RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Chemistry · 10 questions · 24 marks
Why is water bent while carbon dioxide is straight, and why does oxygen stick to a magnet when nitrogen ignores it? This chapter gives you three complementary tools — VSEPR for shape, hybridisation for bonding, and molecular orbital theory for magnetism and bond order — and teaches you when to reach for each one.
The shape and hybridisation of the BrF₃ molecule are:
Answer
T-shaped, sp³d. Bromine has 7 valence electrons; three are used to bond to three fluorine atoms, leaving 2 lone pairs. The total is 3 bond pairs + 2 lone pairs = 5 electron pairs, which demands sp³d hybridisation and a trigonal bipyramidal electron-pair geometry. Because lone pair–lone pair repulsion is the strongest, both lone pairs take equatorial positions where they are 120° apart rather than 90°. The three fluorine atoms then occupy the two axial sites and one equatorial site, giving the molecular shape a T outline with F−Br−F angles slightly under 90°.
According to molecular orbital theory, the bond order of the superoxide ion O₂⁻ is:
Answer
1.5. O₂⁻ has 8 + 8 + 1 = 17 electrons. Filling the molecular orbitals gives 10 electrons in bonding orbitals and 7 in antibonding orbitals. Bond order = (N_b − N_a) ÷ 2 = (10 − 7) ÷ 2 = 1.5. Neutral O₂ has bond order 2, and adding one electron into a π* antibonding orbital lowers it by 0.5, which is why the O−O bond in the superoxide ion is longer and weaker than in oxygen gas.
Which of the following molecules has a zero resultant dipole moment?
Answer
BF₃ has zero dipole moment. Boron in BF₃ is sp² hybridised with no lone pair, so the three highly polar B−F bonds point to the corners of an equilateral triangle at 120° to one another. Their three bond dipoles are equal in magnitude and symmetrically placed, so their vector sum is exactly zero. NF₃ and H₂S are pyramidal and bent respectively, and their lone pairs destroy the symmetry, so a resultant survives. CHCl₃ is tetrahedral but the C−H bond is not equivalent to the three C−Cl bonds, so it too is polar.
The correct order of bond angles is:
Answer
CH₄ > NH₃ > H₂O. All three central atoms are sp³ hybridised with four electron pairs, so the ideal angle is 109.5°. Methane has four bond pairs and no lone pair, so it keeps the ideal 109.5°. Ammonia has one lone pair, which repels the three bonding pairs more strongly than they repel each other, compressing the angle to about 107°. Water has two lone pairs, so the compression happens twice and the angle drops further to about 104.5°. Each extra lone pair narrows the angle, giving the order stated.
Assertion (A): Oxygen gas is paramagnetic, while nitrogen gas is diamagnetic. Reason (R): In molecular orbital theory the last two electrons of O₂ enter two degenerate π* antibonding orbitals singly with parallel spins, whereas every electron in N₂ is paired.
Answer
Both A and R are true and R is the correct explanation of A. A simple Lewis structure O=O shows all electrons paired and predicts oxygen to be diamagnetic, which contradicts the experimental fact that liquid oxygen clings between the poles of a magnet. Molecular orbital theory resolves this. O₂ has 16 electrons, and after filling the bonding orbitals the final two go into the degenerate π*2p_x and π*2p_y orbitals. By Hund's rule they occupy these separately with parallel spins, leaving two unpaired electrons and hence paramagnetism. N₂ has only 14 electrons, which exactly fill through the σ2p_z bonding orbital with none left over for the π* set, so all its electrons are paired and it is diamagnetic. The reason is precisely the explanation of the assertion.
Both CO₂ and SO₂ contain two polar bonds to oxygen, yet CO₂ has a dipole moment of zero while SO₂ has a dipole moment of about 1.6 D. Explain.
Answer
Dipole moment is a vector sum, so the shape decides the outcome. In CO₂ the carbon is sp hybridised with no lone pair, so the molecule is linear and the two C=O bond dipoles are equal in size but exactly opposite in direction. They cancel completely and the resultant is zero. In SO₂ the sulphur carries a lone pair as well as two bonding regions, so it is sp² hybridised and the molecule is bent with an O−S−O angle near 119°. The two S=O bond dipoles no longer oppose each other, and their resultant, reinforced by the lone pair's own contribution, points along the bisector of the angle, giving SO₂ a measurable dipole moment.
