d & f- Block Elements
MCQsSilver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element ?
Answer: Ag has a completely filled 4d orbital (4d10 5s1 in its ground state. Now, silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orboital. However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete (4d9). Hence, it is a transition element.
Answer: Ag has a completely filled 4d orbital (4d10 5s1 in its ground state. Now, silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orboital. However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete (4d9). Hence, it is a transition element.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ/mol. Why ?
Answer: The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: 3d10 4s2), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization.
Answer: The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: 3d10 4s2), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization.
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why ?
Answer: Mn has the maximum number of unpaired electrons present in the d-subshell (five). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7.
Answer: Mn has the maximum number of unpaired electrons present in the d-subshell (five). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7.
How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements ?
Answer: Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as d0, d5, d10. Since these states are exceptionally stable, their ionization enthalpies are very high.
In case of first ionization energy, Cr has low ionization energy because, after losing one electron, it attains the stable configuration (3d5). On the other hand, Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals (3d10 4s2).
Second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out. Also, elements like Cr and Cu have exceptionally high second ionization energies as after losing the first electron, they have attained the stable configuration (Cr+: 3d5 and Cu+: 3d10). Hence, taking out one electron more from this stable configuration will require a lot of energy.
Answer: Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as d0, d5, d10. Since these states are exceptionally stable, their ionization enthalpies are very high.
In case of first ionization energy, Cr has low ionization energy because, after losing one electron, it attains the stable configuration (3d5). On the other hand, Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals (3d10 4s2).
Second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out. Also, elements like Cr and Cu have exceptionally high second ionization energies as after losing the first electron, they have attained the stable configuration (Cr+: 3d5 and Cu+: 3d10). Hence, taking out one electron more from this stable configuration will require a lot of energy.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only ?
Answer: Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.
Answer: Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why ?
Answer: In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more that that experienced by lanthanoids. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.
Answer: In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more that that experienced by lanthanoids. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.
Why are Mn2+compounds more stable than Fe2+ towards oxidation to their +3 state ?
Answer: Electronic configuration of Mn2+is [Ar]18 3d5. Electronic configuration of Fe2+ is [Ar]18 3d6. It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d5 configuration. This is the reason Mn2+shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d6 configuration and by losing one electron, its configuration changes to a more stable 3d5 configuration. Therefore, Fe2+ easily gets oxidized to Fe+3 oxidation state.
Answer: Electronic configuration of Mn2+is [Ar]18 3d5. Electronic configuration of Fe2+ is [Ar]18 3d6. It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d5 configuration. This is the reason Mn2+shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d6 configuration and by losing one electron, its configuration changes to a more stable 3d5 configuration. Therefore, Fe2+ easily gets oxidized to Fe+3 oxidation state.
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: The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting maximum number of oxidation states (+2 to +7). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital. However, Sc does not show +2 oxidation state. Its electronic configuration is 4s2 3d1. It loses all the three electrons to form Sc3+. +3 oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, [Ar]. Ti (+ 4) and V(+5) are very stable for the same reason. For Mn, +2 oxidation state is very stable as after losing two electrons, its d-orbital is exactly half-filled, [Ar] 3d5.
Answer: The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting maximum number of oxidation states (+2 to +7). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital. However, Sc does not show +2 oxidation state. Its electronic configuration is 4s2 3d1. It loses all the three electrons to form Sc3+. +3 oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, [Ar]. Ti (+ 4) and V(+5) are very stable for the same reason. For Mn, +2 oxidation state is very stable as after losing two electrons, its d-orbital is exactly half-filled, [Ar] 3d5.
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: Vanadate,VO3–
Oxidation state of V is + 5.
(ii) Chromate,CrO4–2
Oxidation state of Cr is + 6.
(iii) Permanganate,MnO4–
Oxidation state of Mn is + 7.
Answer: Vanadate,VO3–
Oxidation state of V is + 5.
(ii) Chromate,CrO4–2
Oxidation state of Cr is + 6.
(iii) Permanganate,MnO4–
Oxidation state of Mn is + 7.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction ?
