Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements Text Book solutions

I. Choose the correct answer.

Questions 1.

Sc ( Z=21) is a transition element but Zinc (z=30) is not because ……………..

(a) both Sc³⁺ and Zn²⁺ ions are colourless and form white compounds.

(b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled

(c) last electron as assumed to be added to 4s level in case of zinc

(d) both Sc and Zn do not exhibit variable oxidation states

Answer:

(c) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled

Question 2.

Which of the following d block element has half filled penultimate d sub shell as well as half filled valence sub shell?

(a) Cr

(b) Pd

(c) Pt

(d) none of these

Answer:

(a) Cr

Hint: Cr ? [Ar]3d⁵ 4s¹

Question 3.

Among the transition metals of 3d series, the one that has highest negative (M²⁺/ M) standard electrode potential is ……………..

(a) Ti

(b) Cu

(c) Mn

(d) Zn

Answer:

(a) Ti

Question 4.

Which one of the following ions has the same number of unpaired electrons as present in V³⁺?

(a) Ti³⁺

(b) Fe³⁺

(c) Ni²⁺

(d) Cr²⁺

Answer:

(c) Ni²⁺

Question 5.

The magnetic moment of Mn²⁺ ion is ……………..

(a) 5.92BM

(b) 2.80BM

(c) 8.95BM

(d) 3.90BM

Answer:

(a) 5.92BM

Question 6.

Which of the following compounds is colourless?

(a) Fe³⁺

(6) Ti⁴⁺

(c) CO²

(d) Ni²

Answer:

(b) Ti⁴⁺

Hint: Ti⁴⁺ contains no unpaired electrons in d orbital, hence no d-d transition.

Question 7.

The catalytic behaviour of transition metals and their compounds is ascribed mainly due to ……………..

(a) their magnetic behaviour

(b) their unfilled d orbitals

(c) their ability to adopt variable oxidation states

(d) their chemical reactivity

Answer:

(c) their ability to adopt variable oxidation states

Question 8.

The correct order of increasing oxidizing power in the series ……………..(aVO+2 < Cr2O2-7 < MnO–

Answer:

Question 9.

The alloy of copper that contain Zinc is ……………..

(a) Monel metal

(b) Bronze

(c) bell metal

(d) brass

Answer:

(d) brass

Question 10.

Which of the following does not give oxygen on heating?

(a) K₂Cr₂O₇

(b) (NH₄)₂Cr₂O₇

(c) KClO₃

(d) Zn(ClO₃)₂

Answer:

(b) (NH₄)₂Cr₂O₇

Hint:

Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements-1

Question 11.

In acid medium, potassium permanganate oxidizes oxalic acid to ……………..

(a) Oxalate

(b) Carbon dioxide

(c) acetate

(d) acetic acid

Answer:

(b) Carbon dioxide

Question 12.

Which of the following statements is not true?

(a) on passing H₂S, through acidified K₂Cr₂O₇ solution, a milky colour is observed.

(b) Na₂Cr₂O₇ is preferred over K₂Cr₂O₇ in volumetric analysis

(c) K₂Cr₂O₇ solution in acidic medium is orange in colour

(d) K₂Cr₂O₇ solution becomes yellow on increasing the pH beyond 7

Answer:

(b) Na₂Cr₂O₇ is preferred over K₂Cr₂O₇ in volumetric analysis

Question 13.

Permanganate ion changes to in acidic medium ……………..

(a)

(b) Mn²⁺

(c) Mn³⁺

(d) MnO₂

Answer:

(b) Mn²⁺

Question 14.

A white crystalline salt (A) react with dilute HCl to liberate a suffocating gas (B) and also forms a yellow precipitate . The gas (B) turns potassium dichromate acidified with dil H2SO4 to a green coloured solution(C). A,B and C are respectively ……………..

(a) Na₂SO₃, SO₂, Cr₂(SO₄)₃

(b) Na₂S₂O₃, SO₂, Cr₂(SO₄)₃

(c) Na₂S, SO₂, Cr₂(SO₄)₃

(d) Na₂SO₄, SO₂, Cr₂(SO₄)₃

Answer:

(a) Na₂SO₃, SO₂, Cr₂(SO₄)₃

Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements-2

Question 15.

MnO–4 react with Br in alkaline PH to give ……………..

(a) BrO⁻₃, MnO₂

(b) Br₃, MnO²⁻₄

(c) Br₃, MnO₃

(d) BrO⁻, MnO²⁻₄

Answer:

(a) BrO⁻₃, MnO₂

Question 16.

How many moles of I2 are liberated when 1 mole of potassium dichromate react with potassium iodide?

(a) 1

(b) 2

(c) 3

(d) 4

Answer

(c) 3

Hint: K₂Cr₂O₇ + 6KI + 7H₂SO₄ ? 4K₂SO₄ + Cr₂ (SO₄)₃ + 7H₂O + 3I₂

Question 17.

