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# Maharashtra Board Class 12 Solutions for Chemistry Chapter 8 Transition and Inner Transition Elements - PDF

## Maharashtra Board Class 12 Solutions for Chemistry Chapter 8 Transition and Inner Transition Elements - Download Free PDF with Solution

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### Introduction

Students who are in their class 12 and practising the chapter 8 chapter named Transition and Inner Transition Elements, they can choose the exercise solution available online. The transition and inner transition elements exercise created by chapter 8 offers clear understanding about what the chapter tells us. Students when they are preparing for their class 12 board exam, they find it difficult to handle the pressure and unable to practise well. This is when the Chapter 8 Chemistry Class 12 Exercise Solutions come to the role. Prepared by Vedantu subject experts, students have the option of Transition and Inner Transition Elements Class 12 Notes PDF download. The exercise note provides a complete solution with an easy explanation about what the chapter is all about.

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## Access Maharashtra Board Solution for Class 12 Chemistry Chapter 8 Transition and Inner Transition Elements

### 1. Choose the Most Correct Option.

i. Which one of the following is diamagnetic

1. Cr3⊕

2. Fe3⊕

3. Cu2⊕

4. Sc3⊕

Ans: The correct option is d. Sc3⊕.

It has diamagnetic properties. Diamagnetism is a magnetic field produced by small current loops created by orbital electron mobility. Current loops align with and fight against an imposed external magnetic field. In diamagnetism, the electrons exist in pairs. Magnetic fields repel diamagnetic substances.

ii. Most stable oxidation state of Titanium is

1. +2

2. +3

3. +4

4. +5

Ans: The correct option is c. +4.

Due to its ability to form an octet, the +4 state is the most prevalent and stable. Ti(+3) is oxidised to lose the final electron and transform into Ti (+4), making it a useful reducing agent. The second ionisation energy is the foundation for it.

iii. Components of Nichrome alloy are

1. Ni, Cr, Fe

2. Ni, Cr, Fe, C

3. Ni, Cr

4. Cu, Fe

Ans: The correct option is a. Ni, Cr, Fe.

It mostly consists of 80% Ni, 20% Cr, and a small amount of Fe. Nichrome is a well-liked material for use in heating elements because of its resistance to oxidation and the percentage composition of the elements' constituents. For instance, thick nichrome wire is typically used for heating components in home toasters.

iv. Most stable oxidation state of Ruthenium is

1. +2

2. +4

3. +8

4. +6

Ans: The correct option is b. +4

The region of the big (IV) oxidation state where, once more, ligands based on nitrogen and oxygen predominate. For the +4 oxidising agent, the chemistry, spectroscopy, and structures of these complexes are all briefly reviewed.

v. Stable oxidation states for chromium are

1. +2, +3

2. +3, +4

3. +4, +5

4. +3, +6

Ans: The correct option is d. +3, +6.

Octahedral complexes are a common type of chromium (III). The dark green complex is chromium (III) chloride hydrate, which is marketed. The violet and light green chemicals are closely related. Consequently, they have a comparable radius (63 pm)

vi. Electronic configuration of Cu and Cu+1

1. 3d10, 4s0; 3d9, 4s0

2. 3d9, 4s1; 3d94s0

3. 3d10, 4s1; 3d10, 4s0

4. 3d8, 4s1; 3d10, 4s0

Ans: The correct option is c. 3d10, 4s1; 3d10, 4s0.

Subshells that are partially and completely filled have more stability. Consequently, an electron from 4s2 hops to 3d9. The configuration that results from this is as follows:

1s2, 2s2,2p63s2, 3p6, 3d10,4s1

We take one electron out of 4s1 to make room for the Cu+ ion, leaving the following electrical configuration.

1s2, 2s2,2p63s2, 3p6, 3d10

vii. Which of the following have d0s0 configuration

1. Sc3⊕

2. Ti4⊕

3. V5⊕

4. all of the above

Ans: The correct option is d. all of the above.

None of the aforementioned choices are configured as d0s0.

viii. Magnetic moment of a metal complex is 5.9 B.M. Number of unpaired electrons in the complex is

1. 2

2. 3

3. 4

4. 5

Ans: The correct option is d. 5.

