Lanthanoids and Actinoids Questions in English

(B) The atomic number of Cerium $(Ce)$ is $58$.
Following the Aufbau principle and the stability of $f$-orbitals,the electronic configuration is written as $[Xe] \, 4f^1 \, 5d^1 \, 6s^2$.
This is because the $4f$ and $5d$ energy levels are very close,and the electron enters the $5d$ orbital before filling the $4f$ subshell completely in the case of $Ce$.

(B) The general electronic configuration of lanthanoids (elements with atomic numbers $57$ to $71$) is represented as $[Xe] \, 4f^{1-14} \, 5d^{0-1} \, 6s^2$.
In this configuration,the $4f$ orbitals are progressively filled from $1$ to $14$,while the $5d$ orbital may contain $0$ or $1$ electron,and the $6s$ orbital is completely filled with $2$ electrons.

(B) $1$. $La(OH)_3$ is more basic than $Lu(OH)_3$ because basicity decreases as ionic radius decreases across the lanthanoid series. Thus, option $A$ is incorrect.
$2$. The ionic radius of $Ln^{3+}$ ions decreases across the lanthanoid series due to lanthanoid contraction. This statement is correct.
$3$. $La$ $(Z=57)$ is a $d$-block element, not a $4f$ element, so it is not a lanthanoid. Thus, option $C$ is incorrect.
$4$. The atomic radii of $Zr$ $(160 \ pm)$ and $Hf$ $(159 \ pm)$ are very similar due to lanthanoid contraction. This statement is also correct. However, in standard multiple-choice questions of this type, the most fundamental property described is the decrease in ionic radius.

(D) Due to lanthanoid contraction,the ionic size decreases from $Ce^{3+}$ to $Lu^{3+}$.
As the size of the $M^{3+}$ ion decreases,the covalent character of the $M-OH$ bond increases,which leads to a decrease in basic strength.
Therefore,the basicity order for the given hydroxides is: $Ce(OH)_3 > Eu(OH)_3 > Yb(OH)_3 > Lu(OH)_3$.
Thus,$Ce(OH)_3$ is the most basic.

(D) Mischmetal is an alloy of lanthanoid metals.
It typically consists of approximately $95\%$ lanthanoid metal and $5\%$ iron,along with traces of $S$,$C$,$Ca$,and $Al$.
Therefore,the percentage of iron in pyrophoric mischmetal is approximately $5\%$.

(C) Lanthanoid contraction refers to the steady decrease in the atomic and ionic radii of the lanthanoid elements as the atomic number increases from $La$ $(Z=57)$ to $Lu$ $(Z=71)$.
This is due to the poor shielding effect of the $4f$ electrons,which causes the effective nuclear charge to increase,pulling the valence electrons closer to the nucleus.

(A) In the lanthanide series,as the atomic number increases from $La$ $(Z=57)$ to $Lu$ $(Z=71)$,the ionic radii of $Ln^{+3}$ ions decrease due to lanthanoid contraction.
Therefore,the order of ionic radii for the given ions is $La^{+3} > Ce^{+3} > Pm^{+3} > Yb^{+3}$.
Thus,the increasing order is $Yb^{+3} < Pm^{+3} < Ce^{+3} < La^{+3}$.

(B) Both $Lanthanoids$ and $Actinoids$ exhibit a common oxidation state of $+3$.
While $Lanthanoids$ show limited oxidation states (mainly $+3$),$Actinoids$ show a wider range of oxidation states due to the comparable energies of $5f$,$6d$,and $7s$ orbitals.
However,the $+3$ oxidation state is the most stable and common characteristic shared by both series.

(D) Lanthanoid contraction is primarily caused by the poor shielding effect of the $4f$ electrons.
Because the $4f$ orbitals have a diffuse shape,they are unable to effectively shield the outer electrons from the increasing nuclear charge as we move across the lanthanoid series.

(B) Both lanthanoids and actinoids exhibit a common stable oxidation state of $+3$.

