected to be 2, but observed valency is 4 as in CH4.
(ii) The shapes and geometry of certain molecules:
The valence bond theory cannot explain shapes, geometries and bond angles in certain molecules,
e.g.
(a) Tetrahedral shape of methane molecule.
(b) Bond angles in molecules like NH3 (107°18′) and H2O (104°35′).
However, the valency of the above elements and the observe structural properties of the above molecules can be explained by the concept of hybridization. These are the reasons for need of the concept of hybridization.
In simple words: Hybridization was needed because basic valence bond theory couldn't explain the observed valencies of elements like Beryllium, Boron, and Carbon, nor could it accurately predict the correct shapes and bond angles of many molecules.

🎯 Exam Tip: Hybridization is a fundamental concept that reconciles observed molecular geometries and valencies with atomic orbital theory, especially for polyatomic molecules.

 

Question F. Explain geometry of methane molecule on the basis of Hybridisation.
Answer:
Formation of methane (CH4) molecule on the basis of sp³ hybridization:
(i) Methane molecule (CH4) has one carbon atom and four hydrogen atoms.
(ii) The ground state electronic configuration of C (Z = 6) is 1s² 2px¹ 2py¹ 2pz⁰;
Electronic configuration of carbon:

	1s	2s	2p
Ground state:	`\( \uparrow\downarrow \)`	`\( \uparrow\downarrow \)`	`\( \uparrow \)`	`\( \uparrow \)`	`\( \phantom{\uparrow} \)`
Excited state:	`\( \uparrow\downarrow \)`	`\( \uparrow \)`	`\( \uparrow \)`	`\( \uparrow \)`	`\( \uparrow \)`
sp³ Hybrid orbitals:	`\( \uparrow\downarrow \)` (1s)	`\( \uparrow \uparrow \uparrow \uparrow \)` (four sp³ hybrid orbitals)

(iii) In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp³ hybridization.
(iv) One electron from the 2s orbital of carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, px, py and pz mix and recast to form four new sp³ hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with H-C-H bond angle of 109°28′. Each hybrid orbital contains one unpaired electron.
(v) Each of these sp³ hybrid orbitals with one electron overlap axially with the 1s orbital of hydrogen atom to form one C-H sigma bond. Thus, in CH4 molecule, there are four C-H bonds formed by the sp³-s overlap.
Diagram:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मीथेन (\(CH_4\)) अणु की ज्यामिति को sp³ संकरण के आधार पर समझाता है। केंद्रीय कार्बन परमाणु चार sp³ संकरित ऑर्बिटल बनाता है जो चतुष्फलकीय रूप से व्यवस्थित होते हैं। प्रत्येक sp³ ऑर्बिटल हाइड्रोजन के 1s ऑर्बिटल के साथ ओवरलैप करके एक सिग्मा बॉन्ड बनाता है, जिसके परिणामस्वरूप 109°28′ का H-C-H बॉन्ड कोण होता है।
In simple words: Methane's tetrahedral geometry with 109°28′ bond angles is explained by the central carbon atom undergoing sp³ hybridization, forming four equivalent hybrid orbitals that maximize separation and overlap with hydrogen's 1s orbitals.

🎯 Exam Tip: Remember that sp³ hybridization always leads to a tetrahedral electron geometry, and if all positions are occupied by bonding pairs, the molecular geometry is also tetrahedral with ideal 109°28′ bond angles.

 

Question G. In Ammonia molecule the bond angle is 107°18 and in water molecule it is 104°35′, although in both the central atoms are sp³ hybridized Explain.
Answer:
(i) The ammonia molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 107°28'. It is due to the following reasons.
- One lone pair and three bond pairs are present in ammonia molecule.
- The strength of lone pair-bond pair repulsion is much higher than that of bond pair-bond pair repulsion.
- Due to these repulsions, there is a small decrease in bond angle (~2°) from 109°28′ to 107°18′.
(ii) The water molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 104°35′. It is due to the following reasons.
- Two lone pairs and two bond pairs are present in water molecule.
- The decreasing order of the repulsion is Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-bond pair.
- Due to these repulsions, there is a small decrease in bond angle (~5°) from 109°28′ to 104°35′.
In simple words: Although both ammonia and water have sp³ hybridized central atoms, their bond angles are smaller than the ideal tetrahedral angle (109°28′) because the lone pairs of electrons exert greater repulsive forces on the bonding pairs, with water's two lone pairs causing a larger distortion than ammonia's single lone pair.