The value of m will change in direct proportion to the number of unpaired electrons because the magnetic moment is directly proportional to this quantity. Thus, 5.9 can be rounded up to 5, and there will be 5 unpaired electrons.

ix. In which of the following series all the elements are radioactive in nature

1. Lanthanide

2. Actinoids

3. d-block elements

4. s-block elements

Ans: The correct option is b. Actinoids.

The actinides' radioactivity is essentially why they are valuable. These substances can be employed as energy sources for a variety of purposes, including the creation of electrical energy for equipment on the moon and cardiac pacemakers. Nuclear weapons and nuclear power plants have both used uranium and plutonium in their construction.

x. Which of the following sets of ions contain only paramagnetic ions

1. Sm3⊕, Ho3⊕, Lu3⊕

2. La3⊕, Ce3⊕, Sm3⊕

3. La3⊕, Eu3⊕, Gd3⊕

4. Ce3⊕, Eu3⊕, Yb3⊕

Ans: The correct option is d. Ce3⊕, Eu3⊕, Yb3⊕.

All of the elements in this choice are paramagnetic because they all have unpaired electrons in their orbitals.

xi. Which actinide, other than uranium, occurs in a significant amount naturally?

1. Thorium

2. Actinium

3. Protactinium

4. Plutonium

Ans: The correct option is a. Thorium.

Thorium is a metallic, once-weekly radioactive chemical element with the atomic number 90 and the symbol Th. When exposed to air, thorium tarnishes to a dark grey colour and produces thorium dioxide. When finely divided, it is extremely reactive and can catch fire in the air. Thorium is water resistant and keeps its lustre for a while.

xii. The flux added during extraction of Iron from hematite are its?

1. Silica

2. Calcium carbonate

3. Sodium carbonate

4. Alumina

Ans: The correct option is b. Calcium carbonate,

In a blast furnace, iron is produced from its ore, haematite. When ore is fed into the furnace from the top, calcium carbonate and coke serve as flux. In order to eliminate impurities like silicon dioxide, calcium carbonate transforms into calcium oxide, which then combines with the impurities to generate slag. This facilitates the removal of pure iron.

i. What is the oxidation state of Manganese in (i) MnO42– (ii) MnO4

Ans: According to the question, manganese is in the following oxidation state:

(i) MnO42– = +6

(ii) MnO4 = +7

ii. Give uses of KMnO4.

Ans: The applications of KMnO4 are as follows:

• For the removal of iron and hydrogen sulphide from well water, potassium permanganate is also utilised as a regeneration chemical.

• Additionally used as a disinfectant, this substance treats dermatitis and certain fungal infections of the feet.

• Potassium permanganate has a boiling point of 100°C.

• Potassium permanganate has a melting point of 200°C.

• Potassium permanganate has a density of 2.703g/cm3.

iii. Why salts of Sc3⊕, Ti4⊕, V5⊕ are colourless?

Ans: The explanation for this is that a lot of transition metal complexes have electrodes in the d orbital that can be easily stimulated to higher energy levels by absorbing a lot of visible light. As a result, transition metal compounds have an apparent colour. Sc3+, Ti4+, and V5+ compounds are colourless because of this.

iv. Which steps are involved in the manufacture of potassium dichromate from chromite ore?

Ans: (1) Chromite ore concentration is the first step in producing potassium dichromate from chromite ore.

(2) Making sodium chromate from chromite ore (Na2CrO4).

(3) The creation of sodium dichromate (Na2Cr2O7) from Na2CrO4.

(4) Na2Cr2O7 is transformed into K2Cr2O7.

v. Balance the following equation

1. KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + H2O + O2

2. K2Cr2O7 + KI + H2SO4 → K2SO4 + Cr2(SO4)3 + 7H2O + 3I2

Ans: The balanced nature of the responses is demonstrated as follows:

a. 2KMnO4 + 3H2SO4 + 5H2C2O4 → K2SO4 + 2MnSO4 + 8H2O + 10CO2

b. K2Cr2O7 + 6KI + 7H2SO4 → 4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2

vi. What are the stable oxidation states of plutonium, cerium, manganese, Europium?

Ans: The following are the stable oxidation states of the posed problem:

Plutonium = + 3 to + 7

Cerium = + 3, + 4

Manganese = + 2, + 4, + 6, + 7

Europium = +2, +3

vii. Write the electronic configuration of chromium and copper.

