ADDITIONAL QUESTIONS 2014 – 5 MARK

1. Write :

(a) IUPAC name (b) Central metal ion (c) Ligand (d) Coordination number (e) Geometry of the complex [Co(en)3]Cl3

[Co(en)3]Cl3

(a) IUPAC name of [Co(en)3]Cl3                  Tris(ethylenediamine)cobalt(III) chloride                  

(b) Central metal ion                                       Cobalt(III) / Co+3 / Cobalt

(c) Ligand                                                       en / ethylenediamine / NH2 – CH2 – CH2 – NH2

(d) Coordination number                                6

(e) Geometry of the complex [Co(en)3]Cl3    Octahedral

2. What is the potential of a half-cell consisting of zinc electrode in 0.01 M ZnSO4 solution 25oC [Eo = 0.763 V].

The half-cell reaction is           Zn → Zn2+ + 2e–

The Nernst equation for the oxidation half-cell reaction is

E = Eo – (0.0591/n) log [Zn2+]

The number of electrons transferred n = 2, [Zn2+] = 0.01 = 10–2 and Eo = 0.763 V.

Substituting these values in the Nernst equation we have

E = 0.763 – (0.0591/2) log 10–2

    = 0.763 – (0.0591/2) (–2)log10

    = 0.763 + 0.0591

                                                    = 0.8221 V

3. For the complex [Co(NH3)3(NO2)3] write the following :

(a) IUPAC name (b) Central metal ion (c) Ligands (d) Coordination number (e) Geometry

(a) IUPAC name                                            Triamminetrinitrocobalt(III)              

(b) Central metal ion                                       Cobalt(III) / Co+3 / Cobalt

(c) Ligands                                                     NH3 / ammine & NO2 / nitro

(d) Coordination number                                 6

(e) Geometry                                                  Octahedral

4. Show that for a first order reaction, the time required for 99.9% completion of the reaction is 10 times that required for 50% completion.

5. Write all possible isomers with the molecular formula C₄H₁₀O and name them.

Molecular formula	Ethers	Alcohols
C4H10O	CH3CH2– O –CH2CH3 Diethyl ether / Ethoxy ethane CH3– O –CH2–CH2–CH3 Methyl-n-propyl ether / 1-Methoxy propane                 CH3                  | CH3– O –CH –CH3 Methyl isopropyl ether / 2-Methoxy propane	CH3–CH2–CH2–CH2–OH n-Butyl alcohol / 1-Butanol / Propyl carbinol           CH3            | CH3–CH–CH2OH  iso-Butyl alcohol / 2-Methyl-1-propanol / isopropyl carbinol                     OH                      | CH3–CH2–CH–CH3 sec-Butyl alcohol / 2-Butanol / Ethyl methyl carbinol           CH3            | CH3–C–OH            |           CH3 tert-Butyl alcohol / 2-Methyl-2-propanol / Trimethyl carbinol

6. A zinc rod is placed in 0.095 M zinc chloride solution at 25oC. emf of this half cell is – 0.79 V. Calculate Eo _{Zn2+ /Zn}.

Concentration of Zn2+, [Zn2+] = 0.095 M = 1 x 10 – 1 M

emf of this half cell E_{Zn/ Zn2+}  = – 0.79 V

The number of electrons transferred, n = 2

Standard potential of Zn/ Zn2+ half-cell, Eo_{Zn/ Zn2+}  = ?

The half-cell reaction is: Zn2+ + 2e– → Zn

 E_{Zn/ Zn2+}   = Eo_{Zn/ Zn2+}   – 0.0591/n. log 1 / [Zn2+]

 E_{Zn/ Zn2+}   = Eo_{Zn/ Zn2+}   + 0.0591/2. log 0.095

      – 0.79 = Eo_{Zn/ Zn2+}  + 0.02955 log (1 x 10 – 1)                  [0.095 M = 1 x 10 – 1 M

      – 0.79 = Eo_{Zn/ Zn2+}  –  0.02955

Eo_{Zn/ Zn2+}   = – 0.79 + 0.02955

Eo_{Zn/ Zn2+}   = – 0.76045 V

7. How does Salicylic acid react with the following?

(a) CH3OH (b) (CH3CO)2O (c) Br2/H2O / Bromine water

(a) On heating Salicylic acid with methyl alcohol in presence of conc. H2SO4 a pleasant smelling liquid-methyl salicylate is formed.

 (b) Salicylic acid undergoes acetylation by heating with acetic anhydride to form aspirin or Acetyl salicylic acid.

 (c) Salicylic acid reacts with bromine water gives a white precipitate of tribromo phenol.

 This reaction involves bromination with decarboxylation.

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