Describe the bonding in an ethyne (C₂H₂) molecule in terms of hybridisation, and account for its linear shape and short carbon–carbon bond.
Answer
Each carbon in ethyne uses one 2s and one 2p orbital to form two sp hybrid orbitals, leaving two unhybridised 2p orbitals perpendicular to each other and to the hybrid axis. One sp orbital on each carbon overlaps head-on with the sp orbital of the other carbon to form the C−C sigma bond, and the remaining sp orbital on each carbon overlaps with the 1s orbital of a hydrogen to give a C−H sigma bond. The two unhybridised 2p orbitals on one carbon then overlap sideways with the matching pair on the other carbon, creating two mutually perpendicular pi bonds. Since sp hybrid orbitals lie at 180° to one another, all four atoms fall on a straight line. The carbon–carbon linkage is one sigma plus two pi bonds, giving bond order 3, so the electron density between the nuclei is very high and the bond contracts to about 120 pm, much shorter than the 154 pm single bond in ethane. The 50% s character of sp orbitals also holds the bonding electrons closer to the nuclei, reinforcing the shortening.
Using VSEPR theory, predict the shape, hybridisation and approximate bond angle of each of the following, giving reasons: (i) SF₄, (ii) XeF₄, (iii) ClO₃⁻, (iv) BeCl₂ (gaseous), (v) ICl₄⁻.
Answer
The method in each case is to count sigma bond pairs plus lone pairs on the central atom, use that total to fix the hybridisation and electron-pair geometry, then let lone pairs distort the shape. (i) SF₄: sulphur has 6 valence electrons, 4 used in bonding, so 4 bond pairs + 1 lone pair = 5 pairs, sp³d, trigonal bipyramidal electron geometry. The lone pair takes an equatorial position to minimise repulsion, giving a see-saw shape with angles of about 117° (equatorial) and 173° (axial), both squeezed from the ideal 120° and 180°. (ii) XeF₄: xenon has 8 valence electrons, 4 used in bonding, leaving 2 lone pairs, so 4 + 2 = 6 pairs, sp³d², octahedral electron geometry. The two lone pairs sit trans to each other on opposite axial sites to be 180° apart, leaving the four fluorines in one plane. The molecular shape is square planar with F−Xe−F angles of exactly 90°. (iii) ClO₃⁻: the chlorine has 3 sigma bonds to oxygen and 1 lone pair, so 4 pairs, sp³, tetrahedral electron geometry. The lone pair pushes the three oxygens down into a trigonal pyramidal shape with O−Cl−O angles a little under 109.5°, about 107°. (iv) BeCl₂ in the gas phase: beryllium has only 2 valence electrons and forms 2 bond pairs with no lone pair, so 2 pairs, sp hybridisation. The molecule is linear with a Cl−Be−Cl angle of exactly 180°. This is a classic incomplete-octet species, which is why solid BeCl₂ polymerises through chlorine bridges to complete the octet. (v) ICl₄⁻: iodine has 7 valence electrons plus 1 from the negative charge = 8; four are used in bonding, leaving 2 lone pairs, so 4 + 2 = 6 pairs, sp³d², octahedral electron geometry. As in XeF₄ the lone pairs go trans, and the ion is square planar with 90° angles. The pattern across all five is that the electron-pair geometry follows the total count, while the reported shape names only the positions of the atoms.
(i) State the postulates of molecular orbital theory. (ii) Draw the electron distribution and calculate the bond order for N₂ and for O₂. (iii) Explain, using bond order, why the N≡N bond enthalpy (about 945 kJ/mol) is far higher than that of O₂ (about 498 kJ/mol). (iv) Predict the magnetic behaviour of each. (v) State one observation that valence bond theory fails to explain but molecular orbital theory does.