Ans: Chromium only possesses 4 electrons, leaving it needing 5 to be only partially full. Copper gains one electron from the d- orbital to achieve an entirely filled electrical state. Similar to this, Cu intends to create a more stable, fully filled subshell in 3D. a 3d orbital has less energy than 4s. Cr and Cu both have remarkable electrical configurations as a result.

The electronic configuration of chromium is $[Ar]$4s13d5. The electrical configuration of copper is $[Ar]$4s13d10.

viii. Why is nobelium the only actinoid with +2 oxidation state?

Ans: Nobelium is a man-made chemical element with the atomic number 102 and the symbol No. Both the +3 and +3 oxidation states are possible for it. The electrical configuration of nobelium is $[Rn]$5f147s2.

It loses two electrons and then changes to No2+, which results in a stable structure with a fully filled f-subshell. No other actinoid can reach the orbital configuration with all of its atoms filled.

ix. Explain with the help of a balanced chemical equation, why the solution of Ce(IV) is prepared in acidic medium.

Ans: The lanthanide that is most stable in the +4 oxidation state is cerium. A popular one-electron oxidant is a cerium (IV). Since the hydrated ion {Ce(H2O)n}4+ is rapidly hydrolyzed and loses its oxidising power, cerium (IV) is a relatively powerful Lewis acid. As a result, the {Ce(H2O)n}4+ ion can only be found in concentrated perchloric acid solutions.

x. What is meant by ‘shielding of electrons’ in an atom?

Ans: The outermost electron in an atom is screened or shielded from the nuclear attraction by the inner shell electrons. The shielding effect is the name of this phenomenon. The quantity of inner electrons affects how much shielding is produced.

xi. The atomic number of an element is 90. Is this element diamagnetic or paramagnetic?

Ans: The closest member of the 18th group in this situation is Rn, which belongs to the 6th period and has a fully filled p orbital. It has an electronic arrangement $[Rn]$6d27s2. In the d-subshell, there are two unpaired electrons. The fact that there are two unpaired electrons makes thorium a paramagnetic element.

i. Explain the trends in atomic radii of d-block elements

Ans: The elements in the transition series have decreasing atomic radii from left to right. The nuclear charge rises from left to right as we move. The penultimate (n-1) d subshell is where the final filled electron enters. However, because d orbitals in an atom are more diffused, they have a smaller screening effect than s orbitals. As a result, effective nuclear charge also rises in a transition series as the atomic number does. Consequently, from Cr to Cu, the particle size does not significantly alter.

Moving lower in the group causes the nuclear radius to grow. The group VIII elements in a given series have the lowest nuclear radius values, and from there, it rises toward the end of the series. This is brought on by the additional electrons' strong electrostatic repulsion. Due to the filling of 4f subshells, the atomic radii of the second and third transition elements are quite close.

ii. Name different zones in the Blast furnace. Write the reactions taking place in them.

Ans: The blast furnace has different temperature zones. The furnace's temperature keeps rising from top to bottom.

• Zone of combustion: Coke is oxidised by hot air to produce CO, which is an exothermic reaction that raises the furnace's temperature. CO partially splits into finely divided carbon and oxygen. In the furnace, heated gases containing CO rise to the top and heat the charge as it descends. CO serves as both a fuel and a reducing agent.

• Zone of reduction: CO converts Fe2O3 to spongy or porous iron at a temperature of roughly 900°C. Fe2O3 is also partially reduced by carbon to Fe.

• Zone of slag formation: At 1200 K, the charge of limestone, CaCO3, decomposes to produce a basic flux, CaO, which reacts with gangue (SiO2, Al2O3) at 1500 K to produce a slag of CaSiO3 and Ca3AlO3. Through an outlet, the slag is taken out of the furnace's bottom.

• Zone of fusion: While SiO2 is transformed to Si, impurities like MnO2 and Ca3(PO4)2 in the ore are reduced to Mn and P. As it descends through the furnace, the spongy iron melts in the fusion zone, dissolving impurities like C, Si, Mn, phosphorus, and sulphur. At the furnace's base, molten iron gathers. The lighter slag keeps the molten iron from oxidising by floating on top of it.

iii. What are the differences between cast iron, wrought iron and steel?

