Concepts of Acids and Bases



Solution to Subjective Problems 



  1. H3O+ > H2O > OH
  2. The highly branched tertiary butyl group involve appreciable back – strain (B-strain) when the boron atom changes to pyramidal environment on adduct formation. This destabilizes the adduct. Hence the order is 

B(t–Bu)3 < B(n–Bu)3

  1. II > IV > III > I
  2. BI3 > BBr3 > BCl3 > BF3
  3. Ag+ is stronger Lewis acid due to pseudo inert gas configuration.
  4. Both HCl and SO3 are Lewis acids and can react with amine base to form polar substances which could presumably undergo ionic dissociation in a solvent sufficiently more polar than benzene. The reactioins may be represented as follows (R = C4H9):

Sulphur is a third element  which can expand its octet due to availability of vacant d-orbitals. Thus sulphur expands its number of valance electron by attaching to the lone pair on the nitrogen. The N – S bond will be polar because of the big difference in  electronegativity between N and S.


Here the proton of the HCl attaches to the lone pair on the N. The connection between N – H and Cl is designated by (—-) symbolizing an electron pair on the Cl connected to nitrogen by a hydrogen bond.

  1. d < c < a < b
  2. The order in this case is the reverse of that for BX3. π-conjugation from the halogen p-orbital to the Si-d orbital is not as intense as in the case of BX3 and the order of acidity follows the increase in electron withdrawing power of the halogen from I to F. Hence the order is 

SiI4 < SiBr4 < SiCl4 < SiF4

  1. Water ionises as H2O + H2O H3O+ + OH with strong acid water behaves as a  base and accept the proton gives by the acid e.g.

HOI + H2O Cl + H3O+

While with strong base, water behaves as an acid by liberating a proton e.g.

H2O + NH3 NH4+ + OH

Such solvents are known amphiprotic solvents.

  1. Assume that each has lost a proton. So we get,

OCl, OClO, OClO2, OClO3

The species which have more is than one oxygen atoms show resonance.

That is:

It can be clearly seen that more the no. of oxygen atoms, more the resonance and more is the volume available for the  negative charge.

Therefore  OCl, OClO, OClO2, OClO3

  1. Volume available for the negative charge is increasing from left to right
  2. Charge density is decreasing from left to right.
  3. Basicity of the anion is decreasing from left to right.
  4. Acidity of the conjugate acid is increasing from left to right.

Therefore, the order of acidic strength is :

HOCl < HOClO <  HOClO2  < HOClO3



  1. The oxide with the highest positive oxidation state on the element other than O should be most acidic. Oxidation states  of V in V2O5 and N in N2O5 are both +5. But the electronegativity of N is higher, making N2O5 the most acidic oxide.
  2. OI > OBr > OCl

3. IV > III > I > II

  1. The order of catalytic activity in water is same as the order of acidity. In anhydrous ammonia all three are strong.
  2. CH3 – CH2– CHCl –COOH > CH3 – CHCl – CH2 – COOH > CH2Cl – CH2 – CH2 – COOH > CH3 – CH2 – CH2 – COOH
  3. Reaction SOCl2+ Na2SO3 ⎯→ 2NaCl + 2SO2

Cation SO++

Anion SO32–

  1. To predict their relative acidity predict the relative basicity of their conjugates base. The size of central atom decreases from C to F, at the same time volume available to electron on the central atom also decreases because in CH3 ion only 1/4th of the space is available for electrons. This increasing basic character of these conjugated base is

F < OH < NH2 < CH3

So, increasing acidic character of the corresponding acid is

CH4 < NH3 < H2O < HF


  1. O–hydroxy benzoate anion is stabilized by chelation due to hydrogen bonding which is absent in meta and para isomers.


  1. It can be seen that hydrogens in these molecules are not all bonded to oxygens. It is clear from their structures,

that the number of terminal oxygen atoms is 1 in all three acids. The electronegativities of P and H are almost the same. Thus no much difference in acidity is expected. 


  1. I > V > IV > II > III



  1. Carbon does not have d-orbitals so accommodate the accepted electron pairs while Si and Sn possess d-orbitals and thus accept electron pair electrons.
  2. BF3 is electron deficient compound. It can accept an electron pair. N in NF3 has not present. It cannot accept an electron pair.
  3. As+ ion is soft acid. Its interaction with I ion, a soft base, yields a complex where as its interaction with F ion hard base yield an unstable complex.
  4. HF < HCl < HBr < HI
  5. SiO2 < CO2 < N2O5 < SO3
  6. The pH will shift down below 7. It is clear from the following equation.

CuSO4 + H2O Cu(OH)2 + SO42– + 2H+

  1. The conjugate base of CH3OH – CH3O is stronger than the conjugate base of CH3SH i.e CH3SH i.e. CH3S. Since S is large, the negative charge in CH3S is dispersed over a large. Hence CH3S is the weaker base.
  2. The –CH3 group which is present in acetic acid decrease the acidic character of the –COOH group.
  3. Conjugate base of phenol in resonance stabilized white that of alcohol is not.
  4. Among the halogens, fluorine has the smallest size hence it has availability of electron most.
  5. R2NH > RNH2 > R3N
  6. Acidic nature

CH4 < H2S < HI

Size of the central atoms is greatest in HI hence it is most acidic.

  1. N-alkylated anilines are stronger bases than aniline because of steric effects which decrease the resonance of the lone pair of electron on nitrogen and hence makes it more available for protonation. Ethyl group is bigger than the methyl group so N-ethyl aniline is stronger base than N-methyl aniline.
  2. NH3 > NH2 NH2 > NH2OH
  3. HF > H2O > NH3 > CH4 > BeH2 > LiH