The number of moles of acidified KMnO₄ required to oxidize 1 mole of ferrous oxalate(FeC₂O₄) is …………..

(a) 5

(b) 3

(c) 0.6

(d) 1.5

Answer:

(c) 0.6

Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements-3

Question 18.

When a brown compound of Mn (A) ids treated with HCl, it gives a gas (B). The gas (B) taken in excess reacts with NH₃ to give an explosive compound (C). The compound A, B and C are ……………

(a) MnO₂, Cl₂, NCl₃

(b) MnO, Cl₂, NH₄Cl

(c) Mn₃O₄, Cl₂, NCl₃

(d) MnO₃, Cl₂, NCl₂

Answer:

(a) MnO₂, Cl₂ NCl₃

Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements-4

Question 19.

Which one of the following statements related to lanthanons is incorrect?

(a) Europium shows +2 oxidation state.

(b) The basicity decreases as the ionic radius decreases from Pr to Lu.

(c) All the lanthanons are much more reactive than aluminium.

(d) Ce⁴⁺ solutions are widely used as oxidising agents in volumetric analysis.

Answer:

(c) All the lanthanons are much more reactive than aluminium.

Hint: As we move from La to Lu , their metallic behaviour because almost similar to that of aluminium.

Question 20.

Which of the following lanthanoid ions is diamagnetic?

(a) Eu²⁺

(b) Yb²⁺

(c) Ce²⁺

(d) Sm²⁺

Answer:

(6) Yb²⁺

Question 21.

Which of the following oxidation states is most common among the lanthanoids?

(a) 4

(b) 2

(c) 5

(d) 3

Answer:

(d) 3

Question 22.

Assertion: Ce⁴⁺ is used as an oxidizing agent in volumetric analysis.

Reason: Ce⁴⁺ has the tendency of attaining +3 oxidation state.

(a) Both assertion and reason are true and reason is the correct explanation of assertion.

(b) Both assertion and reason are true but reason is not the correct explanation of assertion.

(c) Assertion is true but reason is false. .

(d) Both assertion and reason are false.

Answer:

(a) Both assertion and reason are true and reason is the correct explanation of assertion.

Question 23.

The most common oxidation state of actinoids is ……………

(a) +2

(b) +3

(c) +4

(d) +6

Answer:

(c) +4

Question 24.

The actinoid elements which show the highest oxidation state of +7 are ……………

(a) Np, Pu, Am

(b) U, Fm, Th

(c) U, Th, Md

(d) Es, No, Lr

Answer:

(a) Np, Pu, Am

Question 25.

Which one of the following is not correct?
(a) La(OH)₂ is less basic than Lu(OH)₃
(b) In lanthanoid series ionic radius of Ln³⁺ ions decreases
(c) La is actually an element of transition metal series rather than lanthanide series
(d) Atomic radii of Zr and Hf are same because of lanthanide contraction
Answer:
(a) La(OH)₂ is less basic than Lu(OH)₃
II. Answer the following Questions:
Question 1.
What are transition metals? Give four examples.
Answer:
Transition metal is an element whose atom has an incomplete d sub shell or which can give rise to cations with an incomplete d sub shell.
They occupy the central position of the periodic table, between s-block and p-block elements.
Their properties are transitional between highly reactive metals of s-block and elements of p-block which are mostly non metals.
Example – Iron, Copper, Tungsten, Titanium.
Question 2.
Explain the oxidation states of 3d series elements.
Answer:
The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by losing electrons from (n-l)d orbital and ns orbital as the energy difference between them is very small.
At the beginning of the 3d series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.
The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases. For example, in 3d series, first element Sc has only one oxidation state +3; the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.
Mn²⁺ (3d⁵) is more stable than Mn⁴⁺ (3d³) is due to half filled stable configuration.
Question 3.
What are inner transition elements?
Answer:
1. The inner transition elements have two series of elements.
Lanthanoids
Actinoids
2. Lanthanoid series consists of 14 elements from Cerium (₅₈Ce) to Lutetium (₇₁Lu) following Lanthanum (₅₇La). These elements are characterised by the preferential filling of 4f orbitals. .
3. Actinoids consists of 14 elements from Thorium (₉₀Th) to Lawrencium (₁₀₃Lr) following Actinium (₈₉Ac). These elements are characterised by the preferential filling of 5f orbital.
Question 4.
Justify the position of lanthanides and actinides in the periodic table.
Answer:
1. In sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these 14 elements following lanthanum show similar chemical properties. Therefore these elements are grouped together and placed at the bottom of the periodic table. This position can be justified as follows.
Lanthanoids have general electronic configuration [Xe] 4f²⁻14 5d0⁻¹ 6s²
The common oxidation state of lanthanoids is +3
All these elements have similar physical and chemical properties.
2. Similarly the fourteen elements following actinium resemble in their physical and chemical properties.
3. If we place these elements after Lanthanum in the periodic table below 4d series and actinides below 5d series, the properties of the elements belongs to a group would be different and it would affect the proper structure of the periodic table.
4. Hence a separate position is provided to the inner transition elements at the bottom of the periodic table.
Question 5.
What are actinoides? Give three examples.
Answer:
The fourteen elements following actinium ,i.e., from thorium (Th) to lawrentium (Lr) are called actinoids.
All the actinoids are radioactive and most of them have short half lives.
The heavier members being extremely unstable and not of natural occurrence. They are produced synthetically by the artificial transformation of naturally occuring elements by nuclear reactions.
Example – Thorium, Uranium, Plutonium, Californium.
Question 6.
Why Gd³⁺ is colourless?
Answer:
Gd – Electronic Configuration: [Xe] 4f⁷ 5d¹ 6s²
Gd³⁺ – Electronic Configuration: [Xe] 4f⁷
In Gd³⁺, no electrons are there in outer d-orbitals. d-d transition is not possible. So it is colourless.
Question 7.
Explain why compounds of Cu²⁺ are coloured but those of Zn²⁺ are colourless.
Answer:
Cu (Z = 29) Electronic configuration is [Ar] 3d¹⁰ 4s¹
Cu²⁺: Electronic configuration is [Ar] 3d⁹.
In Cu²⁺, promotion of electrons take place in outer d-orbital by the absorption of light form visible region involves d-d transition. Due to this Cu²⁺ compounds are coloured. Where in Zn²⁺ electronic configuration is [Ar]3d¹⁰. It has completely filled d-orbital. So there is no chance of d – d transition. So Zn²⁺ compounds are colourless.
Question 8.
Describe the preparation of potassium dichromate.
Answer:
Preparation of potassium dichromate:
1. Potassium dichromate is prepared from chromite ore. The ore FeO. Cr₂O₃ is concentrated by gravity separation process.
2. The concentrated ore is mixed with excess sodium carbonate and lime and roasted in a reverbratory furnace.
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ 900-10000C-?--------- 8 Na₂CrO₄ + 2Fe₂O₃ + 8CO₂?
3. The roasted mass is treated with water to separate soluble sodium chromate from insoluble iron oxide. The yellow solution of sodium chromate is treated with concentrated sulphuric acid which converts sodium chromate into sodium dichromate.
Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements-5