🎯 Exam Tip: When analyzing bond angles, always account for lone pair repulsions; the more lone pairs present, the greater the compression of bond angles from the ideal geometry.

 

Question H. Give reasons for:
(a) Sigma (σ) bond is stronger than Pi (π) bond.
(b) HF is a polar molecule
(c) Carbon is a tetravalent in nature.
Answer:
(a) (i) The strength of the bond depends on the extent of overlap of the orbitals.
(ii) A sigma bond is formed by the coaxial overlap of the atomic orbitals which are oriented along the internuclear axis, hence the extent of overlap is maximum.
(iii) A pi bond is formed by the lateral overlap of the atomic orbitals which are oriented perpendicular to the internuclear axis, hence the extent of orbital overlapping in side wise manner is less.
Hence, sigma bond is stronger than pi bond.
(b) (i) When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges.
(ii) In H-F, fluorine is more electronegative than hydrogen. Therefore, the shared electron pair is pulled towards fluorine and fluorine acquires partial -ve charge and simultaneously hydrogen acquires partial +ve charge. This gives rise to dipole and H-F bond becomes polar. Hence, H-F is a polar molecule.
(c) The electronic configuration of carbon is:
1s² 2s² 2px¹ 2py¹
One electron from '2s' orbital is promoted to the empty '2p' orbital.
Thus, in excited state, carbon has four half-filled orbitals.

	1s	2s	2p
C=	`\( \uparrow\downarrow \)`	`\( \uparrow \)`	`\( \uparrow \)`	`\( \uparrow \)`	`\( \uparrow \)`

Hence, carbon can form 4 bonds and is tetravalent in nature.
In simple words: Sigma bonds are stronger due to greater direct orbital overlap; HF is polar because fluorine's higher electronegativity pulls shared electrons closer to itself, creating a charge separation; carbon is tetravalent as its excited state allows it to form four stable covalent bonds.

🎯 Exam Tip: When justifying bond strength, polarity, or valency, always refer back to fundamental principles like extent of orbital overlap, electronegativity differences, and electron configurations (including excited states for valency).

 

Question I. Which type of hybridization is present in ammonia molecule? Write the geometry and bond angle present in ammonia.
Answer:
The type of hybridization present in ammonia (NH3) molecule is sp³.
Geometry of ammonia molecule is pyramidal or distorted tetrahedral.
Bond angle in ammonia molecule is 107°18′.
In simple words: Ammonia (NH3) has sp³ hybridization, resulting in a pyramidal (distorted tetrahedral) geometry with a bond angle of 107°18′ due to the presence of one lone pair on the nitrogen atom.

🎯 Exam Tip: For sp³ hybridized molecules with lone pairs, remember that the molecular geometry differs from the electron geometry (tetrahedral), and the lone pairs compress bond angles.

 

