d-Block

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d & f-BLOCK ELEMENTS
MULTIPLE CHOICE QUESTIONS

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1. The general outer electronic configuration if transition element is
(a) (n-1)d^1-10ns¹ (b) (n-1)d¹⁰ns² (c) (n-1)d^1-10ns^1-2 (d) (n-1)d⁹ns²
Ans (d) (n-1)d⁹ns²

2. Which Of the following does not show variable oxidation state?
(a) Iron (b) copper
(c) Zinc (d) Manganese
Ans (d) Manganese

3. Which ion gives coloured solution?
(a) Cu⁺
(b) Zn²⁺
(c) Ag⁺
(d) Fe²⁺
Ans (c) Ag⁺

4. Electronic configuration of a transition element X in +2 oxidation state is [Ar]3d¹⁰. The atomic number of the element is …………….
(a) 28
(b) 27
(c) 29
(d) 30
Ans (a) 28

5. Identify the configuration of transition element, which shows the highest magnetic moment.
(a) 3d⁷
(b) 3d⁵
(c) 3d⁸
(d) 3d²
Ans (c) 3d⁸

SHORT ANSWER TYPE QUESTION

i) Why do d-block elements form coloured ions?
Ans i) Because of d-d transition

ii) Why do d-block elements form complex compound?
Ans ii) Because of small size of metallic ionic, high ionic charge of metal ions and the availability of d orbitals for bond formation

iii) Why do d-block elements show catalytic properties?
Ans iii) Because of ability to adopt multiple oxidation States and to form complexes

iv) Why do d-block elements form interstitial compound?
Ans iv) Because of larger size of d-block elements and formation of larger voids

v) Why do d-block elements form alloy?
Ans v) Because of similar atomic size

vi) Explain why Cu⁺ ion is not stable in aqueous solutions?
Ans vi) Hydration enthalpy of Cu²⁺ is larger than ionization enthalpy of Cu⁺.

vii) Iron has higher enthalpy of atomization than that of copper.
Ans vii) Iron has greater number of valence electron. Hence the metallic bond of iron is larger than copper. Thus iron has higher enthalpy of atomization than that of copper.

(viii) Sc³⁺ is colourless in aqueous solution whereas Ti³⁺ is coloured.
Ans (viii) Sc³⁺ is colourless in aqueous solution because of absence of unpaired electron in d sub orbit. d-d transition is not possible. Ti³⁺ contains unpaired electron in d suborbit.

(ix) Silver atom has completely filled d orbitals in its ground State. How can you say that it is a transition element?
Ans (ix) It contains unpaired electron in Ag²⁺

(x) The transition elements are very hard.
Ans (x) Because of strong metallic bond

LONG ANSWER TYPE QUESTION

Q What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Answer : As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron. Also, with the increase in atomic number, the number of electrons in the 4f orbital also increases. The 4f electrons have poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.

Consequences of lanthanoid contraction

There is similarity in the properties of second and third transition series. ii.

Separation of lanthanoids is possible due to lanthanide contraction.

It is due to lanthanide contraction that there is variation in the basic strength of

lanthanide hydroxides. (Basic strength decreases from La(OH)3 to Lu(OH)3.)

Q. What are the characteristics of the transition elements and why are they

called transition elements? Which of the d-block elements may not be

regarded as the transition elements?

Answer : Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between s-block and p-block. Therefore, these are called transition elements.

Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell.

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

Answer: Transition metals have a partially filled d−orbital. Therefore, the electronic configuration of transition elements is (n − 1)d^1-10 ns^0-2.

The non-transition elements either do not have a d−orbital or have a fully filled d−orbital.

Therefore, the electronic configuration of non-transition elements is ns1-2 or ns2 np^1-6.

Q Explain giving reasons:

(i) Transition metals and many of their compounds show paramagnetic behaviour.

(ii) The enthalpies of atomisation of the transition metals are high.

(iii) The transition metals generally form coloured compounds.

(iv) Transition metals and their many compounds act as good catalyst.

Answer

(i) Transition metals show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. However, in the first transition series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is only because of the unpaired electron.

(ii) Transition elements have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals is high.

(iii) Most of the complexes of transition metals are coloured. This is because of the absorption of radiation from visible light region to promote an electron from one of the d−orbitals to another. In the presence of ligands, the d-orbitals split up into two sets of orbitals having different energies. Therefore, the transition of electrons can take place from one set to another. The energy required for these transitions is quite small and falls in the visible region of radiation. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.

(iv) The catalytic activity of the transition elements can be explained by two basic facts.

(a) Owing to their ability to show variable oxidation states and form complexes, transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, Ea, for the reaction.

(b) Transition metals also provide a suitable surface for the reactions to occur.

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

Answer : Transition metals are large in size and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic size), such as H, C, N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.

Q How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

Answer

In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by 1 (Fe2+ and Fe3+; Cu+ and Cu2+). In non-transition elements, the oxidation states differ by 2, for example, +2 and +4 or +3 and +5, etc.

Q Describe the preparation of potassium dichromate from iron chromite ore.

What is the effect of increasing pH on a solution of potassium dichromate?

Answer

Potassium dichromate is prepared from chromite (FeCr₂O₄) ore in the following steps.

