Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 8.1:

Write down the electronic configuration of:

(i) Cr³⁺+ (iii) Cu⁺ (v) Co²+ (vii) Mn²⁺

(ii) Pm³⁺ (iv) Ce⁴⁺ (vi) Lu²⁺ (viii) Th⁴⁺

Answer:

(i) Cr³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³

Or, [Ar]¹⁸ 3d³

(ii) Pm³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f⁴

Or, [Xe]⁵⁴ 3d³

(iii) Cu⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

Or, [Ar]¹⁸ 3d¹⁰

(iv) Ce⁴⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶

Or, [Xe]⁵⁴

(v) Co²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Or, [Ar]¹⁸ 3d⁷

(vi) Lu²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d¹

Or, [Xe]⁵⁴ 2f¹⁴ 3d³

(vii) Mn²⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

Or, [Ar]¹⁸ 3d⁵

(viii) Th⁴⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s² 6s⁶

Or, [Rn]⁸⁶

Question 8.2:

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

Answer:

Electronic configuration of Mn²⁺ is [Ar]¹⁸ 3d⁵.

Electronic configuration of Fe²⁺ is [Ar]¹⁸ 3d⁶.

It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d⁵ configuration. This is the reason Mn²⁺ shows resistance to oxidation to Mn³⁺. Also, Fe²⁺ has 3d⁶ configuration and by losing one electron, its configuration changes to a more stable 3d⁵ configuration. Therefore, Fe²⁺ easily gets oxidized to Fe⁺³ oxidation state.

Question 8.3:

Explain briefly how +2 state becomes more and more stable in the first half

of the first row transition elements with increasing atomic number?

Answer:

The oxidation states displayed by the first half of the first row of transition metals are given in the table below.

+2 oxidation state is attained by the loss of the two 4s electrons by these metals. Since the number of d electrons in (+2) state also increases from Ti(+2) to Mn(+ 2), the stability of +2 state increases (as d-orbital is becoming more and more half-filled). Mn (+2) has d⁵ electrons (that is half-filled d shell, which is highly stable).It can be easily observed that except Sc, all others metals display +2 oxidation state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also increases from 1 to 5.

Question 8.4:

To what extent do the electronic configurations decide the stability of

oxidation states in the first series of the transition elements? Illustrate

your Answer with examples.

Answer:

The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting maximum number of oxidation states (+2 to +7). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital. However, Sc does not show +2 oxidation state. Its electronic configuration is 4s² 3d¹. It loses all the three electrons to form Sc³⁺. +3 oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, [Ar]. Ti (+ 4) and V(+5) are very stable for the same reason. For Mn, +2 oxidation state is very stable as after losing two electrons, its d-orbital is exactly half-filled, [Ar] 3d⁵.

Question 8.5:

What may be the stable oxidation state of the transition element with the

following d electron configurations in the ground state of their atoms : 3d³,

3d⁵, 3d⁸and 3d⁴?

Answer:

Name the oxometal anions of the first series of the transition metals inQuestion 8.6:

which the metal exhibits the oxidation state equal to its group number.

Answer:

(i) Vanadate, 

Oxidation state of V is + 5.

(ii) Chromate, 

Oxidation state of Cr is + 6.

(iii) Permanganate, 

Oxidation state of Mn is + 7.

Question 8.7:

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

(i) There is similarity in the properties of second and third transition series.

2. Separation of lanthanoids is possible due to lanthanide contraction.

(iii) 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.)

Question 8.8:

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.

Question 8.9:

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 ns^1-2 or ns² np^1-6.

Question 8.10:

What are the different oxidation states exhibited by the lanthanoids?

Answer:

In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.

Question 8.11:

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 toanother. 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, E_a, for the reaction.

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

Question 8.12:

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.

Question 8.13:

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.

Question 8.14:

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 ore  in the following steps.

Step (1): Preparation of sodium chromate

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

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

Potassium dichromate being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration.

The dichromate ion  exists in equilibrium with chromate  ion at pH 4. However, by changing the pH, they can be interconverted.

2CrO42- →Acid 2HCrO4-  →Acid 2Cr2O72-Cr2O72- →Base 2HCrO42- →Base 2CrO42-

Question 8.15:

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(i) iodide (ii) iron(II) solution and (iii) H₂S

Answer:

 acts as a very strong oxidising agent in the acidic medium.

 takes up electrons to get reduced and acts as an oxidising agent. The reaction of K₂Cr₂O₇ with other iodide, iron (II) solution, and H₂S are given below.

(i)   oxidizes iodide to iodine.

(ii)   oxidizes iron (II) solution to iron (III) solution i.e., ferrous ions to ferric ions.

(iii)   oxidizes H₂S to sulphur.