Chemical Bonding



Subjective Problems 



  1. Vacant orbital on born on BF3is involved in pπ – pπ bonding with completely filled orbital on fluorine. But this pπ – pπ bonding is not effective in BCl3 due to greater length of B – Cl bond. Less availability of vacant orbital on boron decreases its strength as lewis acid.
  2. I is a better reducing agent than Cl and reduces Cu2+and itself changes to I2.
  3. In both cases S is in its maximum oxidation state (+6) and therefore cannot accommodate six bulky Cl ions around it in octahedral arrangement.
  4. Nitride ion N3– is very difficult to form and is formed by only those metals which can form nitride with high lattice energy. Since lattice energy of Li3N is much higher than Na3N formation of former is more favourable.
  5. Boron does not have d-orbital to accommodate electrons provided by 6OH ions.
  6. This is because the resultant dipole moment of two C – Cl bonds cancels that of other two chlorine which is not possible in case of CH2Cl2, whree even though resultant of C – H and C – Cl cancels each other but that of C – H and C – H and that of C – Cl and C – Cl can not, thus some dipole moment is retained.
  7. CaCO3 reacts with HCl to form CaCl2 whose lattice energy is less than hydration energy, so it does not dissolve in water.
  8. The number of H–bonds  broken per mole of HF on vapourisation is much less than in the case of H2O. Each HF forms  two bonds, while each H2O molecule forms 4 bonds. In case of water vapours at boiling pt. contains essentially monomers while HF contains polymers i.e., all H–bonds are not  broken on vapourisation whereas all 4-H bonds are broken per molecule of water.
  9. In the ground state both fluorine and chlorine has got the unpaired electron and hence exhibits a valency of one. But in the excited state transfer of s-electrons into the vacant d-orbtals of Cl is possible and therefore exhibit valencies of four and six.


  1. A substance is said to be soluble in water if it is capable if of forming H-bonding with water molecule. In onitrophenol due to chelation in OH group is not available to form hydrogen bond with water hence it is sparingly soluble in water. On the other hand the OH group is available in p-nitrophenol to form H-bond with water and hence it is more soluble compared to the O is answer.
  2. Boron in BF3 is sp2 hybridised giving a planar geometry whereas nitrogen in NF3 is sp3 hybridised with a lone pair, which account for its pyramidal geometry.
  3. Dipole moment μ = e × d coulomb metre

For KCl d = 2.6 × 10–10 m

For complete separation of unit charge

e = 1.602 × 10–19 C

Hence μ = 1.602 × 10–19 × 2.6 × 10–10 = 4.1652 × 10–29 Cm

μKCl = 3.336 × 10–29 Cm

% ionic character of KCl = = 80.09%

  1. Pb(+IV) is less table than Pb(+II) due to inert pair effect and therefore Pb(+IV) is reduced to Pb(+II) by I which changes to I2(I is a good reducing agent)
  2. The shape of PCl4+ is tetrahedral while that of PCl6 is octahedral. Now conduction of electricity is possible whenever the ions can move. Here due to strong  coulombic attraction between the tetrahedra and octahedra the ions cannot move and hence do not conduct electricity.



  1. Nitrogen cannot show pentavalency is it lacks d-orbital. In PCl5, central phosphorus shows pentavalecny as electron from 3s orbital is promoted to vacant d-orbital, giving rise to five unpaired electrons.

In bismuth due to pronounced inert pair effect 5s2 electrons do not prefer un-pair making Bi trivalent only.

  1. Loss of H+ from HI results in I which due its larger surface area disperses negative charge in a better way than in Cl ion.
  2. HF reacts with SiO2 present in glass forming fluoro silicate ions which is soluble in water.

SiO2 + 6HF ⎯→ [SiF6]2– + 2H+ + 2H2O

  1. Due to greater lone pair lone pair repulsion in F2 which arises due to shorter bond length of F – F bond) results in decrease in bond strength of F2.
  2. The Lewis structure for POCl3, would show some double bond character between P and O. (P is allowed to exceed the octet because of the availability of 3d orbitals). The increased electron density in the P = O bond would make the intensive repulsion between the P = O bond and a P – Cl bond greater than between the P – Cl bonds. Thus the Cl – P – Cl angle is lowered and the Cl – P = O angle raised, as compared with a regular tetrahedron.
  3. Because of its very high solvation due to its large charge density, its effective size increases and its mobility decreases.

In Cl2O delocalisation of lone pair of  electrons of oxygen to chlorine decrease the repulsion by lone pairs and increases the repulsion between bond pairs. The bond angle thus becomes larger compared to F2O.

