Lanthanoids and Actinoids Questions in English

(D) The rare-earth metals are defined as the $15$ lanthanoids (from $Lanthanum$ to $Lutetium$) plus $Scandium$ and $Yttrium$. $Lanthanum$ $(Z = 57)$ is the first element of the lanthanoid series and is considered the first element of the rare-earth metals.

(C) The lanthanides are the elements with atomic numbers $57$ to $71$.
$Cerium$ $(Ce)$ has an atomic number of $58$,which places it within the lanthanide series.
$Cadmium$ $(Cd)$ is a transition metal ($Group$ $12$).
$Californium$ $(Cf)$ is an actinide ($atomic$ $number$ $98$).
$Cesium$ $(Cs)$ is an alkali metal ($Group$ $1$).
Therefore,the correct option is $C$.

(C) The term $ \text{rare-earth elements} $ typically refers to the $15$ lanthanides $( \text{Lanthanum to Lutetium} )$ plus $ \text{Scandium} $ and $ \text{Yttrium} $,making a total of $17$ elements. However,in many contexts,the $14$ elements from $ \text{Cerium} $ $(Z=58)$ to $ \text{Lutetium} $ $(Z=71)$ are specifically referred to as the lanthanide series. Given the standard classification,the total count is $17$.

(D) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] \,4f^1 \,5d^1 \,6s^2$.
Since the differentiating electron enters the $4f$ orbital,$Ce$ is classified as an $f$-block element.

(B) In lanthanoids,electrons are progressively filled in $4f$ orbitals.
The general electronic configuration for lanthanoids is $[Xe] \, 4f^{1-14} \, 5d^{0-1} \, 6s^2$.

(B) The correct answer is $(B)$. The ionic radii of trivalent lanthanides do not increase with an increase in the atomic number; instead,they show a steady decrease,a phenomenon known as lanthanoid contraction.

(D) $(d)$ The trivalent ion having the largest size in the lanthanide series is $La^{3+}$.
As we move from $La^{3+}$ to $Lu^{3+}$, the ionic radius decreases due to lanthanide contraction, which is the steady decrease in the size of lanthanide ions with an increase in atomic number.

(D) The elements provided include lanthanoids ($Eu$,$La$,$Gd$) and an actinoid $(Am)$.
Actinoids exhibit a greater range of oxidation states compared to lanthanoids due to the comparable energies of $5f$,$6d$,and $7s$ orbitals.
$Am$ (Americium) exhibits oxidation states ranging from $+2$ to $+6$ (and sometimes $+7$ in specific conditions).
In contrast,lanthanoids like $La$,$Eu$,and $Gd$ primarily show the $+3$ oxidation state,with $Eu$ also showing $+2$ and $Gd$ showing $+3$ as its most stable state.
Therefore,$Am$ shows the maximum number of different oxidation states.

(A) The ionic size depends on the charge and the shell number.
$Sn^{4+}$ has a configuration of $[Kr] 4d^{10}$,which is a pseudo-noble gas configuration,and it is in the $5th$ period,making it larger than the lanthanide ions.
$Ce^{4+}$ $(Z=58)$ is a lanthanide ion with a $4f^0$ configuration.
$Yb^{3+}$ $(Z=70)$ and $Lu^{3+}$ $(Z=71)$ are lanthanide ions.
Due to lanthanide contraction,the ionic size decreases as the atomic number increases from $Ce$ to $Lu$.
Thus,the order is $Sn^{4+} > Ce^{4+} > Yb^{3+} > Lu^{3+}$.

(A) The $f$-block elements (lanthanoids and actinoids) form hydrides with limiting compositions of $MH_2$ and $MH_3$.

(A) The element californium $(Cf)$ has an atomic number of $98$.
It is a member of the $5f$ block elements,which are known as the actinide series.

(A) Misch metal is an alloy consisting of a significant proportion of rare earth elements (lanthanoids).
The typical composition is:
Rare earth metals: $94-95\%$
Iron $(Fe)$: $5\%$
Traces of $S, C, Ca, Al$.
Among the given options,$La$ (Lanthanum) is a rare earth metal (lanthanoid) which is the primary component of misch metal.

(D) Lanthanum $(Z = 57)$ has the electronic configuration $[Xe] 5d^1 6s^2$.
Although it does not have any electrons in the $4f$ orbital in its ground state,it is considered the first member of the lanthanoid series because its chemical properties are very similar to the elements of the $4f$ block (lanthanoids).
Therefore,it is grouped with the $f-$block elements.