4. The above solution is concentrated to remove less soluble sodium sulphate. The resulting solution is filtered and concentrated. It is cooled to get the crystals of Na₂SO₂.2H₂O.

5. The saturated solution of sodium dichromate in water is mixed with KCl and then concentrated to get crystals of NaCl. It is filtered while hot and the filtrate is cooled to obtain K₂Cr₂O₇ crystals.
Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements-6

Question 9.
What is lanthanide contraction and what are the effects of lanthanide contraction?
Answer:
As we move across 4f series, the atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. This decrease in ionic size is called lanthanoid contraction.
Effects (or) Consequences of lanthanoid contraction:
1. Basicity differences: As we move from Ce³⁺ to Lu³⁺ , the basic character of Ln³⁺ ions decrease. Due to the decrease in the size of Ln³⁺ ions, the ionic character of Ln – OH bond decreases (covalent character increases) which results in the decrease in the basicity.

2. Similarities among lanthanoids – In the complete f-series only 10 pm decrease in atomic radii and 20 pm decrease in ionic radii is observed. Because of this very small change in radii of lanthanoids, their chemical properties are quite similar.

The elements of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction. For example,

4d series – Zr – Atomic radius 145 pm
5d series – Hf – Atomic radius 144 pm
Question 10.
Complete the following
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements-7

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Question 11.
What are interstitial compounds?
Answer:

An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice.
They are usually non-stoichiometric compounds.
Transition metals form a number of interstitial compounds such as TiC, ZrH₁.₉₂, Mn₄N etc.
The elements that occupy the metal lattice provide them new properties.
They are hard and show electrical and thermal conductivity
They have high melting points higher than those of pure metals
Transition metal hydrides are used as powerful reducing agents
Metallic carbides are chemically inert.
Question 12.
Calculate the number of unpaired electrons in Ti³⁺, Mn²⁺ and calculate the spin only magnetic moment.
Answer:
Ti³⁺:
Ti (Z = 22). Electronic configuration [Ar] 3d² 4s²
Ti³⁺ – Electronic configuration [Ar] 3d¹
So, the number of unpaired electrons in Ti³⁺ is equal to 1.

Spin only magnetic moment of Ti³⁺ = 1(1+2)-------v = 3–v = 1.73µB

Mn²⁺:
Mn (Z = 25). Electronic configuration [Ar] 3d⁵ 4s²
Mn²⁺ – Electronic configuration [Ar] 3d⁵
Mn²⁺ has 5 unpaired electrons.
Spin only magnetic moment of Mn²⁺ = 5(5+2)-------v = 35--v = 5.91µB

Question 13.
Write the electronic configuration of Ce⁴⁺ and CO²⁺.
Answer:
Ce (Z = 58) ? Ce⁴+4e⁻
Ce⁴⁺ – Is² – 2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶ 5s²4d¹⁰ 5p⁶
CO²⁺ – Is²2s²2p⁶3s²3p⁶4s²3d⁵.

