Chemical Bonding and Molecular Structure Class 11 Chemistry NCERT Solutions

Step-by-step solution

1. 1Atoms tend to acquire the electron configuration of the nearest noble gas because a filled valence shell gives a lower-energy, more stable state.
2. 2As atoms approach, attraction between each nucleus and the electrons of the other atom competes with repulsion between the two nuclei and between the two electron clouds.
3. 3At the equilibrium bond distance, the net attraction stabilises the pair and lowers the potential energy of the system.
4. 4If electron transfer produces stable ions, the resulting electrostatic attraction is an ionic bond; if valence electrons are shared, the resulting electron density between nuclei gives a covalent bond.
5. 5Thus bond formation is the energetic stabilisation that accompanies electron rearrangement, whether the bonding is ionic, covalent or metallic.

Step-by-step solution

1. 1A Lewis symbol shows only the valence-shell electrons of a free atom: Mg has 2, Na has 1, B has 3, O has 6, N has 5 and Br has 7 valence electrons.
2. 2Place these electrons beside the element symbol, pairing them where convenient: Mg has two single dots; Na has one; B has three; O has six; N has five; and Br has seven.
3. 3The exact left–right placement of equivalent dots is only a drawing convention; the essential Lewis information is the number and pairing of the valence electrons around each atom.

Step-by-step solution

1. 1S has 6 valence electrons. In S²⁻, two electrons have been gained, so the ion has 8 valence electrons, arranged as four lone pairs, and its formal charge is −2.
2. 2Al has 3 valence electrons. In Al³⁺, all three have been removed, so the ion has no valence electrons to display and its formal charge is +3.
3. 3H has 1 valence electron. In H⁻, one electron has been gained, so the ion has 2 electrons, forming one lone pair and completing the duet; its formal charge is −1.
4. 4The cation symbols show no dots because the displayed Lewis symbol contains the remaining valence electrons, while the charge identifies electron loss.

Step-by-step solution

1. 1Count valence electrons: H₂S has 8, SiCl₄ has 32, BeF₂ has 16, CO₃²⁻ has 24 and HCOOH has 18 valence electrons.
2. 2H₂S: S is central and forms two S–H single bonds. Sulfur retains 2 lone pairs, so both H atoms and sulfur satisfy the octet.
3. 3SiCl₄: Si forms 4 Si–Cl single bonds. Silicon has no lone pair and each chlorine has 3 lone pairs.
4. 4BeF₂: Be forms 2 Be–F single bonds and has no lone pair. Each F has 3 lone pairs; Be retains an incomplete duet, a recognised exception to the octet rule.
5. 5CO₃²⁻: carbon is bonded to 3 oxygen atoms. One contributor has one C=O bond and two C–O single bonds; the double-bonded O has 2 lone pairs, each singly bonded O has 3 lone pairs and carries −1, giving total charge −2.
6. 6HCOOH: the connectivity is H–C(=O)–O–H. The carbonyl O and hydroxyl O each have 2 lone pairs, and every atom has an octet except hydrogen, which has a duet; all formal charges are zero.
7. 7
   $$\mathrm{HCOOH}:4+2(1)+2(6)=18e^{-}$$

Step-by-step solution

1. 1The octet rule states that atoms tend to combine by transfer or sharing of valence electrons so that each atom attains the electron configuration of the nearest noble gas.
2. 2Its significance is that it provides a simple qualitative test for stable Lewis structures, common formulas and whether a bond is ionic or covalent.
3. 3An incomplete octet is possible in species such as BeCl₂, BF₃ and AlCl₃, so the rule is not universal for electron-deficient compounds.
4. 4Odd-electron species such as NO and NO₂ cannot give every atom an octet.
5. 5Period-3 and heavier atoms can exceed an octet, as in PF₅ and SF₆, and noble gases such as Xe and Kr can form compounds such as XeF₂ and KrF₂.
6. 6Finally, the rule does not by itself predict molecular shape or relative stability; those require VSEPR, resonance or valence-bond and molecular-orbital reasoning.