Ans: The differences between cast iron, wrought iron and steel is as follows-

 Cast Iron Wrought Iron Steel Is rough and fragile. Very gentle It's not too stiff or too soft. Includes 4% carbon. Less than 0.2% carbon is present. 0.2 to 2% carbon content. Is used in the production of pipes, vehicle parts, pots, pans, and kitchenware. Is used to create pipes, stay bolt bar stock, engine bolts, and rivets. Is used in tools, ships, automobiles, guns, buildings, and other structures.

iv. Iron exhibits +2 and +3 oxidation states. Write their electronic configuration. Which will be more stable? Why?

Ans: Fe2+ has the following electronic configuration: 1s22s22p63s23p63d6. Fe3+ has the following electrical configuration: 1s22s22p63s23p63d5. Iron reaches a 3+ oxidation state as a result of losing one electron from the 3d-orbital. Fe3+ is more stable than Fe2+ because the 3d-orbital in that compound is half-filled, which adds additional stability.

v. Give the similarities and differences in elements of 3d, 4d and 5d series.

Ans: Similarities:

• The metals in the d block are all electropositive.

• They have varied valencies and produce colourful complexes and salts.

• They are good reducing agents.

• The majority of the metals are powerful catalysts that speed up biological reactions and have excellent malleability and tensile strength.

• With both transitional and non-transitional elements, they can create alloys.

Differences:

• Lanthanide and actinoid contraction is shown in the 4d and 5d series. Atomic size variations are less noticeable in the 3d series.

• When compared to 3d elements, 4d and 5d elements have higher coordination numbers.

• While the 3d series has different qualities from the 4d and 5d series,

vi. Explain trends in ionisation enthalpies of d-block elements.

Ans:

• Transitional elements have quite high ionisation enthalpies that fall between those of 5-block and p-block elements. This is thus because p-block and 5-block elements have nuclear charges and atomic radii that are higher than those of transition elements.

• Even if the increase is irregular, the first ionisation enthalpy rises as the atomic number of transition elements rises along the period and along the group.

• If the first, second, and third ionisation enthalpies of the transition components are IE1, IE2, and IE3, respectively, then IE1<IE2<IE3.

• The last differentiating electron in the transition elements enters the (n - 1) d orbital, protecting the valence electrons from the nuclear attraction. This results in the (n - 1) d-electrons' screening effect.

• The screening action of the (n - 1) d electrons causes a sluggish and irregular increase in ionisation enthalpy.

vii. What is meant by diamagnetic and paramagnetic metal? Give one example of diamagnetic and paramagnetic transition metal and lanthanoid metal.

Ans: Paramagnetic substances are those that, when a magnetic field is applied, are drawn toward the magnetic field. Ni2+ and Pr4+ as examples.

Diamagnetic compounds are those that are attracted to magnetic fields when a magnetic field is applied. Example: Zn2+ and La3+.

Ferromagnetic compounds are those that are highly attracted to a magnetic field when it is applied. These materials are magnetizable. Fe, Co, and Ni are ferromagnetic examples.

viii. Why are the ground-state electronic configurations of gadolinium and lawrencium different than expected?

Ans: The intended electronic configuration for gadolinium is $[Xe]$4f86s2, whereas the actual configuration is $[Xe]$4f75d16s2. This is because, after receiving one electron from 5d, the 4f orbital will reach its fully filled state, which will increase the atom's stability. Because the orbitals have almost the same energy, this is possible.

The electrical configuration of Lawrencium is also $[Rn]$5f146d17s2. The electron enters the 5f orbital, filling it entirely and increasing its stability.

ix. Write steps involved in the metallurgical process

Ans: The following are the numerous procedures and ideas involved in extracting pure metals from their ores:

(1) Ore concentration after removing gangue contaminants.

(2) The transformation of ores into metal oxides or other reducible chemicals.

(3) Ore reduction to produce crude metals

(4) Metals are refined to produce pure metals.

x. Cerium and Terbium behave as good oxidising agents in +4 oxidation state. Explain.

Ans: Lanthanides are most stable in the +3 oxidation state. Therefore, Ce4+ (cerium) and Tb4+ (terbium) are more likely to achieve the more stable +3 oxidation state. In the +4 oxidation state, they are effective oxidising agents because they decrease by receiving electrons.

xi. Europium and Ytterbium behave as good reducing agents in +2 oxidation states.

Ans: Lanthanides are most stable in the +3 oxidation state. As a result, by losing one electron, Eu2+ and Yb2+ typically transition to +3 oxidation states. They are effective reducing agents in the +2 oxidation state because they become oxidised.

