//M1//QN1//SUB//DL0

Write down the electronic configuration of: (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+ (v) Co2+ (vi) Lu2+ (vii) Mn2+ (viii) Th4+

//X

(i) Cr3+ : 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3
(ii) Pm3+ : 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6 4f 4
: [Xe]54 4 f 4
(iii) Cu+ : 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10
: [Ar]18 3 d 10
(iv) Ce4+ : 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6
: [Xe]54
(v) Co2+ : 1s 2 2s 2 2p 6 3s 2 3p 6 3d 7
(vi) Lu2+ : 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6 4f 14 5d 1
: [Xe]54 4f 14 5d 1
(vii) Mn2+ : 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5
(viii) Th4+ : 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 4f 14 5s 2 5p 6 5d 10 6s 2 6s 6
: [Rn]86

//M2//QN2//SUB//DL0

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

//X

Electronic configuration of Mn2+ is [Ar]183d 5.
Electronic configuration of Fe2+ is [Ar]183d 6.
It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d 5 configuration.
This is the reason Mn2+ shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d 6 configuration and by losing one electron, its configuration changes to a more stable 3d 5 configuration. Therefore, Fe2+ easily gets oxidized to Fe3+ oxidation state.

//M2//QN3//SUB//DL0

Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

//X

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.
Sc (+2) = d 1
Ti (+2) = d 2
V (+2) = d 3
Cr (+2) = d 4
Mn (+2) = d 5
+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 d 5 electrons (that is half-filled d-shell, which is highly stable).

//M2//QN4//SUB//DL0

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.

//X

If an element contain d 0, d 5, d 10 electronic configuration it become more stable.
Example,
Sc3+ is more stable than Sc+ because of d 0 configuration.
V5+ is more stable than V3+ because of d 0 configuration.
Mn2+ is more stable than Mn3+ because of
d 5 configuration.
Fe3+ is more stable than Fe2+ because of
d 5 configuration.
Zn2+ is more stable than Zn+ because of
d 10 configuration.

//M1//QN5//SUB//DL0

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

d electron configurations in the ground state of their atoms: 3d 3, 3d 5, 3d 8 and 3d 4?

//X

(no answer)

//M0//QN6//SUB//DL0

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.

//X

(1) Vanadate ion (VO–3 )
Oxidation state of vanadium is +5. which is equal to its group number 5.
(2) Dichromate (Cr2O2– 7 ) and chromate (CrO2–4 )
Oxidation state of chromium in both oxoanions is +6. which is equal to its group number 6.
(3) Permanganate ion (MnO4– )
Oxidation state of manganese is +7. which is equal to its group number 7.

//M0//QN7//SUB//DL0

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

//X

Refer Que. 15 (Page No.240)

//M3//QN8//SUB//DL0

What are the characteristics of the transition elements and why are they called transition elements? Which d-block elements may not be regarded as the transition elements?

//X

Refer Que. 9 (Page No.238)

//M2//QN9//SUB//DL0

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

[Topic 4.0] [2 Marks]

//X

Refer Que. 10 (Page No.238)

//M0//QN10//SUB//DL0

What are the different oxidation states exhibited by the lanthanoids?

//X

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.

//M2//QN11//SUB//DL0//EQ

Explain giving reasons: (i) Transition metals and many of their compounds show paramagnetic behaviour. (ii) The enthalpies of atomization 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.

//X

(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.
Magnetic moment is calculated by following formula.
µ = B.M.
where, n = No. of 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 in 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.
Example, V2O5 in contact process, Fe in Haber's process.

//M0//QN12//SUB//DL0

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

//X

Refer Que. 26 (Page No.244)

//M0//QN13//SUB//DL0

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

//X

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.

//M4//QN14//SUB//DL0

Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate? [Topic 4.4]

//X

Dichromates are generally prepared from chromate. which in turn are obtained by the fusion of chromite ore (FeCr2O4) with sodium or potassium carbonate in free access of air.
The reaction with sodium carbonate occurs as follows:
4FeCr2O4 + 8Na2CO3 + 7O2 →
8Na2CrO4 + 2Fe2O3 + 8CO2
The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7 . 2H2O can be crystallised.
2Na2CrO4 + 2 H+ → Na2Cr2O7 + 2Na+ + H2O
Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.
Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl
Orange crystals of potassium dichromate crystallise out.
The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution.
2CrO2–4 + 2H+ → Cr2O2–7 + H2O
Cr2O2–7 + 2OH– → 2CrO2–4 + H2O

//M0//QN15//SUB//DL0//EQ

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.