Step-by-step solution

1. 1Formation of an ionic compound requires energy to remove electrons from a metal and add them to a non-metal, followed by energy released when the resulting ions form a crystal lattice.
2. 2A low ionisation enthalpy makes electron loss from the metal inexpensive, while a high magnitude of exothermic electron-gain enthalpy makes electron addition favourable.
3. 3A large lattice enthalpy stabilises the ionic solid; it is especially large for highly charged ions and small ionic radii.
4. 4
   $$\Delta H_{\mathrm{formation}}\approx \sum IE-\sum EAH+\text{lattice formation enthalpy}$$
5. 5Therefore, easy cation formation, easy anion formation and strong electrostatic attraction in the lattice all favour ionic bonding.

Step-by-step solution

1. 1Use steric number = number of σ bonds + number of lone pairs on the central atom, then assign hybridisation and geometry.
2. 2BeCl₂: steric number = 2 + 0 = 2, so it is sp hybridised, linear, with a Cl–Be–Cl angle of 180°.
3. 3BCl₃: steric number = 3 + 0 = 3, so it is sp² hybridised, trigonal planar, with bond angles of 120°.
4. 4SiCl₄: steric number = 4 + 0 = 4, so it is sp³ hybridised, tetrahedral, with bond angles of 109.5°.
5. 5AsF₅: steric number = 5 + 0 = 5, so it is sp³d hybridised and trigonal bipyramidal. Equatorial–equatorial angles are 120°, axial–equatorial angles are 90° and the axial–axial angle is 180°.
6. 6H₂S: steric number = 2 σ bonds + 1 lone pair = 3, so it is sp³ hybridised. The electron arrangement is trigonal planar, but molecular shape is bent with an H–S–H angle of about 92°.
7. 7PH₃: steric number = 3 σ bonds + 1 lone pair = 4, so it is sp³ hybridised, trigonal pyramidal, with an H–P–H angle of about 93.5°.
8. 8Lone-pair repulsion compresses the bond angles of H₂S and PH₃ below the ideal tetrahedral value of 109.5°.

Step-by-step solution

1. 1For NH₃, steric number = 3 σ bonds + 1 lone pair = 4; the electron-pair geometry is tetrahedral and the atom is sp³ hybridised.
2. 2For H₂O, steric number = 2 σ bonds + 2 lone pairs = 4; its electron-pair geometry is also tetrahedral and oxygen is sp³ hybridised.
3. 3VSEPR repulsion follows lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.
4. 4Water has one lone-pair–lone-pair interaction and each lone pair also compresses the O–H bonds. Ammonia has only one lone pair, so its H–N–H bonds are compressed less.
5. 5The resulting bond angles are approximately 104.5° in H₂O and 107.8° in NH₃.

Step-by-step solution

1. 1In molecular-orbital theory, bond order measures the net number of bonding interactions and is a dimensionless quantity.
2. 2
   $$\text{Bond order}=\frac{N_b-N_a}{2}$$
3. 3Here Nᵦ is the number of electrons in bonding orbitals and Nₐ is the number in antibonding orbitals.
4. 4Removing an electron from an antibonding orbital or adding one to a bonding orbital raises bond order and usually strengthens and shortens the bond; the reverse lowers it.

Step-by-step solution

1. 1Bond length is the equilibrium internuclear distance between two bonded atoms in a molecule or ion.
2. 2It is commonly expressed in picometres or ångströms.
3. 3
   $$1\text{A˚}=10^{-10}\mathrm{m}=100\mathrm{pm}$$
4. 4For an ionic bond, the distance is approximately the sum of the cation and anion radii; for a covalent bond it is approximately the sum of the bonded atoms' covalent radii.
5. 5Bond lengths are obtained from diffraction, rotational spectroscopy and other structural measurements. A shorter bond is generally associated with stronger bonding, although atom sizes must also be considered.