Question J. Identify the type of orbital overlap present in
(a) H2
(b) F2
(c) H-F molecule.
Explain diagramatically.
Answer:
(i) s-s σ overlap:
(a) The overlap between two half-filled s orbitals of two different atoms containing unpaired electrons with opposite spins is called s-s overlap.
e.g. Formation of H2 molecule by s-s overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹. The 1s¹ orbitais of two hydrogen atoms overlap along the internuclear axis to form a σ bond between the atoms in H2 molecule.
(b) Diagram for H2:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र हाइड्रोजन अणु (\(H_2\)) में s-s सिग्मा ओवरलैप को दर्शाता है। इसमें दो हाइड्रोजन परमाणुओं के 1s ऑर्बिटल एक-दूसरे के साथ अक्षीय रूप से ओवरलैप करते हुए दिखाए गए हैं, जिससे एक सिग्मा बंधन बनता है। यह इलेक्ट्रॉन घनत्व को अंतरनाभिकीय अक्ष पर केंद्रित करता है।
(ii) p-p σ overlap:
(a) This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis.
e.g. Formation of F2 molecule by p-p overlap:
Fluorine atom (Z = 9) has electronic configuration 1s² 2s² 2px² 2py² 2pz¹. During the formation of F2 molecule, half-filled 2pz orbital of one F atom overlaps with similar half-filled 2pz orbital containing electron with opposite spin of another F atom axially and a p-p σ bond is formed.
(b) Diagram for F2:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फ्लोरीन अणु (\(F_2\)) में p-p सिग्मा ओवरलैप को दर्शाता है। इसमें दो फ्लोरीन परमाणुओं के 2pz ऑर्बिटल एक-दूसरे के साथ अक्षीय रूप से ओवरलैप करते हुए दिखाए गए हैं, जिससे एक मजबूत सिग्मा बंधन बनता है। यह इलेक्ट्रॉन घनत्व को अंतरनाभिकीय अक्ष पर केंद्रित करता है।
(iii) s-p σ overlap:
(a) In this type of overlap one half filled s orbital of one atom and one half filled p orbital of another orbital overlap along the internuclear axis.
e.g. Formation of HF molecule by s-p overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹ and fluorine atom (Z = 9) has electronic configuration 1s² 2s² 2px² 2py² 2pz¹. During the formation of HF molecule, half-filled 1s orbital of hydrogen atom overlaps coaxially with half-filled 2pz orbital of fluorine atom with opposite electron spin and an s-p σ bond is formed.
(b) Diagram for HF:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र हाइड्रोजन फ्लोराइड अणु (HF) में s-p सिग्मा ओवरलैप को दर्शाता है। इसमें हाइड्रोजन परमाणु का 1s ऑर्बिटल और फ्लोरीन परमाणु का 2pz ऑर्बिटल एक-दूसरे के साथ अक्षीय रूप से ओवरलैप करते हुए दिखाए गए हैं, जिससे एक सिग्मा बंधन बनता है। यह इलेक्ट्रॉन घनत्व को अंतरनाभिकीय अक्ष पर केंद्रित करता है।
In simple words: H2 forms via s-s overlap, F2 via p-p overlap, and H-F via s-p overlap, all resulting in sigma bonds from axial (head-on) overlap of their respective half-filled atomic orbitals.

🎯 Exam Tip: Be able to identify and illustrate the different types of sigma orbital overlap (s-s, s-p, p-p) as these foundational concepts underpin covalent bond formation.

 

Question K. F-Be-F is a liner molecule but H-O-H is angular. Explain.
Answer:
(i) In the BeF2 molecule, the central beryllium atom undergoes sp hybridization giving rise to two sp hybridized orbitals placed diagonally opposite with an angle of 180°. Thus, F-Be-F is a linear molecule.
(ii) In the H2O molecule, the central oxygen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. There are two lone pairs of electrons in two of the sp³ hybrid orbitals of oxygen. The lone pair-lone pair repulsion distorts the structure. Hence, H-O-H is angular or V-shaped.
In simple words: BeF2 is linear due to sp hybridization and no lone pairs on beryllium, while H2O is angular (V-shaped) despite sp³ hybridization because two lone pairs on oxygen cause greater repulsion, compressing the H-O-H bond angle.

🎯 Exam Tip: Molecular geometry is determined by electron domains (bonding pairs and lone pairs) and their repulsion; linear shapes result from two electron domains, while lone pairs lead to bent or angular shapes in otherwise tetrahedral arrangements.

 

Question L. BF3 molecule is planar but NH3 pyramidal. Explain.
Answer:
(i) In the BF3 molecule, the central boron atom undergoes sp² hybridization giving rise to three sp² hybridized orbitals directed towards three comers of an equilateral triangle. Thus, the geometry is trigonal planar.
(ii) In the NH3 molecule, the central nitrogen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. The expected geometry of NH3 molecule is regular tetrahedral with bond angle 109°28′. There is one lone pair of electrons in one of the sp³ hybrid orbitals of nitrogen. The lone pair-bond pair repulsion distorts the bond angle. Hence, the structure of NH3 is distorted and it has pyramidal geometry.
In simple words: BF3 is trigonal planar because boron undergoes sp² hybridization with no lone pairs, whereas NH3 is pyramidal due to sp³ hybridization and the presence of one lone pair on nitrogen, which repels the bonding pairs and distorts the ideal tetrahedral shape.