Step (1): Preparation of sodium chromate

4 FeCr₂O₄+ 16 NaOH + 7O2 → 8Na₂Cr₂O₄+ 2Fe₂O₃ + 8H₂O

Step (2): Conversion of sodium chromate into sodium dichromate

2Na₂Cr₂O₄+ Conc. H₂SO₄ →Na₂Cr₂O₄+Na₂SO₄+H₂O

Step(3): Conversion of sodium dichromate to potassium dichromate

Na₂Cr₂O₇+ KCl → K₂Cr₂O₇+ 2NaCl

Potassium chloride being less soluble than sodium chloride is obtained in the form of

orange coloured crystals and can be removed by filtration.

The dichromate $\displaystyle \left( {C{{r}_{2}}O_{7}^{{2-}}} \right)$ ion exists in equilibrium with chromate $\displaystyle \left( {CrO_{4}^{{2-}}} \right)$ ion at pH 4.

However, by changing the pH, they can be interconverted.

Q Describe the preparation of potassium permanganate. How does the acidified

permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid?

Write the ionic equations for the reactions.

Answer

Potassium permanganate can be prepared from pyrolusite (MnO2). The ore is fused with

KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO3

or KClO4, to give K2MnO4.

2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O

The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution. Electrolytic oxidation

K₂MnO₄ ⇌ 2K⁺ + $\displaystyle MnO_{4}^{{2-}}$

H₂O ⇌ H⁺ + OH^–

At anode, manganate ions are oxidized to permanganate ions.

$\displaystyle MnO_{4}^{{2-}}\rightleftarrows MnO_{4}^{-}+{{e}^{-}}$

Oxidation by chlorine

2K₂MnO₄ + Cl₂→2KMnO₄+ 2KCl

$\displaystyle 2MnO_{4}^{{2-}}+C{{l}_{2}}\to 2MnO_{4}^{-}+2C{{l}^{-}}$

Oxidation by ozone

2K2MnO4 + O3 + H2O → 2KMnO4 + 2KOH + O2

$\displaystyle 2MnO_{4}^{{2-}}+{{O}_{3}}+{{H}_{2}}O\to 2MnO_{4}^{{2-}}+2O{{H}^{-}}+{{O}_{2}}$

(i) Acidified KMnO4 solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to ferric

ions.

$\displaystyle MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{{2+}}}+4{{H}_{2}}O$

$\displaystyle \left[ {F{{e}^{{2+}}}\to F{{e}^{{3+}}}+{{e}^{-}}} \right]\times 5$

$\displaystyle MnO_{4}^{-}+5F{{e}^{{2+}}}+8{{H}^{+}}\to M{{n}^{{2+}}}+5F{{e}^{{3+}}}+4{{H}_{2}}O$

(ii) Acidified potassium permanganate oxidizes SO2 to sulphuric acid.

$\displaystyle \left[ {MnO_{4}^{-}+6{{H}^{+}}+5{{e}^{-}}\to M{{n}^{{2+}}}+3{{H}_{2}}O} \right]\times 2$

$\displaystyle \left[ {2{{H}_{2}}O+2S{{O}_{2}}+{{O}_{2}}\to 5{{H}^{+}}+2SO_{4}^{{2-}}+2{{e}^{-}}} \right]\times 5$

$\displaystyle 2MnO_{4}^{-}+10S{{O}_{2}}+5{{O}_{2}}+4{{H}_{2}}O\to 2M{{n}^{{2+}}}+10SO_{4}^{{2-}}+8{{H}^{+}}$

(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

$\displaystyle \left[ {MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to 2M{{n}^{{2+}}}+4{{H}_{2}}O} \right]\times 2$

$\displaystyle \left[ {{{C}_{2}}O_{4}^{{2-}}\to 2C{{O}_{2}}+2{{e}^{-}}} \right]\times 5$

$\displaystyle 2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{{2-}}+16{{H}^{+}}\to 2M{{n}^{{2+}}}+10C{{O}_{2}}+8{{H}_{2}}O$

Q Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(i) electronic configuration (iii) oxidation state

(ii) atomic and ionic sizes and (iv) chemical reactivity.

Answer

(i) Electronic configuration

The general electronic configuration for lanthanoids is [Xe]54 4f0-14 5d0-1 6s2 and that for actinoids is [Rn]86 5f1-14 6d0-1 7s2. Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Oxidation states

The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of + 2 and + 4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies. Again, (+3) is the principal oxidationstate for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.

(iii) Atomic and lonic sizes

Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of 5f orbitals.

iv. Chemical reactivity

In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

Q Give examples and suggest reasons for the following features of the transition metal chemistry:

(i)The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.

(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Answer

(i) In the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base. On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding and so, they are unavailable. There is also a high effective nuclear charge. As a result, it can accept electrons and behave as an acid.

For example, MnO is basic and Mn₂O₇ is acidic.

(ii) Oxygen and fluorine act as strong oxidising agents because of their high

electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits higher oxidation states in oxides and fluorides.

For example, in OsF6 and V2O5, the oxidation states of

Os and V are +6 and +5 respectively.

(iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size. So, oxo-anions of a metal have the highest oxidation state.

For example, in $\displaystyle MnO_{4}^{-}$ , the

oxidation state of Mn is +7.

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

Answer

An alloy is a solid solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to possess different physical properties than those of the component elements. An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94−95%), iron

(5%), and traces of S, C, Si, Ca, and Al.

Uses

(1) Mischmetal is used in cigarettes and gas lighters.

(2) It is used in flame throwing tanks.

(3) It is used in tracer bullets and shells.

Q Compare the chemistry of the actinoids with that of lanthanoids with reference to:

(i) electronic configuration (ii)oxidation states and (iii) chemical reactivity.

Answer

Electronic configuration

Oxidation states

The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of + 2 and + 4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.

Chemical reactivity

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