Step-by-step solution

1. 1Carbonate has 4 + 3(6) + 2 = 24 valence electrons and a trigonal-planar arrangement with three equivalent C–O connections.
2. 2Contributor I places the C=O bond on one oxygen and single bonds to the other two oxygens, which each carry −1.
3. 3Contributor II places the C=O bond on a second oxygen, while contributor III places it on the third oxygen. Carbon carries +1 in each expanded-octet contributor, and the remaining two oxygens carry −1 each.
4. 4The nuclei remain in the same positions in all three contributors; only the distribution of π electrons and formal charges changes, so they are canonical forms rather than separate structures that interconvert.
5. 5The real carbonate ion is the resonance hybrid of the three equivalent contributors. Delocalisation lowers its energy and makes all three C–O bonds equal, with bond order 1⅓ and length intermediate between a C–O single and C=O double bond.
6. 6
   $$\mathrm{C}{\mathrm{O}}_3^{2-}:\pi \text{-delocalisation over all three C-O links}$$

Step-by-step solution

1. 1Canonical resonance contributors must have exactly the same positions of nuclei and differ only in electron placement.
2. 2The two displayed forms change which hydrogen is attached directly to phosphorus and therefore change atomic connectivity, not merely π-electron placement.
3. 3They are distinct structural arrangements rather than resonance contributors.
4. 4The conventional connectivity of phosphorous acid is H–P(=O)(OH)₂; its two equivalent P–OH bonds can participate in resonance, but the displayed I and II cannot be paired as I and II resonance forms.

Step-by-step solution

1. 1SO₃: describe three equivalent principal contributors. In turn, choose a different oxygen for the S=O bond; the other two oxygens are singly bonded and each carries −1, while S carries +1. Each double-bonded O carries 0 and has 2 lone pairs; each O⁻ has 3 lone pairs.
2. 2A less important all-single-bond contributor has three S–O single bonds, S carrying +2 and each O carrying −1, but the three principal expanded-octet forms account for the observed equivalence of the S–O bonds.
3. 3NO₂: there are 17 valence electrons and two equivalent contributors. In one, N=O and N–O⁻; in the other, the N=O bond is on the other oxygen. N has formal charge +1, the singly bonded O has −1, and N retains one unpaired electron.
4. 4NO₃⁻: there are 24 valence electrons and three equivalent contributors, each placing the N=O bond on a different oxygen. N has +1; two singly bonded O atoms each have −1; the double-bonded O has 0.
5. 5In every case, the nuclei stay fixed while π electrons are delocalised, and the true species is the lower-energy resonance hybrid rather than any single contributor.

Step-by-step solution

1. 1K has configuration 2, 8, 8, 1 and S has 2, 8, 6. Each K transfers one electron to an S atom, forming K⁺ and S⁻, both with the argon configuration; two K⁺ ions are required per S²⁻ in K₂S.
2. 2Ca has configuration 2, 8, 8, 2 and O has 2, 6. Ca transfers two electrons to O, forming Ca²⁺ and O²⁻, each with the argon configuration; their formula is CaO.
3. 3Al has configuration 2, 8, 3 and N has 2, 5. Al transfers three electrons to N, forming Al³⁺ and N³⁻, each with the neon configuration; their formula is AlN.
4. 4In each transfer, both ions gain stable noble-gas configurations, and electrostatic attraction between the oppositely charged ions stabilises the compound.

Step-by-step solution

1. 1Oxygen is more electronegative than carbon, so each C=O bond in CO₂ is polar and directed from carbon toward oxygen.
2. 2CO₂ has steric number = 2 + 0 = 2, is sp hybridised and is linear at 180°. The two equal C=O bond dipoles oppose one another and cancel, so μ(CO₂) = 0 D.
3. 3Oxygen is also more electronegative than hydrogen, so each O–H bond in water is polar toward oxygen.
4. 4H₂O has steric number = 2 + 2 = 4, is sp³ hybridised and is bent with an H–O–H angle of 104.5°. The two equal O–H bond dipoles do not cancel because they are not opposite.
5. 5Their vector resultant points along the molecular bisector and has a magnitude of 1.84 D for water.

