//M2//QN1//SUB//DL0

What are d-block elements? Explain classification of d-block elements.

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//M0//QN2//SUB//DL0

What transition elements? Explain with examples.

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

What is ƒ-block element?

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

Explain position of d-block in periodic table.

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//M0//QN5//SUB//DL0

Electronic configuration of first transition metal.

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

Electronic configuration of second transition metal.

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

Electronic configuration of third transition metal.

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//M0//QN8//SUB//DL0

Why the chromium and copper have exceptional electronic configuration?

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//M3//QN9//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?

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//M2//QN10//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]

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//M0//QN11//SUB//DL0

Explain physical properties of transition elements.

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//M2//QN12//SUB//DL0

Aqueous solution of TiCl3 is coloured where as aqueous solution of TiCl4 is colourless. Give reason.

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//M2//QN13//SUB//DL0//EQ

Why the tansition elements are hard and have high melting and boiling points?

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The high melting points of these metals are attributed to the involvement of greater number of electrons from (n-1)d in addition to the ns electrons in the interatomic metallic bonding.

In any row the melting points of these metals rise to a maximum at d5 except for anomalous values of Mn and Tc and fall regularly as the atomic number increases.

Zn, Cd, and Hg are soft and volatile and they have low melting point because all these elements have no unpaired electrons.

//M0//QN14//SUB//DL0//EQ

Write note on enthalpy of atomization of transition metals.

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The transition elements have high enthalpy of atomization because of presence of strong bonds between the atoms.

The enthalpy of atomization increases with the increase in number of unpaired electrons which results in formation of strong bonds due to strong interatomic interactions.

The metals of the second and third series have greater enthalpies of atomization than the corresponding elements of the first series; this is an important factor in accounting for the occurrence of much more frequent metal–metal bonding in compounds of the heavy transition metals.

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

What is lanthanoid contraction? What are the consequences of lanthanoid contraction? or Explain atomic radii of transition element along the period.

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In a given series decrease in radius with increasing atomic number.

This is because the new electron enters a
d-orbital each time the nuclear charge increases by unity. Hence, the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases.

The same trend is observed in the atomic radii of a given series. However, the variation within a series is quite small.

The curves in Fig. show an increase from the first (3d) to the second (4d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series.

This phenomenon is associated with the intervention of the 4ƒ-orbitals which must be filled before the 5d series of elements begin.

The filling of 4f before 5d-orbital results in a regular decrease in atomic radii called lanthanoid contraction.

The net result of the lanthanoid contraction is that the second and the third d series exhibit similar radii (e.g. Zr-160 pm, Hf-159 pm) and have very similar physical and chemical properties much more than that expected on the basis of usual family relationship.

//M0//QN16//SUB//DL0

Explain the variation in ionization enthalpies of transition element in 3d-series.

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//M3//QN17//SUB//DL0

Write about oxidation states of transition series?

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//M0//QN18//SUB//DL0//EQ

Explain trends in M2+/M standard electrode potentials.

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An element in M2+ state in aqueous medium is more stable if the electrode potential (M2+/M) value is more negative.

The general trend towards less negative Eo values across the series is related to the general increase in the sum of the first and second ionization enthalpies.

The unique behaviour of Cu, having a positive E°/V, accounts for its inability to liberate H2 from acids. Only oxidising acids (nitric and hot concentrated sulphuric) react with Cu, the acids being reduced.

It is interesting to note that the value of E°/V for Mn, Ni and Zn are more negative than expected from the trend.

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

Write a note on halide compounds of Transition elements.

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//M2//QN20//SUB//DL0

Explain chemical reactivity of transition metals.

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//M3//QN21//SUB//DL0//EQ

Explain magnetic properties of transition elements.

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When a magnetic field is applied to substances, mainly two types of properties magnetic behaviour are observed: diamagnetism and paramagnetism.

Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted.

Substances which are attracted very strongly are said to be ferromagnetic.

In fact, ferromagnetism is an extreme form of paramagnetism. Many of the transition metal ions are paramagnetic.

Paramagnetism arises from the presence of unpaired electrons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum.

For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the 'spin-only' formula,
i.e., m =

Where, n = The number of unpaired electrons µ = Magnetic moment in units of Bohr
magneton (BM).

The magnetic moment increases with the increasing number of unpaired electrons.

Calculated and Observed Magnetic Moments (BM)
Ion	Configu-ration	Unpaired electron(s)	Magnetic moment
			Calculated	Observed
Sc3+ Ti3+ Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+	3d 0 3d 1 3d 2 3d 3 3d 4 3d 5 3d 6 3d 7 3d 8 3d 9 3d 10	0 1 2 3 4 5 4 3 2 1 0	0 1.73 2.84 3.87 4.90 5.92 4.90 3.87 2.84 1.73 0	0 1.75 2.76 3.86 4.80 5.96 5.3 - 5.5 4.4 - 5.2 2.9 - 3, 4 1.8 - 2.2 0

//M3//QN22//SUB//DL0//EQ//PYQ

Calculate the magnetic moment of a tivalent ion in aqueous solution if its atomic number

is 26.[May 2021] [3 Marks]

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26M: [Ar] 3d5 4s2

M3+: [Ar] 3d5

Unpaired electrons = 5

∴ Magnetic moment (µ)=

=

=

= 5.92 B.M

//M3//QN23//SUB//DL0

Why transition elements form a coloured ions?

