d-and f-block elements

Q.1.Why are d-block elements called transition elements?

Ans.The position of d-block elements in the periodic table is between s-block elements and p-block elements. They exhibit transitional behaviour between s-block and p-block elements,so they are known as transition elements.

Q.2.Write different  types of transition elements.

Ans.There are mainly three types of transition series.

1.1^st transition series-  3d series (Sc to Zn).

2. 2^nd transition series-4d series (Y to Cd)

3. 3^rd transition series-5d series (La to Hg)

Q.3.Zinc, cadmium and mercury are not considered as transition elements why?

Ans.They have completely filled d orbital(d¹⁰) in their ground state as well as in their oxidation state(M²⁺) hence they are not considered as transition elements.

Q.4.What is the general electronic configuration of d-block elements?

Ans.(n-1)d^1–10ns^1–2 where n is outermost shell.

Q.5.Write the electronic configuration of Cr and Cu.

Ans.Cr(24)- [Ar] 3d⁵ 4s¹

         Cu(29)-[Ar] 3d¹⁰4s¹

Q.6.Transition metals are hard ,have high melting points and have low volatility why?

Ans. Transition metals have strong interatomic metallic bonding due to the involvement of greater number of electrons from (n-1)d in addition to the ns electrons because the energy difference between ns electrons and (n-1)d electrons is very small.

Q.7.Zn, Cd and Hg are not hard why?

Ans.They have completely filled d orbital(d¹⁰) hence the interatomic metallic bonding is weak. Thus Zn, Cd and Hg are not hard.

Q.8.In any row of transition series the melting points of transition elements rise to a maximum at middle except for anomalous values of Mn and Tc and fall regularly as the atomic number increases why?

Ans.The strength of interatomic metallic bonding increases and becomes maximum at middle(d⁵) because number of unpair electron increases up to middle. From middle to end strength of metallic bonding decreases as number of unpair electron decreases. Thus in any row the melting points of transition elements  rise to a maximum at middle except for anomalous values of Mn and Tc and fall regularly as the atomic number increases.

Q.9.Enthalpies of atomisation of transition elements is very high why?

Ans.It is due to the presence of large number of unpaired electrons in their atoms they have strong interatomic metallic bonding and hence higher enthalpies of atomisation.

Q.10.In 1^st transition series the enthalpy of atomization of zinc is the lowest why?

Ans. Zinc have completely filled d orbital(d¹⁰) hence the interatomic metallic bonding is weak.

Q.11.Atomic and ionic size of of transition element decreases with increasing atomic number why?

Ans. As atomic number increases the new electron enters a d orbital each time the nuclear charge increases by unity and also the shielding effect of a d electron is very low,hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic /atomic radius decreases.

Q.12.The variation in atomic /ionic size within a series is quite small why?

Ans.It is due to the presence of d electrons and shielding effect of these electrons is very low.

Q.13.Define lanthanoid contraction.

Ans.The filling of 4f before 5d orbital results in a regular decrease in atomic radii called lanthanoid contraction which essentially compensates for the expected increase in atomic size with increasing atomic number

Q.14.The radii of the 3^rd (5d) series are nearly same with those of the corresponding members of the 2^nd series.

Ans.It is due to lanthanoid contraction. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called lanthanoid contraction which essentially compensates for the expected increase in atomic size with increasing atomic number.

Q.15.Zr and Hf  have very similar physical and chemical properties why?

Ans.Zr and Hf have similar atomic radii due to lanthanoid contraction. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called lanthanoid contraction which essentially compensates for the expected increase in atomic size with increasing atomic number.

Q.16.There is an increase in ionisation enthalpy along each series of the transition elements from left to right why?

Ans. Atomic and ionic size of transition element decreases with increasing atomic number due to poor shielding effect d electrons.

Q.17.Variation in ionization enthalpies of transition element is small why?

Ans. Because variation in atomic /ionic size within a series is quite small.