Step-by-step solution

1. 1Dipole moment is the product of charge magnitude and the separation of the centres of positive and negative charge; it is a vector directed from positive to negative charge.
2. 2
   $$\mu =qr,1\mathrm{D}=3.33564\times 10^{-30}\mathrm{C}\mathrm{m}$$
3. 3Its SI unit is C·m; the commonly used debye unit D is exactly equivalent to the value shown.
4. 4A nonpolar molecule has μ = 0 D because its bond dipoles cancel, whereas a polar molecule has a nonzero vector resultant. Thus dipole moment distinguishes isomers with different symmetry.
5. 5Bond dipole data help identify molecular geometry, bond polarity and the percentage ionic character of a bond. For a diatomic bond, μ(observed) divided by μ(ionic) provides the percentage ionic character; for polyatomic molecules, the vector sum must be used.

Step-by-step solution

1. 1Electronegativity is the ability of an atom in a bond to attract the shared electron pair toward itself.
2. 2It is a relative property, is not directly measurable as an absolute quantity and can vary with the atom or environment to which a given element is bonded.
3. 3Electron gain enthalpy is the enthalpy change when one mole of electrons is added to one mole of a neutral gaseous atom to form a gaseous anion.
4. 4
   $$\mathrm{X}(\mathrm{g})+{\mathrm{e}}^{-}⟶{\mathrm{X}}^{-}(\mathrm{g})$$
5. 5Electron gain enthalpy is experimentally measurable, is a property of an isolated gaseous species, and may be exothermic or endothermic; electronegativity is dimensionless and comparative.

Step-by-step solution

1. 1A polar covalent bond forms when two bonded atoms have different electronegativities, so the shared electron pair is displaced toward the more electronegative atom.
2. 2For HCl, chlorine is more electronegative than hydrogen: the dimensionless Pauling values are approximately 3.16 and 2.20, respectively.
3. 3The shared pair therefore lies closer to chlorine, producing partial charges rather than complete ions.
4. 4
   $${\mathrm{H}}^{\delta +}-\mathrm{C}{\mathrm{l}}^{\delta -}$$
5. 5HCl is therefore covalent because the pair is shared, but polar because the sharing is unequal.

Step-by-step solution

1. 1Ionic character generally increases as the electronegativity difference between bonded atoms increases.
2. 2
   $$\%\text{ ionic character}\propto |{\chi }_A-{\chi }_B|$$
3. 3The N–N bond in N₂ joins identical atoms, so Δχ = 0 and it is nonpolar covalent.
4. 4S–O and Cl–F are polar covalent bonds; their electronegativity differences are similar, with Cl–F conventionally placed after S–O for the NCERT qualitative comparison.
5. 5K–O has a much larger difference and is strongly ionic, while Li–F has the largest difference in the set and is the most ionic.
6. 6The order of increasing ionic character is therefore N₂ < SO₂ < ClF₃ < K₂O < LiF.

Step-by-step solution

1. 1Count the valence electrons: 2(4) + 4(1) + 2(6) = 24.
2. 2Keep the skeleton CH₃–C–O–H and provide carbon with four bonds in total.
3. 3Use the carboxyl carbon to form a C=O double bond, a C–OH single bond and a C–CH₃ single bond; the methyl carbon forms three C–H single bonds and the hydroxyl oxygen forms one O–H single bond.
4. 4Place 2 lone pairs on the carbonyl oxygen and 2 lone pairs on the hydroxyl oxygen. No atom then exceeds an octet, hydrogen has a duet and all formal charges are zero.
5. 5
   $$\mathrm{C}{\mathrm{H}}_3-\mathrm{C}(=\mathrm{O})-\mathrm{O}-\mathrm{H}$$

Step-by-step solution

1. 1In CH₄, carbon has steric number = 4 σ bonds + 0 lone pairs = 4, so its four bonding domains are sp³ hybridised.
2. 2The four equivalent sp³ orbitals point toward the corners of a tetrahedron, giving an H–C–H angle of 109.5°.
3. 3A square-planar arrangement would require 90° separations and stronger electron-pair repulsions; the tetrahedral arrangement minimises these repulsions.
4. 4It would also formally require dsp²-type participation from energetically high d orbitals, which is not competitive with sp³ bonding for carbon in methane.
5. 5Both valence-bond and VSEPR reasoning therefore select the tetrahedral structure rather than a square plane.

