Lanthanoids and Actinoids Questions in English

(C) The $+3$ oxidation state of lanthanoids is the most stable. Therefore,lanthanoids in the $+4$ oxidation state have a strong tendency to gain an electron and convert into the $+3$ state,acting as strong oxidizing agents.
For example,$Ce^{+4}$ in $CeO_{2}$ is used to oxidize alcohols,aldehydes,and ketones.
Lanthanoids in the $+2$ oxidation state have a strong tendency to lose an electron and convert into the $+3$ oxidation state,thus acting as strong reducing agents.
Therefore,$EuSO_{4}$ (containing $Eu^{2+}$) acts as a strong reducing agent.
Both statements are true.

(D) Most lanthanoids form $M_{2}O_{3}$ type oxides.
However,$Ce$,$Pr$,$Nd$,$Tb$,and $Dy$ can form $MO_{2}$ type oxides.
$Yb$ does not form $MO_{2}$ type oxide; it typically forms $Yb_{2}O_{3}$.

(NONE) The atomic number of $Np$ is $93$.
The ground state electronic configuration of $Np$ is $[Rn] 5f^{4} 6d^{1} 7s^{2}$.
The $Rn$ core $(Z=86)$ contains $14$ electrons in the $4f$ subshell.
In the $5f$ subshell,there are $4$ electrons.
Total number of $f$ electrons = $14$ $(4f)$ + $4$ $(5f)$ = $18$.

(B) The electronic configuration of $Yb$ $(Z=70)$ is $[Xe] 4f^{14} 6s^2$.
In the $+2$ oxidation state,$Yb^{2+}$ has the configuration $[Xe] 4f^{14}$.
Since all $4f$ orbitals are completely filled,there are no unpaired electrons,making it diamagnetic.

(C) The atomic number of Europium $(Eu)$ is $63$.
The ground state electronic configuration of $Eu$ is $[Xe] 4f^{7} 6s^{2}$.
When $Eu$ forms the $Eu^{2+}$ ion,it loses two electrons from the $6s$ orbital.
Therefore,the electronic configuration of $Eu^{2+}$ is $[Xe] 4f^{7}$.
This configuration is highly stable due to the half-filled $f$-orbital $(f^{7})$,which makes $Eu^{2+}$ a strong reducing agent as it tends to lose an electron to reach the even more stable $Eu^{3+}$ state $(4f^{6})$.

(B) The correct answer is $B$.
Most trivalent Lanthanoid ions $(Ln^{3+})$ are colored in both solid and aqueous states due to $f-f$ transitions.
Actinoid contraction is indeed greater than Lanthanoid contraction because the $5f$ electrons provide poorer shielding than $4f$ electrons.
Lanthanoids are typical metallic elements and are good conductors of heat and electricity.
Actinoids are highly reactive metals,especially when finely divided,as they form a protective oxide layer slowly.

(A) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$.
By losing four electrons,$Ce$ attains the stable noble gas configuration of Xenon $([Xe])$,resulting in the $+4$ oxidation state.
Therefore,$Ce^{4+}$ is a stable ion.

(A) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$.
After losing $3$ electrons,$Ce^{3+}$ has the configuration $[Xe] 4f^1$.
However,$Ce$ can easily lose one more electron to form $Ce^{4+}$,which has the configuration $[Xe] 4f^0 5d^0$.
This $Ce^{4+}$ ion achieves a stable noble gas configuration,making it easy for $Ce$ to deviate from the $+3$ oxidation state.

(C) The electronic configuration of $Tb$ $(Z=65)$ is $[Xe] 4f^9 6s^2$. Therefore,$Tb^{4+}$ has the configuration $[Xe] 4f^7$,which is half-filled.
The electronic configuration of $Yb$ $(Z=70)$ is $[Xe] 4f^{14} 6s^2$. Therefore,$Yb^{2+}$ has the configuration $[Xe] 4f^{14}$,which is completely filled.
Thus,$Tb^{4+}$ and $Yb^{2+}$ correspond to half-filled and completely filled $f$ orbitals,respectively.