🎯 Exam Tip: Differentiate between electron geometry and molecular geometry; the former is based on all electron domains, while the latter only considers atom positions, with lone pairs causing distortions from ideal angles.

 

Question M. In case of bond formation in Acetylene molecule :
(a) How many covalend bonds are formed ?

 

Question N. Define :
a. Bond Enthalpy
b. Bond Length
Answer:
a. Bond Enthalpy:
Bond enthalpy is defined as the amount of energy required to break one mole of a bond of one type, present between two atoms in a gaseous state.
b. Bond Length:
Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.
In simple words: Bond enthalpy is the energy needed to break a chemical bond. Bond length is the distance between the nuclei of two bonded atoms at equilibrium.

🎯 Exam Tip: These definitions are fundamental to understanding chemical bonding energetics and molecular structure.

 

Question O. Predict the shape and bond angles in the following molecules:
a. CF4
b. NF3
c. HCN
d. H2S
Answer:
a. CF4: There are four bond pairs on the central atom. Hence, shape of CF4 is tetrahedral and F-C-F bond angle is 109° 28'.
b. NF3: There are three bond pairs and one lone pair on the central atom. Hence, shape of NF3 is trigonal pyramidal and F-N-F bond angle is less than 109° 28'.
c. HCN: There are two bond pairs on the central atom. Hence, shape of HCN is linear and H-C-N bond angle is 180°.
d. H2S: There are two bond pairs and two lone pairs on the central atom. Hence, shape of H2S is bent or V-shaped and H-S-H bond angle is slightly less than 109° 28'.
In simple words: CF4 is tetrahedral (109°28'), NF3 is trigonal pyramidal (less than 109°28'), HCN is linear (180°), and H2S is bent/V-shaped (slightly less than 109°28'). These shapes are determined by the number of bond pairs and lone pairs around the central atom.

🎯 Exam Tip: VSEPR theory is crucial for predicting molecular geometry and bond angles. Remember that lone pair-lone pair repulsions are stronger than lone pair-bond pair, which are stronger than bond pair-bond pair repulsions, affecting bond angles.

 

4. Using data from the Table, answer the following :

Examples	C2H6 Ethane	C2H4 Ethene	C2H2 Ethyne
Structure	H3C-CH3	H2C=CH2	HC≡CH
Type of bond between carbons	single	double	triple
Bond length (nm)	0.154	0.134	0.120
Bond Enthalpy kJ mol-1	348	612	837

a. What happens to the bond length when unsaturation increases?
b. Which is the most stable compound?
c. Indicate the relation between bond strength and Bond enthalpy.
d. Comment on overall relation between Bond length, Bond Enthalpy and Bond strength and stability.
Answer:
a. When unsaturation increases, the bond length decreases.
b. The stable compound is ethyne (C2H2).
c. Bond strength \( \propto \) Bond enthalpy
Larger the bond enthalpy, stronger is the bond.
d. As bond length decreases, bond enthalpy, bond strength and stability increase.
In simple words: As the number of bonds between carbons increases (more unsaturation), the bond length gets shorter, and the bond becomes stronger and more stable, requiring more energy (enthalpy) to break.

🎯 Exam Tip: Understand the inverse relationship between bond order and bond length, and the direct relationship between bond order, bond strength, and bond enthalpy. This table clearly illustrates these concepts.

