Bihar Board - Class 12 Chemistry - Chapter 8: The D And F Block Elements Short Answer Question

Short Question Answer

Q. 1. Why do transition metals show variable oxidation state ?

Ans⇒Transition elements show variable state oxidation in their compounds because there is a very small energy difference in between (n-1)d and ns orbitals. As a result, electrons of (n-1)d orbitals as well as ns-orbitals take part in bond formation.

Q. 2. Only Xe  forms a chemical compound among inert gasses.

Ans⇒ Xe The atom has a larger radius, therefore the electron attraction to the nucleus is weaker in comparison to the other noble gasses. It has also d-sub shell comparison. So unpaired electron d-sub shell comparison. So an unpaired electron of d-subshell can pair to another electron of non-metal atom and form a bond. Thus noble gas xenon forms real chemical compounds.

Q.3. Write down the electronic configuration of

(a) Cr³⁺    (b) Pm³⁺    (c) Cu⁺   (d) Ce⁴⁺  (e) Co²⁺  (f) Lu²⁺ (g) Mn²⁺  (h) Th⁴⁺

Ans⇒

(a) Cr³⁺  = [Ar]¹⁸ 3d³       

(b) Pm³⁺= [Xe]⁵⁴ 4f⁵

(c) Cu⁺   = [Ar]¹⁸ 3d¹⁰              

(d) Ce⁴⁺   = [Xe]⁵⁴            

(e) Co²⁺   = [Ar]¹⁸ 3d⁷        

(f) Lu²⁺     = [Xe]⁵⁴ 4f¹⁴ 5d¹       

(g) Mn²⁺    = [Ar]¹⁸ 3d⁵          

(h) Th⁴⁺   = [Rn]⁸⁶         

Q. 4. In what way is the electronic configuration of the transition elements different from that of the non-transition elements ?

Ans⇒ Transition elements contain partially filled d subshell, i.e., their electronic configuration is (n-1)d^1-10 ns^0-2 where as non-transition elements have no d subshell or their d-the subshell is completely filled. s-Block elements have their inner d-subshells empty and p-block elements have their inner d-subshells filled (d¹⁰). Non- transition elements have ns^1-2 or ns₂ np^1-6  in their valence shells.

Q. 5. What are the different oxidation states exhibited by the lanthanides ?

Ans⇒ The most common oxidation state shown by all lanthanides is +3. Apart from +3, they also show oxidation states of +2 and +4 .

Q. 6. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why ?

Ans⇒ Cu  which has the electronic configuration of 3d¹⁰ 4s¹ is the metal which exhibits the +1 oxidation state most frequently after the removal of 1 electron form 1s¹, inner d subshell is fully filled, i.e., 3d¹⁰.

Q. 7. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Ans⇒  An alloy is a homogeneous mixture of two or more metals or metals and non-metals. An important alloy containing lanthanoid metals is MISCH metal which contains 95% lanthanoid metal and iron is 5% along with traces of S , C , Ca and Al. It is used in Mg based alloy to produce bullets, shells and lighter flints. (2) Magnesium mixed with 3% misch metal (to increase its strength) is used in making jet engine parts

Q. 8. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why ?

Ans⇒ Mn (Z=25) shows the maximum number of oxidation states because it utilizes all the seven valence electrons (3d⁵ 4s²) for bond formations in different compounds like MnCl₂ , Mn₂O₃ , MnO₂,K₂MnO₄,KMnO₄, manganese shows oxidation states of +2 , +3 , +4 , +6 , +7 respectively.

Q. 9. Why is the highest oxidation state of a metal exhibited in its oxides or fluoride only ?

Ans⇒ The highest oxidation state of a metal is exhibited in oxides or fluorides only because of their small size and high electronegativity; both of them can oxidize the metal to its highest oxidation state.

Q. 10. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Ans⇒

(i) Vanadate, VO₃^-

Oxidation state of V is +.

(ii) Chromate, CrO₄^2-

Oxidation state of Cr is +6 .

(iii) Permanganate,  MnO₄^-

Oxidation state of Mn is +7.

Q. 11. Actinoid contraction is greater from element to element than lanthanoid contraction. Why ?

Ans⇒ Though the actinoid contraction is like lanthanoid contraction, this contraction is greater from element to element resulting from poor shielding by 5f electrons as compared to 4f electrons of lanthanides.

Q. 12. Why are the IES of 5d elements greater than 3d elements ?

Ans⇒ In the  5d series, after  La(57) There is lanthanide contraction. As a result, in each group the atomic size of  5d The element is small and its nuclear charge is large. Hence the ionization energies of  5d elements are larger than  3d elements.