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//M2//QN24//SUB//DL0

Why the transition element form a large number of complex compounds?

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//M2//QN25//SUB//DL0

Explain Catalytic property of transition elements.

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

What are interstitial compounds? Why are such compounds well known for transition metals?or What are interstitial compounds. Write its any two characteristics. or Write note on interstitial compounds.

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//M0//QN27//SN//DL0

Write short note on alloys.

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

Write note on oxides and oxoanions of 3d transition metals.

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

Discuss structures and magnetic properties of manganate and permanganate ions.

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The manganate and permanganate ions are tetrahedral.

The green manganate is paramagnetic with one unpaired electron but the permanganate is diamagnetic.

The π-bonding takes place by overlap of
p-orbitals of oxygen with d-orbitals of manganese.

//M2//QN30//SUB//DL0//EQ

Give the structure of chromate and dichromate ions, and explain interconvertible property of these ions.

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The structures of chromate ion, CrO2–4 and the dichromate ion, Cr2O2–7 are shown below.

The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedral sharing one corner with Cr–O–Cr bond angle of 126°.

The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The oxidation state of chromium in chromate and dichromate is the same.

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

Explain the chemical properties of KMnO4.

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KMnO4 act as strong oxidising agent in neutral, alkaline and in acidic medium.

(I) In Acidic Solution:

(a) Iodine is liberated from potassium iodide:

(b) Fe2+ ion (green) is converted to Fe3+ (yellow):

(c) Oxalate ion or oxalic acid is oxidised at 333K :

(d) Hydrogen sulphide is oxidised, sulphur being precipitated:

H2S → 2H+ + S2–

5S2– + 2MnO–4 + 16H+ → 2Mn2+ + 8H2O + 5S

(e) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid:

5SO2–3 + 2MnO–4 + 6H+ → 2Mn2+ + 3H2O + 5SO2–4

(f) Nitrite is oxidised to nitrate:

5NO–2 + 2MnO–4 + 6H+ → 2Mn2+ + 5NO–3 + 3H2O

(2) In neutral or faintly alkaline solutions:

(a) A notable reaction is the oxidation of iodide to iodate:

2MnO–4 + H2O + I– → 2MnO2 + 2OH– + IO–3

(b) Thiosulphate is oxidised almost quantitatively to sulphate:

8MnO–4 + 3S2O2–3 + H2O → 8MnO2 + 6SO2–4 + 2OH–

(c) Manganous salt is oxidised to MnO2; the presence of zinc sulphate or zinc oxide catalyses the oxidation:

2MnO–4 + 3Mn2+ + 2H2O → 5MnO2 + 4H+

//M2//QN32//SUB//DL0

Write down the uses of KMnO4.

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//M3//QN33//SUB//DL0//EQ

Complete the reactions and balance them. (i) MnO2 + KOH + O2 → (ii) MnO–4 + H2O + I– (iii) MnO–2 + S2O2–3 + H2O

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

Write note on inner transition elements or ƒ-block elements.

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

Write note on physical properties of lanthanoids.

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//M0//QN36//SUB//DL0

Write down uses of lanthanoids and its compounds.

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//M3//QN37//SUB//DL0//EQ

Discuss chemical properties of the lanthanoids.

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In their chemical behaviour, in general, the earlier members of the series are quite reactive similar to calcium but, with increasing atomic number, they behave more like aluminium.

The value of E° of half reaction

+ 3e– Ln(s) range from – 2.2 V to 2.4 V except for Eu that has – 2.0 V.

Lanthanoids heated with carbon forms carbides Ln3C, Ln2C3 and LnC2.

Lanthanoids form trihalides LnX3 with halogen and liberates hydrogen gas when treated with dilute acids.

These metals combine with oxygen to form oxides of type Ln2O3.

Oxides are basic in nature and when dissolved in water they form hydroxide Ln(OH)3.

//M3//QN38//SUB//DL0

Write down electronic configuration of all lanthanoids elements. [3 Marks]

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//M3//QN39//SUB//DL0//EQ

Explain the lanthanoid contraction.

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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 4ƒ-orbital also increases. The 4ƒ 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.
• Separation of lanthanoids is possible due to lanthanoid contraction.
• It is due to lanthanoid contraction that there is variation in the basic strength of lanthanoid hydroxides. (Basic strength decreases from La(OH)3 to Lu(OH)3.)

//M0//QN40//SUB//DL0

Write down general electronic configuration of elements of actinoids. Why there are irregularities in the electronic configurations?

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//M0//QN41//SUB//DL0

Discuss comparison between lathanoids and actinoids.

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//M0//QN42//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.

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//M0//QN43//SUB//DL0

Discuss applications of d and ƒ-block elements.

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//M3//QN44//SUB//DL0//PYQ

Give reason:[March 2023] [3 Marks]

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