//X

Potassium dichromates are strong oxidising agents
In acidic solution, its oxidising action can be represented as follows
Cr2O2–7 + 14H+ + 6e– → 2Cr3+ + 7H2O
(1) Acidified potassium dichromate will oxidise iodides to iodine,
(2) Acidified potassium dichromate will oxidise iron (II) solution to iron (III) solution.
(3) Acidified potassium dichromate will oxidise H2S to sulphur.

//M3//QN16//SUB//DL0//EQ

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.

OR State the balance reaction equations of acidic permanganate ion with (i) Fe 2+ (ii) C 2 O 2–4 (iii) SO 2–3 [June 2024] [3 Marks]

//X

Preparation:
Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate.
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
3MnO2–4 + 4H+ → 2MnO–4 + MnO2 + 2H2O
Commercially it is prepared by the alkaline oxidative fusion of MnO2 followed by the electrolytic oxidation of manganate (VI).
In the laboratory, a manganese (II) ion salt is oxidised by peroxodisulphate to permanganate.
2Mn2+ + 5S2O2–8 + 8H2O → 2MnO–4 + 10SO2–4 + 16H+
In Acidic Medium oxidising action of KMnO4
+ 8H+ + 5e– → Mn2+ + 4H2O
(i) Acidified KMnO4 solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to ferric ions.
(ii) Acidified potassium permanganate oxidizes SO2 to sulphuric acid.
(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

//M0//QN17//SUB//DL0

For M2+/M and M3+/M2+ systems the Eo values for some metals are as follows: Cr 2+/Cr –0.9 V Cr3/Cr2+: –0.4 V Mn 2+/Mn –1.2 V Mn3+/Mn2+ +1.5 V Fe 2+/Fe –0.4 V Fe3+/Fe2 +0.8 V Use this data to comment upon: (i) The stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+. (ii) The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

//X

(i) The E° value for Fe3+/Fe2+ is higher than that for Cr3+/Cr2+ and lower than that for
Mn3+/Mn2+. So, the reduction of Fe3+ to Fe2+ is easier than the reduction of Mn3+ to Mn2+, but not as easy as the reduction of Cr3+ to Cr2+. Hence, Fe3+ is more stable than Mn3+, but less stable than Cr3+. These metal ions can be arranged in the increasing order of their stability as: Mn3+ < Fe3+ < Cr3+.
(ii) The reduction potentials for the given pairs increase in the following order:
Mn2+ / Mn < Cr2+ / Cr < Fe2+ / Fe
So, the oxidation of Fe to Fe2+ is not as easy as the oxidation of Cr to Cr2+ and the oxidation of Mn to Mn2+. Thus, these metals can be arranged in the increasing order of their ability to get oxidised as: Fe < Cr < Mn.

//M2//QN18//SUB//DL0

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

//X

Only the ions that have electrons in d-orbital and in which d-d transition is possible will be coloured.
The ions in which d-orbitals are empty or completely filled will be colourless as no d-d transition is possible in those configurations.

Element | Atomic number | Ionic state | Electronic configuration in ionic state | Number of unpaired electrons
Ti | 22 | T1 3+ | [Ar] 3 d 1 | 1
V | 23 | V 3+ | [Ar] 3 d 2 | 2
Cu | 29 | Cu+ | [Ar] 3 d 10 | 0
Sc | 21 | Sc 3+ | [Ar] | 0
Mn | 25 | Mn 2+ | [Ar] 3 d 5 | 5
Fe | 26 | Fe 3+ | [Ar] 3 d 5 | 5
Co | 27 | Co 2+ | [Ar] 3 d 7 | 3

From the above table, it can be easily observed that only Sc3+ has an empty d-orbital and Cu+ has completely filled d-orbitals. All other ions, except Sc3+ and Cu+, will be coloured in aqueous solution because of d-d transitions.

//M0//QN19//SUB//DL0

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

//X

The number of oxidation states increases on moving from Sc to Mn. On moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons.
The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d-orbital.

//M0//QN20//SUB//DL0

Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) Electronic configuration, (ii) Oxidation state, (iii) Atomic and ionic sizes and (iv) Chemical reactivity.