(C) The stability of the divalent state in lanthanoids is related to the electronic configuration and the reduction potential $E^0_{M^{3+}/M^{2+}}$.
$Eu^{2+}$ has a stable half-filled $f$-orbital configuration $([Xe] 4f^7)$.
$Yb^{2+}$ has a stable fully-filled $f$-orbital configuration $([Xe] 4f^{14})$.
However,comparing the standard reduction potentials: $E^0_{Eu^{3+}/Eu^{2+}} = -0.35 \ V$ and $E^0_{Yb^{3+}/Yb^{2+}} = -1.05 \ V$.
$A$ less negative reduction potential indicates that $Eu^{3+}$ is more easily reduced to $Eu^{2+}$ than $Yb^{3+}$ is to $Yb^{2+}$,making $Eu^{2+}$ the most stable among the given options.

(B) For the $3^{rd}$ ionisation energy,the electron must be removed from the $+2$ oxidation state of the given metals.
The electronic configuration of $Ce^{2+}$ is $4f^{1} 5d^{1}$.
The electronic configuration of $Nd^{2+}$ is $4f^{3}$.
The electronic configuration of $Eu^{2+}$ is $4f^{7}$.
The electronic configuration of $Dy^{2+}$ is $4f^{10}$.
$Eu$ will have the maximum $3^{rd}$ ionisation enthalpy because the electron has to be removed from the stable half-filled $4f^{7}$ configuration of $Eu^{2+}$.
$Ce$ will have the lowest $3^{rd}$ ionisation enthalpy because the electron has to be removed from the $5d$-orbital,which is more shielded and further from the nucleus compared to the $4f$-orbitals.

(D) In the lanthanoid series,the most stable oxidation state is $+3$.
$Ce^{4+}$ is a strong oxidizing agent because it has a strong tendency to revert to the more stable $+3$ oxidation state $(Ce^{4+} + e^- \rightarrow Ce^{3+})$.
Similarly,$Tb^{4+}$ also acts as a good oxidizing agent as it tends to gain an electron to reach the stable $+3$ state $(Tb^{4+} + e^- \rightarrow Tb^{3+})$.
Therefore,both $Ce^{4+}$ and $Tb^{4+}$ are good oxidizing agents.

(B) The atomic number of Neodymium $(Nd)$ is $60$.
The ground state electronic configuration of $Nd$ is $[Xe] 4f^4 6s^2$.
When $Nd$ forms the $Nd^{2+}$ ion,it loses two electrons from the outermost $6s$ orbital.
Therefore,the electronic configuration of $Nd^{2+}$ is $[Xe] 4f^4$.

(A) The electronic configurations of the given lanthanoids in their ground state are as follows:
$1.$ ${}_{62}Sm: [Xe] 4f^6 6s^2$
$2.$ ${}_{63}Eu: [Xe] 4f^7 6s^2$
$3.$ ${}_{64}Gd: [Xe] 4f^7 5d^1 6s^2$
$4.$ ${}_{65}Tb: [Xe] 4f^9 6s^2$
$5.$ ${}_{61}Pm: [Xe] 4f^5 6s^2$
$A$ half-filled $f$-orbital contains $7$ electrons $(f^7)$.
From the configurations above,$Eu$ $(4f^7 6s^2)$ and $Gd$ $(4f^7 5d^1 6s^2)$ both possess a half-filled $4f$ subshell.
Therefore,the correct elements are $B$ $(Eu)$ and $D$ $(Gd)$.

(C) The decrease in atomic size across the actinoid series is more pronounced than in the lanthanoid series.
This is because the $5f$ orbitals have a poorer shielding effect compared to the $4f$ orbitals.
Due to the poor shielding by $5f$ electrons,the effective nuclear charge $(Z_{eff})$ increases more significantly across the actinoid series.
Since atomic size is inversely proportional to $Z_{eff}$ $(Size \propto \frac{1}{Z_{eff}})$,the size decreases more rapidly.

(B) The $4f$ orbitals are deeply buried within the atom,shielded by outer shells,which limits their participation in chemical bonding.
In contrast,the $5f$ orbitals are less buried and extend further from the nucleus compared to $4f$ orbitals.
Because the $5f$ electrons experience less nuclear attraction and are more accessible,they can participate in bonding to a far greater extent than $4f$ electrons.
Therefore,both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(B) The third ionization energy $(IE_3)$ is high for elements that have stable electronic configurations in their $+2$ oxidation state,as removing the third electron disrupts this stability.
$Eu$ $(Z=63)$ has the configuration $[Xe] 4f^7 6s^2$. Its $Eu^{2+}$ ion is $[Xe] 4f^7$,which is a stable half-filled configuration.
$Yb$ $(Z=70)$ has the configuration $[Xe] 4f^{14} 6s^2$. Its $Yb^{2+}$ ion is $[Xe] 4f^{14}$,which is a stable fully-filled configuration.
Therefore,both $Eu$ and $Yb$ exhibit high third-ionization energy.