### Do You Know? (Textbook Page No 165)

1. In which block of the modern periodic table are the transition and inner transition elements placed?

Ans: The transition elements are arranged in the d-block of the modern periodic table, while inner transition elements are arranged in the f-block.

### Use Your Brain Power! (Textbook Page No 167)

1. Fill in the blanks with correct outer electronic configurations.

Ans: The filled table with correct outer electronic configurations is,

 2nd Series Y Zr Nb Mo Te Ru Rh Pd Ag Cd Z 57 72 73 74 75 76 77 78 79 80 Valence Electron Shell Configuration 4d15s2 4d25s2 4d45s1 4d55s1 4d65s1 4d75s1 4d85s1 4d105s0 4d105s1 4d105s2 3rd Series La Hf Ta W Re Os Ir Pt Au Hg Z 39 40 41 42 43 44 45 46 47 48 Valence Electron Shell Configuration 5d16s2 5d26s2 5d36s2 5d46s2 5d56s2 5d66s2 5d76s2 5d96s2 5d106s1 5d106s2 4th Series Ac Rf Db Sg Bh Hs Mt Ds Rg Uvb Z 89 104 105 106 107 108 109 110 111 112 Valence Electron Shell Configuration 6d17s2 6d27s2 6d37s2 6d47s2 6d57s2 6d67s2 6d77s2 6d87s2 6d107s1 6d107s2

### Try This. (Textbook Page No 168)

1. Write the electronic configuration of Cr and Cu.

Ans: Chromium and copper are configured electronically as follows:

• 24Cr → $[Ar]$3d54s1

• 30Cu → $[Ar]$3d104s1

### Can You Tell? (Textbook Page No 168)

1. Which of the first transition series elements shows the maximum number of oxidation states and why?

Ans: The reasons are as follows,

• Manganese exhibits the most oxidation states overall.

• J-subshell is only partially filled with manganese.

• Manganese can lose or share electrons from both orbitals due to the minimal energy difference between the 3d and 4s orbitals.

• The oxidation states of manganese range from +2 to +7.

2. Which elements in the 4d and 5d-series will show maximum number of oxidation states?

Ans: For Ruthenium $[Ru]$, the greatest number of oxidation states in the 4d series are + 2, +3, +4, +6, +7, + 8. Osmium $[Os]$ can exist in a maximum of +2 to +8 oxidation states in the 5d-series.

### Try This. (Textbook Page No 168)

1. Write the electronic configuration of Mn6+, Mn4+, Fe4+, Co5+, Ni2+.

Ans: The table shown below comprises the electronic configuration of the given ions.

 Ions Electronic configuration Mn6+ $[Ar]$3d1 Mn4+ $[Ar]$3d3 Fe4+ $[Ar]$3d4 Co5+ $[Ar]$3d4 Ni2+ $[Ar]$3d8

### Try This. (Textbook Page No 171)

1. Pick up the paramagnetic species from the following : Cu1+, Fe3+, Ni2+, Zn2+, Cd2+, Pd2+.

Ans: Fe3+, Ni2+, and Pd2+ are the paramagnetic ions.

### Try This. (Textbook Page No 171)

1. What will be the magnetic moment of transition metal having 3 unpaired electrons?

1. Equal to 1.73 B.M.?

2. Less than 1.73 BM.

3. More than 1.73 B.M.?

Ans: The answer is c. more than 1.73 B.M.

The formula used:

$\mu = \sqrt{n(n+2)}$

Calculation:

$\mu = \sqrt{3(3+2)}$

$\mu = \sqrt{3(5)}$

$\mu = 3.87 B.M.$

### Use Your Brain Power! (Textbook Page No 171)

1. A metal ion from the first transition series has two unpaired electrons. Calculate the magnetic moment.

Ans: The formula used:

$\mu = \sqrt{n(n+2)}$

Calculation:

$\mu = \sqrt{2(2+2)}$

$\mu = \sqrt{2(4)}$

$\mu = \sqrt{8}$

$\mu = 2.84 B.M.$

### Try This….. (Textbook Page No 172)

1. Calculate the spin-only magnetic moment of a divalent cation of element Slaving atomic number 27.

Ans: The divalent ion of an element with the atomic number 27 has the electrical configuration $[Ar]$3d7.