(D) The general electronic configuration of $f-$ block elements is $ns^2(n-1)d^{0-1}(n-2)f^{1-14}$. However,the given configuration $ns^2(n-1)d^{1-10}(n-2)f^{1-14}$ is a generalized representation often used to identify the filling of the $f-$ orbital,which characterizes the $f-$ block elements (Lanthanoids and Actinoids).

(D) is the correct answer.
Terbium $(Tb)$ is a lanthanide element with atomic number $65$,belonging to the $4f$-series with the electronic configuration $[Xe] 4f^9 6s^2$.
Curium $(Cm)$,Californium $(Cf)$,and Uranium $(U)$ are all members of the $5f$-series,which are known as actinides.

(B) The chemistry of the lanthanides differs from main group elements and transition metals because of the nature of the $4f$ orbitals.
These orbitals are buried inside the atom and are shielded from the atom's environment by the $5s$,$5p$,and $5d$ electrons.
As a consequence,the ionic radii of trivalent lanthanides steadily decrease with an increase in atomic number,a phenomenon known as the lanthanide contraction.
Therefore,statement $B$ is false because the radii decrease,not increase.
All lanthanide elements exhibit the $+3$ oxidation state as their most typical state.
Due to the lanthanide contraction,these elements have high density.

(D) . Europium is an $f-$block element as it follows the general electronic configuration of the $f-$block elements $(4f^{1-14}5d^{0,1}6s^2)$.
$Eu = [Xe] \, 4f^7 6s^2$

(D) The general oxidation state of lanthanoids is $+III$. However,some elements exhibit $+II$ or $+IV$ oxidation states due to stable electronic configurations (like half-filled or fully-filled $f$-orbitals).
Europium ($Eu$,atomic number $63$) has the electronic configuration $[Xe] 4f^7 6s^2$.
It easily loses two electrons from the $6s$ orbital to form the $Eu^{2+}$ ion,which has a stable half-filled $4f^7$ configuration.
It also commonly exists in the $+III$ oxidation state $(Eu^{3+})$.
Therefore,$+II$ and $+III$ oxidation states are common for $Eu$.

(D) Lanthanide contraction occurs because $f-$orbital electrons are poor shielders of nuclear charge.
Due to the poor shielding effect of $f-$electrons,the effective nuclear charge increases as we move across the lanthanide series,which results in a gradual decrease in the atomic and ionic radii.

(A) The rare-earth elements,also known as lanthanides,consist of elements with atomic numbers from $58$ to $71$.
These elements are characterized by the filling of the $4f$ orbitals.

(C) The $f-$block elements consist of two series: the Lanthanoids (atomic numbers $58$ to $71$) and the Actinoids (atomic numbers $90$ to $103$).
Therefore,both $(a)$ and $(b)$ are correct.

(C) The phenomenon where there is a regular decrease in the atomic and ionic radii of lanthanoids with an increase in atomic number is known as lanthanide contraction. This is a characteristic feature of the $f-$ block elements.

(B) Inner transition elements are defined as the $f-$block elements.
In these elements,the three outermost shells are incomplete,specifically the $(n-2)f$,$(n-1)d$,and $ns$ orbitals.
Therefore,the number of incomplete orbitals is $3$.

(C) Cerium ($Ce$,atomic number $58$) is a lanthanoid element.
Its electronic configuration is $[Xe] 4f^1 5d^1 6s^2$.
It commonly exhibits $+3$ oxidation state,which is the characteristic oxidation state of lanthanoids.
It also exhibits $+4$ oxidation state because,in the $+4$ state,it achieves a stable noble gas configuration $([Xe])$.
Therefore,the most common oxidation states of $Ce$ are $+3$ and $+4$.

(A) Cerium $(Z = 58)$ has an electronic configuration of $[Xe] 4f^1 5d^1 6s^2$.
It exhibits both $+3$ and $+4$ oxidation states.
The $+4$ oxidation state of cerium is well-known and stable in aqueous solutions,where it acts as a strong oxidizing agent.
Therefore,the statement that the $+4$ oxidation state is not known in solutions is incorrect.

(A) is the correct option.
Lanthanoids consist of $14$ elements starting from Cerium $(Z = 58)$ to Lutetium $(Z = 71)$.
These elements are characterized by the progressive filling of the $4f$ orbitals.
They are placed in the sixth period of the periodic table and are shown in a separate row at the bottom of the table.

(B) Both lanthanides and actinides exhibit a common stable oxidation state of $+3$.
The chemical properties of actinides are very similar to those of lanthanides when both are in the $+3$ oxidation state.

