Question:

What are the different oxidation states exhibited by the lanthanoids?

Answer:

Answer:

1. Lanthanoids exhibit oxidation states ranging from +2 to +4.
2. The common oxidation states of the lanthanoids are +2, +3, and +4.
3. The +3 oxidation state is the most common oxidation state exhibited by lanthanoids.
4. Some lanthanoids can also exhibit +1 and +5 oxidation states, but these are less common.

Question:

Explain given reasons: (i) Transition metals and many of their compounds show paramagnetic behaviour. (ii) The enthalpies of atomisation of the transition metals are high. (iii) The transition metals generally form coloured compounds. (iv) Transition metals and their many compounds act as good catalyst.

Answer:

(i) Transition metals and many of their compounds show paramagnetic behaviour because they have unpaired electrons in their outermost shells. This means that they are attracted to an external magnetic field and can become magnetised.

(ii) The enthalpies of atomisation of the transition metals are high because they have strong metallic bonds between their atoms. This makes them more difficult to break apart, resulting in higher enthalpies of atomisation.

(iii) The transition metals generally form coloured compounds because they have partially filled d-orbitals. These orbitals are able to absorb light of specific wavelengths, which gives the compounds their colour.

(iv) Transition metals and their many compounds act as good catalysts because they can easily form coordination complexes. These complexes can then act as intermediates in chemical reactions, allowing them to take place more quickly and efficiently.

Question:

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

Answer:

i) Electronic configurations: Compare the electronic configurations of the first series of transition metals with those of the second and third series metals in the respective vertical columns. The first series of transition metals have a general electronic configuration of [Ar] 3d^n, where n is an integer from 4 to 10, while the second and third series of transition metals have a general electronic configuration of [Ar] 4s² 3d^n, where n is an integer from 5 to 10.

ii) Oxidation states: Compare the oxidation states of the first series of transition metals with those of the second and third series metals in the respective vertical columns. The first series of transition metals generally have oxidation states of +2, +3 and +4, while the second and third series of transition metals generally have oxidation states of +2, +3, +4, +5 and +6.

iii) Ionisation enthalpies: Compare the ionisation enthalpies of the first series of transition metals with those of the second and third series metals in the respective vertical columns. The first series of transition metals generally have ionisation enthalpies that decrease from left to right across the periodic table, while the second and third series of transition metals generally have ionisation enthalpies that increase from left to right across the periodic table.

iv) Atomic sizes: Compare the atomic sizes of the first series of transition metals with those of the second and third series metals in the respective vertical columns. The first series of transition metals generally have atomic sizes that decrease from left to right across the periodic table, while the second and third series of transition metals generally have atomic sizes that increase from left to right across the periodic table.

Question:

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?

Answer:

1. Calculate the number of unpaired electrons in Mn³⁺: Mn³⁺ has 7 electrons, so it has 3 unpaired electrons.

2. Calculate the number of unpaired electrons in Cr³⁺: Cr³⁺ has 6 electrons, so it has 3 unpaired electrons.

3. Calculate the number of unpaired electrons in V³⁺: V³⁺ has 5 electrons, so it has 1 unpaired electron.

4. Calculate the number of unpaired electrons in Ti³⁺: Ti³⁺ has 4 electrons, so it has 0 unpaired electrons.

5. Which one of these is the most stable in aqueous solution? Ti³⁺ is the most stable in aqueous solution, since it has no unpaired electrons.

Question:

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

Answer:

(i) The stability of Fe³⁺ in acid solution is greater than that of Cr³⁺ or Mn³⁺. This is because the E^⊖ value for Fe³⁺/Fe²⁺ is +0.8V, while the E^⊖ values for Cr³⁺/Cr²⁺ and Mn³⁺/Mn²⁺ are -0.4V and +1.5V respectively.

(ii) The ease with which iron can be oxidised is greater than the ease with which chromium or manganese can be oxidised. This is because the E^⊖ value for Fe²⁺/Fe is -0.4V, while the E^⊖ values for Cr²⁺/Cr and Mn²⁺/Mn are -0.9V and -1.2V respectively.

Question:

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.

Answer:

1. Determine the first series of transition metals.
2. Identify the oxidation state of the metals in the first series of transition metals.
3. Name the oxometal anions of the first series of transition metals in which the metal exhibits the oxidation state equal to its group number.

