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Q1. What are interstitial compounds? Give their characteristics.

Solution

Interstitial compounds are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals. The atoms in interstitial compounds are usually in non stoichiometric ratio and are neither typically ionic nor covalent. Characteristics: (i) They have high melting points. (ii) They are very hard. (iii) They retain metallic conductivity. (iv) They are chemically inert.

Q2. Why the melting points of transition metals are very high and what is the trend of melting point across the period of 3d series?

Solution

The high melting points of transition metals are due to the involvement of greater number of electrons of (n-1)d in addition to the ns electrons in the interatomic metallic bonding. Across a period of 3d series, the melting points of these metals increases to a maximum at d⁵ except for anomalous values of Mn and Tc decreases regularly as the atomic number increases.

Q3. Why Cr²⁺ is reducing agent and Mn³⁺ an oxidizing agent although both have d⁴ configuration?

Solution

Cr²⁺ is reducing agent as its configuration changes from d⁴ to d³, when it is oxidized to Cr³⁺ .The d³ configuration have a half-filled t_2g level which is very stable. On the other hand, the reduction of Mn³⁺ to Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability hence Mn³⁺ acts as oxidizing agent.    

Q4. Why do the transition elements exhibit higher enthalpies of atomisation and why is enthalpy of atomisation of zinc  lowest in the 3d series?

Solution

The value of enthalpy of atomisation increases as the interatomic interaction increases. Further the interatomic interaction is affected by the number of unpaired electrons. The greater the number of unpaired electrons greater is interatomic interaction and greater will be the enthalpy of atomisation. Since transition elements have unpaired electrons they have greater interatomic interaction and exhibit higher enthalpies of atomisation. Zinc contains no unpaired electrons hence enthalpy of atomisation of zinc is the lowest.

Q5. Why scandium (Z = 21) is a transition element but zinc (Z = 30) is not?

Solution

The outer electronic configuration of scandium ( Z = 21) is 3d¹4s² while that of zinc( Z = 30) is 3d¹⁰4s². Since scandium has incompletely filled d orbital in its ground state it is regarded as transition metal while zinc do not have incompletely filled d-orbital in either ground state or any of its oxidized state so it is not regarded as transition metal.

Q6. What is mischmetall and what is its use?

Solution

Mischmetall is an alloy of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. A large quantity of mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint.

Q7. Write down the most common oxidation state for lanthanoids.

Solution

The most stable oxidation state of Lanthanides is +3.        

Q8. Give reaction of KMnO₄ with KI in (i) Acidic medium,(ii) Alkaline medium.

Solution

Acidic medium: Iodine is liberated from potassium iodide 10I^– + 2MnO₄^- + 16H⁺  2Mn²⁺ + 8H₂O + 5I₂   Alkaline medium: Iodide is oxidized to iodate 2MnO₄^– + H₂O + I^–     2MnO₂ + 2OH^– + IO₃^–

Q9. Why it is very difficult to study the chemistry of Actinoid?

Solution

The actinoids are radioactive elements having relatively short half-lives. The elements of actinoid series could be prepared only in nanogram quantities. This is the reason why their study is more difficult.

Q10. What are alloys and how these are formed by the elements of transition metals?

Solution

Alloys are homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other. Alloys are formed by atoms with almost similar metallic radii. Transition metals readily form alloys because of similar radii. The alloys so formed are hard, have high melting points and improved quality than original metal.

Q11. Give the method of preparation of potassium dichromate.

Solution

Dichromates are generally prepared from chromate, which is obtained by the fusion of chromite ore (FeCr₂O₄) with sodium or potassium carbonate in free access of air. 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂  8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂   Solution of sodium chromate is filtered and acidified with sulphuric acid to give solution from which Na₂Cr₂O₇. 2H₂O can be crystallized. 2Na₂CrO₄ + 2H⁺   Na₂Cr₂O₇ + 2Na⁺ + H₂O   Potassium dichromate can be prepared by treating the solution of sodium dichromate with potassium chloride. Na₂Cr₂O₇ + 2KCl  K₂Cr₂O₇ + 2NaCl

Q12. What is Lanthanoid contraction and what is the factor responsible for that?

Solution

There is an increase in radii from the first (3d) to the second (4d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is called Lanthanoid contraction. The possible reason for that is the intervention of the 4f orbitals which must be filled before the 5d series of elements begins. The imperfect shielding of one 4f electron by another result in the decrease in the size of the entire 4f ⁿ orbitals as the nuclear charge increases along the series.

Q13. Which transition element of the 3d series   exhibit the largest number of oxidation states and why?

Solution

Manganese exhibit the largest number of oxidation states. It shows the oxidation states +2, +3, +4, +5 ,+6,  and + 7.The reason for that is the maximum number of unpaired electrons present in its outermost shell i.e. 3d⁵4s².

Q14. Why Ce⁴⁺ behaves as good oxidising agent although being a stable species?

Solution

The formation of Ce⁴⁺ is favoured by its noble gas configuration, but it is a strong oxidant reverting to the common +3 state because the E^o value for Ce⁴⁺/ Ce³⁺ is    + 1.74 V which is positive.

Q15. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: (i)  Iodide (ii) Iron (II) solution and (iii) H₂S.

Solution

In acidic solution, the oxidising action of Potassium dichromate can be represented as follows: Cr₂O₇^2– + 14H⁺ + 6e^–  2Cr³⁺ + 7H₂O (i)    Cr₂O₇^2– + 14H⁺ + 6 I^–  2Cr³⁺ + 7H₂O+ 3I₂  (ii) Cr₂O₇^2– + 8H⁺ + 3 H₂S  2Cr³⁺  + 7H₂O + 3S (iii) Cr₂O₇^2– + 14H⁺ + 6 Fe²⁺  2Cr³⁺ + 6 Fe³⁺ + 7 H₂O

Q16. Why do Zr and Hf occur together in nature?

Solution

The atomic radii of Zr (160 pm) and Hf (159 pm) is almost identical due to lanthanoid contraction. So they occur together in nature.

Q17. In Tc ( Z = 43) and Tb( Z = 65) which one is inner transition metal and which one is transition metal and why?

Solution

The outer Electronic configuration of Tc (Z = 43) is 4d⁶5s¹ and that of Tb (Z = 65) is 4f⁹5d⁰6s². Since in Tc there is incompletely filled d orbitals hence it is Transition metal or d-block element and in Tb there is incompletely filled f orbitals hence it is inner transition metal or f-block element.

Q18. Write the electronic configuration of the following elements of lanthanoid and their possible oxidation state other than +3: (a) Eu and (b) Yb.

Solution

(a) Eu (63): 4f⁷6s² the possible oxidation state other than +3 is +2 having 4f⁷ configuration. (b) Yb (70): 4f¹⁴6s² the possible oxidation state other than +3 is +2 having 4f¹⁴ configuration.

Q19. Among oxygen and fluorine which one has the greater stability to stabilize high oxidation states?

Solution

The ability of oxygen to stabilize high oxidation states is greater than that of fluorine.

Q20. Silver atom has completely filled d orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?

Solution

Silver Ag(Z = 47) has outer electronic configuration 5s¹4d¹⁰  in its ground state showing that it is not transition metal in first instance but in +2 oxidation state its electronic configuration is 5s⁰4d⁹, proving that it is transition metal with incompletely filled d orbital.

Q21. Why the elements of 3d series forms coloured ions?

Solution

When an electron of a lower energy d orbital of the transition metal ion is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed. Since this frequency lies in the visible region hence the corresponding complementary color of the absorbed light can be observed.