(A) The atomic number of Neodymium $(Nd)$ is $60$.
The nearest noble gas is Xenon $(Xe)$ with atomic number $54$.
The remaining $6$ electrons are filled in the $4f$ and $6s$ orbitals.
According to the Aufbau principle and experimental observations for lanthanoids,the configuration is $[Xe] 4f^4 6s^2$.

(B) Statement $I$ is true: The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$. Upon losing four electrons,$Ce^{+4}$ attains the stable noble gas configuration of $[Xe]$.
Statement $II$ is true: $Ce^{+4}$ has a high reduction potential $(E^{\circ} = +1.74 \ V)$ for the reaction $Ce^{+4} + e^- \rightarrow Ce^{+3}$. Due to this,it readily gains an electron to revert to the more stable $+3$ oxidation state,making it a strong oxidizing agent.

(C) Lanthanides generally exhibit a stable $+3$ oxidation state.
The electronic configuration of $Eu$ $(Z=63)$ is $[Xe] 4f^7 6s^2$.
Therefore,$Eu^{2+}$ has the configuration $[Xe] 4f^7$.
Although the $4f^7$ configuration is half-filled and stable,$Eu^{2+}$ tends to lose one more electron to achieve the most stable $+3$ oxidation state $(Eu^{3+})$ common to lanthanides.
Thus,$Eu^{2+}$ acts as a strong reducing agent.
In contrast,$Ce^{4+}$ $([Xe] 4f^0)$ tends to gain an electron to reach the $+3$ state,making it a strong oxidizing agent.

(B) species is diamagnetic if it has no unpaired electrons ($4f^0$ or $4f^{14}$ configuration).
$La$ $(Z=57)$ has the electronic configuration $[Xe] 5d^1 6s^2$. Thus,$La^{3+}$ is $[Xe] 4f^0$,which is diamagnetic.
$Ce$ $(Z=58)$ has the electronic configuration $[Xe] 4f^1 5d^1 6s^2$. Thus,$Ce^{4+}$ is $[Xe] 4f^0$,which is diamagnetic.
Therefore,both $La^{3+}$ and $Ce^{4+}$ are diamagnetic.

(C) Lanthanoids are the elements with atomic numbers $57$ to $71$.
$Eu$ $(Z=63)$,$Er$ $(Z=68)$,$Tb$ $(Z=65)$,$Yb$ $(Z=70)$,and $Lu$ $(Z=71)$ are all lanthanoids.
$Cm$ $(Z=96)$ is an actinide.
Therefore,only $1$ element $(Cm)$ does not belong to the lanthanoid series.

(B) The colour of lanthanoid ions depends on the presence of unpaired $f$-electrons ($f-f$ transitions).
Electronic configurations of the given ions are:
$La^{3+} = [Xe] 4f^0$ (No unpaired electrons)
$Nd^{3+} = [Xe] 4f^3$ ($3$ unpaired electrons)
$Sm^{3+} = [Xe] 4f^5$ ($5$ unpaired electrons)
$Eu^{3+} = [Xe] 4f^6$ ($6$ unpaired electrons)
$Lu^{3+} = [Xe] 4f^{14}$ (No unpaired electrons)
Ions with $f^0$ or $f^{14}$ configurations are colourless because they lack unpaired electrons for $f-f$ transitions.
Thus,$La^{3+}$ and $Lu^{3+}$ are colourless.
The total number of colourless ions is $2$.

(B) The atomic number of Einsteinium $(Es)$ is $99$.
The electronic configuration is based on the noble gas Radon ($Rn$,atomic number $86$).
The remaining $13$ electrons are filled in the $5f$,$6d$,and $7s$ orbitals.
The correct configuration is $[Rn] 5f^{11} 6d^0 7s^2$.