(B) The lanthanide contraction refers to the steady decrease in the atomic and ionic radii of lanthanoids as the atomic number increases from $La$ $(Z=57)$ to $Lu$ $(Z=71)$.
This contraction is observed in both the neutral atoms and their corresponding $M^{3+}$ ions.

(D) The atomic number of Cerium $(Ce)$ is $58$.
The electronic configuration of $Ce$ is $[Xe] 4f^1 5d^1 6s^2$.
Since the last electron enters the $4f$ orbital,it belongs to the $f-$ block elements (Lanthanoids).

(A) Unlike lanthanoids,actinoids exhibit a wider range of oxidation states (from $+3$ to $+7$) because the energy gap between the $5f$,$6d$,and $7s$ subshells is very small.
In contrast,the $4f$ orbitals in lanthanoids are more deeply buried and have a larger energy gap relative to the $5d$ and $6s$ orbitals,which limits their oxidation states primarily to $+3$.

(A) In the lanthanide series,as the atomic number increases from $La$ $(Z=57)$ to $Lu$ $(Z=71)$,the ionic radii of the trivalent ions $(Ln^{3+})$ decrease steadily due to lanthanoid contraction.
Since $La^{3+}$ is the first ion in the series,it possesses the largest ionic radius among the given options.
Therefore,the correct option is $A$.

(A) Lanthanoid contraction is primarily caused by the poor shielding effect of $4f$ electrons.
As we move across the lanthanoid series,the nuclear charge increases by one unit at each step,while the additional electron enters the $4f$ subshell.
The $4f$ orbitals have a very diffuse shape,which results in poor shielding of the nuclear charge for the outer electrons.
Consequently,the effective nuclear charge increases,causing the atomic and ionic radii to decrease gradually.

(B) The elements are classified as follows:
$Ce$ $(Cerium)$ is a lanthanide.
$Cf$ $(Californium)$ is an actinide.
$Ca$ $(Calcium)$ is an alkaline earth metal.
$Cs$ $(Cesium)$ is an alkali metal.
Therefore,the correct option is $B$.

(A) In the lanthanoid series,as the atomic number increases from $La$ $(Z=57)$ to $Lu$ $(Z=71)$,the ionic radii of the trivalent ions $(Ln^{+3})$ decrease steadily due to the phenomenon known as lanthanoid contraction.
This contraction occurs because the $4f$ electrons provide poor shielding for the increasing nuclear charge.
Therefore,the order of ionic radii for the given ions is $La^{+3} > Ce^{+3} > Pm^{+3} > Yb^{+3}$.
In increasing order,this is $Yb^{+3} < Pm^{+3} < Ce^{+3} < La^{+3}$.

(D) The lanthanoids are the elements from $Z = 57$ to $71$.
$La$ $(Z = 57)$ is the first element of the $4f$ series and is often considered the prototype for lanthanoid chemistry due to its $3+$ oxidation state and ionic radius similarity.
$Lu$ $(Z = 71)$ is the last element of the lanthanoid series.
$Ce$ $(Z = 58)$ is a lanthanoid itself.
$Lr$ $(Z = 103)$ is an actinoid.
Among the given options,$La$ is the most characteristic element representing the lanthanoid series properties.

(B) As we move across the lanthanoid series from $La$ to $Lu$,the ionic radius of the $Ln^{3+}$ ions decreases due to lanthanoid contraction.
According to Fajan's rule,as the ionic radius decreases,the covalent character of the $Ln-OH$ bond increases.
Due to the increase in covalent character,the tendency to release $OH^-$ ions decreases.
Therefore,the basicity of lanthanoid hydroxides decreases from $La(OH)_3$ to $Lu(OH)_3$.

(B) The separation of $Lanthanoid$ elements is difficult due to the $Lanthanoid$ contraction,which results in very similar ionic radii and chemical properties for all $Lanthanoid$ elements. However,they can be separated based on the slight differences in their basicity,which arises from the gradual decrease in ionic radii across the series.

(D) The lanthanoids typically exhibit a $+3$ oxidation state.
However,some elements show $+2$ or $+4$ oxidation states due to stable electronic configurations (like half-filled or fully-filled $f$-orbitals).
$Eu$ ($Europium$,atomic number $63$) has the electronic configuration $[Xe] 4f^7 6s^2$.
By losing two electrons from the $6s$ orbital,it forms the $Eu^{2+}$ ion,which has a stable half-filled $4f^7$ configuration.
Thus,$Eu$ commonly exhibits both $+2$ and $+3$ oxidation states.

