Lanthanoids and Actinoids Questions in English

(B) In the lanthanide series,the atomic radius generally decreases as the atomic number increases due to the poor shielding effect of $4f$ electrons,a phenomenon known as lanthanide contraction.
However,elements like $Eu$ $(Europium)$ and $Yb$ $(Ytterbium)$ show an anomaly in this trend.
$Eu$ has a stable half-filled $4f^7$ configuration,which results in a larger atomic radius compared to the expected trend because the metallic bonding is weaker due to the involvement of fewer electrons in the bonding process.
Therefore,the element $\underline{X}$ is $Eu$.

(B) In lanthanides,the $4f$ orbitals have a poor shielding effect.
As a result,the atomic radius generally decreases with an increase in atomic number from $Ce$ to $Lu$ due to lanthanide contraction.
However,$Eu$ $(Europium)$ is an exception,as it exhibits a larger atomic radius compared to its neighbors due to its stable $f^7$ configuration and metallic bonding characteristics.

(C) The atomic numbers of the given lanthanides are: $Ce = 58$,$Eu = 63$,and $Lu = 71$.
As we move across the lanthanide series from $Ce$ to $Lu$,the ionic radius decreases due to lanthanide contraction.
This decrease in ionic radius leads to an increase in the covalent character of the $M-OH$ bond according to Fajan's rule.
Consequently,the basic strength of the hydroxides decreases as the atomic number increases.
Therefore,the correct order of basic strength is $Lu(OH)_3 < Eu(OH)_3 < Ce(OH)_3$.

(D) Statement $I$: The electron gain enthalpy becomes more negative as we move across a period. The order of electronegativity is $P < S < Cl < F$. Thus, $P$ has the least negative electron gain enthalpy. Statement $I$ is correct.
Statement $II$: In $Eu$ $([Xe] 4f^7 6s^2)$ and $Yb$ $([Xe] 4f^{14} 6s^2)$, the atomic radii do not show the expected decrease due to lanthanoid contraction because of the stable half-filled and fully-filled $4f$ subshells, which provide better shielding. Thus, Statement $II$ is correct.
Statement $III$: Among lanthanoid hydroxides, basicity decreases as the ionic radius decreases (lanthanoid contraction). Since $Ce^{3+}$ has the largest ionic radius, $Ce(OH)_3$ is the most basic. Statement $III$ is correct.
Statement $IV$: The radius of a cation is always smaller than its parent atom. $Na^+$ $(95 \ pm)$ is smaller than $Na$ $(186 \ pm)$. The statement incorrectly assigns the values. Statement $IV$ is incorrect.
Therefore, statements $I, II,$ and $III$ are correct.

(C) The electronic configuration of Lanthanoids is $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$.
For $Eu$ $(Z=63)$,the configuration is $[Xe] 4f^7 6s^2$.
When $Eu$ forms $Eu^{2+}$ ion,it loses two $6s$ electrons,resulting in the configuration $[Xe] 4f^7$.
$Pr^{3+}$ $(Z=59)$ has configuration $[Xe] 4f^2$.
$Lu^{3+}$ $(Z=71)$ has configuration $[Xe] 4f^{14}$.
$Ce^{4+}$ $(Z=58)$ has configuration $[Xe] 4f^0$.
Therefore,the ion with $4f^7$ configuration is $Eu^{2+}$.

(C) The general electronic configuration of lanthanoids is $[Xe] 4f^{0-14} 5d^{0-1} 6s^2$.
Elements with a $5d^1$ electron in their ground state are $Ce (Z=58)$,$Gd (Z=64)$,and $Lu (Z=71)$.
Their configurations are:
$Ce = [Xe] 4f^1 5d^1 6s^2$
$Gd = [Xe] 4f^7 5d^1 6s^2$
$Lu = [Xe] 4f^{14} 5d^1 6s^2$
Thus,the correct set is $Ce, Gd, Lu$.

(B) The lanthanides that exhibit the $+4$ oxidation state in their oxides are $Pr, Nd, Tb,$ and $Dy$.
These elements show stability in the $+4$ state due to configurations approaching the stable $4f^0$ or $4f^7$ configurations.
Specifically,$PrO_2, NdO_2, TbO_2,$ and $DyO_2$ are known oxides where these elements exhibit the $+4$ oxidation state.
Therefore,the total number of such elements is $4$.

(B)

Element	Maximum Oxidation State
$U$	$+6$
$Np$	$+7$
$Am$	$+6$
$Pa$	$+5$

Among the given actinoids,Neptunium $(Np)$ exhibits the highest oxidation state of $+7$.

(C) Actinides exhibit a larger number of oxidation states than lanthanides because the energy gap between the $5f$,$6d$,and $7s$ subshells is very small.
This allows the electrons to be easily excited to higher energy levels,leading to variable oxidation states.
In contrast,the energy gap between $4f$,$5d$,and $6s$ subshells in lanthanides is relatively larger.
Therefore,the Assertion is correct,but the Reason is incorrect because the energy gap is small,not large.
Thus,the correct option is $(C)$.

(D) $(I)$ The ionic radius of lanthanoids decreases with an increase in atomic number due to lanthanoid contraction. The order is $Pr^{3+} (1.013 \ \mathring{A}) > Sm^{3+} (0.964 \ \mathring{A}) > Dy^{3+} (0.908 \ \mathring{A})$. Thus,statement $(I)$ is incorrect.
$(II)$ $Eu^{2+}$ $(Z = 63)$ has a stable $4f^7$ configuration but easily oxidizes to $Eu^{3+}$ $(4f^6)$,making it a strong reducing agent. Thus,statement $(II)$ is correct.
$(III)$ $Pu$ $(Z = 94)$ has the electronic configuration $[Rn] 5f^6 6d^0 7s^2$. It can exhibit oxidation states up to $+7$ by utilizing its $5f, 6d,$ and $7s$ electrons. Thus,statement $(III)$ is correct.
Therefore,the correct option is $(D)$.