Question 14.
Explain briefly how +2 states becomes more and more stable in the first half of the first row transition elements with increasing atomic number.
First transition series.
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements-8

Question 15.
Which is more stable? Fe3+³⁺ or Fe²⁺ – explain.
Answer:
Fe (Z = 26)
Fe ? Fe²⁺ + 2e⁻
Fe ? Fe³⁺ + 3e⁻
Fe²⁺ [Number of electrons 24]
Electronic configuration = [Ar]3d⁶
Fe³⁺ [Number of electrons 23]
Electronic configuration = [Ar]3d⁵
Among Fe³⁺ and Fe²⁺, Fe³⁺ is more stable due to half filled d-orbital. This can be explained by Aufbau principle. Half filled and completely filled d-orbitals are more stable than partially filled d-orbitals. So Fe³⁺ is more stable than Fe²⁺.

Question 16.
Explain the variation in E⁰M²⁺/M³⁺3d series.
Answer:
1. In transition series, as we move down from Ti to Zn, the standard reduction potential E⁰M²⁺/M³ value is approaching towards less negative value and copper has a positive reduction potential, i.e. elemental copper is more stable than Cu²⁺.

2. E⁰M²⁺/M value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d⁵ configuration in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺.

3. The standard electrode potential for the M³⁺ / M²⁺ half cell gives the relative stability between M³⁺ and M²⁺.

4. The high reduction potential of Mn³⁺ / Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺.

5. For Fe³⁺ / Fe²⁺ the reduction potential is 0.77 V, and this low value indicates that both Fe³⁺ and Fe²⁺ can exist under normal condition.

6. Mn³⁺ has a 3d² configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible [E° = +1.51 V]

Question 17.
Compare lanthanides and actinides.
Answer:
Lanthanoids:

Differentiating electron enters in 4f orbital
Binding energy of 4f orbitals are higher
They show less tendency to form complexes
Most of the lanthanoids are colourless
They do not form oxo cations
Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases.
Actinoids:

Differentiating electron enters in 5f orbital
Binding energy of 5f orbitals are lower
They show greater tendency to form complexes
ost of the actinoids are coloured.
E.g : U³⁺ (red), U⁴⁺ (green), UO²⁺₂(yellow)
They do form oxo cations such as UO²⁺₂ NpO²⁺₂ etc.
Besides +3 oxidation states actinoids show higher oxidation states such as +4, +5, +6 and +7

Question 18.
Explain why Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidizing.
Answer:
Cr²⁺ is strong reducing while Mn³⁺ is strongly oxidising.
E⁰Cr³⁺/Cr²⁺ is -0.41 V
Cr²⁺ + 2e⁻ → Cr E° = – 0.91V.
If the standard electrode potential E° of a metal is large and negative, the metal is a powerful reducing agent because it loses electrons easily.
Mn ? Mn³⁺ + 3e⁻
Mn³⁺ + e⁻ → Mn²⁺

Mn³⁺ [Ar]3d⁴
E° = + 1.51 V
If the standard electrode potential E° of a metal is large and positive, the metal is a powerful oxidising agent because it gains electrons easily.

Question 19.
Compare the ionization enthalpies of first series of the transition elements. Ionization enthalpies of first transition series:
Answer:

Ionization energy of transition element is intermediate between those of s and p block elements.
As we move from left to right in a transition metal series, the ionization enthalpy increases as expected. This is due to increase in nuclear charge corresponding to the filling of d electrons.
The increase in first ionisation enthalpy with increase in atomic number along a particular series is not regular. The added electron enters (n-l)d orbital and the inner electrons act as a shield and decrease the effect of nuclear charge on valence ‘ns’ electrons. Therefore, it leads to variation in the ionization energy values.
Question 20.
Actinoid contraction is greater from element to element than the lanthanoid contraction, why?
Answer:
Actinoid contraction is greater from element to element than lanthanoid contraction. The 5f orbitals in Actinoids have a very poorer shielding effect than 4f orbitals in lanthanoids.
Thus, the effective nuclear charge experienced by electron in valence shells in case of actinoids is much more than that experienced by lanthanoids.
In actinoids, electrons are shielded by 5d, 4f, 4d and 3d whereas in lanthanoids, electrons are shielded by 4d, 4f only.
Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.
SamacheerKalvi.Guru

Question 21.
Out of LU(OH)³ and La(OH)³ which is more basic and why?
Answer:

