Questions: Chemical Bonding and Molecular Structure
Chemical Bonding and Molecular Structure25 min readLesson 6 of 7
Q1 State the octet rule and give two of its limitations.
Answer (English): The octet rule says that atoms combine until each atom in the resulting molecule has eight electrons in its valence shell, matching a noble gas configuration. Hydrogen and helium are satisfied with two. Two limitations are that the rule cannot explain molecules with an odd number of electrons, such as nitric oxide, which has eleven valence electrons and cannot give every atom an octet. It also cannot explain molecules such as sulfur hexafluoride, in which the central sulphur atom carries twelve electrons around it and is stable.
Hinglish: Octet rule kehta hai atom aise combine karte hain ki valence shell mein 8 electrons ho jaayein. Hydrogen/helium ke liye 2 kaafi hain. Limit: odd electron molecules jaise NO explain nahi hote, aur SF6 jaisi species mein sulphur ke paas 12 electrons hote hain phir bhi molecule stable hota hai.
Q2 Draw the Lewis structure of CO2 and calculate the formal charge on each atom.
Answer (English): Carbon contributes four valence electrons and each oxygen six, so the total is sixteen. Connect carbon to both oxygens and complete the octets of the oxygens with double bonds. The structure is O double bond C double bond O, with two lone pairs on each oxygen. For carbon, formal charge is 4 minus 0 minus 4 = 0. For each oxygen, formal charge is 6 minus 4 minus 2 = 0. Every atom carries zero formal charge, which is why this is the preferred structure. The molecule is linear and its bond order is two.
Hinglish: CO2 mein 4 + 6 + 6 = 16 valence electrons. Structure O=C=O hai, har oxygen par do lone pairs. Carbon ka formal charge 4 - 0 - 4 = 0 aur oxygen ka 6 - 4 - 2 = 0. Sab formal charges zero hain, isliye yahi best structure hai; molecule linear hai aur bond order 2 hai.
Q3 Give the shape of H2O, NH3, CH4, PCl5 and SF6 according to VSEPR theory, with the number of lone pairs on the central atom.
Answer (English): Water has two bond pairs and two lone pairs on oxygen, so its four domains are tetrahedral and the molecular shape is bent. Ammonia has three bond pairs and one lone pair, giving a tetrahedral electron geometry and a trigonal pyramidal shape. Methane has four bond pairs and no lone pairs, so it is tetrahedral. Phosphorus pentachloride has five bond pairs and no lone pairs, so it is trigonal bipyramidal. Sulfur hexafluoride has six bond pairs and no lone pairs, so it is octahedral.
Hinglish: H2O: 2 bond pairs + 2 lone pairs → electron geometry tetrahedral, shape bent. NH3: 3 + 1 → trigonal pyramidal. CH4: 4 + 0 → tetrahedral. PCl5: 5 + 0 → trigonal bipyramidal. SF6: 6 + 0 → octahedral. Lone pairs se bond angle ghatta hai.
Q4 Find the hybridisation of the central atom in CH4, C2H4, C2H2, BF3 and PCl5.
Answer (English): Count the domains in each case. Methane has four sigma bonds and no lone pair on carbon, so carbon is sp3. Each carbon of ethene has three sigma bonds and no lone pair, so both are sp2. Each carbon of ethyne has two sigma bonds and no lone pair, so both are sp. Boron trifluoride has three sigma bonds and no lone pair, so boron is sp2. Phosphorus pentachloride has five sigma bonds and no lone pairs, so phosphorus is sp3d. The domain count gives the hybridisation directly.
Hinglish: CH4 mein 4 domains → sp3; C2H4 mein har carbon ke 3 sigma bonds → sp2; C2H2 mein 2 sigma bonds → sp; BF3 mein 3 domains → sp2; PCl5 mein 5 domains → sp3d. Yaad rakho: domains count = sigma bonds + lone pairs.
Q5 Explain the relation between bond order, bond length and bond energy with an example.
Answer (English): Bond order is the number of electron pairs shared between two atoms. A higher bond order puts more electron density between the nuclei, so the nuclei are held closer and the bond length decreases. More energy is then needed to break the bond, so the bond energy increases. The carbon-carbon bonds make the pattern easy to see: the single bond has order one and length 154 pm, the double bond has order two and length 134 pm, and the triple bond has order three and length 120 pm. Bond energy rises in the same order.
Hinglish: Bond order badhne par zyada electron density nuclei ke beech hoti hai, isliye bond chhota aur strong ho jaata hai. C-C single (154 pm), double (134 pm) aur triple (120 pm) bond isi pattern ko dikhate hain; bond energy bhi single se triple tak badhti hai.
Q6 Use molecular orbital theory to explain why O2 is paramagnetic and place O2+, O2, O2- and O2^2- in order of increasing bond length.