Question:

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

Answer:

Answer:

1. Element with atomic number 61: Electronic configuration = [Kr] 4d¹⁰ 5s² 5p³

2. Element with atomic number 91: Electronic configuration = [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁴

3. Element with atomic number 101: Electronic configuration = [Md] 5f¹⁴ 6d¹⁰ 7s² 7p⁵

4. Element with atomic number 109: Electronic configuration = [Mt] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

Question:

What is meant by disproportionation? Give two examples of disproportionation reaction in aqueous solution.

Answer:

1. Disproportionation is a type of redox reaction in which an element is both oxidized and reduced simultaneously.

2. Examples of disproportionation reactions in aqueous solution include the oxidation of sulfite ions (SO₃ ^2-) to sulfate ions (SO₄ ^2-) and the reduction of nitrate ions (NO₃ ^-) to nitrite ions (NO₂ ^-).

Question:

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: (i) iodide (ii) iron(II) solution and (iii) H₂S

Answer:

(i) Oxidising action of potassium dichromate: Potassium dichromate (K₂Cr2O7) is a strong oxidising agent. It can oxidise iodide ions (I-) to iodine (I₂) in acidic solution. The ionic equation for this reaction is:

2K₂Cr2O7 + 6I- + 14H+ → 4Cr³⁺ + 2K+ + 6I₂ + 7H₂O

(ii) Oxidising action of potassium dichromate: Potassium dichromate (K₂Cr2O7) is a strong oxidising agent. It can oxidise iron(II) ions (Fe2+) to iron(III) ions (Fe3+) in acidic solution. The ionic equation for this reaction is:

2K₂Cr2O7 + 6Fe2+ + 14H+ → 4Cr³⁺ + 2K+ + 6Fe3+ + 7H₂O

(iii) Oxidising action of potassium dichromate: Potassium dichromate (K₂Cr2O7) is a strong oxidising agent. It can oxidise hydrogen sulfide (H₂S) to sulfur (S) and water (H₂O) in acidic solution. The ionic equation for this reaction is:

2K₂Cr2O7 + H₂S + 14H+ → 4Cr³⁺ + 2K+ + S + 7H₂O

Question:

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

Answer:

1. The +2 state of the transition elements with increasing atomic number becomes more and more stable due to the increase in nuclear charge.

2. As the atomic number increases, the number of protons in the nucleus also increases, which increases the nuclear charge.

3. The increased nuclear charge attracts the electrons more strongly, leading to a higher effective nuclear charge, which makes the +2 state more stable.

4. This stability is further increased by the shielding effect of the inner electrons, which decreases the effective nuclear charge and further stabilizes the +2 state.

Question:

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.

Answer:

1. Preparation of Potassium Permanganate: Potassium permanganate is prepared by the reaction of manganese dioxide with potassium hydroxide in aqueous solution. The reaction equation is:

MnO2 + 2KOH → K₂MnO4 + H₂O

2. Reaction of Acidified Permanganate Solution with Iron(II) Ions: When an acidified solution of potassium permanganate is added to a solution containing iron(II) ions, a red-brown precipitate of iron(III) hydroxide is formed. The ionic equation for the reaction is:

2Fe2+ + 2H+ + MnO4− → Fe2O₃ + Mn2+ + 2H₂O

3. Reaction of Acidified Permanganate Solution with SO2: When an acidified solution of potassium permanganate is added to a solution containing SO2, a yellow precipitate of manganese sulfite is formed. The ionic equation for the reaction is:

MnO4− + SO2 + H+ → MnSO₃ + H₂O

4. Reaction of Acidified Permanganate Solution with Oxalic Acid: When an acidified solution of potassium permanganate is added to a solution containing oxalic acid, a brown precipitate of manganese dioxide is formed. The ionic equation for the reaction is:

2MnO4− + 5H+ + 2C2O4H₂ → 2MnO2 + 4CO2 + 6H₂O

Question:

Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity

Answer:

(i) Electronic Configuration: Actinoids: The electronic configuration of the actinoids is [Rn] 5f1-14 6d1-2 7s².

Lanthanoids: The electronic configuration of the lanthanoids is [Xe] 4f1-14 5d1-2 6s².

(ii) Oxidation States: Actinoids: The actinoids typically exhibit oxidation states of +3, +4, +5, +6, and +7.

Lanthanoids: The lanthanoids typically exhibit oxidation states of +2, +3, +4, and +6.

(iii) Chemical Reactivity: Actinoids: The actinoids are generally more reactive than the lanthanoids due to their higher oxidation states and larger atomic radii.

