Important Questions Class 12 Chemistry Chapter 8

Important Questions Class 12 Chemistry Chapter 8

Important Questions for CBSE Class 12 Chemistry Chapter 8 – The d and f Block Elements

Class 12 Chemistry Chapter 8 Important Questions for Chapter 8 “The d and f Block Elements” of Class 12 Chemistry, are provided by Extramarks. These vital questions are based on CBSE curriculum and on the most recent Class 12 Chemistry Syllabus.

CBSE Class 12 Chemistry Chapter-8 Important Questions

Study Important Question for class 12 Chemistry Chapter 8 – The d and f Block Elements

A sample of Important Questions Class 12 Chemistry Chapter 8 is given below:

Very Short Answer Questions: 1 Mark

Q1. What is meant by lanthanide contraction?

Ans. The steady decrease in the ionic radius from La3+ to Lu3+ is termed as lanthanoid contraction.

Q2. Write the electronic configuration of Cr3 + ion (atomic number of Cr=24 )

Ans. The atomic number of chromium is 24 , hence the electronic configuration of Cr3 + (21) will be 1s2 2s2 2p6 3s2 3p6 4s0 3d3.

Q3. Why is the third ionisation energy of Manganese (Z = 25 ) unexpectedly high?

Ans. Manganese has an atomic number of 25 and an electronic configuration of [Ar]4s2,3d5. The loss of two electrons transforms Mn2 + into [Ar]3d5, a structure with a half-filled d-orbital that is exceedingly stable. As the third electron must be removed from the stable configuration of Mn2 + , the third ionisation energy is extremely large.

Q4. Why does vanadium pentoxide act as a catalyst?

Ans. Vanadium is a transition metal that quickly switches between oxidation states. It creates unstable intermediates in one oxidation state and then quickly converts to products by gaining a new stable oxidation state, opening up a new reaction pathway.

Q5. What are interstitial compounds?

Ans. Interstitial compounds are formed when very small atoms, such as hydrogen, nitrogen and carbon, become trapped inside the crystal lattices of metals. These chemicals are usually non-stoichiometric and not ionic or covalent.

Q6. Transition metals and their compounds are known for their catalytic activity. Give two specific reasons to justify the statement.

Ans. Transition metals have a wide range of oxidation states and are capable of forming complexes. They produce unstable intermediates. They open up a novel pathway with lower reaction activation energy. They also provide an appropriate surface for the reaction to take place on.

Q7. Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number. 

Permanganate ion, i.e., MnO4– with oxidation number +7.

Q8. Write the names of the catalyst used in the: 

(a) Manufacture of sulphuric acid by contact process.

Ans: Vanadium oxide is the catalyst used in the manufacture of sulphuric acid by contact process.

(b) Manufacture of polythene.

Ans: Ziegler Nata catalyst which is the combination of titanium tetrachloride and trimethyl aluminium is used in the manufacture of polythene.

Q9. Mention the name of the element among lanthanoids known to exhibit a +4 oxidation state.

Ans. Cerium is the element in the lanthanoid series which is known to exhibit +4 oxidation state.

Q10. Name one ore each of manganese and chromium.

Ans. The ore of manganese is pyrolusite and chromite is known as the ore of chromium.

Short Answer Type Questions – 2 Mark

Q1. Why does copper not replace hydrogen from acids?

Ans. Copper does not replace hydrogen from acids because Cu has a positive E∘ value, i.e., it is less reactive than hydrogen, which has an electrode potential of 0.0 V. So, Cu cannot replace hydrogen from acids.

Q2. Transition elements show high melting points. Why?

Ans. The high melting points of transition metals are due to the involvement of a greater number of electrons of (n-1)d in addition to the ns electrons in the interatomic metallic bonding.

Q3. Give reason for:

(a) In permanganate ions, all bonds formed between manganese and oxygen are covalent. 

Ans: MnO4− in the highest oxidation state, +7. Transition metals produce covalent bonds in high oxidation states (according to Fajan’s rules, as the oxidation state increases, ionic character decreases).

(b) Permanganate titrations in presence of hydrochloric acid are unsatisfactory.

Ans. Because hydrochloric acid is oxidised to chlorine, permanganate titrations in the presence of the acid are unreliable. In preparative organic chemistry, this is a preferred oxidant.

Q4. Answer the following:

(a) Why do transition metals show high melting points? 

Ans: Transition metals have a high density, as well as high melting and boiling temperatures. Metallic bonding via delocalised d electrons causes these characteristics, which leads to increased cohesion as the number of shared electrons grows.

(b) Out of Fe and Cu, which one would exhibit a higher melting point?

Ans: The melting point of Fe is greater than that of Cu. This is since iron contains four unpaired electrons in the 3d-subshell, but copper only has one electron in the 4s-subshell. As a result, metallic connections of iron are significantly stronger than copper.