//X

Refer Que. 42 (Page No.250)

//M0//QN21//SUB//DL0//EQ

How would you account for the following: (i) Cr2+ is strongly reducing while manganese (III )is strongly oxidising. (ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii) The d 1 configuration is very unstable in ions.

//X

(i) Cr2+ is strongly reducing in nature. It has a d 4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d 3). This d 3 configuration can be written as configuration, which is a more stable configuration. In the case of Mn3+ (d 4), it acts as an oxidizing agent and gets reduced to Mn2+ (d 5). This has an exactly half-filled d-orbital and is highly stable.
(ii) Co (II) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co (III). Although the 3rd ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.
(iii) The ions in d 1 configuration tend to lose one more electron to get into stable d 0 configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.

//M2//QN22//SUB//DL0//EQ

What is meant by 'disproportionation'? Give an example of disproportionation reaction in aqueous solution.

//X

It is found that sometimes a relatively less stable oxidation state undergoes an oxidation – reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation.
e.g. 3 + 4H+ → 2 + MnO2 + 2H2O

//M0//QN23//SUB//DL0

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

//X

In the first transition series, Cu exhibits
+1 oxidation state very frequently. It is because Cu (+1) has an electronic configuration of [Ar] 3d 10. The completely filled d-orbital makes
it highly stable.

//M0//QN24//SUB//DL0//EQ

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?

| Gaseous ions | Number of unpaired electrons
(i) | Mn3+: [Ar] 3d 4 | 4
(ii) | Cr3+: [Ar] 3d 3 | 3
(iii) | V3+: [Ar] 3d 2 | 2
(iv) | Ti3+: [Ar] 3d 1 | 1

//X

Cr3+ is the most stable in aqueous solutions owing to a configuration.

//M3//QN25//SUB//DL0

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.

//X

(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.
e.g.: MnO = Basic, MnO 2 = Amphoteric,
Mn 2 O 7 = 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 MnO–4, the oxidation state of Mn is +7.

//M3//QN26//SUB//DL0

Indicate the steps in the preparation of: (i) K2Cr2O7 from chromite ore. (ii) KMnO4 from pyrolusite ore.

//X

(i) K 2 Cr 2 O 7 from chromite ore
4FeCr 2 O 4 + 8Na 2 CO 3 + 7O 2
→ 8Na 2 CrO 4 + 2Fe 2 O 3 + 8CO 2
2Na 2 CrO 4 + 2H+ → Na 2 Cr 2 O 7 + 2Na+ + H 2 O
Na 2 Cr 2 O 7 + 2KCl → K 2 Cr 2 O 7 + 2NaCl
(ii) KMnO 4 from pyrolusite ore
Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3.
This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate.
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
3MnO2–4 + 4H+ → 2MnO–4 + MnO2 + 2H2O

//M0//QN27//SUB//DL0

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

//X

“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:

- Mischmetal is used in cigarettes and gas lighters.
- It is used in flame throwing tanks.
- It is used in tracer bullets and shells.

//M0//QN28//SUB//DL0

What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.

//X

Inner transition metals are those elements in which the last electron enters the ƒ-orbital.
The elements in which the 4ƒ-orbitals and the 5ƒ-orbitals are progressively filled are called ƒ-block elements.
Among the given atomic numbers, the atomic numbers of the inner transition elements are
59, 95, and 102.

//M2//QN29//SUB//DL0

The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.

//X

Lanthanoids primarily show three oxidation states (+2, +3, +4).
Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4ƒ, 5d, and 6s-orbitals is quite large.
On the other hand, the energy difference between 5ƒ, 6d, and 7s-orbitals is very less. Hence, actinoids display a large number of oxidation states.
For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, +7. The most common oxidation state in case of actinoids is also +3.

//M1//QN30//SUB//DL0

Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

//X

The last element in the actinoid series is lawrencium, Lr. Its atomic number is 103 and its electronic configuration is [Rn]5ƒ 14. 6d 1. 7s 2. The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable f 14 configuration.

//M0//QN31//SUB//DL0//EQ

Use Hund's rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.

//X

Electronic configuration of Ce: [xe] 4f 1 5d 1 6s 2
Ce3+: [xe] 4ƒ 1 (n = 1)
µ = B.M.
=
=
= 1.73 B.M.

//M0//QN32//SUB//DL0

Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.

//X

The lanthanoids that exhibit +2 and +4 states are shown in the given table. The atomic numbers of the elements are given in the parenthesis.