 

5. Complete the flow chart

ℹ️ चित्र व्याख्या (Diagram Explanation): यह फ्लोचार्ट बेरिलियम क्लोराइड (BeCl2), कार्बन डाइऑक्साइड (CO2) और एथाइन (C2H2) अणुओं की ज्यामिति और संबंधित बंध कोण को दर्शाता है। प्रत्येक अणु के लिए, यह केंद्रीय परमाणु के आसपास की संरचना, ज्यामिति (जैसे रैखिक) और बंध कोण (जैसे 180°) को स्पष्ट करता है, जिससे छात्रों को इन सरल अणुओं की स्थानिक व्यवस्था को समझने में मदद मिलती है।
Answer:

BeCl2	Cl-Be-Cl	Linear	180°
CO2	O=C=O	Linear	180°
C2H2	H-C≡C-H	Linear	180°

In simple words: The completed flowchart shows that BeCl2, CO2, and C2H2 all have linear geometries with 180° bond angles due to the arrangement of their atoms.

🎯 Exam Tip: Linear geometry with a 180° bond angle is characteristic of molecules where the central atom has two electron domains (e.g., sp hybridization) and no lone pairs, like BeCl2, CO2, and alkynes such as C2H2.

 

6. Complete the following Table

Molecule	Type of Hybridisation	Type of bonds	Geometry	Bond angle
CH4			Tetrahedral
NH3	sp3
H2O			angular	104.5°
BF3	sp2			120°
C2H4
BeF2			Linear
C2H2	sp

Answer:

Molecule	Type of hybridization	Type of bonds	Geometry	Bond angle
CH4	sp3	4 C-H 4 \( \sigma \) bonds	Tetrahedral	109.5°
NH3	sp3	3 N-H 3 \( \sigma \) bonds 1 lone pair	Pyramidal	107°18'
H2O	sp3	2 O-H, 2 \( \sigma \) bonds 2 lone pairs	Angular	104.5°
BF3	sp2	3 B-F, 3 \( \sigma \) bonds, One empty p orbital	Trigonal planar	120°
C2H4	sp2	(5 \( \sigma \) + 1 \( \pi \)) 1 C-C \( \sigma \), 4 C-H \( \sigma \), 1 C-C \( \pi \)	Trigonal planar	120°
BeF2	sp	2 Be-F 2 \( \sigma \) bonds	Linear	180°
C2H2	sp	(3 \( \sigma \) + 2 \( \pi \)) 1 C-C \( \sigma \) 2 C-H \( \sigma \) 2 C-C \( \pi \)	Linear	180°

In simple words: This table summarizes the hybridization, bond types, geometry, and bond angles for common molecules like methane, ammonia, water, boron trifluoride, ethene, beryllium fluoride, and ethyne, demonstrating how hybridization influences molecular structure.

🎯 Exam Tip: Mastering the correlation between hybridization type (sp, sp2, sp3), the number of bond pairs and lone pairs, and the resulting molecular geometry and bond angle is crucial for questions on molecular structure.

 

7. Answer in one sentence:

Question A. Indicate the factor on which stability of ionic compound is measured?
Answer: The stability of an ionic compound is measured by the amount of energy released during lattice formation.
In simple words: Ionic compound stability is determined by the energy released when the crystal lattice forms.

🎯 Exam Tip: Lattice enthalpy (or lattice energy) is a key concept for understanding the stability of ionic compounds. Higher lattice enthalpy indicates greater stability.

 

Question B. Arrange the following compounds on the basis of lattice energies in decreasing (descending) order: BeF2, AlCl3, LiCl, CaCl2, NaCl.
Answer: AlCl3 > BeF2 > CaCl2 > LiCl > NaCl
In simple words: The lattice energies decrease in the order: Aluminum Chloride, Beryllium Fluoride, Calcium Chloride, Lithium Chloride, and Sodium Chloride.

🎯 Exam Tip: Lattice energy generally increases with higher charge and smaller ionic radii. Consider the charge density of the ions for accurate ranking.

 

Question C. Give the total number of electrons around sulphur (S) in SF6 compound.
Answer: The total number of electrons around sulphur (S) in SF6 is 12.
In simple words: Sulfur in SF6 has 12 electrons around it, demonstrating an expanded octet.

🎯 Exam Tip: SF6 is a classic example of an expanded octet, where the central atom (sulfur) accommodates more than eight valence electrons by utilizing its empty d-orbitals.