Answer: As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron. Also, with the increase in atomic number, the number of electrons in the 4f orbital also increases. The 4f electrons have poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.
Consequences of lanthanoid contraction:
1. There is similarity in the properties of second and third transition series.
2. Separation of lanthanoids is possible due to lanthanide contraction.
3. It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides. (Basic strength decreases from La(OH)3 to Lu(OH)3.)
Answer: As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron. Also, with the increase in atomic number, the number of electrons in the 4f orbital also increases. The 4f electrons have poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.
Consequences of lanthanoid contraction:
1. There is similarity in the properties of second and third transition series.
2. Separation of lanthanoids is possible due to lanthanide contraction.
3. It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides. (Basic strength decreases from La(OH)3 to Lu(OH)3.)
What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
Answer: Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between s-block and p-block. Therefore, these are called transition elements. Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell.
Answer: Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between s-block and p-block. Therefore, these are called transition elements. Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell.
What are the different oxidation states exhibited by the lanthanoids ?
Answer: In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.
Answer: In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.
What are interstitial compounds? Why are such compounds well known for transition metals ?
Answer:Transition metals are large in size and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic size), such as H, C, N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.
Answer:Transition metals are large in size and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic size), such as H, C, N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.
How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Answer: In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by 1 (Fe2+ and Fe3+; Cu+ and Cu2+). In non-transition elements, the oxidation states differ by 2, for example, +2 and +4 or +3 and +5, etc.
Answer: In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by 1 (Fe2+ and Fe3+; Cu+ and Cu2+). In non-transition elements, the oxidation states differ by 2, for example, +2 and +4 or +3 and +5, etc.
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. Cr2+is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d3). This d3 configuration can be written as t2g3 configuration, which is a more stable configuration. In the case of Mn3+(d4), it acts as an oxidizing agent and gets reduced to Mn2+(d5). This has an exactly half-filled d-orbital and is highly stable.
ii. Co(II) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co (III). Although the 3rd ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.
iii. The ions in d1configuration tend to lose one more electron to get into stable d0 configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.
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. Cr2+is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d3). This d3 configuration can be written as t2g3 configuration, which is a more stable configuration. In the case of Mn3+(d4), it acts as an oxidizing agent and gets reduced to Mn2+(d5). This has an exactly half-filled d-orbital and is highly stable.
ii. Co(II) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co (III). Although the 3rd ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.
iii. The ions in d1configuration tend to lose one more electron to get into stable d0 configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why ?
Answer: In the first transition series, Cu exhibits +1 oxidation state very frequently. It is because Cu ( +1) has an electronic configuration of [Ar] 3d10. The completely filled d-orbital makes it highly stable.
Answer: In the first transition series, Cu exhibits +1 oxidation state very frequently. It is because Cu ( +1) has an electronic configuration of [Ar] 3d10. The completely filled d-orbital makes it highly stable.
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Answer: An alloy is a solid solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to possess different physical properties than those of the component elements. An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94-95%), iron (5%), and traces of S, C, Si, Ca, and Al.
Uses:
1. Mischmetal is used in cigarettes and gas lighters.
2. It is used in flame throwing tanks.
3. It is used in tracer bullets and shells.
Answer: An alloy is a solid solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to possess different physical properties than those of the component elements. An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94-95%), iron (5%), and traces of S, C, Si, Ca, and Al.
Uses:
1. Mischmetal is used in cigarettes and gas lighters.
2. It is used in flame throwing tanks.
3. It is used in tracer bullets and shells.
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: Inner transition metals are those elements in which the last electron enters the f-orbital. The elements in which the 4f and the 5f orbitals are progressively filled are called f-block elements. Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95, and 102.
Answer: Inner transition metals are those elements in which the last electron enters the f-orbital. The elements in which the 4f and the 5f orbitals are progressively filled are called f-block elements. Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95, and 102.
The chemistry of the actinoid elements is not so smooth as that of the Lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
Answer: Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states. For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, and +7. The most common oxidation state in case of actinoids is also +3.
Answer: Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states. For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, and +7. The most common oxidation state in case of actinoids is also +3.
Share