(D) substance is diamagnetic if it has no unpaired electrons $(n = 0)$.
Electronic configurations of the given ions:
$Ce^{4+} \Rightarrow [Xe] 4f^0$ (No unpaired electrons,diamagnetic)
$Yb^{2+} \Rightarrow [Xe] 4f^{14}$ (All electrons paired,diamagnetic)
$Ce^{3+} \Rightarrow [Xe] 4f^1$ (One unpaired electron,paramagnetic)
$Eu^{2+} \Rightarrow [Xe] 4f^7$ (Seven unpaired electrons,paramagnetic)
$Gd^{3+} \Rightarrow [Xe] 4f^7$ (Seven unpaired electrons,paramagnetic)
$Eu^{3+} \Rightarrow [Xe] 4f^6$ (Six unpaired electrons,paramagnetic)
$Pm^{3+} \Rightarrow [Xe] 4f^4$ (Four unpaired electrons,paramagnetic)
$Sm^{3+} \Rightarrow [Xe] 4f^5$ (Five unpaired electrons,paramagnetic)
Thus,the pair $Ce^{4+}$ and $Yb^{2+}$ is diamagnetic.

(A) The electronic configurations of the given ions are:
$Eu^{2+} (Z=63): [Xe] 4f^7 6s^0$
$Gd^{3+} (Z=64): [Xe] 4f^7 5d^0 6s^0$
$Eu^{3+} (Z=63): [Xe] 4f^6 6s^0$
$Tb^{3+} (Z=65): [Xe] 4f^8 6s^0$
$Sm^{2+} (Z=62): [Xe] 4f^6 6s^0$
Thus,only $Eu^{2+}$ and $Gd^{3+}$ have a $4f^7$ configuration.

(C) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$. $Ce^{4+}$ is stable due to noble gas configuration,making $Ce^{3+}$ a reducing agent.
$Tb$ $(Z=65)$ has configuration $[Xe] 4f^9 6s^2$. $Tb^{4+}$ has a $4f^7$ configuration,which is half-filled and stable,but $Tb^{4+}$ itself is a strong oxidizing agent because it readily gains an electron to achieve the stable $4f^7$ configuration from a $4f^8$ state (if it were $Tb^{3+}$) or simply acts as a strong oxidant to return to the stable $Tb^{3+}$ state.
Among the given options,$Tb^{4+}$ is the strongest oxidizing agent because it has a high tendency to be reduced to $Tb^{3+}$.

(A) The basicity of lanthanoid oxides decreases as the ionic radius decreases from $La^{3+}$ to $Lu^{3+}$ due to lanthanoid contraction.
Since $Pr$ (Praseodymium) appears earliest in the lanthanoid series among the given options,it has the largest ionic radius.
Therefore,$Pr_2O_3$ is the most basic oxide among the choices provided.

(A) The actinoids exhibit a wide range of oxidation states due to the comparable energies of $5f$,$6d$,and $7s$ orbitals.
$Np$ (Neptunium) and $Pu$ (Plutonium) are known to exhibit oxidation states ranging from $+3$ to $+7$.
$U$ (Uranium) typically shows $+3$ to $+6$,while $Am$ (Americium) shows $+3$ to $+6$.
Therefore,$Np$ and $Pu$ are the elements that can form compounds in the $+3$ to $+7$ oxidation states.

(B) The general oxidation state of lanthanoids is $+3$.
However,some lanthanoids exhibit $+2$ or $+4$ oxidation states due to stable electronic configurations (half-filled or fully-filled $f$-orbitals).
$Eu$ $(Z = 63)$ has the configuration $[Xe] 4f^7 6s^2$. By losing two electrons,it forms $Eu^{2+}$ $(4f^7)$,which is stable due to the half-filled $f$-subshell.
$Yb$ $(Z = 70)$ has the configuration $[Xe] 4f^{14} 6s^2$. By losing two electrons,it forms $Yb^{2+}$ $(4f^{14})$,which is stable due to the fully-filled $f$-subshell.
Therefore,the pair showing $+2$ oxidation state is $_{63}Eu$ and $_{70}Yb$.

(D) $1$. Lanthanides are elements from $Z = 57$ to $71$.
$2$. All lanthanides are non-radioactive except Promethium $(Pm)$,which is radioactive. So,statement $A$ is correct.
$3$. All lanthanides have a $6s^2$ configuration in their outermost shell. So,statement $B$ is correct.
$4$. The most common and stable oxidation state for all lanthanides is $+3$. So,statement $C$ is correct.
$5$. Lanthanum $(La, Z = 57)$ has the electronic configuration $[Xe] 5d^1 6s^2$. It does not have any electrons in the $4f$ orbital. Therefore,the statement that Lanthanum has $[Xe] 4f^1 6s^2$ is incorrect.