Lanthanoids: The lanthanoids are generally less reactive than the actinoids due to their lower oxidation states and smaller atomic radii.

Question:

Indicate the steps in the preparation of: (i) K₂​Cr2​O7​ from chromite ore. (ii) KMnO4​ from pyrolusite ore.

Answer:

(i) K₂​Cr2​O7​ from chromite ore Step 1: Obtain chromite ore and grind it into a fine powder.

Step 2: Heat the powder in an electric furnace.

Step 3: Add potassium carbonate to the heated powder and mix it thoroughly.

Step 4: Heat the mixture until it melts and then cool it.

Step 5: Filter the cooled mixture and collect the solid residue.

Step 6: Dissolve the solid residue in water and filter it.

Step 7: Heat the filtered solution in an electric furnace.

Step 8: Add potassium chloride to the heated solution and mix it thoroughly.

Step 9: Heat the mixture until it melts and then cool it.

Step 10: Filter the cooled mixture and collect the solid residue.

Step 11: The solid residue is K₂​Cr2​O7​.

(ii) KMnO4​ from pyrolusite ore Step 1: Obtain pyrolusite ore and grind it into a fine powder.

Step 2: Heat the powder in an electric furnace.

Step 3: Add potassium hydroxide to the heated powder and mix it thoroughly.

Step 4: Heat the mixture until it melts and then cool it.

Step 5: Filter the cooled mixture and collect the solid residue.

Step 6: Dissolve the solid residue in water and filter it.

Step 7: Heat the filtered solution in an electric furnace.

Step 8: Add potassium permanganate to the heated solution and mix it thoroughly.

Step 9: Heat the mixture until it melts and then cool it.

Step 10: Filter the cooled mixture and collect the solid residue.

Step 11: The solid residue is KMnO4​.

Question:

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

Answer:

1. Transition elements are elements in the d-block of the periodic table that have partially filled d orbitals. They are also known as transition metals. They are called transition elements because they have a wide range of oxidation states due to the ability to move electrons between the d orbitals.

2. The elements in the d-block that may not be regarded as transition elements are the lanthanides and actinides. These elements have completely filled d orbitals and do not have the ability to move electrons between them.

Question:

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

Answer:

1. Interstitial compounds are chemical compounds formed when an atom of a transition metal is squeezed into the space between other atoms in a crystal lattice structure.

2. These compounds are well known for transition metals because the atoms of transition metals have a larger atomic radius than the atoms of other elements, which allows them to fit into the spaces between other atoms more easily. This creates a strong bond between the atoms, resulting in a more stable compound.

Question:

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

Answer:

1. The actinoids are a series of elements in the periodic table that includes elements 89 (actinium) to 103 (lawrencium).

2. The lanthanoids are a series of elements in the periodic table that includes elements 57 (lanthanum) to 71 (lutetium).

3. The chemistry of the actinoids is not as smooth as that of the lanthanoids because the actinoids have a greater range of oxidation states than the lanthanoids.

4. For example, actinium has an oxidation state of +3, +4, and +7, while lanthanum has an oxidation state of +2 and +3.

5. Another example is that protactinium has an oxidation state of +3, +4, +5, and +7, while cerium has an oxidation state of +2 and +3.

6. This shows that the chemistry of the actinoids is not as smooth as that of the lanthanoids because of their greater range of oxidation states.

Question:

How would you account for the following: (i) Of the d4 species, 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 d1 configuration is very unstable in ions

Answer:

(i) This is due to the fact that Cr2+ has two electrons in its d orbitals, which makes it more susceptible to reduction reactions. Manganese(III) has three electrons in its d orbitals, which makes it more prone to oxidation reactions.

(ii) Cobalt(II) is stable in aqueous solution because it has two electrons in its d orbitals, which makes it more stable and less prone to oxidation. However, in the presence of complexing reagents, the electrons in the d orbitals become more easily accessible, which makes it easier to oxidise.

(iii) The d1 configuration is very unstable in ions because it only has one electron in its d orbitals. This makes it very susceptible to oxidation and reduction reactions, making it very reactive and unstable.

Question:

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

Answer:

1. First, identify the elements of the first transition series. These are elements from scandium (Sc) to zinc (Zn).

2. Next, research the +2 oxidation state for each of these elements.

3. Analyze the data to determine the stability of each element’s +2 oxidation state.

4. Compare the stability of each element’s +2 oxidation state to one another.

5. Finally, draw a conclusion about the stability of the +2 oxidation state for the elements of the first transition series.