Answer (English): Oxygen has twelve valence electrons in its molecular orbitals, and the highest two occupy separate pi antibonding orbitals with parallel spins. The presence of unpaired electrons makes the molecule paramagnetic. Bond order is half the difference between bonding and antibonding electrons: O2+ is 2.5, O2 is 2.0, O2- is 1.5 and O2^2- is 1.0. Bond length increases as bond order falls, so the order of increasing bond length is O2+ below O2 below O2- below O2^2-.
Hinglish: O2 ke highest do electrons alag pi antibonding orbitals mein unpaired rehte hain, isliye paramagnetic hai. Bond order: O2+ 2.5, O2 2.0, O2- 1.5, O2^2- 1.0. Bond order ghatne par bond length badhti hai, isliye order O2+ < O2 < O2- < O2^2- hai.
Q7 Define hydrogen bonding. Distinguish between intermolecular and intramolecular hydrogen bonding with one example each.
Answer (English): A hydrogen bond is the electrostatic attraction between a hydrogen atom bonded to a strongly electronegative atom such as F, O or N and a lone pair on another such atom. When the two groups belong to different molecules the bond is intermolecular; water and hydrogen fluoride show this, and it raises their boiling points above those of their group neighbours. When both groups are in the same molecule the bond is intramolecular, as in o-nitrophenol, where the hydroxyl hydrogen reaches the nitro group on the same ring. Intermolecular bonding raises the boiling point and viscosity, while intramolecular bonding usually lowers them compared with an isomer that cannot form the internal bond.
Hinglish: Hydrogen bond H (jo F/O/N se juda ho) aur doosre F/O/N ke lone pair ke beech electrostatic attraction hai. Alag molecules ke beech ho to intermolecular (paani, HF), boiling point badhta hai. Ek hi molecule ke andar ho to intramolecular (o-nitrophenol), yeh aksar boiling point kam kar deta hai.
Q8 What is lattice enthalpy? Which two factors increase its magnitude?
Answer (English): Lattice enthalpy is the enthalpy change when one mole of an ionic solid forms from its separated gaseous ions. Its magnitude measures how strongly the ions are bound in the crystal. Two factors control it. First, the charges on the ions: the attraction grows as the product of the charges, so magnesium oxide with 2+ and 2- ions has a much larger lattice enthalpy than sodium chloride with 1+ and 1-. Second, the size of the ions: smaller ions can approach more closely, and the attraction grows strongly as the distance falls, so lithium fluoride has a larger lattice enthalpy than caesium iodide. A large lattice enthalpy is one of the factors that makes an ionic compound stable.
Hinglish: Lattice enthalpy ek mole ionic solid ke gaseous ions se banne par enthalpy change hai. Do factors magnitude badhate hain: ions par charge ka product zyada hona (MgO > NaCl) aur ions ka size chhota hona (LiF > CsI). Bada lattice enthalpy ionic compound ko stable banata hai.
Q9 Both CO2 and H2O contain polar bonds, yet CO2 has zero dipole moment while H2O has a large one. Explain.
Answer (English): Individual bond moments add like vectors to give the molecular moment. Carbon dioxide is linear, so the two C-O bond moments point in exactly opposite directions and cancel, leaving a net moment of zero. Water is bent with an angle of about 104.5 degrees, so the two O-H bond moments do not cancel and add to a net moment of about 1.85 D. The same bond polarity therefore gives opposite molecular results, and the difference comes entirely from the shape.
Hinglish: CO2 linear hai, isliye dono C-O bond moments opposite direction mein cancel ho jaate hain aur net μ zero hai. H2O bent hai (~104.5°), isliye dono O-H moments cancel nahi hote aur net μ ≈ 1.85 D banta hai. Bond polarity same hai, farq sirf shape ka hai.
Q10 Count the sigma and pi bonds in ethyne and ethene, and state the hybridisation of the carbon atoms in each.
Answer (English): Ethyne is H-C triple bond C-H. The two C-H bonds are sigma, the triple bond contains one sigma and two pi bonds, so the total is three sigma and two pi bonds. Each carbon uses two sp hybrids for the sigma bonds and keeps two p orbitals for the two pi bonds. Ethene is H2C double bond CH2. The four C-H bonds and the C-C sigma bond make five sigma bonds, and the double bond contributes one pi bond, so the total is five sigma and one pi. Each carbon is sp2 hybridised and the molecule is planar.
Hinglish: Ethyne (C2H2) mein 3 sigma aur 2 pi bonds hain; har carbon sp hai. Ethene (C2H4) mein 5 sigma aur 1 pi bond hain; har carbon sp2 hai aur molecule planar hai. Triple bond = 1 sigma + 2 pi; double bond = 1 sigma + 1 pi.
That closes the chapter questions. Try the quick quiz below before moving on to the previous year questions.