Answer
(i) Postulates. Atomic orbitals of comparable energy and correct symmetry combine to form the same number of molecular orbitals; each combination gives one bonding molecular orbital of lower energy and one antibonding molecular orbital of higher energy. Molecular orbitals belong to the molecule as a whole, not to individual atoms. They are filled following the Aufbau principle, the Pauli exclusion principle and Hund's rule, exactly as atomic orbitals are. (ii) For N₂ (14 electrons) the filling order up to nitrogen puts σ2p_z above the π2p pair, giving σ1s² σ*1s² σ2s² σ*2s² π2p_x² π2p_y² σ2p_z². Bonding electrons N_b = 10, antibonding N_a = 4, so bond order = (10 − 4) ÷ 2 = 3. For O₂ (16 electrons) the σ2p_z now lies below the π2p set, giving σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹. Here N_b = 10 and N_a = 6, so bond order = (10 − 6) ÷ 2 = 2. (iii) Bond enthalpy rises steeply with bond order because a higher bond order means more electron density concentrated between the two nuclei and less in antibonding regions that push the nuclei apart. Nitrogen's bond order of 3 means three shared electron pairs holding the nuclei together with no electrons in antibonding π* orbitals at all. Oxygen's bond order of 2 comes with two electrons already occupying antibonding orbitals, which partly cancels the bonding interaction. Hence the nitrogen bond needs nearly twice the energy to break, and it is also shorter (110 pm against 121 pm). (iv) N₂ has every electron paired, so it is diamagnetic and is weakly repelled by a magnetic field. O₂ has two unpaired electrons in the degenerate π* orbitals, so it is paramagnetic and is attracted into a magnetic field — liquid oxygen visibly bridges the poles of a magnet. (v) The paramagnetism of O₂ is the standard example. A valence bond Lewis structure writes O=O with all electrons neatly paired and therefore predicts diamagnetic behaviour, contradicting experiment. Only molecular orbital theory, by placing the last two electrons singly in degenerate antibonding orbitals, predicts the two unpaired spins correctly. It also accounts naturally for species such as He₂⁺ and O₂⁻, where fractional or unusual bond orders arise.
A student notices that a bottle of hydrogen fluoride boils at 19.5 °C, while hydrogen chloride, a heavier molecule, boils at −85 °C. She also observes that ice cubes float on the water in her glass, and that ethanol mixes with water in all proportions while petrol does not. Answer: (a) Explain the boiling point anomaly of HF. (b) Why is ice less dense than liquid water? (c) Account for the miscibility of ethanol with water and the immiscibility of petrol. (d) Would you expect H₂S to show the same anomaly as HF within group 16? Justify.
Answer
(a) Fluorine is the most electronegative element, so the H−F bond is strongly polar and the hydrogen carries a large partial positive charge on a very small atom. Neighbouring HF molecules therefore link through strong intermolecular hydrogen bonds, F−H···F. Boiling requires these hydrogen bonds to be broken in addition to the ordinary dispersion forces, so a great deal of energy is needed and the boiling point is abnormally high. Chlorine is much less electronegative and larger, so HCl has only weak dipole–dipole and dispersion forces despite its greater molar mass, and it boils far lower. (b) In liquid water each molecule hydrogen bonds to about three or four neighbours in a constantly rearranging network, and the molecules can pack fairly close. On freezing, each oxygen forms exactly four hydrogen bonds directed tetrahedrally, producing an open cage-like lattice with substantial empty space at the centre of each ring. The same mass now occupies a larger volume, so ice has a lower density than water at 0 °C and floats — a fact that lets fish survive under a frozen lake surface. (c) Ethanol has an O−H group, so it can both donate and accept hydrogen bonds. When ethanol is added to water the hydrogen bonds broken between water molecules are replaced by comparable ethanol–water hydrogen bonds, so mixing costs almost no energy and the two are miscible in all proportions. Petrol is a mixture of hydrocarbons with only weak dispersion forces and no polar site. Inserting them between water molecules would break strong hydrogen bonds and pay nothing back, so the hydrocarbons are excluded and a separate layer forms — the like dissolves like principle. (d) No. Hydrogen bonding needs hydrogen bonded to a small, highly electronegative atom, effectively only F, O or N. Sulphur has an electronegativity of about 2.5, close to hydrogen's 2.1, and is a much larger atom, so the H−S bond is only weakly polar and the charge is spread over a bigger volume. H₂S therefore forms no significant hydrogen bonds and boils at about −60 °C, far below water's 100 °C. Within group 16 it is water that is the anomaly, and H₂S follows the normal trend of rising boiling point with molar mass.
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