(B) The lanthanoids exhibit a $+3$ oxidation state as their most stable state.
$1$. The ionic radii of $Ln(III)$ ions decrease with an increase in atomic number due to lanthanoid contraction.
$2$. $Ln(III)$ compounds are often colored due to $f-f$ transitions,unlike the statement that they are mostly colorless.
$3$. $Ln(III)$ hydroxides are basic in nature,and their basicity decreases as the ionic radius decreases.
$4$. Due to the large size of $Ln(III)$ ions,they form compounds that are predominantly ionic in nature.
Therefore,the statement that $Ln(III)$ compounds are mostly colorless is incorrect.

(C) The phenomenon of lanthanoid contraction refers to the steady decrease in the ionic radii of lanthanoids from $La^{3+}$ $(Z=57)$ to $Lu^{3+}$ $(Z=71)$.
As the atomic number increases,the shielding effect of $4f$ electrons is poor,leading to an increase in the effective nuclear charge,which pulls the electrons closer to the nucleus.
Since $_{57}La^{3+}$ has a radius of $1.06 \ \mathring{A}$,the radius of $_{71}Lu^{3+}$ must be smaller than $1.06 \ \mathring{A}$ due to lanthanoid contraction.
Among the given options,$0.85 \ \mathring{A}$ is the only value smaller than $1.06 \ \mathring{A}$.

(D) The electronic configuration of $Ce$ $(Z = 58)$ is $[Xe] 4f^1 5d^1 6s^2$.
By losing four electrons,it achieves the stable noble gas configuration of $Xe$,thus exhibiting a $+4$ oxidation state.
$La$ $(Z = 57)$ typically shows a $+3$ state.
$Eu$ $(Z = 63)$ and $Gd$ $(Z = 64)$ typically show $+3$ states,with $Eu$ also showing $+2$.

(B) In the $f$-block elements,specifically the lanthanoids,the atomic and ionic radii decrease with an increase in atomic number. This phenomenon is known as lanthanoid contraction. This occurs because the $4f$ electrons provide poor shielding for the nuclear charge,causing the outer electrons to be pulled closer to the nucleus.

(B) Mischmetal is an alloy of lanthanoid metals.
It typically consists of approximately $50\%$ Cerium $(Ce)$,$40\%$ Lanthanum $(La)$,$7\%$ Iron $(Fe)$,and $3\%$ other trace metals (such as $Al, Ca, C, S, Si$).
This composition is widely used in the manufacture of pyrophoric alloys for lighter flints.

(A) The correct statement is that $La(OH)_3$ is more basic than $Lu(OH)_3$.
As we move from $La^{3+}$ to $Lu^{3+}$,the ionic size decreases due to lanthanoid contraction,which increases the covalent character and decreases the basicity.
Therefore,$La(OH)_3$ is the most basic and $Lu(OH)_3$ is the least basic.
Option $A$ is incorrect.

(A) Thorium $(Th)$ has an atomic number of $90$.
Its electronic configuration is based on the $Rn$ $(Z=86)$ core.
Although $Th$ is often placed in the actinoid series,its ground state electronic configuration is $[Rn] \, 5f^0 \, 6d^2 \, 7s^2$ because the $5f$ orbitals are not yet occupied in the neutral atom.

(B) Statement $(i)$ is incorrect because $La(OH)_3$ is the most basic lanthanide hydroxide due to the largest ionic radius of $La^{3+}$. As we move from $La$ to $Lu$,the ionic radius decreases,leading to a decrease in basicity.
Statement $(ii)$ is correct because $Zr^{4+}$ and $Hf^{4+}$ have almost identical ionic radii due to lanthanoid contraction.
Statement $(iii)$ is correct because $Ce^{4+}$ has a strong tendency to gain an electron to form the stable $Ce^{3+}$ $(4f^1)$ configuration,thus acting as a strong oxidizing agent.
Therefore,statements $(ii)$ and $(iii)$ are correct.

(B) Lanthanoid contraction is the steady decrease in the atomic and ionic radii of lanthanoids with an increase in atomic number.
Due to the poor shielding effect of $4f$ electrons,the effective nuclear charge increases,which causes the size of the atoms to decrease.
This phenomenon is responsible for the fact that the second and third transition series elements have almost identical atomic and ionic radii.
Specifically,$Zr$ ($4d$ series) and $Hf$ ($5d$ series) exhibit almost identical covalent and ionic radii due to lanthanoid contraction.