  1. This is because, as the size of anion increases, its plolarisatibility increases and hence % of covalent character increases.
  2. SnCl4 is more covalent in nature compared to SnF4 due to the high polarisability of Cl ion and hence boiling is less.
  3. Cl – Cl bond is stronger. It’s because of smaller size of F which brings about its high electron density causing greater repulsion between bonded F atoms making the bonding weaker.
  4. In case of nitrogen intramolecular multiplicity is possible leading to completion of octet via π bond formation. Hence octet of N2, gets fulfilled by  N N. In case of P intramolecular multiplicity is ruled out because of longer P – P single bond. Hence P has to complete its octet by undergoing sigma bond formation with other phosphorus. Hence formula is P4.
  5. The order is

BI3 > BBr3 > BCl3 > BF3

This is because, extent of back donation of electron decreases in the order

BF3 > BCl3 > BBr3 > BI3

due to increase in the size of p-orbitals which causes poorer overlap.

  1. Since Al – O and O – H bond strength are almost equal, there exist almost equal probability of their breaking making them amphoteric in nature.
  2. Because of one lone pair, which repel the four Cl atoms slightly up, making them and the central I, non-coplanar.
  3. A substance is said to be soluble in water if it is capable of forming H-bonding with water molecule. In o-hydroxy benzaldehyde due to intra-molecular chelation the –OH group is not available to form hydrogen bond with water hence it is sparingly soluble in water. On the other hand the –OH group is available in p-hydroxy benzaldehyde to form H-bond with water and hence it is more soluble compared to the o-isomer. 



  1. In (SiH3)3N pπ-dπ bonding exists between fully filled p-orbital of nitrogen and vacant d-orbital of silicon which requires planarity in molecule that is possible only when nitrogen is sp2 hybridised.
  2. dπ-pπ bonding is possible in case of H3SiNCO but it is not possible in case of H3CNCO as carbon lacks d-orbital, dπ-pπ bonding is most effective when molecule is linear.
  3. Aqua-regia is 1 part HNO3 and 3 parts HCl. HNO3 oxidizes Au to Au3+ and Cl forms a complex ion [AuCl4] with Au3+ enhancing its solubility.
  4. Orange colour in SnI4 is caused by absorption of blue light, the reflected light thus containing a higher proportion of red and orange. The energy absorbed in this way causes transfer of an electron from I to Sn resulting in temporary reduction of Sn(IV) to Sn(+III). Since transferring of an electron is transferring of charge such spectra are called charge transfer spectra. This occurs in SnI4 because both Sn and I atoms have similar energy levels.
  5. CO forms a dative bond with metal (M C O+). This dative bond is unstable and is strengthened by back bonding which involves overlap of filled dx–y orbital of metal and vacant π*2py orbital of carbon, as a result M = C = O bond is formed. Thus in metal carbonyl placement  of electrons in anti-bonding orbital of CO reduces its bond order from three to two and thus increasing C – O bond length from 1.12Å to 1.15Å.
  6. FeCl2 is having greater melting point since it is more ionic than FeCl3. This can be explained on the basis of Fajan’s rule which states that a cation having higher magnitudes of positive charge has got more polarising power and forms compound having more covalent character.
  7. P-dimethyoxy benzene can exist in two following forms.

Because of (II) structure there exist some dipole moment.

  1. It’s because of intra molecular hydrogen bonding in ortho-hydroxy benzoate ion, which impart stability to it but not in p-hydroxy benzoate ion.
  2. Because of larger size of ‘Si’ it cannot form strong π bond with oxygen as is done by ‘C’ in CO2. So it satisfies it octet by forming giant three dimensional structure using σ bond.
  3. In trimethylamine, N  is sp3hybridised and in trisilylamine N is sp2 hybridised nitrogen. In trisilyl amine the vacant d orbitals of Si can undergo overlap with the filled p-orbitals of nitrogen and for this geometry becomes planar. (As delocalisation is possible in planar geometry). This delocalisation is not possible in trimethyl amine as C has got no vacant d-orbitals.
  4. Lithium due to its small size and high charge density prevents the electron cloud of oxygen to spread to other oxygen. But due to large size of sodium, it fails to prevent spreading of electron cloud to other oxygen due to which peroxide results.
  5. Because of its very high degree of covalent character due to large size I which can thus be polarised easily.
  6. NCl3 + 4H2O ⎯→ 3HOCl + NH4OH

PCl3 + 3H2O ⎯→ 3HCl + H3PO3

In NCl3, H2O attacks vacant d-orbital of Cl where as in PCl3, H2O attacks the vacant d-orbital of P.

  1. It is because when S combines with strongly electronegative atoms like ‘F’ their d-orbital contracts in size and becomes lower in energy and thereafter they undergo hybridization with s and p-orbital to give sp3d2 hybridized orbitals. But with ‘H’ it can not occur.
  2. Since ‘F’ is having very high electronegativity it can form strong hydrogen bond with HF as F Hδ+ –– Fδ but due to lower electronegativity ‘Cl’, HCl cannot do it with Cl can not do it.