(A) As we move from $Ce$ to $Lu$ in the lanthanoid series,the atomic and ionic radii decrease due to lanthanoid contraction.
$Z_{eff}$ increases,which leads to a decrease in the size of the ions.
Due to the decrease in size,the polarizing power of the ions increases,which increases the covalent character of halides.
As the size decreases,the ionic character of the oxides decreases,meaning the basic nature of the oxides decreases.
Therefore,the basic nature of oxides decreases from $Ce$ to $Lu$.

(C) The Assertion is True: The ionisation enthalpies of early actinoids are lower than those of early lanthanoids because the $5f$ orbitals are more extended in space than $4f$ orbitals,making the $5f$ electrons less tightly bound to the nucleus.
The Reason is False: The shielding effect of $5f$ electrons is actually poorer than that of $4f$ electrons. Because $5f$ electrons are less effective at shielding,the effective nuclear charge experienced by the outer electrons is higher,but the spatial extension of $5f$ orbitals dominates,leading to lower ionisation enthalpies for actinoids compared to lanthanoids.

(D) The highest oxidation state in plutonium (Atomic number = $94$) is $+7$.
This is because the $5f$,$6d$,and $7s$ energy levels have comparable energies,allowing for a wide range of oxidation states.
Plutonium $(Pu)$ exhibits oxidation states of $+3$,$+4$,$+5$,$+6$,and $+7$ in its various compounds.

(B) In the lanthanoid series,as the atomic number $(Z)$ increases,the atomic radius decreases due to lanthanoid contraction.
Since the atomic numbers are $Ce$ $(58)$,$Pm$ $(61)$,$Tm$ $(69)$,and $Yb$ $(70)$,the decreasing order of their atomic radii is $Ce > Pm > Tm > Yb$.

(D) In the lanthanide series,the ionic radius decreases from $La^{3+}$ to $Lu^{3+}$ due to lanthanide contraction.
As the ionic radius decreases,the covalent character of the $M-OH$ bond increases,which leads to a decrease in basic strength.
Since $La^{3+}$ has the largest ionic radius among the given options,$La(OH)_3$ is the most basic hydroxide.

(A) Inner transition elements are the elements in which the last electron enters the $f$-orbital. These include the lanthanoids $(58-71)$ and actinoids $(90-103)$.
$Cm$ ($Curium$,atomic number $96$) is an actinoid,which is a type of inner transition element.
$W$ $(Tungsten)$,$Mo$ $(Molybdenum)$,and $Ru$ $(Ruthenium)$ are transition elements belonging to the $d$-block.

(A) The atomic number of Lutetium $(Lu)$ is $71$.
The electronic configuration of $Lu$ is $[Xe] 4f^{14} 5d^1 6s^2$.
In the $+3$ oxidation state,$Lu$ loses three electrons (two from $6s$ and one from $5d$).
The electronic configuration of $Lu^{3+}$ is $[Xe] 4f^{14}$.
Since the $4f$ subshell is completely filled with $14$ electrons,there are no unpaired electrons present.
Therefore,the number of unpaired electrons is $0$.

(A) The electronic configuration of lanthanoids follows the general pattern $(Xe)4f^{n} 5d^{0-1} 6s^{2}$.
$A$ half-filled $f$-orbital corresponds to $f^{7}$.
Looking at the provided table:
- $Ce$ $(Z=58)$: $(Xe)4f^{1} 5d^{1} 6s^{2}$
- $Pm$ $(Z=61)$: $(Xe)4f^{5} 6s^{2}$
- $Sm$ $(Z=62)$: $(Xe)4f^{6} 6s^{2}$
- $Eu$ $(Z=63)$: $(Xe)4f^{7} 6s^{2}$
Since $Eu$ has $4f^{7}$,it possesses a half-filled $f$-orbital. This configuration is stable and matches the expected filling order for this element.

(C) Promethium is a chemical element with the symbol $Pm$ and atomic number $61$.
All of its isotopes are radioactive.
It is extremely rare,with only about $500-600 \ g$ naturally occurring in Earth's crust at any given time.

(B) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$.
Cerium $(Ce)$ exhibits $+3$ and $+4$ oxidation states.
The $+3$ state is the most stable,while the $+4$ state is also common due to the attainment of a stable noble gas configuration $([Xe])$ upon losing four electrons.