(B) The electronic configuration of lanthanoids is $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$,while for actinoids it is $[Rn] 5f^{1-14} 6d^{0-1} 7s^2$.
In actinoids,the energy difference between the $5f$ and $6d$ orbitals is much smaller than the energy difference between the $4f$ and $5d$ orbitals in lanthanoids.
Additionally,because the $5f$ orbitals are further from the nucleus,they are less tightly held,allowing for greater participation in bonding and the exhibition of a wider range of oxidation states.

(B) The actinoid series consists of elements with atomic numbers $89$ to $103$.
These elements are $Actinium$ ($Ac$,$Z=89$) through $Lawrencium$ ($Lr$,$Z=103$).
Therefore,the correct range is $Ac$ to $Lr$.

(B) The flint used in gas lighters is made of a pyrophoric alloy known as mischmetal.
It consists of a lanthanoid metal (about $95\%$) along with iron (about $5\%$) and traces of $S$,$C$,$Ca$,and $Al$.
This alloy is pyrophoric,meaning it produces sparks when struck or scraped,which ignites the gas.

(B) The lanthanoids are a series of $14$ elements from Cerium ($Ce$,$Z=58$) to Lutetium ($Lu$,$Z=71$).
All lanthanoids exhibit a common oxidation state of $+3$ due to the loss of two $6s$ electrons and one $4f$ or $5d$ electron.
While some elements show $+2$ or $+4$ oxidation states in specific compounds,$+3$ is the most stable and characteristic oxidation state for the entire series.

(D) The actinoids are the series of elements from $Actinium$ $(Z = 89)$ to $Lawrencium$ $(Z = 103)$.
$Uranium$ $(Z = 92)$,$Curium$ $(Z = 96)$,and $Californium$ $(Z = 98)$ are all members of the actinoid series.
$Erbium$ $(Z = 68)$ is a lanthanoid element,belonging to the $4f$ series,not the $5f$ actinoid series.
Therefore,the correct answer is $Erbium$.

(A) Among the lanthanoids,all elements are non-radioactive except for Promethium $(Pm)$.
Promethium is the only synthetic radioactive element in the lanthanoid series with atomic number $61$.

(D) Due to the $Lanthanoid \ contraction$,the atomic and ionic radii of lanthanoids decrease regularly from $La^{3+}$ to $Lu^{3+}$.
As the atomic number increases from $57$ to $71$,the effective nuclear charge increases,which pulls the electrons closer to the nucleus.
Therefore,the ionic radius of $Lu^{3+}$ must be smaller than that of $La^{3+}$ $(1.06 \ \mathop{A}\limits^o)$.
Among the given options,$0.85 \ \mathop{A}\limits^o$ is the closest and most reasonable value.

(A) The $f$-block elements are divided into two series: the lanthanides ($4f$-series) and the actinides ($5f$-series).
Because both series involve the filling of $f$-orbitals and exhibit similar chemical properties,the actinides are considered to be analogous or homologous to the lanthanides.

(B) The most common lanthanide is $Cerium$ $(Ce)$. It is the most abundant of the lanthanide series in the Earth's crust.

(B) The actinoids exhibit a greater range of oxidation states compared to lanthanoids because the $5f$,$6d$,and $7s$ energy levels are very close in energy.
This small energy gap allows electrons from these subshells to participate in bonding,making excitation easier.

(C) The earlier members of the lanthanoid series are quite reactive,similar to $Ca$. However,with an increase in atomic number,their reactivity decreases,and they behave more like $Al$. Therefore,the statement that all lanthanons are much more reactive than $Al$ is incorrect.

(B) The electronic configuration of Lanthanoids follows the general pattern $[Xe] \, 4f^{n} \, 5d^{0-1} \, 6s^2$.
For $Eu$ $(Z=63)$: The configuration is $[Xe] \, 4f^7 \, 6s^2$.
For $Gd$ $(Z=64)$: The configuration is $[Xe] \, 4f^7 \, 5d^1 \, 6s^2$ (due to the stability of the half-filled $f$-orbital).
For $Tb$ $(Z=65)$: The configuration is $[Xe] \, 4f^9 \, 6s^2$.
Thus,the correct option is $B$.