Step-by-step solution

1. 1Hydrogen is more electronegative than beryllium, so each Be–H bond is polar and its bond dipole points from Be toward H.
2. 2The central Be has steric number = 2 σ bonds + 0 lone pairs = 2, so it is sp hybridised and gaseous BeH₂ is linear with a 180° H–Be–H angle.
3. 3The two Be–H bond dipoles have equal magnitude and point in exactly opposite directions.
4. 4
   $${\vec{\mu }}_1+{\vec{\mu }}_2=0$$
5. 5Vector cancellation makes the molecular dipole zero even though each individual bond is polar.

Step-by-step solution

1. 1Both central N atoms have steric number = 3 σ bonds + 1 lone pair = 4, so both molecules are sp³ hybridised and trigonal pyramidal.
2. 2Nitrogen is more electronegative than hydrogen, so the three N–H bond dipoles point toward N and reinforce the nitrogen lone-pair contribution.
3. 3Fluorine is more electronegative than nitrogen, so the three N–F bond dipoles point from N toward F and oppose the nitrogen lone-pair contribution; consequently much of the vector sum cancels.
4. 4The observed values are μ(NH₃) = 1.46 D and μ(NF₃) = 0.24 D.
5. 5Therefore, despite a larger bond-polarity difference in N–F, the molecular dipole moment of NH₃ is much greater.

Step-by-step solution

1. 1Hybridisation is the mathematical mixing of atomic orbitals of similar energy on the same atom to produce an equal or equivalent set of directed hybrid orbitals.
2. 2The number of hybrid orbitals equals the number of electron domains, that is σ bonds plus lone pairs on the central atom.
3. 3sp hybridisation mixes one s and one p orbital to form 2 linear sp orbitals directed 180° apart.
4. 4sp² hybridisation mixes one s and two p orbitals to form 3 coplanar sp² orbitals directed 120° apart in a trigonal plane.
5. 5sp³ hybridisation mixes one s and three p orbitals to form 4 equivalent sp³ orbitals directed toward the vertices of a tetrahedron with bond angles of 109.5°.
6. 6
   $$\mathrm{sp}:180^{\circ },\mathrm{s}{\mathrm{p}}^2:120^{\circ },\mathrm{s}{\mathrm{p}}^3:109.5^{\circ }$$

Step-by-step solution

1. 1In monomeric AlCl₃, Al has steric number = 3 σ bonds + 0 lone pairs = 3.
2. 2The three bonding domains are sp² hybridised, so isolated AlCl₃ is trigonal planar with 120° bond angles.
3. 3The electron pair on Cl⁻ is donated into the empty orbital on Al, creating a coordinate Al–Cl bond.
4. 4AlCl₄⁻ then has steric number = 4 σ bonds + 0 lone pairs = 4, so Al changes from sp² to sp³ and the geometry changes from trigonal planar to tetrahedral with 109.5° angles.
5. 5
   $$\mathrm{AlC}{\mathrm{l}}_3:s{p}^2⟶AlC{l}_4^{-}:s{p}^3$$

Step-by-step solution

1. 1In BF₃, B has steric number = 3 σ bonds + 0 lone pairs = 3 and is sp² hybridised with an empty unhybridised 2p orbital.
2. 2The nitrogen lone pair in NH₃ is donated into that empty 2p orbital to form a coordinate B←N bond.
3. 3In the adduct, B has steric number = 4 σ bonds + 0 lone pairs = 4, so B changes from sp² to sp³ and becomes approximately tetrahedral.
4. 4N in NH₃ already has steric number = 3 σ bonds + 1 lone pair = 4 and is sp³ hybridised; donating the pair to form a fourth σ bond does not increase its steric number, so N remains sp³.