Q. 13. The melting and boiling points of  Zn , Cd and  Hg are low. Why ?

Ans⇒ In Zn , Cd and  Hg all the electrons in the d subshell are paired. Hence the metallic bonds present in them are weak. That is why they have low melting and boiling points.

Q. 14. Explain why transition elements have many irregularities in their electronic configuration ?

Ans⇒ In the transition elements, the (n-1)d subshell and ns subshell have very small energies. The incoming electron may enter into ns or (n-1)d  subshell. Hence they show irregularities in their electronic configurations.

Q. 15. Chromium is a typical hard metal while mercury is a liquid. Why ?

Ans⇒ Chromium (Cr) has five unpaired d electrons in the d subshell (3d⁵  4s¹). Hence metallic bonds are strong. In mercury all the d-orbitals are fully filled (3d¹⁰  4s²). Hence the metallic bonding is weak.

Q. 16. Why are the properties of the third transition series very similar to the second transition series ?

Ans⇒ In the 3rd transition series, after lanthanum, there is lanthanide contraction. Due to this contraction, the size of any atom of the third series is almost the same as that of the element lying just above it in the second transition series. This leads to similarity in their properties.

Q. 17. What are interstitial compounds? Why are such compounds well known for transition metals?

Ans⇒ Interstitial compounds are those in which small atoms occupy the interstitial sites in the crystal lattice. Interstitial compounds are well known for transition metals because small-sized atoms of H , B, C, N etc., can easily occupy position in the voids present in the crystal lattices of transition metals.

Q. 18. Describe the oxidizing action of potassium dichromate and write the ionic equations for its reaction with

(i) iodide

(ii) iron (H) solution and

(iii) H₂S

Ans⇒

(i) Cr₂O₇^2-+ 14H⁺+ 6I ^- - 2Cr³⁺ + 7H₂O+3I₂

(ii) Cr₂O₇^2- + 14H⁺ + 6Fe^2 + - 2Cr₃⁺ + 7H₂O+6Fe³⁺ 

(iii) Cr2O₇^2- + 8H⁺ + 3H₂SCr³⁺ + 7H₂O+3S

Q. 19. Predict which of the following will be coloured in aqueous solution? Ti³⁺ , V³⁺ ,Cu⁺ ,Sc³⁺ ,Mn²⁺ ,Fe³⁺ and Co²⁺. Give reasons for each.

Ans⇒ Only those ions will be coloured which have incompletely filled d-orbitals. Those with fully-filled or empty d-orbitals are colorless. Due to d-d transition, Ti³⁺ , V³⁺,Mn²⁺ ,Fe³⁺ and Co²⁺ are coloured.

Q. 20. Compare the stability of +2 oxidation state for the elements of the first transition series.

Ans⇒ The decreasing negative electrode potentials of M₂+/M+1 in the first transition series shows that in general, the stability of +2 oxidation state decrease from left-to right (exception being Mn and Zn). The decrease in the negative electrode potentials is due to increase in the sum IE₂+IE₂.

Q.21. Calculate the number of unpaired electrons in the following gaseous ions: Mn³⁺ ,Cr³⁺ ,V³⁺ and Ti³⁺. Which one of these is the most stable in aqueous solution?

Ans⇒

Mn³⁺=3d⁴= 4 unpaired electrons

Cr³⁺=3d³= 3 unpaired electrons

V³⁺=3d²= 2 unpaired electrons

Ti³⁺=3d¹= 1 unpaired electron

Out of these species Cr³⁺ is the most stable in aqueous solution due to its tendency of complex formation.

Q. 22. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Ans⇒ An alloy is a homogeneous mixture of two or more metals, or metals and non-metals. An important alloy containing lanthanoid metals is misch metal which contains 95 % lanthanide metals and 5% iron along with traces of S , C , Ca and Al. It is used in Mg-based alloy to produce bullets, shells and lighter flints.

Q.23. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Ans⇒ Oxygen and fluorine have small size and high electronegativity. Hence, they can oxidize the metal to the highest oxidation state.

Q. 24 . Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Ans⇒ In actinides, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanide). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinides is much more than that experienced by lanthanides. Hence, the size contraction in actinoids is greater as compared to that in lanthanide.

Q. 25. How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?

Ans⇒ Irregular variation of ionization enthalpies is mainly attributed to varying degree of stability of different 3d configuration (e.g., d⁰ , d⁵ , d¹⁰  are exceptionally stable).