(A) Lanthanoid contraction is the regular decrease in atomic and ionic radii of lanthanides.
This is due to the imperfect shielding (or poor screening effect) of $f$-orbitals.
Due to their diffused shape,$f$-orbitals are unable to effectively counterbalance the effect of the increased nuclear charge.
Hence,the net result is a contraction in the size of lanthanoids.

(B) lanthanoid ion with no unpaired electrons is diamagnetic in nature.
$Ce_{58} = [Xe] 4f^{2} 5d^{0} 6s^{2} \implies Ce^{2+} = [Xe] 4f^{2}$ (two unpaired electrons)
$Sm_{62} = [Xe] 4f^{6} 5d^{0} 6s^{2} \implies Sm^{2+} = [Xe] 4f^{6}$ (six unpaired electrons)
$Eu_{63} = [Xe] 4f^{7} 5d^{0} 6s^{2} \implies Eu^{2+} = [Xe] 4f^{7}$ (seven unpaired electrons)
$Yb_{70} = [Xe] 4f^{14} 5d^{0} 6s^{2} \implies Yb^{2+} = [Xe] 4f^{14}$ (no unpaired electrons)
Because of the absence of unpaired electrons,$Yb^{2+}$ is diamagnetic.

(C) Actinium $(Ac)$ has the atomic number $89$.
The electronic configuration of $Ac$ is $[Rn] \ 6d^1 \ 7s^2$.
Due to the stability of its electronic configuration,it exhibits only a $+3$ oxidation state.
In contrast,$Th$ $(Z=90)$ exhibits $+3$ and $+4$,$Pa$ $(Z=91)$ exhibits $+3, +4, +5$,and $U$ $(Z=92)$ exhibits $+3, +4, +5, +6$ oxidation states.

(D) The most common oxidation state exhibited by lanthanoids is $+3$.

(B) The regular decrease in the radii of lanthanide ions from $La^{3+}$ to $Lu^{3+}$ is known as lanthanide contraction.
It is due to the poor shielding effect of $4f$ electrons,which allows the increased nuclear charge to pull the electrons closer to the nucleus.
As a result of lanthanide contraction,the atomic radii of elements of the $4d$ and $5d$ series become very similar,leading to similarities in their chemical properties.
Therefore,the statement that $4d$ and $5d$ series have no similarities is incorrect.

(C) The actinoids exhibit a greater number of oxidation states compared to the lanthanoids.
This is primarily because the energy difference between the $5f$ and $6d$ orbitals in actinoids is smaller than the energy difference between the $4f$ and $5d$ orbitals in lanthanoids.
Consequently,the $5f$ electrons can participate in bonding more easily than the $4f$ electrons.

(B) The main reason for lanthanoid contraction is the poor shielding of the outer shell electrons by $4f$ electrons.
Because the $4f$ orbitals have a diffuse shape,their shielding effect is imperfect.
Consequently,the effective nuclear charge experienced by the outer electrons increases,causing the nucleus to attract the outer shell electrons more strongly,which leads to a decrease in atomic and ionic radii across the lanthanoid series.

(A) The $5f$ orbitals extend further from the nucleus compared to the $4f$ orbitals.
Due to this,the $5f$ electrons are less tightly held by the nucleus and are more available for bonding.
Consequently,the actinoids exhibit a greater variety of oxidation states compared to the lanthanoids.

(B) The main reason for actinoids exhibiting a larger number of oxidation states compared to lanthanoids is the smaller energy difference between the $5f$ and $6d$ orbitals compared to the energy difference between the $4f$ and $5d$ orbitals.
Because the $5f$,$6d$,and $7s$ orbitals in actinoids have comparable energies,electrons from these orbitals can participate in bonding,leading to a wider range of oxidation states.

(B) The chemistry of lanthanoids is dominated by the $+3$ oxidation state.
Most $Ln^{3+}$ ions contain unpaired $f$-electrons,which undergo $f-f$ transitions,making them coloured.
Therefore,the statement that $Ln(III)$ compounds are generally colourless is incorrect.
$La^{3+}$ $(f^0)$ and $Lu^{3+}$ $(f^{14})$ are colourless,but most others are coloured.