As we move from Ce³⁺ to Lu³⁺, the basic character of Lu³⁺ ions decreases.
Due to the decrease in the size of Lu³⁺ ions, the ionic character of Lu – OH bond decreases, covalent character increases which results in the decrease in the basicity.
Hence, La(OH)³ is more basic than Lu(OH)³.
Question 22.
Why europium (II) is more stable than Cerium (II)?
Answer:
Eu (Z = 63) – Electronic configuration – [Xe] 4f⁷ 5d° 6s²
Eu²⁺ – Electronic configuration Electronic 6s¹
Ce (Z = 58) – configuration – [Xe] 4f²⁺ 5d° 6s²⁺
Ce²⁺ – Electronic confluration – [Xe] 4f² 5d°
According to Aufbau principle, half filled and completely filled d (or) f orbitals are more stable than partially filled f orbitals.
Hence Eu²⁺ [Xe] 4f⁷ 5d° is more stable than Ce²⁺ [Xe] 4f² 5d°

Question 23.
Why do zirconium and Hafnium exhibit similar properties?
Answer:

The element of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction.
e.g., Zr – 4d series -Atomic radius 145 pm Hf – 5d series – Atomic radius 144 pm
The radii are very similar even though the number of electrons increases.
Zr and Hf have very similar chemical behaviour, having closely similar radii and electronic configuration.
Radius dependent properties such as lattice energy, solvation energy are similar.
Thus lanthanides contraction leads to formation of pair of elements and those known as chemical twins, e.g., Zr – Hf
SamacheerKalvi.Guru

Question 24.
Which is stronger reducing agent Cr²⁺ or Fe²⁺ ?
Answer:
Cr²⁺ and Fe²⁺
Cr (Z = 24) – Electronic configuration – [Ar] 3d⁵ 4s¹
Cr2+ Electronic configuration – [Ar] 3d⁴ 4s⁰
Fe (Z = 26) – Electronic configuration – [Ar] 3d⁶ 4s²
Fe²⁺ Electronic confimiration – [Ar] 3d⁶ 4s⁰

If standard electrode potential (E°) of a metal is large and negative, the metal is a powerful reducing agent

Cr²⁺ + 2e– → Cr
Fe²⁺+2e⁻ → Fe
E° = – 0.91V
E°= – 0.44V

By comparing the above equation, Cr²⁺ is a powerful reducing agent.

Question 25.
The E⁰M²⁺/M value for copper is positive. Suggest a possible reason for this.
Answer:
Copper has a positive reduction potential. Elemental copper is more stable than Cu2+.
Copper having positive sign for electrode potential merely means that copper can undergo reduction at faster rate than reduction of hydrogen.
The electron giving reaction (oxidation) of copper is slower than that of hydrogen. It is determined from the result of S.H.E (Standard Hydrogen Electrode) potential value experiment.
Question 26.
Predict which of the following will be coloured in aqueous solution Ti²⁺ , V³⁺, Sc⁴⁺, Cu⁺, SC³⁺, Fe³⁺, Ni²⁺ and CO³⁺

Answer:
Among Ti²⁺ , V³⁺, Sc⁴⁺, Cu⁺, Sc³⁺, Fe³⁺, Ni²⁺ and CO³⁺ in aqueous solution state.
Ti (Z = 22) – Ti²⁺ – Electronic configuration is [Ar] 3d²
V (Z = 23) – V³⁺ – Electronic configuration is [Ar] 3d²
Sc (Z = 21) – SC⁴⁺ – Electronic configuration is [Ar] 1s² 2s² 2p⁶ 3s² 3p⁵
Cu (Z = 29) – Cu⁺ – Electronic configuration is [Ar] 3d¹⁰
Sc (Z = 21) – SC³⁺ – Electronic configuration is [Ar] 3d°4s°
Fe (Z = 26) – Fe³⁺ – Electronic configuration is [Ar] 3d⁵
Ni (Z = 28) – Ni²⁺ – Electronic configuration is [Ar] 3d⁵
CO (Z = 27) – CO³⁺ – Electronic configuration is [Ar] 3d⁶
A transition metal ion is coloured if it has one or more unpaired electrons in (n-l)d orbital, i.e. 3d orbitals in the case of first transition series, when such species are exposed to visible radiation, d – d transition take place and the species are coloured.
Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements-9

Question 27.
Describe the variable oxidation state of 3d series elements.
Answer:
1. The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by loosing electrons from (n-l)d orbital and ns orbital as the energy difference between them is very small.

2. At the beginning of the 3d series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.

3. The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases. For example, in the 3d series, first element Sc has only one oxidation state +3 the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.

4. Mn²⁺ (3d⁵) is more stable than Mn⁴⁺ (3d³) is due to half filled stable configuration.

Question 28.
Which metal in the 3d series exhibits +1 oxidation state most frequently and why?
Answer:
1. The first transition metal copper exhibits only +1 oxidation state.