Question:

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

Answer:

1. Transition metals typically have more variable oxidation states than non-transition metals. This is because transition metals have partially filled d-orbitals which can be easily filled or emptied to create different oxidation states.

2. For example, the transition metal iron (Fe) has an oxidation state of +2, +3, and +4, while the non-transition metal magnesium (Mg) only has an oxidation state of +2.

3. Another example is the transition metal copper (Cu), which has an oxidation state of +1, +2, and +3, while the non-transition metal chlorine (Cl) only has an oxidation state of -1.

Question:

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

Answer:

Step 1: Understand Hund’s Rule Hund’s Rule states that electrons fill orbitals of the same energy in a way that maximizes the number of unpaired electrons.

Step 2: Determine the electronic configuration of Ce3+ ion Ce3+ has an atomic number of 58, which means it has 58 electrons. The electronic configuration is 1s² 2s² 2p6 3s² 3p6 3d¹⁰ 4s² 4p6 4d¹⁰.

Step 3: Determine the number of unpaired electrons The 4s² orbital has two electrons, which are paired, so there are no unpaired electrons in the 4s² orbital. The 4p6 orbital has six electrons, three of which are paired and three of which are unpaired. The 4d¹⁰ orbital has ten electrons, five of which are paired and five of which are unpaired.

Step 4: Calculate the magnetic moment The magnetic moment of the Ce3+ ion is calculated using the spin-only formula, which is μ = √(n(n+2)), where n is the number of unpaired electrons. The Ce3+ ion has 8 unpaired electrons, so the magnetic moment is μ = √(8(8+2)) = 6.4 Bohr magnetons.

Question:

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.

Answer:

1. Inner transition elements are a group of elements in the periodic table that are located between the s and p blocks of elements.

2. The atomic numbers of the inner transition elements are 59, 74, 95, and 102.

Question:

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

Answer:

1. Mn²⁺ compounds are more stable than Fe²⁺ compounds because Mn has a higher oxidation state than Fe.

2. This is because Mn has more electrons in its outermost shell than Fe, giving it greater stability against oxidation.

3. The extra electrons make it more difficult for the Mn²⁺ compound to be oxidized to its +3 state, since it would require the addition of three electrons to the outermost shell.

4. In contrast, the Fe²⁺ compound has fewer electrons in its outermost shell, making it easier to be oxidized to its +3 state since it would only require the addition of two electrons to the outermost shell.

Question:

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 be occupied for these hydrated ions (octahedral).

Answer:

Ti²⁺: 3d electrons = 4 V²⁺: 3d electrons = 3 Cr³⁺: 3d electrons = 3 Mn²⁺: 3d electrons = 4 Fe²⁺: 3d electrons = 4 Fe³⁺: 3d electrons = 3 Co²⁺: 3d electrons = 4 Ni²⁺: 3d electrons = 4 Cu²⁺: 3d electrons = 3

For the hydrated octahedral ions, the five 3d orbitals are expected to be filled as follows:

Ti²⁺: 3d1², 3d2², 3d3², 3d4⁰ V²⁺: 3d1², 3d2², 3d3¹, 3d4⁰ Cr³⁺: 3d1², 3d2², 3d3¹, 3d4⁰ Mn²⁺: 3d1², 3d2², 3d3², 3d4⁰ Fe²⁺: 3d1², 3d2², 3d3², 3d4⁰ Fe³⁺: 3d1², 3d2², 3d3¹, 3d4⁰ Co²⁺: 3d1², 3d2², 3d3², 3d4⁰ Ni²⁺: 3d1², 3d2², 3d3², 3d4⁰ Cu²⁺: 3d1², 3d2², 3d3¹, 3d4⁰

Question:

What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d³,3d⁵,3d⁸ and 3d⁴?

Answer:

Answer:

1. The oxidation state of a transition element is determined by the number of unpaired electrons in its d orbital.
2. For the given d electron configurations in the ground state of their atoms (3d³,3d^5,3d^8 and 3d^4), the number of unpaired electrons is 3, 1, 0 and 2, respectively.
3. Therefore, the stable oxidation states of the transition element are +3, +1, 0 and +2, respectively.

Question:

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

Answer:

Step 1: Identify the first series of transition metals.

Answer: The first series of transition metals are scandium (Sc), titanium (Ti), vanadium (V), chromium (Cr), manganese (Mn), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu), and zinc (Zn).

Step 2: Determine which metal exhibits +1 oxidation state most frequently.