The formula used:

$\mu = \sqrt{n(n+2)}$

Calculation:

$\mu = \sqrt{5(5+2)}$

$\mu = \sqrt{5(7)}$

$\mu = \sqrt{35}$

$\mu = 5.92B.M.$

### Can you tell? (Textbook Page No 172)

1. Compounds of s and p-block elements are almost white. What could be the absorbed radiation? (uv or visible)?

Ans: A compound's white colour denotes the absorption of ultraviolet light.

### Can You Tell? (Textbook Page No 181)

1. Why are f-block elements called inner transition metals?

Ans: Since their f-orbitals are significantly closer together than those of transition metals, f-block elements are also known as inner transition elements. They have an f-orbital with 1 to 14 electrons.

2. Are there any similarities between transition and inner transition metals?

Ans: Transitional and inner transition metals share some characteristics.

• They sit between the s-block and p-block components.

• They have inner subshells filling in their electrical architecture, making them metals.

• They exhibit slates with varying oxidation.

• They manifest magnetism.

• It results in colored compounds.

• They are capable of catalysis.

### Use Your Brain Power! (Textbook Page No 185)

1. Do you think that lanthanide complex would show magnetism?

Ans: Magnetism may be present in lanthanide complexes.

2. Can you calculate the spin only magnetic moment of lanthanide complexes using the same formula that you used for transition metal complexes?

Ans: The magnetic moment of lanthanoid complexes cannot be calculated using a spin-only calculation because orbital momentum must also be taken into account.

3. Calculate the spin only magnetic moment of La3+. Compare the value with that given in the table.

Ans: No electrons are unpaired in the La3+ ion.

The formula used:

$\mu = \sqrt{n(n+2)}$

Calculation:

$\mu = \sqrt{0(0+2)}$

$\mu = 0B.M.$

La3+ ion's magnetic moment is zero.

## Introduction to Transition and Inner Transition Elements

The chapter Transition and Inner Transition Elements speaks about different elements and their grouping. It talks about Inner Transition elements and components that make up the f-block are those where the 4f and 5f orbitals gradually fill up. Although they are officially members of group 3, these elements are displayed separately in the f-block of the periodic table. Inner transition elements are another name for the f-constituent block's parts.

Lanthanoids and actinoids are the names for the two series of inner transition elements that are 4f and 5f series, respectively. The inner transition elements U, Th, and Pa are excellent nuclear energy sources. The transitional parts of the d block have (n-1) d-orbitals that are only partially filled. An element's placement in the periodic table strongly reflects both its characteristics and nature.

The Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements give students a complete insight about the chapter. There are examples and formulas that can make you understand better.

## Benefits of using Chapter 8 Chemistry Class 12 Exercise Solutions

One of the benefits of using Maharashtra Board Class 12 Chemistry Solutions Chapter 8 Transition and Inner Transition Elements is the easy language under which notes are prepared. It covers different questions and their solutions which are easy to understand and saves time. Practising for the class 12 board is quite a pressure moment for students. They cannot waste their time on looking for solutions to certain chapters. By using the solution, students can-

• Practice before the exam without referring to huge booklets or notes. Students can download notes and start practising.

• Students can save enough time by using this solution for the practice purpose. This is because class 12 preparation matters the most and students must manage time.

## FAQs on Maharashtra Board Class 12 Solutions for Chemistry Chapter 8 Transition and Inner Transition Elements - PDF

1. What do D Block Elements in Modern Periodic Table Consists?

The elements in the group of 3 to 12 make up the d block position in the periodic table. The d orbitals in this group fill up one at a time. The transition elements in the contemporary periodic table are those that exist between the s and p block elements. The (n-1) d-orbitals of transition elements are typically partially filled.

2. What is the first transition series?

The first transition series elements often fall into this series if their atomic numbers fall between 21 and 30. Ten elements—Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, and Zn—are part of this series. They always match the third subshell's filling.

3. What is another name for inner transition metals?

The lanthanides and actinides are groupings of elements in the periodic table. They are the substances that are occasionally mentioned after the primary part of the periodic table. There are a total of thirty elements in lanthanides and actinides.

4. Why are transition metals different?

The peculiarity of the transition elements is that they could have an unfinished internal subshell that permits valence electrons to be located in a shell other than the outer shell. Only valence electrons exist in the outer shell of some elements.