2. It is unique in 3d series having a stable +1 oxidation state.
Cu (Z = 29) Electronic configuration is [Ar] 3d¹⁰ 4s²

3. So copper element only can have +1 oxidation state.

Question 29.
Why first ionization enthalpy of chromium is lower than that of zinc?
Answer:
The first ionization enthalpy of chromium is lower than that of zinc. Cr (Z = 24) Electronic configuration [Ar] 3d⁵ 4s¹. In the case of Cr, first electron has to be removed easily from 4s orbital to attain the more stable half filled configuration. So Cr has lower ionization enthalpy. But in the case of Zinc (Z = 30), electronic configuration [Ar] 3d¹⁰ 4s². The first electron has to be removed from the most stable fully filled electronic configuration becomes difficult and it requires more energy.

Question 30.
Transition metals show high melting points why?
Answer:

All the transition metals are hard.
Most of them are hexagonal close packed, cubic close packed (or) body centered cubic which are characteristics of true metals.
The maximum melting point at about the middle of transition metal series indicates that d5 configuration of favourable for strong inter atomic attraction.
Due to the strong metallic bonds, atoms of the transition elements are closely packed and held together. This leads to high melting point and boiling point.

Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements Evaluate yourself

Question 1.
Compare the stability of Ni⁴⁺ and Pt⁴⁺ from their ionisation enthalpy values
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements-10
Pt⁴⁺ compounds are stable than Ni⁴⁺ compounds because the energy needed to remove 4 electrons in Pt is less than that of Ni.

Question 2.
Why iron is more stable in +3 oxidation state than in +2 and the reverse is true for Manganese?
Answer:
Fe (Z = 26). Electronic configuration [Ar] 3d⁶ 4s²
Fe → Fe³⁺ + 3e⁻
Fe3+ Electronic configuration p° [Ar]3d³.
If’d’ orbital is half filled, it is more stable than . Fe²⁺ where it is [Ar]3d⁶.
Mn (Z = 25). Electronic configuration [Ar]3d⁵ 4s²
Mn ? Mn²⁺ + 2e– . By the loss of 2e⁻, Mn²⁺ is more stable due to half filled configuration. Mn → Mn³⁺ + 3e⁻.
Mn³⁺ Electronic configuration [Ar]3d⁴ 4s°.
Among this Fe³⁺ is more stable than Fe²⁺ and the Mn²⁺ is more stable than Mn³⁺.

Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements Additional Questions
Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements 1 Mark Questions and Answers

I. Choose the correct answer.

Question 1.
The elements whose atom has incomplete d sub-shell are called …………..
(a) s-block element
(b) Alkali metals
(c) transition elements
(d) Representative elements
Answer:
(c) transition elements

2. Which one of the following is the other name of d-block elements?
(a) Chalcogens
(b) Halogens
(c) Inner-transition elements
(d) Transition elements
Answer:
(d) Transition elements

Question 3.
Which metals play an important role in the development of human civilization?
(a) Al and Mg
(b) Na and K
(c) Fe and Cu
(d) Mn and Ni
Answer:
(c) Fe and Cu

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Question 4.
Which transition element is used in light bulb filaments?
(a) Al
(b) Ni
(c) W
(d) Fe
Answer:
(c) W

Question 5.
Which metal is used in manufacturing artificial joints?
(a) Molybdenum
(6) Titanium
(c) Tungsten
(d) Iron
Answer:
(b) Titanium

Question 6.
Which transition metal is applied in the manufacturing of boiler plants?
(a) Iron
(b) Copper
(c) Aluminium
(d) Molybdenum
Answer:
(d) Molybdenum

Question 7.
Identify the transition metal present in Hemoglobin …………..
(a) Cobalt
(b) Iron
(c) Manganese
(d) Copper
Answer:
(b) Iron

Question 8.
Which of the following transition metal is present in Vitamin B₁₂?
(a) Cobalt
(b) Platinum
(c) Copper
(d) Iron
Answer:
(a) Cobalt

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Question 9.
The metal cobalt is present in
(a) Vitamin-A
(b) Vitamin-B
(c) Vitamin-B₁₂
(d) Vitamin-B₆
Answer:
(c) Vitamin-B₁₂

Question 10.
Consider the following statements.
Answer:
(i) Transition metals occupy group-3 to group-12 of the modem periodic table.
(ii) Representative elements occupy group-3 to group-12 of the modem periodic table.
(iii) Except group-11 elements of all transition metals are hard.
(iv) d-block elements are mostly non-metals.

Question 11.
Consider the following statements.
Answer:
(i) d-block elements composed of 3d series, Sc to Zn (4th period).
(ii) 4d series composed of Y to Cd.
(iii) 5d series composed of La, Hf to Mercury.
(iv) d-block elements composed of 4d series Y to Cd.