 

Question D. Covalent bond is directional in nature. Justify.
Answer: Covalent bond is formed by the overlap of two half-filled atomic orbitals. The atomic orbitals are oriented in specific directions in space (except s-orbital which is spherical). Hence, covalent bond is directional in nature.
In simple words: Covalent bonds are directional because they form from the overlap of specific, spatially oriented atomic orbitals (like p or d orbitals), which gives molecules defined shapes.

🎯 Exam Tip: Contrast covalent bonds with ionic bonds, which are non-directional as electrostatic forces act equally in all directions. The directional nature of covalent bonds is foundational to molecular geometry.

 

Question E. What are the interacting forces present during formation of a molecule of a compound ?
Answer:
a. Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
b. Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).
In simple words: During molecule formation, there's a balance of attractive forces (nuclei to opposite electrons) and repulsive forces (electrons to electrons, nuclei to nuclei).

🎯 Exam Tip: The interplay between these attractive and repulsive forces determines whether a stable bond forms and at what internuclear distance, leading to an overall decrease in energy.

 

Question F. Give the type of overlap by which pi (\( \pi \)) bond is formed.
Answer: The type of overlap by which pi (\( \pi \)) bond is formed is p-p lateral overlap.
In simple words: Pi bonds form from the side-by-side (lateral) overlap of parallel p-orbitals.

🎯 Exam Tip: Remember that sigma (\( \sigma \)) bonds form from head-on (axial) overlap, while pi (\( \pi \)) bonds form from lateral overlap. This difference in overlap dictates their strength and electron density distribution.

 

Question G. Mention the steps involved in Hybridization.
Answer: The steps involved in hybridization are:
- formation of the excited state and
- mixing and recasting of orbitals.
In simple words: Hybridization involves promoting electrons to a higher energy level (excited state) and then mixing atomic orbitals to form new hybrid orbitals.

🎯 Exam Tip: Hybridization is a theoretical concept that helps explain observed molecular geometries and equivalent bond strengths that pure atomic orbitals cannot account for.

 

Question H. Write the formula to calculate bond order of molecule.
Answer: Bond order of a molecule = \( \frac{N_b - N_a}{2} \)
where, Nb is the number of electrons present in bonding MOs and Na is the number of electrons present in antibonding MOs.
In simple words: Bond order is calculated by subtracting the number of electrons in antibonding molecular orbitals from those in bonding molecular orbitals, and then dividing the result by two.

🎯 Exam Tip: A higher bond order generally corresponds to greater bond strength and shorter bond length. Bond order can be fractional.

 

Question I. Why is O2 molecule paramagnetic?
Answer: The electronic configuration of O2 molecule is \( (\sigma1s)^2 (\sigma*1s)^2 (\sigma2s)^2 (\sigma*2s)^2 (\sigma2pz)^2 (\pi2px)^2 (\pi2py)^2 (\pi*2px)^1 (\pi*2py)^1 \)
Since the oxygen molecule contains two unpaired electrons, it is paramagnetic.
In simple words: Oxygen is paramagnetic because its molecular orbital diagram shows two unpaired electrons in the antibonding pi orbitals.

🎯 Exam Tip: Molecular Orbital Theory (MOT) is essential for explaining magnetic properties like paramagnetism (presence of unpaired electrons) and diamagnetism (all electrons paired) in molecules like O2, which Valence Bond Theory struggles to explain.

 

Question J. What do you mean by formal charge ? Explain its significance with the help of suitable example.
Answer: Formal charge is the charge assigned to an atom in a molecule, assuming that all electrons are shared equally between atoms, regardless of their relative electronegativities.
Structure (I):

Number of atom	Total number of electrons in free atom (V.E.)	Total number of non-bonding electrons (N.E.)	Total number of shared electrons in bond (B.E.)	Formal charge F.C = (V.E.) - (N.E.) - \( \frac{1}{2} \) (B.E.)
1	6	4	4	F.C = 6 - 4 - \( \frac{1}{2} \)(4) = 0
2	4	0	8	F.C = 4 - 0 - \( \frac{1}{2} \)(8) = 0
3	6	4	4	F.C = 6 - 4 - \( \frac{1}{2} \)(4) = 0