Q.18.Zinc has high 1^st ionization enthalpy why?

Ans.Because electron is removing from stable 4s² configuration.

Q.19.Zinc has lower 2^nd ionization enthalpy.

Ans.After removal of an electron from 4s¹ electronic configuration of d orbital remain unchanged.

Q.20.What is the lowest common oxidation state of elements of 1^st transition series.

Ans.The lowest common oxidation state of elements of 1^st transition series is +2 when 4s electrons are used.

Q.21.Second ionisation enthalpy of Cr and Cu has unusual high values why?

Ans.After removal of one electron from Cr and Cu it converts into stable Cr⁺(3d⁵)configuration and Cu⁺(3d¹⁰) configuration respectively hence second ionisation enthalpy of Cr and Cu has unusual high values.

Q.22.The third ionization enthalpy of Mn is high why.

Ans After removal of 2 electrons from Mn it converts into  stable Mn²⁺(3d⁵)configuration.

Q.23.Transition elements show variable oxidation state why?

Ans.Transition elements have large number of unpair electrons and also these elements use not only ns electrons but also( n-1)d electrons because energy difference between ns and (n-1)d electrons are very small.

Q.24.Scandium does not exhibit variable oxidation states why?

Ans .Sc has only one electron in d orbital hence it can show only one oxidation state +3.

Q.25.Manganese  shows largest number of oxidation states that is from +2 to 7 why?

Ans. Mn  atoms have maximum number of unpaired electrons (3d⁵ configuration).

Q.26.The first and second  ionization enthalpies in the first series of the transition elements are irregular why?

Ans.Ionisation enthalpy depends not only on the size of elements but also stability of electronic configurations .Some electronic configurations like d⁰.d⁵,d¹⁰ have extra stability hence the first and second  ionization enthalpies in the first series of the transition elements are irregular.

Q.27.The E⁰(Cu²⁺/Cu) value for copper is positive (+0.34V) why?

Ans. E⁰ value depends on following factors.  

   i.Enthalpy of atomization.

   ii.Ionisation enthalpy.

   iii.Hydration enthalpy.

Since enthalpy of atomization of Cu is high and hydration enthalpy of Cu²⁺ is low hence, the E⁰(Cu²⁺/Cu) value for copper is positive (+0.34V).

Q.28.Cu is not able  to liberate H₂ from acids why?

                    Or

Only oxidising acids (nitric and hot concentrated sulphuric) react with Cu.

Ans. The E⁰ (Cu²⁺/Cu) value for copper is positive (+0.34V) due to low hydration enthalpy of Cu²⁺ and high enthalpy of atomization of Cu.

Q.29.E⁰value for Mn, Ni and Zn are more negative than expected from the trend why?

Ans.It is due to stable half filled electronic configuration of Mn²⁺ and fully filled stable electronic configuration of Zn²⁺ and high hydration enthalpy of Ni²⁺.

Q.30.Mn³⁺ is oxidising in nature but Cr²⁺ reducing in nature why?

Ans.Both Mn³⁺and Cr²⁺ have d⁴ configuration but Mn³⁺ is reduced into Mn²⁺ (from d⁴ to d⁵) and Cr²⁺ is oxidized into Cr³⁺ (from d⁴ to d³).Both Mn²⁺ and Cr³⁺ are stable due to different reasons. Mn²⁺ has stable half filled d⁵ configuration and Cr³⁺ has stable half filled t_2g configuration,hence Mn³⁺ is oxidising but Cr²⁺ reducing in nature.

Q.31.E⁰(M^3+/M²⁺)value for Mn is high why?

Ans.It is due to high ionization enthalpy is required to remove an electron from half filled stable electronic configuration of Mn²⁺( d⁵), E⁰(M^3+/M²⁺) value for Mn is high.

Q.32.Copper (I) compounds are unstable in aqueous solution and undergo disproportionation?