Q5. Transition metals show low oxidation states with carbon monoxide. 

Ans. Synergic bonds enable the formation of complexes by transition elements in zero oxidation states. In metal carbonyls, the metal-carbon bond has both s and p characters. The M-C sigma bond is created when a lone pair of electrons from the carbonyl carbon are donated to a metal’s empty orbital. The M-C bond is created when a pair of electrons from a metal’s full d orbital is donated to unoccupied antibonding pi* orbital of carbon monoxide. The synergic action of metal to ligand bonding enhances the connection between carbon monoxide and the metal.

Short Answer Type Questions – 3 marks

Q1. Account for the following: 

(a) La(OH)3 is more basic than Lu(OH)3 

Ans. The most basic is La(OH)3 while the least basic is Lu(OH)3. The covalent nature of the hydroxides rises as the size of lanthanide ions falls from La3 +  to Lu3 +  , and therefore the basic strength lowers.

(b) Zn2 + salts are white. 

Ans. There are no unpaired electrons in the electronic arrangement of Zn2 + (3d10) salts. The presence of an unpaired electron in metal ions allows for electron transition in the visible range. This is why Zn2 + ion salts are white in colour.

(c) Cu (I) compounds are unstable in an aqueous solution and undergo disproportionation.

Ans. Cu +  is more unstable in aqueous solution than Cu2 +  because, while copper’s 2nd  I.E. is considerable, Cu2 +  hydration enthalpy is significantly lower than Cu +  , and so it more than compensates for copper’s 2nd I.E. As a result, many Cu +   complexes in aqueous solution are unstable and disproportionate.

Q2. Answer the following questions:

(a) Deduce the number of 3d electrons in the following ions: Fe3 + , Cu2 + and Sc3 +  . 

Ans. The electronic configuration of Fe3 + is 1s2 2s2 2p6 3s2 3p6 3d5. So, Fe3 + has 5 electrons in 3d-orbital. The electron configuration of Cu2 + is 1s2 2s2 2p6 3s2 3p6 3d9. So, Cu2 +  has 9 electrons in 3d-orbital. The electronic configuration of Sc3 +  is 1s2 2s2 2p6 3s2 3p6 3d0. So, Sc 3 +  has no electrons in 3d-orbitals.

(b) Why do transition metals form alloys?

Ans. Transition metals have atomic sizes that are extremely close to one another. Due to their comparable atomic sizes, one metal may easily replace the other in its lattice and create a solid solution, which is the alloy. This is why, in the molten state, transition metals create homogenous mixes with one another.

(c) Write any two characteristics of interstitial compounds.

Ans. The melting temperature of interstitial compounds is higher than that of pure metals, and they are tougher and more corrosion resistant.

Long Answer Type Questions – 5 mark 

Q1. Assign reasons for the following: 

(a) There are no regular trends in E° values of M2 + /M  systems in 3d  series. 

Ans. Due to uneven electronic configurations from left to right, irregular change in ionisation enthalpies (sum of 1st and 2nd ionisation enthalpies), heat of sublimation, and enthalpy of hydration from left to right in the period. As a result, irregular variations in Eo(M2 + /M)

values for ionisation metals are caused by irregular variations in ionisation enthalpies, and heat of sublimation and hydration enthalpy.

(b) There is a gradual decrease in the ionic radii of M2 +  ion in the 3d series. 

Ans. The ionic radii get smaller as the nuclear charge gets higher. Because incoming electrons enter the inner (n−1)d’orbitals as nuclear charge rises, this reduction occurs. When the oxidation state goes from +2 to+3 , the ionic radii gradually decrease.

(c) The majority of transition metals form complexes. 

Ans. Transition metals are the d-block elements having the outer shell electronic configuration as ns2(n−1)d1 – 10 . These metal ions can easily form complexes with a group of negative ions or neutral molecules that have lone pairs of electrons. This is due to: the metals’ small size and strong nuclear charge. Availability of empty d-orbitals with sufficient energy to accept ligand-donated lone pairs of electrons.

(d) Ce3 + can be easily oxidised to Ce4 +  

Ans.  The outer shell electronic configuration of Ce3 +  is 4f1 5d0 6s0 . Ce3 +  can easily lose one electron from the 4f-orbital and forms Ce4 +  with an outer shell electronic configuration of 4f0 5d0 6s0. Thus, Ce3 +  gets easily oxidised to Ce4 + .

(e) Tantalum and palladium metals are used to electroplate coinage metals.

Ans. At a bare minimum, the coinage metals are those metallic chemical elements that have historically been utilised as components in coin alloys. Tantalum and palladium are valuable materials for electroplating coinage because of their chemical inertness.