+2 | +4 Oxidation State
Nd (60) | Ce (58)
Sm (62) | Pr (59)
Eu (63) | Nd (60)
Tm (69) | Tb (65)
Yb (70) | Dy (66)

Ce after forming Ce4+ attains a stable electronic configuration of [Xe].
Tb after forming Tb4+ attains a stable electronic configuration of [Xe] 4f 7.
Eu after forming Eu2+ attains a stable electronic configuration of [Xe] 4f 7.
Yb after forming Yb2+ attains a stable electronic configuration of [Xe] 4f 14.

//M0//QN33//SUB//DL0

Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) Electronic configuration (ii) Oxidation states and (iii) Chemical reactivity.

//X

Refer Que. 42 (Page No.250)

//M0//QN34//SUB//DL0

Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

//X

61Pm (Promethium) = [xe] 4ƒ 5 5d 0 6s 2
91 Pa (Protactinium) = [Rn] 5 ƒ 2 6 d 1 7 s 2
101Md (Mendelevium) = [Rn] 5ƒ 13 6d 0 7s 2
109 Mt (Meitnerium) = [Rn] 5 ƒ 14 6 d 7 7 s 2

//M0//QN35//SUB//DL0

Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points: (i) Electronic configurations, (ii) Oxidation states, (iii) Ionisation enthalpies and (iv) Atomic sizes.

//X

(i) In the 1st, 2nd and 3rd transition series, the 3d, 4d and 5d-orbitals are respectively filled.
(ii) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends.
However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states.
The stability of the +2 and +3 oxidation states decreases in the second and the third transition series, wherein higher oxidation states are more important.
(iii) In each of the three transition series, the first ionization enthalpy increases from left to right. However, there are some exceptions.
The first ionization enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of
4f electrons in the third transition series.
(iv) Atomic size generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the second transition series are greater than those of the elements corresponding to the same vertical column in the first transition series.
However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

//M3//QN36//SUB//DL0

Write down the number of 3d electrons in each of the following ions: Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to the occupied for these hydrated ions (octahedral).

//X

Ions | Electronic Configuration | Number of d-electron | Filling of d-electron
Ti2+ | [Ar] 3d 2 | 2 | t 2 g 2 eg0
V2+ | [Ar] 3d 3 | 3 | t 2g 3 eg0
Cr3+ | [Ar] 3d 3 | 3 | t 2g 3 eg0
Mn2+ | [Ar] 3d 5 | 5 | t 2g 3 eg2
Fe2+ | [Ar] 3d 6 | 6 | t 2g 4 eg2
Fe3+ | [Ar] 3d 5 | 5 | t 2g 3 eg2
Co2+ | [Ar] 3d 7 | 7 | t 2g 5 eg2
Ni2+ | [Ar] 3d 8 | 8 | t 2g 6 eg2
Cu2+ | [Ar] 3d 9 | 9 | t 2 g 6 eg3

//M0//QN37//SUB//DL0

Elements of the first transition series possess many properties different from those of heavier transition elements.

//X

The properties of the elements of the first transition series differ from those of the heavier transition element in many ways.
(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2nd and 3rd transition series).
However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.
(ii) +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.
(iii) The enthalpies of atomionization of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.
(iv) The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of stronger metallic bonding
(M-M bonding).
(v) The elements of the first transition series form
low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.

//M0//QN38//SUB//DL0//EQ

What can be inferred from the magnetic moment values of the following complex species?

Example | Magnetic moment (B.M.)
(i) | K 4[Mn(CN)6] | 2.2
(ii) | [Fe(H 2 O)6]2+ | 5.3
(iii) | K 2[MnCl 4] | 5.9

//X

(i) K4[Mn(CN)6]
For in transition metals, the magnetic moment is calculated from the spin-only formula.
Therefore,
µ = = 2.2 B.M.
We can see from the above calculation that the given value is closest to n = 1. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.
Hence, we can say that CN– is a strong field ligand that causes the pairing of electrons.
(ii) [Fe(H2O)6]2+
µ = = 5.3 B.M.

- We can see from the above calculation that the given value is closest to n = 4. Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 6 electrons in the d-orbital.
- Hence, we can say that H2O is a weak field ligand and does not cause the pairing of electrons.

(iii) K2[MnCl4]
µ = = 5.9 B.M.
We can see from the above calculation that the given value is closest to n = 5. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.
Hence, we can say that Cl– is a weak field ligand and does not cause the pairing of electrons.