Ans.

of Cu²⁺ is higher than that of Cu⁺ hence 2^nd ionization enthalpy of  cu is much more compensated by hydration enthalpy of Cu²⁺ .

Q.33.The ability of oxygen to stabilise higher oxidation states exceeds that of fluorine why?

                                                 Or

The highest Mn fluoride is MnF₄ whereas the highest oxide is Mn₂O₇ why?

Ans. It is due to ability to form multiple bonds by oxygen atom.

Q.34.Compare reducing strength between Cr²⁺ and Fe²⁺.

Ans.

since 3d³ configuration is more stable than 3d⁵ configuration in aq solution hence E⁰ value of Cr³⁺/Cr²⁺ is lesser than Fe³⁺/Fe²⁺ thus Cr²⁺ is more reducing than  Fe²⁺.

Q.35.In the first row transition metals show irregular values  of E^o(M²⁺/M)  as we go from left to right why?

Ans. E⁰ value depends on following factors.  

   i.Enthalpy of atomization.

   ii.Ionisation enthalpy.

   iii.Hydration enthalpy

 Since first and  second  ionization enthalpies in the first series of the transition elements are irregular and sublimation enthalpy of manganese and vanadium is much less hence  the first row transition metals show irregular values  of E^o(M²⁺/M)  as we go from left to right.

Q.36.E⁰(Mn³⁺/Mn²⁺) value is  more positive than Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺?

Due to very high ionization enthalpy in removal of an electron from half filled stable electronic configuration of Mn²⁺( d⁵) E⁰(Mn³⁺/Mn²⁺) value is  more positive than Cr³⁺/Cr²⁺ or Fe³⁺/Fe².

Q.37.The highest oxidation state of a metal present in its oxide or fluoride only why?

Ans.It is due to high electronegativity of oxygen and fluorine which can stabilize the large oxidation state of metal.

Q.38.Many of the transition metal ions are paramagnetic why?

Ans.Most transition element ions having a magnetic moment due to presence of unpaired electrons hence they are paramagnetic.

Q.39.Sc³⁺ and  Zn²⁺ are diamagnetic why?

Ans. These ions have no magnetic moment due to presence paired electrons hence they are diamagnetic.

Q.40.What is ‘spin-only’ formula and how magnetic moment is determined by this formula.

n= number of unpaired electrons.Higher the number of unpair electrons,higher the magnetic moment and thus higher the paramagnetism.

Q.41.Why are transitional metal ions coloured?

Ans.It is due to d-d transition ,in presence of ligands splitting of d orbitals are occurred and electrons of lower d orbitals are excited to higher d orbitals by absorbing light in visible region,and thus new color of solution is complementary colour of the light absorbed.

Q.42.Sc³⁺ and  Zn²⁺ are colorless why?

Ans. d-d transition is not possible on both cases because Sc³⁺ has no d electron and Zn²⁺  has completely filled d orbitals.

Q.43.The transition metals form a large number of complex compounds why?

Ans.It is due to following reasons.

   i. Smaller sizes of the metal ions.

    ii. High ionic charges.

   iii.Availability of d orbitals for bond formation.  

Q.44.The transition metals are good catalyst why?

Ans.It is due to the following reasons.

i. During catalytic reaction they can adopt multiple oxidation states.

ii.They can form complexes.

Q.45.Give examples of some reactions in which transition metal can acts like catalyst.

Ans. I. Vanadium(V) oxide in contact Process for manufacture of sulphuric acid.

       II. Finely divided iron in Haber’s Process for manufacturing ammonia.

      III. Nickel in Catalytic Hydrogenation of alkenes.

Q.46.Define interstitial compounds.

Ans.A compound of transition metals and hydrogen, carbon ,nitrogen etc when these smaller metals are trapped into interstitial of crystal lattice.

Examples: example, TiC, Mn₄N, Fe₃H

Q.47.What are the properties of interstitial compounds.