Answer: Copper (Cu) exhibits +1 oxidation state most frequently.

Step 3: Explain why copper (Cu) exhibits +1 oxidation state most frequently.

Answer: Copper (Cu) exhibits +1 oxidation state most frequently because it has one valence electron in its outer shell, making it easier for it to lose that electron and form a +1 oxidation state.

Question:

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

Answer:

1. First, it is important to understand what an electronic configuration is. An electronic configuration is the arrangement of electrons in an atom or molecule according to a set of rules.

2. The electronic configuration of the transition elements is different from that of the non transition elements in that the transition elements have partially filled d-orbitals. This is due to the fact that transition elements have more than one possible oxidation state and the electrons in the d-orbitals are responsible for this.

3. Non transition elements, on the other hand, have a complete outer shell and do not have partially filled d-orbitals. This is because they only have one possible oxidation state and the electrons in the outer shell are responsible for this.

4. Therefore, the electronic configuration of the transition elements is different from that of the non transition elements because the transition elements have partially filled d-orbitals while the non transition elements do not.

Question:

Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (ii) atomic and ionic sizes and (iii) oxidation state (iv) chemical reactivity.

Answer:

(i) Electronic Configuration: Actinoids and lanthanoids both have similar electronic configurations in the outermost shell. Both of them have a half-filled f-shell, which is the last electron shell. Actinoids have a few extra electrons in the d-shell, which is the penultimate electron shell.

(ii) Atomic and Ionic Sizes: Actinoids have larger atomic radii than lanthanoids. This is due to the extra electrons present in the d-shell of actinoids. Similarly, the ionic radii of actinoids are larger than those of lanthanoids.

(iii) Oxidation State: The oxidation states of actinoids and lanthanoids are similar. Both of them can have oxidation states of +2, +3, +4, and +6.

(iv) Chemical Reactivity: Actinoids are more reactive than lanthanoids due to the extra electrons present in the d-shell of actinoids. This makes actinoids more reactive than lanthanoids.

Question:

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.

Answer:

(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.

Examples: The lowest oxide of iron is FeO, which is basic, whereas the highest oxide of iron is Fe2O₃, which is amphoteric.

Reason: The basicity of an oxide of a transition metal increases as the oxidation state of the metal decreases. As the oxidation state increases, the oxide becomes more acidic.

(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.

Examples: Cobalt exhibits its highest oxidation state of +7 in the oxide Co2O7 and the fluoride CoF₃.

Reason: Oxides and fluorides are both stable compounds that can accommodate a high oxidation state. Oxides contain oxygen, which has a high electronegativity, allowing it to easily accept electrons from the metal, while fluorides contain fluorine, which has a high electron affinity, allowing it to easily donate electrons to the metal.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Examples: Iron exhibits its highest oxidation state of +8 in the oxoanion FeO4^2-.

Reason: Oxoanions are highly charged species that can accommodate a high oxidation state. The oxygen atoms in the oxoanion can easily accept electrons from the metal, allowing the metal to reach a high oxidation state.

Question:

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

Answer:

1. Preparation of Potassium Dichromate from Iron Chromite Ore: a. The iron chromite ore is first crushed and mixed with concentrated sulfuric acid. b. The mixture is then heated, causing the formation of chromium sulfate. c. The chromium sulfate is then reacted with potassium hydroxide, forming potassium dichromate.

2. Effect of Increasing pH on a Solution of Potassium Dichromate: a. As the pH of the solution increases, the potassium dichromate molecules become more basic, leading to the formation of chromate and dichromate ions. b. The chromate ions are more stable at higher pH values, while the dichromate ions are more stable at lower pH values. c. As the pH increases, the amount of chromate ions increases, while the amount of dichromate ions decreases.

Question:

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

Answer:

1. Alloys are mixtures of two or more metals, or a metal and a non-metal.

2. An important alloy which contains some of the lanthanoid metals is called Magnalium.

3. Magnalium is a combination of magnesium and aluminum and contains some of the lanthanoid metals.

4. Magnalium is used in the aerospace industry, for the production of lightweight components and for the manufacturing of aircraft parts. It is also used in the automotive industry for the production of lightweight components. Additionally, Magnalium is used in the electronics industry for the production of circuit boards.

Question:

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.

Answer:

Answer: The last element in the series of actinoids is Lawrencium (Lr). The electronic configuration of Lawrencium is [Rn] 5f¹⁴ 7s². The possible oxidation state of Lawrencium is +3 due to its electronic configuration.