Structure (II):

Number of atom	Total number of electrons in free atom (V.E.)	Total number of non-bonding electrons (N.E.)	Total number of shared electrons in bond (B.E.)	Formal charge F.C = (V.E.) - (N.E.) - \( \frac{1}{2} \) (B.E.)
1	6	2	6	F.C = 6 - 2 - \( \frac{1}{2} \)(6) = +1
2	4	0	8	F.C = 4 - 0 - \( \frac{1}{2} \)(8) = 0
3	6	6	2	F.C = 6 - 6 - \( \frac{1}{2} \)(2) = -1

Structure (III):

Number of atom	Total number of electrons in free atom (V.E.)	Total number of non-bonding electrons (N.E.)	Total number of shared electrons in bond (B.E.)	Formal charge F.C = (V.E.) - (N.E.) - \( \frac{1}{2} \) (B.E.)
1	6	6	2	F.C = 6 - 6 - \( \frac{1}{2} \)(2) = -1
2	4	0	8	F.C = 4 - 0 - \( \frac{1}{2} \)(8) = 0
3	6	2	6	F.C = 6 - 2 - \( \frac{1}{2} \)(6) = +1

While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. The structure having the lowest formal charge has the lowest energy.
In structure (I), the formal charge on each atom is 0 while in structures (II) and (III) formal charge on carbon is 0 while oxygens have formal charge -1 or +1. Hence, the possible structure with the lowest energy will be structure (I). Thus, formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.
In simple words: Formal charge is a hypothetical charge on an atom in a molecule, assuming equal sharing of electrons, and helps identify the most stable (lowest energy) Lewis structure by minimizing charges.

🎯 Exam Tip: Calculate formal charges to evaluate the stability of resonance structures. The most stable Lewis structure generally has formal charges closest to zero on all atoms, or negative formal charges on more electronegative atoms.

 

11th Chemistry Digest Chapter 5 Chemical Bonding Intext Questions and Answers

Question 1. Why are atoms held together in chemical compounds?
Answer: Atoms are held together in chemical compounds due to chemical bonds.
In simple words: Atoms are connected in compounds by chemical bonds.

🎯 Exam Tip: The formation of chemical bonds lowers the overall energy of the system, making the compound more stable than individual atoms.

 

Question 2. How are chemical bonds formed between two atoms?
Answer: There are two ways of formation of chemical bonds:
1. by loss and gain of electrons
2. by sharing a pair of electrons between the two atoms.
In either process of formation of chemical bond, each atom attains a stable noble gas electronic configuration.
In simple words: Chemical bonds form when atoms either transfer electrons (ionic bonding) or share electrons (covalent bonding) to achieve a stable electron configuration like noble gases.

🎯 Exam Tip: Emphasize the octet rule (or duplet rule for hydrogen/helium) as the driving force behind bond formation, leading to increased stability.

 

Question 3. Which electrons are involved in the formation of chemical bonds?
Answer: The electrons present in the outermost shell of an atom are involved in the formation of a chemical bond.
In simple words: Valence electrons, those in the outermost shell, are the ones involved in forming chemical bonds.

🎯 Exam Tip: Valence electrons dictate an atom's chemical reactivity and bonding behavior. Inner-shell electrons are generally not involved in bond formation.

 

Internet my friend (Textbook Page No. 55)

Question 1. Search more atoms, which complete their octet during chemical combinations.
Answer: In compounds like KCl, MgCl2, CaO, NaF, etc, the constituent atoms complete their octet by lose or gain of electrons.
e.g. K \( \longrightarrow \) K+ + e-
Cl + e- \( \longrightarrow \) Cl-
K+ + Cl- \( \longrightarrow \) KCl
[Note: Students are expected to search more atoms on their own.]
In simple words: Atoms like potassium, chlorine, magnesium, calcium, sodium, and fluorine achieve an octet by losing or gaining electrons to form ionic bonds.

🎯 Exam Tip: Focus on main group elements, especially those in Groups 1, 2, 16, and 17, as they readily form ionic bonds to satisfy the octet rule.