Step-by-step solution

1. 1In C₂H₄, each carbon has steric number = 3 σ bonds + 0 lone pairs = 3, so both carbons are sp² hybridised and locally trigonal planar.
2. 2An sp² orbital on each C forms the C–C σ bond. Each carbon uses its other two sp² orbitals to form two C–H σ bonds with 1s orbitals of H.
3. 3One unhybridised 2p orbital on each carbon overlaps side by side to form one π bond; hence the C=C bond contains 1 σ + 1 π bond.
4. 4In C₂H₂, each C has steric number = 2 σ bonds + 0 lone pairs = 2, so both are sp hybridised and the molecule is linear.
5. 5One sp orbital on each C forms the C–C σ bond, and the other forms a C–H σ bond with H 1s. Two unhybridised p orbitals on each C overlap laterally to form two π bonds, so C≡C contains 1 σ + 2 π bonds.
6. 6
   $$\mathrm{C}=\mathrm{C}:1\sigma +1\pi ,C\equiv C:1\sigma +2\pi$$

Step-by-step solution

1. 1Every atom-to-atom connection contains exactly one σ bond; a double bond adds one π bond and a triple bond adds two π bonds.
2. 2C₂H₂ has 2 C–H σ bonds and 1 σ component in C≡C, giving 3 σ bonds; the two remaining components of the triple bond are 2 π bonds.
3. 3C₂H₄ has 4 C–H σ bonds and 1 σ component in C=C, giving 5 σ bonds; the second component of the double bond is 1 π bond.
4. 4
   $${\mathrm{C}}_2{\mathrm{H}}_2:3\sigma +2\pi ;{\mathrm{C}}_2{\mathrm{H}}_4:5\sigma +1\pi$$

Step-by-step solution

1. 1A σ bond forms by head-on overlap along the internuclear axis.
2. 21s–1s, 1s–2pₓ and 1s–2s pairs can overlap head-on along the x-axis and form σ bonds.
3. 3The two 2pᵧ orbitals are perpendicular to the x-axis. If they are parallel, their overlap is lateral rather than head-on.
4. 4Lateral overlap of parallel 2pᵧ orbitals forms a π bond, not a σ bond.

Step-by-step solution

1. 1Count four electron domains around each carbon: each σ bond and each lone pair contributes one; a multiple bond counts as only one domain.
2. 2(a) In CH₃–CH₃, each C has 4 σ bonds and 0 lone pairs, so both C atoms are sp³ hybridised.
3. 3(b) In CH₃–CH=CH₂, C₁ has 4 σ domains and is sp³, while C₂ and C₃ each have 3 σ domains and are sp².
4. 4(c) In CH₃–CH₂–OH, each C has 4 σ bonds and no lone pair, so both C atoms are sp³.
5. 5(d) In CH₃–CHO, the methyl C is sp³ and the carbonyl C has 3 σ domains and is sp².
6. 6(e) In CH₃COOH, the methyl C is sp³ and the carboxyl C has 3 σ domains and is sp²; the π component of C=O does not add a fourth hybrid domain.

Step-by-step solution

1. 1A bond pair is a shared pair of valence electrons that forms a covalent σ bond between two atoms.
2. 2In C₂H₆, all 14 valence electrons are used in bonding: there are 1 C–C bond and 6 C–H bonds, so 7 bond pairs and no lone pairs on carbon.
3. 3A lone pair is a pair of valence electrons localised on one atom and not shared in a bond.
4. 4In H₂O, oxygen forms 2 O–H bond pairs and retains 2 lone pairs; its steric number is therefore 2 + 2 = 4.
5. 5Bond pairs determine connections between atoms, while lone pairs influence molecular shape and bond angles through electron-domain repulsion.