(A) The electronic configuration of $Cerium$ ($Ce$,$Z=58$) is $[Xe] 4f^1 5d^1 6s^2$.
Due to the loss of three electrons,it forms a stable $+3$ oxidation state.
It also exhibits a $+4$ oxidation state because the removal of four electrons results in a stable noble gas configuration $([Xe])$.

(A) The lanthanoids are the elements with atomic numbers $57$ to $71$ ($La$ to $Lu$).
$Er$ (Erbium) has an atomic number of $68$,which falls within the range of the lanthanoid series.
$Am$ (Americium),$Np$ (Neptunium),and $Lr$ (Lawrencium) are actinoid elements.

(D) The electronic configuration of $Tb$ $(Z = 65)$ is $[Xe] 4f^9 6s^2$.
In the $+4$ oxidation state,$Tb^{4+}$ loses four electrons,resulting in the configuration $[Xe] 4f^7$.
This $f^7$ configuration is half-filled and therefore highly stable.
Thus,$Tb$ exhibits a $+4$ oxidation state with an $f^7$ configuration.

(C) The most common and stable oxidation state for lanthanoid elements $(Ln)$ is $+3$.
Therefore,when a lanthanoid reacts with the hydroxide ion $(OH^-)$,it forms a compound where the charge is balanced.
Since the hydroxide ion has a charge of $-1$,three hydroxide ions are required to balance the $+3$ charge of the $Ln^{3+}$ ion.
Thus,the general formula for lanthanoid hydroxide is $Ln(OH)_3$.

(C) In the lanthanoid series,the ionic radii of $Ln^{3+}$ ions decrease regularly with an increase in atomic number due to lanthanoid contraction.
Since $La$ $(Z=57)$ has the lowest atomic number among the given elements $(La, Pr, Sm, Yb)$,its $La^{3+}$ ion will have the largest ionic size.
The order of ionic size is: $La^{3+} > Pr^{3+} > Sm^{3+} > Yb^{3+}$.

(B) The effective magnetic moment $(\mu_{eff})$ is calculated using the formula $\mu_{eff} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For a lanthanoid to exhibit zero effective magnetic moment in the $+3$ state,it must have zero unpaired electrons $(n=0)$.
$Lu$ (Lutetium) has the atomic number $71$ and its electronic configuration is $[Xe] \ 4f^{14} \ 5d^1 \ 6s^2$.
In the $+3$ oxidation state,$Lu^{3+}$ loses three electrons to form $[Xe] \ 4f^{14}$.
Since the $4f$ subshell is completely filled,there are no unpaired electrons $(n=0)$,resulting in a magnetic moment of $0 \ BM$.

(D) The first ionisation enthalpy $(IE_1)$ of lanthanoids generally increases across the series due to the lanthanoid contraction,which leads to a decrease in atomic size and an increase in effective nuclear charge.
Among the given options,$Yb$ (Ytterbium,$Z=70$) has the highest atomic number.
$Yb$ has a stable electronic configuration of $[Xe] 4f^{14} 6s^2$.
Due to the completely filled $4f$ subshell and the high effective nuclear charge,$Yb$ exhibits a very high first ionisation enthalpy compared to the other listed lanthanoids ($Nd$,$Dy$,$Tm$).

(A) In the lanthanoid series,as we move from $La$ $(Z = 57)$ to $Lu$ $(Z = 71)$,the ionic radius of the $M^{3+}$ ions decreases regularly. This phenomenon is known as lanthanoid contraction.
Lanthanoid contraction occurs due to the poor shielding effect of $4f$ electrons,which causes the effective nuclear charge to increase,pulling the electrons closer to the nucleus.
Therefore,the ionic size decreases as the atomic number increases.
Among the given elements,$Lu$ $(Z = 71)$ has the highest atomic number,so $Lu^{3+}$ has the smallest ionic size.

(C) The electronic configuration of Cerium ($Ce$,atomic number $58$) is $[Xe] 4f^1 5d^1 6s^2$.
When $Ce$ loses four electrons to form $Ce^{4+}$,it loses the two $6s$,one $5d$,and one $4f$ electron.
The resulting configuration of $Ce^{4+}$ is $[Xe] 4f^0$,which is a stable noble gas configuration.
Therefore,$Ce$ is the lanthanoid that exhibits a $+4$ oxidation state with an $f^0$ configuration.