 

Use your brainpower. (Textbook Page No. 60)

Question 1. Which atom in NH4+ will have formal charge +1?
Answer: In NH4+, nitrogen atom (N) will have formal charge of+1.
In simple words: In the ammonium ion (NH4+), the central nitrogen atom carries a formal charge of +1.

🎯 Exam Tip: To verify, calculate the formal charge on nitrogen: Valence electrons (5) - Non-bonding electrons (0) - 1/2 * Bonding electrons (8) = 5 - 0 - 4 = +1.

 

Question 1. How many electrons will be around I in the compound IF7?
Answer: Lewis structure of IF7 is:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र IF7 अणु की लुईस संरचना को दर्शाता है, जिसमें केंद्रीय आयोडीन (I) परमाणु सात फ्लोरीन (F) परमाणुओं से घिरा हुआ है। प्रत्येक फ्लोरीन परमाणु एकल बंध के माध्यम से आयोडीन से जुड़ा है, और प्रत्येक फ्लोरीन पर तीन एकाकी इलेक्ट्रॉन युग्म (lone pairs) होते हैं। यह चित्र आयोडीन परमाणु के विस्तारित अष्टक को स्पष्ट रूप से चित्रित करता है।
In IF7, iodine (I) atom will be surrounded by 14 electrons.
In simple words: The central iodine atom in IF7 is surrounded by 14 electrons, forming an expanded octet.

🎯 Exam Tip: IF7 is an example of hypervalency, where the central atom (iodine) can accommodate more than eight valence electrons due to the availability of d-orbitals, leading to a pentagonal bipyramidal geometry.

 

Question 2. Why is H2 stable even though it never satisfies the octet rule?
Answer: The valence shell configuration of hydrogen atom is 1s1. Two hydrogen atoms approach each other and share their valence electrons. By having two electrons in its valence shell, H atom attains the nearest noble gas configuration of He. H2 molecule attains stability due to duplet formation. Hence, H2 is stable even though it never satisfies the octet rule.
In simple words: H2 is stable because each hydrogen atom achieves a stable duplet configuration (like helium) by sharing two electrons, even though it doesn't follow the octet rule.

🎯 Exam Tip: For elements like hydrogen and helium, the duplet rule (having two valence electrons) is the criterion for stability, not the octet rule.

 

Question 1. Lowering of energy takes during bond formation. How does this happen?
Answer:
i. When two combining atoms approach each other to form a covalent bond, the following interacting forces come into play.
- Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
- Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).
ii. The balance between attractive and repulsive forces decide whether the bond will be formed or not.
iii. When the magnitude of attractive forces is more than the magnitude of repulsive forces, the energy of the system decreases and a covalent bond is formed.
iv. When the magnitude of repulsive forces becomes more than that of attraction, the total energy of the system increases and a covalent bond is not formed. Hence, lowering of energy takes during bond formation.
In simple words: Bond formation occurs when attractive forces between nuclei and electrons outweigh repulsive forces, leading to a net decrease in the system's energy and increased stability.

🎯 Exam Tip: The potential energy curve illustrates this process: energy decreases as atoms approach, reaching a minimum at the equilibrium bond length, and then increases sharply due to strong nuclear repulsion upon closer approach.

 

Can you tell? (TextBook Page No. 76)

Question 1. Which molecules are polar?
H-I, H-O-H, H-Br, Br2, N2, I2, NH3
Answer:
i. H-I: Polar
ii. H-O-H: Polar
iii. H-Br: Polar
iv. Br2: Nonpolar
v. N2: Nonpolar
vi. I2: Nonpolar
vii. NH3: Polar
In simple words: H-I, H-O-H, H-Br, and NH3 are polar molecules because they have uneven electron distribution, while Br2, N2, and I2 are nonpolar due to symmetrical electron distribution.

🎯 Exam Tip: Polarity depends on two factors: bond polarity (difference in electronegativity) and molecular geometry. Even if a molecule has polar bonds, if its geometry is symmetrical (e.g., CO2, CCl4), the bond dipoles can cancel out, resulting in a nonpolar molecule.