Step-by-step solution

1. 1A σ bond forms by head-on overlap along the internuclear axis; a π bond forms by lateral overlap of parallel orbitals perpendicular to that axis.
2. 2σ bonds can arise from s–s, s–p, p–p or hybrid-orbital overlap, whereas the π bonds considered here arise from lateral p-orbital overlap.
3. 3The electron density of a σ bond is cylindrically symmetric about the internuclear axis; a π bond has two lobes on opposite sides of that axis and a nodal plane containing it.
4. 4A σ bond is generally stronger than a π bond, and rotation about an isolated σ single bond is comparatively free, whereas a π bond restricts rotation because sideways overlap would be lost.
5. 5A double bond contains one σ and one π component, while a triple bond contains one σ and two π components.

Step-by-step solution

1. 1Each H atom has one electron in a spherical 1s orbital. As the atoms approach, attraction develops between each nucleus and the electron of the other atom, while repulsion develops between the nuclei and between the electrons.
2. 2The total potential energy initially falls because nucleus–electron attraction dominates, allowing the two 1s orbitals to overlap.
3. 3The electrons pair with opposite spins, and constructive overlap places increased electron density between the two nuclei.
4. 4This internuclear electron density lowers the energy of the pair. At the equilibrium distance, attractive and repulsive effects balance and the H–H σ bond is formed.
5. 5
   $$1s(A)+1s(B)⟶{\sigma }_{1s}(H-H)$$
6. 6The equilibrium H–H bond length is about 74 pm and the bond dissociation energy is about 436 kJ mol⁻¹.

Step-by-step solution

1. 1The atomic orbitals to be combined must have the same or nearly the same energy; large energy mismatch gives little mixing.
2. 2They must have compatible symmetry and proper orientation so that their overlap is constructive and appreciable.
3. 3The extent of overlap along or across the internuclear axis must be large; distant or poorly oriented orbitals interact weakly.
4. 4The two atomic orbitals combine linearly to give one bonding and, where non-degenerate conditions permit, one corresponding antibonding molecular orbital.
5. 5Because molecular orbitals extend over the whole molecule, an electron in a bonding orbital stabilises both nuclei, whereas an electron in an antibonding orbital reduces the net bond.

Step-by-step solution

1. 1Each Be atom has the configuration 1s² 2s², so Be₂ has 8 valence and total electrons.
2. 2The molecular-orbital configuration is obtained by pairing 1s and 2s atomic orbitals.
3. 3
   $$({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2$$
4. 4There are 4 bonding and 4 antibonding electrons, including the cancelling 1s core orbitals.
5. 5
   $$\text{Bond order}=\frac{4-4}{2}=0$$
6. 6A bond order of 0 gives no net lowering from the separated atoms. Be₂ is therefore not bound as a discrete molecule, although bulk beryllium is metallic and held by delocalised electrons.

Step-by-step solution

1. 1Bond order is one half of bonding electrons minus antibonding electrons; higher bond order means greater stability.
2. 2The 1s core orbitals cancel. The relevant valence configuration common to the oxygen species is σ(2s)², σ*(2s)², σ(2p_z)², π(2pₓ)², π(2pᵧ)², followed by electrons in the degenerate π* orbitals.
3. 3O₂ has two electrons in separate π* orbitals: Nᵦ = 8, Nₐ = 4, bond order = 2 and 2 unpaired electrons, so it is paramagnetic.
4. 4O₂⁺ has one π* electron: Nᵦ = 8, Nₐ = 3, bond order = 2.5 and 1 unpaired electron, so it is paramagnetic.
5. 5O₂⁻ has three π* electrons: Nᵦ = 8, Nₐ = 5, bond order = 1.5 and 1 unpaired electron, so it is paramagnetic.
6. 6O₂²⁻ has four π* electrons: Nᵦ = 8, Nₐ = 6, bond order = 1 and no unpaired electron, so it is diamagnetic.
7. 7Thus the relative stability follows O₂⁺ > O₂ > O₂⁻ > O₂²⁻ because their bond orders are 2.5, 2, 1.5 and 1.
8. 8
   $${\mathrm{O}}_2^{+}:2.5,{\mathrm{O}}_2:2,{\mathrm{O}}_2^{-}:1.5,{\mathrm{O}}_2^{2-}:1$$