(i) High melting points

(ii) Very hard.

(iii) Metallic conductor.

(iv) Chemically inert.

Q.48.Explain disproportionation reaction with example .

Ans.In a particular reaction when same compound is reduced and oxidized both, the reaction is called disproportionation reaction.

In above example oxidation number of Mn increases from 6 to 7 and also decreases from 6 to 4.

Q.49.Inner transition elements are called f-block elements why?

Ans. The last electron of these elements enters in the ‘f’’ orbital that is why they are called are called f-block elements

Q.50.Why f-block elements are called inner transition elements.

Ans d-block elements are called transition elements and their two shells
(last and the penultimate shells)are incomplete and since last  electrons of f-block elements are filled in f orbitals  which are inner orbitals of penultimate shell that is why f-block elements are called inner transition elements.

Q.51.How many types of f-block elements.

Ans. Lanthanoids (The fourteen elements after lanthanum).

        Actinoids  (the fourteen elements after actinium).

Q.52.General electronic configuration of f-block elements.

Ans.(n-2)f^0-14 (n-1)^0-1ns²

Where n is outermost shell.

For Lanthanoids n=6

For Actinoids n=7

Q.53.What is the common oxidation state of lanthanoids.

Ans. +3 is the common oxidation state of lanthanoids.

Q.54.The first and second  ionization enthalpies of the innertransition elements are irregular why?

Ans.It is due to the extra stability of f⁰,f⁷,f¹⁴ electronic configuration.

Q.55.Ce can exhibit +4 oxidation state why?

Ans.Due to noble gas electronic  configuration of Ce in +4 oxidation state it can show this oxidation state.

Q.56.Name the lanthanoids ions which do not show paramagnetic behaviour.

Ans. La³⁺ and Ce⁴⁺(due to f⁰ configuration)

        Yb²⁺ and Lu³(due to f¹⁴ configuration)

Q.57. Name the lanthanoids ions which do not show color in aq solution.

Ans. La³⁺ and Lu³+ due to f⁰ and  f¹⁴configuration configuration respectively.

Q.58.Most of the trivalent lanthanoid ions are coloured both in the solid state and in aqueous solution why?

Ans.Due to presence of unpair f electrons in these trivalent ions they show colour.

Q.59.Explain the abnormally low value of the third ionisation enthalpy of lanthanum, gadolinium and lutetium in lanthanoids series of elements.

Ans.It is due to half filled configuration (f⁷) of La³⁺ and Gd³⁺ and fully filled configuration  (f¹⁴) of Lu⁺³.

Q.60.What are the uses of lanthanoids.

Ans. A lanthanoid metal (almost 95%), iron (almost 5%) and traces of S, C, Ca and Al are used to make an alloy named mischmetall. Mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint.

Q.61.Are all the actinoids are radioactive elements.

Ans, Yes all actinoids are radioactive elements.

Q.62.Define actinoid contraction.

Ans.The gradual decrease in the size of atoms or M³⁺ ions across the actinoids series is called actinoid contraction.

Q.63.Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Ans.In lanthanoids electrons are filled in 4f orbitals but in actinoids electrons are filled in 5f orbitals since shielding effect of 5f electrons are lower than 4f electrons hence actinoid contraction is greater from element to element than lanthanoid contraction.

Q.64.Why actinoids show greater range of oxidation in comparision with lanthanoids .

Ans.The energy difference between 5f, 6d and 7sorbitals in case of actinoids are lesser than that of 4f ,5d and 6s orbitals in case of lanthanoids hence actinoids show greater range of oxidation in comparison with lanthanoids .

Q.65.Describe the reactivity of actinoids.

Ans.1 All actinoids are highly reactive.

       2.The metal readily tarnish in air.

       3.They can react with boiling water and dilute acids and can liberate hydrogen gas.

       4.They form oxides and hydrides and other products when combine with nonmetals dircetly.