(D) Lanthanoids exhibit a common $+3$ oxidation state.
The $4f$ electrons are deeply buried and shielded by $5s$,$5p$,and $5d$ orbitals,making them less available for bonding compared to $d$-block elements.
While some lanthanoids like $Ce$ $(+4)$ and $Tb$ $(+4)$ show $+4$ oxidation states due to stable electronic configurations (like $f^0$ or $f^7$),it is not a property of all members of the series.
Therefore,the statement that all members form compounds in the $+4$ state is incorrect.

(D) Promethium $(Pm)$ is the only element among the lanthanoids that is radioactive in nature.
Promethium is relatively unstable,and all its isotopes are radioactive.
Thus,the correct answer is option $D$.

(A) The ionic radii of lanthanides decrease as the atomic number increases from $La$ $(Z=57)$ to $Lu$ $(Z=71)$.
This phenomenon is known as lanthanoid contraction.
Therefore,the ionic radii order is $La^{3+} > Ce^{3+} > Pm^{3+} > Yb^{3+}$.
In increasing order,this is $Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$.

(D) To find the number of unpaired $f-$electrons,we write the electronic configuration of the lanthanide ions:
$1$. $Tm^{+3} (Z = 69)$: The neutral $Tm$ is $[Xe] 4f^{13} 6s^2$. $Tm^{+3}$ is $[Xe] 4f^{12}$. In $4f^{12}$,there are $2$ unpaired electrons.
$2$. $Nd^{+3} (Z = 60)$: The neutral $Nd$ is $[Xe] 4f^4 6s^2$. $Nd^{+3}$ is $[Xe] 4f^3$. In $4f^3$,there are $3$ unpaired electrons.
$3$. $Dy^{+3} (Z = 66)$: The neutral $Dy$ is $[Xe] 4f^{10} 6s^2$. $Dy^{+3}$ is $[Xe] 4f^9$. In $4f^9$,there are $5$ unpaired electrons.
$4$. $Tb^{+3} (Z = 65)$: The neutral $Tb$ is $[Xe] 4f^9 6s^2$. $Tb^{+3}$ is $[Xe] 4f^8$. In $4f^8$,there are $6$ unpaired electrons.
Thus,$Tb^{+3}$ has $6$ unpaired $f-$electrons.

(D) Due to lanthanide contraction, the atomic radii of $Zr$ $(160 \ pm)$ and $Hf$ $(159 \ pm)$ are almost identical.
Because of this similarity in atomic sizes, $Zr$ and $Hf$ exhibit very similar chemical properties and are difficult to separate.
Therefore, both statements $(a)$ and $(b)$ are correct.

(A) The lanthanoids exhibit a common oxidation state of $+3$.
However,$Ce$ (Cerium,$Z=58$) has an electronic configuration of $[Xe] 4f^1 5d^1 6s^2$.
Upon losing four electrons,it attains the stable noble gas configuration of $Xe$.
Thus,$Ce^{4+}$ is a stable ion in an aqueous medium,making it a strong oxidizing agent.
Other elements like $Eu^{2+}$ and $Yb^{2+}$ are stable,while $Tl$ is a post-transition metal that typically shows $+1$ and $+3$ oxidation states.

(A) . Due to lanthanide contraction,the ionic radius decreases from $La^{3+}$ to $Lu^{3+}$. According to Fajan's rule,as the size decreases,the covalent character increases,which leads to a decrease in basic strength. Thus,$La(OH)_3$ is more basic than $Lu(OH)_3$. Therefore,the statement in option $A$ is $INCORRECT$.
$B$. Lanthanide carbides $(LnC_2)$ contain the $C_2^{2-}$ ion,which reacts with water to produce acetylene $(C_2H_2)$.
$C$. Cerium sulfide $(CeS)$ is highly stable and is used in the manufacture of crucibles.
$D$. Misch-metal,an alloy of lanthanoids (approx. $95\%$) and iron (approx. $5\%$) with traces of $S, C, Ca,$ and $Al$,is used in cigarette lighters.