Step-by-step solution

1. 1A molecular orbital is represented by a spatial wavefunction, and the plus or minus sign specifies the algebraic sign, or phase, of that wavefunction in a region of space.
2. 2Lobes on opposite sides of a node can have opposite signs even though the electron density, proportional to the square of the wavefunction, is positive in both lobes.
3. 3The overall sign of a normalised orbital can be reversed without changing the physical orbital, so a single plus or minus has no independent physical meaning by itself.
4. 4During overlap, lobes of the same phase combine constructively to give a bonding orbital; lobes of opposite phase combine destructively to give an antibonding orbital.
5. 5The signs therefore indicate wavefunction phase and overlap, not positive and negative electrical charge.

Step-by-step solution

1. 1The central P has steric number = 5 σ bonds + 0 lone pairs = 5.
2. 2In the valence-bond description, excited phosphorus uses one 3s, three 3p and one 3d orbital to form five sp³d hybrid orbitals directed toward a trigonal bipyramid.
3. 3Three P–Cl bonds lie in the equatorial plane and are separated by 120°; two lie above and below that plane, at 90° to it and 180° from each other.
4. 4Each axial position has three 90° interactions with equatorial bond pairs, whereas each equatorial position has only two 90° interactions with axial bond pairs.
5. 5The greater repulsion experienced by the axial bond pairs produces longer axial P–Cl bonds than equatorial P–Cl bonds.

Step-by-step solution

1. 1A hydrogen bond is an electrostatic attraction between H bonded to a highly electronegative atom, commonly N, O or F, and a lone pair on an electronegative atom of a neighbouring molecule or group.
2. 2
   $$\mathrm{X}-\mathrm{H}\cdots \mathrm{Y},\mathrm{X},\mathrm{Y}=\mathrm{N},\mathrm{O},\mathrm{or}\mathrm{F}$$
3. 3The X–H bond is strongly polar, so H carries δ+ and attracts the δ− lone-pair region on Y. The hydrogen bond is therefore a strongly oriented dipole–dipole attraction.
4. 4It is stronger than ordinary van der Waals dispersion forces because it requires both a large bond dipole and a specific donor–acceptor orientation.
5. 5Hydrogen bonding may be intermolecular, as in water, or intramolecular, as in suitable ortho-substituted compounds. It is generally weaker than an ordinary covalent or ionic bond and is strongest in the solid state because molecules are more constrained.
6. 6A water molecule can form four hydrogen bonds: two as donor through its H atoms and two as acceptor through its oxygen lone pairs.

Step-by-step solution

1. 1Bond order is one half of the number of electrons in bonding molecular orbitals minus the number in antibonding molecular orbitals; it is dimensionless.
2. 2
   $$\text{Bond order}=\frac{N_b-N_a}{2}$$
3. 3N₂ has 8 bonding and 2 antibonding valence electrons after the cancelling 1s core orbitals are removed, so its bond order is (8 − 2)/2 = 3.
4. 4O₂ has 8 bonding and 4 antibonding valence electrons, so its bond order is (8 − 4)/2 = 2.
5. 5O₂⁺ is formed by removing one electron from an antibonding π* orbital of O₂. It has 8 bonding and 3 antibonding electrons, so its bond order is (8 − 3)/2 = 2.5.
6. 6O₂⁻ is formed by adding one electron to an antibonding π* orbital. It has 8 bonding and 5 antibonding electrons, so its bond order is (8 − 5)/2 = 1.5.
7. 7
   $${\mathrm{N}}_2:3,{\mathrm{O}}_2:2,{\mathrm{O}}_2^{+}:2.5,{\mathrm{O}}_2^{-}:1.5$$