8. THE D- AND f-BLOCK ELEMENTS

8. THE D- AND f-BLOCK ELEMENTS 

Q. 1. How are xenon fluorides XeF₂, XeF₄ and XeF₆ prepared ?

Ans. All three binary fluorides of Xe are formed by direct union of elements under appropriate experimiental conditions. XeF₂ can also be prepared by irradiating a mixture of Xenon and fluorine with sunlight or light from a pressure mercurry arc lamp.

Xe (g) + Fe (g) XeF₂ (g)

1 Vol  2 Vol                                   (xenon difluoride)

Xe(g) + 2FE₂ (g) XeF₄ (s)

Xe (g) + 3F₂ (g) XeF₆ (s)

Q. 2. Describe briefly the following physico- chemical properties of transition metals :
(i) Ionic radii.
(ii) Magnetic behaviour.
(iii) Formation of interstitial compounds.

Ans. (i) Ionic radii : For ions carrying identical charges the ionic radii decreases slowly with increase in the atomic numbers across a given series of the transition elements. The decrease in the size is due to the increase nuclear charge across the series while the electrons being added into the d-orbitals of the penultimate shell
                         (ii) Magnetic properties : Most of the compounds of transition elements are paramagnetic in nature and are attracted by the magnetic field.
The transition elements involve the partial filling of d-subshells. Most of the transition metal ions or their compounds have unpaired electrons in d-subshell and therefore they give rise to paramagnetic character. The larger the number of unpaired electrons in a substance, the greater in the paramagnetic character.
                        (iii) Interstitial compounds : Transition metals form ‘interstitial compounds with lighter elements such as hydrogen, boron, carbon and nitrogen. In these compounds small atoms of these elements occupy the interstitial spaces of the metal lattices. These are non stoichiometric compound and cannot be given any definite formulae. They have essentially the same chemical properties as the parent metals but differ in physical properties such as density, and hardness.

Q. 3. Why do transition metals show variable oxidation state ?

Ans. Transitron elements show variable oxidation state due to the prosena of ‘ns’and (n – 1)d elecrons in bonding when ‘ns’ electron participate in bonding if generally exhibit lower oxidation state, and higher oxidation states are show when ‘ns’ and (n-1) d electrons take part in bonding.

Q. 4. Only Xe forms chemical compound among inert gases.

Ans. Xe atom has larger radie, therefore the electron attraction to the nucleus is weaker in comparison to the another noble gases. It has also d-sub shell comparison. So unpaired electron d-sub shell comparision. So unpaired electron of d-sub shell can pair to another electron of non-metal atom and form bond. Thus noble gas xenon forms real chemical compounds.

Q.5. 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. 6. Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their + 3 state ?

Ans.

Mn²⁺              =                [Ar]3d5
Fe²⁺                =                [Ar]3d6

Mn² has exactly half-filled electronic configuration in its 3d-subshell which is extremely stable. Therefore, 3rd ionisation enthalpy is very high, i.e., 3rd e^– cannot be lost easily.

∴ Mn² has no tendencey to lose an extra electron to undergo oxidation as compared to Fe²⁺ which has one electron extra (than the exactly half-filled configuration) which it can lose readily to undergo oxidation to Fe³⁺.

Fe²⁺ → Fe³⁺ + e^–

Q. 7. In what way is the electronic configuration of the transition elements different form 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-subsell 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. 8. What are the different oxidation states exhibited by the lanthanoids ?

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

Q. 9. 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 3d¹⁰4s¹ is the metal which exhibits the + 1 oxidation state most frequently after the removal of 1 electron form ls¹, inner d subshell is fully filled, i.e., 3d¹⁰.

Q. 10. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Ans. An alloy is a homegeneous 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 flinits. (2) Mangnersium mixed with 3% misch metal (to increse its strength) is used in making jet engine parts.

Q. 11. Which of the 3d series of the transition metals exihibits the largest number of oxidation state and why ?

Ans. Mn (Z = 25) shows the maximum number of oxidation states because it utilizes all the seven valence electrons (3d⁵4s²) for bond formations in different compounds like MnCl₂, Mn₂O₃ , MnO₂, K₂MnO₄, KMnO₄, manganese shows oxidation states of +2, +3, +4, +6, + 7 respectively.

Q. 12. 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 electro negativity both of them can oxidise the metal to its highest oxidation state.

Q. 13. Which is stronger reducing agent Cr²⁺ or Fe²⁺ and why ?

Ans. The reduction potentials for Cr²⁺ and Fe²⁺ are as given

Cr³⁺ e- + Cr²⁺;                                     E° = -0.41 V

Fe³⁺ + e^– → Fe²⁺;                                E° = + 0.77 V

Cr²⁺ → Cr³⁺ + e^–                                 E_x = + 0.41V

or, Fe²⁺ → Fe³⁺ + e^–;                         E_x = -0.77 V

             On the basis of oxidation potentials above Cr²⁺ is more likely to get oxidized to Cr³⁺ than Fe²⁺ which has a negative oxidation potenatial. Hence Cr²⁺ is more reducing in nature than Fe²⁺. [3rd I.E. required to convert Fe²⁺ (3d⁶) to Fe³⁺ (3d⁵-stable electronic configuration) is very small]. In a medium (like water) d³ is more stable as compared to d⁵.

Q. 14. Calculate the “spin only’ magnetic moment of m²⁺ (aq) ion (Z = 27).

Ans. With atomic number 27, the divalention in aqueous solution will have d⁷ configuration (three unpaired electrons)

M = 27 = ls², 2s²2P⁶, 3s²3P⁶3d⁷ 4s²

M²⁺ = ls², 2s²2P⁶, 3s²3P6, 3d⁷

∴ The magnetic moment μ is (spin only)

= B.M. = 3.87 B.M.

Q. 15. Explain why Cu+ ion is not stable in aqueous solutions ?

Ans. The oxidation state of a metal in a solven depends upon the nature of the solvent. The metal in a particular oxidation state may underg dation state may undergo oxidatin or

reduction in the solvent under appropriate conditions. For example, Cu+ is unstable in water as it undergoes oxidatin to Cu2+ and reduction to Cu, i.e., it undergoes disproportionation.

2 Cu+ (aq) → Cu2+ (aq) + Cu (s)
The E॰ value for this reaction is favorable.

Q. 16. 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 lanthanoids.

Q. 17. 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 element is small and its nuclear charge is large. Hence the ionisation energies of 5d elements are larger than 3d elements.

Q. 18. The melting and boiling points of Zn, Cd, and Hg are low. Why ?

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

Q. 19. 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 in energies. The incoming electron may enter into ns or (n − 1 )d subshell. Hence they show irregularities in their electronic configurations.

Q. 20. Chromium is a typical hard metal while mercury is a liquid. Why ?

Ans. Chromium (Cr) has five unparired d electrons in the d subshell (3d54s1). Hence metallic bonds are strong. In mercury all the d-orbitals are fully filled (3d104s2). Hence the metallic bonding is weak.

Q. 21. Why the properties of third transition series are very similar to 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.

Class 12th Chemistry Short Type Question English