Which of the above statements is/are incorrect
(a) (i) and (iv)
(b) (i), (ii) and (iii)
(c) (iii) and (iv)
(d) (iv) only
Ans.
(b) (i), (ii) and (iii)

Question 12.
Which of the following is the correct electronic configuration of Sc (Z = 21)?
(a) [Ar] 3d³
(b) [Ar] 3d’ 4s²
(c) [Ar] 3d² 4s¹
(d) [Ar] 4s² 4p’
Answer:
(b) [Ar] 3d’ 4s²

Question 13.
The correct electronic configuration of Cr is …………..
(a) [Ar] 3d⁴ 4s²
(b) [Ar] 3d⁵
(c) [Ar] 3d⁵ 4s¹
(d) [Ar] 3d⁶
Answer:
(c) [Ar] 3d⁵ 4s¹

Question 14.
Which of the following is the correct electronic configuration of copper?
(a) [Ar] 3d⁵ 4s¹
(b) [Ar] 3d¹⁰ 4s¹
(c) [Ar] 3d⁹ 4s²
(d) [Ar] 3d⁸ 4s² 4p¹
Answer:
(b) [Ar] 3d¹⁰ 4s¹
Question 15.
Which one of the following is the general electronic configuration of transition elements?
(a) [Noble gas] ns² np⁶
(b) [Noble gas] ( n – 2 ) f1-¹⁻¹⁴(n-l)d¹⁻¹⁰ ns²
(c) [Noble gas] ( n – 1 ) d¹⁻¹⁰ (n-l)f¹⁻¹⁴ ns²
(d) [Noble gas] ( n – 1 ) d¹⁻¹⁰ ns²
Answer:
(d) [Noble gas] ( n – 1 ) d¹⁻¹⁰ ns²

Question 16.
Which of the following d-block elements has the highest electrical conductivity at room temperature?
(a) Copper
(b) Silver
(c) Aluminium
(d) Tungsten
Answer:
(b) Silver

Question 17.
Consider the following statements.
(i) The melting point decreases from Scandium to Vanadium in 3d series.
(ii) In 3d transition series, atomic radius decreases from Sc to V and upto copper atomic radius nearly remains the same.
(iii) As we move down in 3d transition series, atomic radius increases.

Which of the above statements is/are incorrect?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(a) (i) only

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Question 18.
Which of the following transition element exhibit only +3 oxidation state?
(a) Cu
(b) Sc
(c) Mn
(d) Cr
Answer:
(b) Sc

Question 19.
Which one of the following transition element has maximum oxidation states?
(a) Manganese
(b) Copper
(c) Scandium
(d) Titanium
Answer:
(a) Manganese

Question 20.
Consider the following statements.
(i) In 3d series, the middle element Mn has +2 to +7 oxidation states.
(ii) The oxidation state of Ru and Os is +8.
(iii) Scandium has six different oxidation states.

Which of the above statements is/are not correct?
(a) (i) and (ii)
(b) (ii) only
(c) (i) only
(d) (iii) only
Answer:
(d) (iii) only

Question 21.
Which one of the following elements show high positive electrode potential?
(a) Ti⁺
(b) Mn²⁺
(c) CO²⁺
(d) Cr²⁺
Ans.wer:
(c) CO²⁺

Question 22.
Which one of the following elements show high negative electrode potential?
(a) Copper
(b) Manganese
(c) Cobalt
(d) Zinc
Answer:
(d) Zinc

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Question 23.
Which one of the following is diamagnetic in nature?
(a) Ti³⁺
(b) Cu²⁺
(c) Zn²⁺
(d) V³⁺
Answer:
(c) Zn²⁺

Question 24.
Which one of the following is paramagnetic in nature?
(a) Sc³⁺
(b) Ti⁴⁺
(c) V⁵⁺
(d) Cu²⁺
Answer:
(d) Cu²⁺

Question 25.
Which of the following pair has maximum number of unpaired electrons?
(a) Mn²⁺, Fe³⁺
(b) CO³⁺, Fe²⁺
(c) Cr³⁺, Mn⁴⁺
(d) Ti²⁺, V³⁺
Answer:
(a) Mn²⁺, Fe³⁺

Question 26.
Which of the following pair has d10 electrons?
(a) Ti³⁺, V⁴⁺
(b) CO³⁺, Fe²⁺
(c) Cu⁺ , Zn²⁺
(d) Mn²⁺, Fe³⁺
Answer:
(c) Cu⁺, Zn²⁺

Question 27.
Which of the following is used as a catalyst in the manufacture of sulphuric acid form SO₃.
(a) V₃O₅
(b) Rh-Ir
(c) Ni
(d) Fe
Answer:
(a) V₃O₅

Question 28.
Which one of the following is Zeigler – Natta catalyst?
(a) CO₂(CO)₈
(b) Rh/Ir complex
(c) TiCl₄ + Al(C₂H₅)₃
(d) Fe / Mo
Answer:
(c) TiCl₄ + Al(C₂H₅)₃

Question 29.
Which one of the following is used as a catalyst in the polymerisation of propylene?
(a) V₂O₅
(b) Pt
(c) TiCl₄ + Al(C₂H₅)₃
(d) Fe / Mo
Answer:
(c)TiCl₄ + Al(C₂H₅)₃

Question 30.
Consider the following statements.
Answer:
(i) Transition metal hydrides are used as powerful oxidising agents.
(ii) Metallic carbides are chemically active.
(iii) Interstitial compounds are hard and show electrical and thermal conductivity.