(B) The lanthanoid series consists of $14$ elements starting from Cerium ($Ce$,atomic number $58$) and ending at Lutetium ($Lu$,atomic number $71$).
Lanthanum ($La$,atomic number $57$) is the element preceding the lanthanoids and is often studied alongside them due to its similar properties,but it is technically a $d$-block element.
Therefore,the lanthanoid series is defined as $Ce$ to $Lu$.

(D) The electronic configuration of $Eu$ is $[Xe] 4f^7 6s^2$.
In the $+2$ oxidation state,$Eu^{2+}$ has the configuration $[Xe] 4f^7$.
Since the $4f$ subshell is half-filled,it is highly stable.
Therefore,$Eu$ exhibits both $+2$ and $+3$ oxidation states.

(D) The general electronic configuration of actinoids (elements with atomic numbers $89$ to $103$) is represented as $[Rn] \, 5f^{0-14} \, 6d^{0-2} \, 7s^2$.
This configuration accounts for the filling of the $5f$ orbitals,the occasional occupancy of the $6d$ orbitals,and the filled $7s$ subshell.

(D) The electronic configuration of lanthanoids follows the general pattern $[Xe]4f^{1-14} 5d^{0-1} 6s^2$.
For Gadolinium ($Gd$,atomic number $64$),the configuration is $[Xe]4f^7 5d^1 6s^2$.
This stability arises because the $4f$ subshell is half-filled $(4f^7)$,which provides extra stability to the atom.

(A) In the lanthanoid series,the ionic radii of $Ln^{3+}$ ions decrease from $La^{3+}$ to $Lu^{3+}$ due to lanthanoid contraction.
Therefore,the order of ionic radii is $La^{3+} > Ce^{3+} > Pm^{3+} > Yb^{3+}$.
Arranging these in increasing order of ionic radii gives: $Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$.

(B) The lanthanoid series is characterized by the filling of $4f$ orbitals.
All lanthanoids exhibit a common $+3$ oxidation state.
Due to the lanthanoid contraction,the ionic radii of the members decrease gradually with an increase in atomic number.
Because of their similar chemical and physical properties,the separation of lanthanoids from one another is very difficult.
Statement $B$ is incorrect because while some lanthanoids exhibit a $+4$ oxidation state (e.g.,$Ce^{4+}$),it is not a property of all members of the series.

(A) The basic character of lanthanoid hydroxides,$Ln(OH)_3$,decreases as the ionic radius of the lanthanoid ion decreases due to lanthanoid contraction.
Lanthanoid hydroxides are less basic than alkaline earth metal hydroxides like $Ca(OH)_2$ but are more basic than $Al(OH)_3$,which is amphoteric.
Therefore,the basicity follows the order: $Ca(OH)_2 > Ln(OH)_3 > Al(OH)_3$.

(A) The lanthanoids are a series of $14$ elements starting from cerium ($Ce$,atomic number $58$) to lutetium ($Lu$,atomic number $71$).
They are often represented by the general symbol $Ln$,where $Ln$ stands for any lanthanoid element.

(D) In the lanthanoid series,as the atomic number increases from $La$ $(Z=57)$ to $Lu$ $(Z=71)$,the ionic radii of the $M^{3+}$ ions decrease due to lanthanoid contraction.
Since the atomic numbers are $La (57) < Ce (58) < Pm (61) < Yb (70)$,the ionic radii follow the order: $La^{3+} > Ce^{3+} > Pm^{3+} > Yb^{3+}$.
Therefore,the correct increasing order is $Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$.

(C) Lanthanoid contraction is the steady decrease in the atomic and ionic radii of lanthanoids with an increase in atomic number.
This is primarily due to the poor shielding effect of $4f$ electrons.
As the atomic number increases,the nuclear charge increases by one unit at each step,but the additional electron enters the $4f$ orbital.
Since $4f$ electrons have a very poor shielding effect,the effective nuclear charge experienced by the outer electrons increases,causing the electron cloud to be pulled closer to the nucleus,resulting in a decrease in size.

(D) The lanthanoids generally exhibit a $+3$ oxidation state. However,some elements show $+2$ or $+4$ oxidation states due to stable electronic configurations (like half-filled or fully-filled $f$-orbitals).
Europium ($Eu$,atomic number $63$) has the electronic configuration $[Xe] 4f^7 6s^2$.
It loses two electrons from the $6s$ orbital to form $Eu^{2+}$ $(4f^7)$,which is a stable half-filled configuration.
It can also lose one more electron to form $Eu^{3+}$ $(4f^6)$.
Therefore,$Eu$ exhibits both $+2$ and $+3$ oxidation states.