Question:

Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

Answer:

1. The first transition series is a group of elements found in the periodic table that includes elements such as scandium, titanium, vanadium, chromium, manganese, iron, cobalt, nickel, copper, and zinc.

2. These elements possess many properties that are distinct from the heavier transition elements, such as lanthanides and actinides.

3. For example, the first transition series elements are generally more reactive and have higher melting points than the heavier transition elements.

4. Additionally, the first transition series elements tend to form compounds with higher oxidation states than the heavier transition elements.

5. Furthermore, the first transition series elements are generally more abundant in the Earth’s crust than the heavier transition elements.

6. In conclusion, it is clear that the elements of the first transition series possess many properties that are different from those of the heavier transition elements.

Question:

To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Answer:

1. The electronic configurations of the elements in the first series of the transition elements play a major role in determining the stability of their oxidation states.

2. This is because the outermost electrons of the elements in the first series of transition elements are the valence electrons, which determine the oxidation state of the element.

3. For example, the element Iron (Fe) has an electronic configuration of [Ar]3d6 4s². This means that the outermost electrons are in the 3d orbital, and the oxidation state of Iron is +2 or +3.

4. Similarly, the element Cobalt (Co) has an electronic configuration of [Ar]3d7 4s². This means that the outermost electrons are in the 3d orbital, and the oxidation state of Cobalt is +2 or +3.

5. Thus, the electronic configurations of the elements in the first series of the transition elements determine the stability of their oxidation states.

Question:

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

Answer:

(i) Cr³⁺: [Ar] 3d³ (ii) Pm³⁺: [Xe] 4f⁶ (iii) Cu⁺: [Ar] 3d¹⁰ (iv) Ce⁴⁺: [Xe] 4f¹ (v) Co²⁺: [Ar] 3d⁷ (vi) Lu²⁺: [Xe] 4f¹⁴ (vii) Mn²⁺: [Ar] 3d⁵ (viii) Th⁴⁺: [Rn] 5f²

Question:

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

Answer:

Ti³⁺: No color, because titanium is a transition metal and does not form colored ions in aqueous solution.

V³⁺: No color, because vanadium is a transition metal and does not form colored ions in aqueous solution.

Cu⁺: Blue color, because copper forms a blue colored ion in aqueous solution.

Sc³⁺: No color, because scandium is a transition metal and does not form colored ions in aqueous solution.

Mn²⁺: Pink color, because manganese forms a pink colored ion in aqueous solution.

Fe³⁺: Brownish-yellow color, because iron forms a brownish-yellow colored ion in aqueous solution.

Co²⁺: Pink color, because cobalt forms a pink colored ion in aqueous solution.

Question:

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.

Answer:

1. The elements that exhibit +4 oxidation states are Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb and Lu.

2. The elements that exhibit +2 oxidation states are La, Ce, Pr, Nd, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb and Lu.

3. The electronic configurations of these elements are such that they have 4f orbitals which are not filled with electrons. This means that they have a tendency to lose electrons to achieve a stable configuration, thus leading to the formation of +4 oxidation states.

4. The elements that exhibit +2 oxidation states have their 4f orbitals filled with electrons. This means that they have a tendency to gain electrons to achieve a stable configuration, thus leading to the formation of +2 oxidation states.

Question:

What is meant by lanthanoid contraction? The actinoid contraction is greater than that of lanthanoid contraction. Why?

Answer:

1. Lanthanoid contraction is the decrease in atomic radii of the elements in the lanthanoid series of the periodic table.

2. Actinoid contraction is the decrease in atomic radii of the elements in the actinoid series of the periodic table.

3. The actinoid contraction is greater than that of lanthanoid contraction because the actinoid series has higher nuclear charge than the lanthanoid series, which results in a greater electrostatic force between the nucleus and the electrons, causing a greater decrease in atomic radii.

Question:

What can be inferred from the magnetic moment values of the following complex species? Example Magnetic Moment (BM) K₄[Mn(CN)₆]².2 [Fe(H₂O)6​]²⁺ 5.3 K₂​[MnCl₄​] 5.9

Answer:

1. The complex species K₄[Mn(CN)₆]².2 has a magnetic moment of 0 BM.
2. The complex species [Fe(H₂O)₆]²⁺ has a magnetic moment of 5.3 BM.
3. The complex species K₂[MnCl₄] has a magnetic moment of 5.9 BM.
4. The magnetic moments of the three complex species differ from each other, indicating that the number of unpaired electrons in each complex species is different.