Which of the above statements is/are incorrect?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) only
(d) (i) only
Answer:
(a) (i) and (ii)

Question 31.
Which one of the following oxide is covalent?
(a) Cr₂O₃
(b) CrO
(c) Mn₂O₇
(d) Na₂O
Answer:
(c) Mn₂O₇

Question 32.
Which one of the following oxide is amphoteric in nature?
(a) CrO
(b) Cr₂O₃
(c) Mn₂O₇
(d) MnO
Answer:
(b) Cr₂O₃

Question 33.
The oxidation state of Chromium in CrO²⁻₄ and in Cr₂O²⁻₇ are …………..
(a) +3, +6
(b) +7, +4
(c) +6, +6
(d) +4, +6
Answer:
(c) +6, +6

Question 34.
Which one of the following is used to identify chloride ion in inorganic qualitative analysis?
(a) Barium chloride test
(b) Chromyl chloride test
(c) Brown ring test
(d) Ammonium molybdate test
Answer:
(b) Chromyl chloride test

II. Fill in the blanks.

Transition elements occupy the central position of the periodic table between …………. elements.
Except …………. elements, all transition metals are hard and have very high melting point.
The metal present in Vitamin ⁻B₁₂ is ………….
…………. metal is used in manufacture of artificial joints.
The extra stability of Cr and Cu is due to …………. of electrons and exchange energy.
Of all the known elements …………. has the highest electrical conductivity at room temperature.
The maximum melting point at about the middle of transition metal series indicates that …………. configuration is favourable for strong attraction.
The atomic radius of 5d elements and 4d elements are nearly same due to ………….
Ni (II) compounds are thermodynamically …………. than Pt (II) compounds.
The first transition metal …………. exhibits only +3 oxidation state.
The middle transition element …………. has six different oxidation states.
The oxidation state of Ru and Os is ………….
…………. is unique in 3d series having a stable +1 oxidation state.
The substance which is oxidised is a …………. agent and the one which is reduced is an …………. agent.
The oxidising and reducing power of an element is measured in terms of ………….
If the E° of a metal is large and negative, the metal is a ………….
The species with all paired electrons exhibit ………….
The magnetic moment of an ion is given by ………….
Many industrial processes use …………. or their as catalyst.
In the catalytic hydrogenation of an alkene …………. is used as catalyst.
In the preparation of acetic acid from acetaldehyde the catalyst used in ………….
The catalyst used in the hydroformylation of olefins is ………….
…………. catalyst is used in polymerization of propylene.
Metallic carbides are chemically ………….
Except Scandium all other 3d series transition elements form …………. metal oxides.
Cr₂O₃…………. is and CrO is …………. in nature.
Mn₂O₇ dissolves in water to give ………….
On heating potassium dichromate, it decomposes to give …………. and molecular oxygen.
Potassium dichromate is a powerful …………. agent in acidic medium.
………….is used in leather tanneries for chrome tanning.
Potassium dichromate is used in quantitative analysis for the estimation of …………. and ………….
Permanganate ion has …………. geometry in which Mn⁺⁷ is …………. hybridised.
Cold dilute alkaline KMnO₄ is known as ………….

…………. is used for the treatment of skin infections and fungal infections of the foot.
Baeyer’s reagent is used for detecting …………. in an organic compounds.
The equivalent weight of KMnO₄ in neutral medium is ………….
The common oxidation state of lanthanoids is ………….
Due to the decrease in the size of Ln³⁺ ions, the ionic character of Ln – OH bond decreases which results in the ………….
All the actinoids are …………. and most of them have …………. half lives.
………….do not form oxo cations.
Answer:
sandpblock
group-II
cobalt
Titanium
symmetrical distribution
silver
d⁵, inter atomic
lanthanoid contraction
more stable
Scandium
Manganese
+ 8
Copper
reducing, oxidising
Standard electrode potential
powerful reducing agent
diamagnetism
µ=gS(S+1)-------vµB
transition metals, compounds
Nickel
Rh/Ir complex
CO₂(CO)₈
Zeigler – Natta (or) TiCl₄ + Al(C₂H₅)₃
inert
ionic
amphoteric, basic
permanganic acid (HMnO₄)
Chromium (III) oxide – Cr₂O₃
Potassium dichrom.ate
iron compounds, iodides
tetrahedral,sp3
Baeyer’s reagent
Potassium permanganate
unsaturation
52.67
+ 3
decrease in the basicity
radioactive. shoit
Lanthanoids

Kalvikavi

10,11,12th Revision Question 2025