(B) The electronic configuration of Cerium $(Ce)$ is $[Xe] 4f^1 5d^1 6s^2$.
Upon losing four electrons,it achieves the stable noble gas configuration of Xenon $([Xe])$,forming $Ce^{+4}$ $([Xe] 4f^0 5d^0 6s^0)$.
Thus,Cerium is well known for exhibiting the $+4$ oxidation state.

(B) Statement $(i)$: $Eu^{2+}$ $(4f^7)$ and $Yb^{2+}$ $(4f^{14})$ are stable,but they can be oxidized to $Eu^{3+}$ $(4f^6)$ and $Yb^{3+}$ $(4f^{13})$ respectively. Thus,they act as reducing agents.
Statement $(ii)$: The atomic number of $Pr$ is $59$. The electronic configuration of $Pr$ is $[Xe] 4f^3 6s^2$. Therefore,$Pr^{3+}$ is $[Xe] 4f^2$. The statement is incorrect.
Statement $(iii)$: $La^{3+}$ has a $4f^0$ configuration. Since there are no unpaired electrons,the aqueous solution of $LaCl_3$ is colourless. The statement is correct.
Therefore,statements $(i)$ and $(iii)$ are correct.

(B) The actinides exhibit a range of oxidation states due to the comparable energies of $5f$,$6d$,and $7s$ orbitals.
While most actinides show a $+3$ oxidation state,some early actinides exhibit higher oxidation states.
Specifically,$Np$ (Neptunium) and $Pu$ (Plutonium) are known to exhibit the $+7$ oxidation state in certain compounds,such as $NpO_2(OH)_3$ and $PuO_2(OH)_3$.

(C) Paramagnetism is exhibited by ions that have unpaired electrons in their electronic configuration.
$1$. $Lr^{3+}$ $(Z=103)$: The electronic configuration of $Lr$ is $[Rn] 5f^{14} 6d^1 7s^2$. $Lr^{3+}$ is $[Rn] 5f^{14}$,which has no unpaired electrons (diamagnetic).
$2$. $Ac^{3+}$ $(Z=89)$: The electronic configuration of $Ac$ is $[Rn] 6d^1 7s^2$. $Ac^{3+}$ is $[Rn]$,which has no unpaired electrons (diamagnetic).
$3$. $Th^{3+}$ $(Z=90)$: The electronic configuration of $Th$ is $[Rn] 6d^2 7s^2$. $Th^{3+}$ is $[Rn] 5f^1$,which has $1$ unpaired electron (paramagnetic).
$4$. $Lu^{3+}$ $(Z=71)$: The electronic configuration of $Lu$ is $[Xe] 4f^{14} 5d^1 6s^2$. $Lu^{3+}$ is $[Xe] 4f^{14}$,which has no unpaired electrons (diamagnetic).
Therefore,$Th^{3+}$ is the only paramagnetic ion among the given options.

(C) Misch metal is an alloy consisting of a lanthanide metal (about $95\%$),iron (about $5\%$),and traces of sulfur,carbon,calcium,and aluminum.
Therefore,the percentage of lanthanides and iron in misch metal is approximately $95\%$ and $5\%$,respectively.

(C) Isoelectronic species are those that have the same number of electrons.
$(A)$ $Pr^{3+}$: Number of electrons $= 59 - 3 = 56$. $Nd^{3+}$: Number of electrons $= 60 - 3 = 57$. Thus,they are not isoelectronic.
$(B)$ $Tb^{3+}$: Number of electrons $= 65 - 3 = 62$. $Dy^{2+}$: Number of electrons $= 66 - 2 = 64$. Thus,they are not isoelectronic.
$(C)$ $Eu^{2+}$: Number of electrons $= 63 - 2 = 61$. $Gd^{3+}$: Number of electrons $= 64 - 3 = 61$. Since both have $61$ electrons,they are isoelectronic.
$(D)$ $Pr^{3+}$: Number of electrons $= 59 - 3 = 56$. $Ce^{4+}$: Number of electrons $= 58 - 4 = 54$. Thus,they are not isoelectronic.

(B) The general electronic configuration of $f$-block elements is $(n-2)f^{0-14} (n-1)d^{0-1} ns^2$.
Lanthanoids involve the filling of $4f$ orbitals,with the general configuration $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$.
Actinoids involve the filling of $5f$ orbitals,with the general configuration $[Rn] 5f^{1-14} 6d^{0-1} 7s^2$.
Therefore,$4f$ and $5f$ orbitals are progressively filled in $f$-block elements.

(A) The electronic configuration of $M^{3+}$ is given as $[Xe] \ 4f^3$.
We analyze the $M^{3+}$ configurations for the given options:
$(A)$ $Nd^{3+} (Z=60): [Xe] \ 4f^3$
$(B)$ $Pr^{3+} (Z=59): [Xe] \ 4f^2$
$(C)$ $Sm^{3+} (Z=62): [Xe] \ 4f^5$
$(D)$ $Dy^{3+} (Z=66): [Xe] \ 4f^9$
Comparing these with the given configuration,$M^{3+}$ corresponds to $Nd^{3+}$.

(B) $I$. $Ce^{4+}$ $(4f^0)$ and $Tb^{4+}$ $(4f^7)$ have a strong tendency to gain an electron to reach stable configurations,thus acting as strong oxidising agents. This statement is correct.
$II$. $Eu^{2+}$ $(4f^7)$ and $Yb^{2+}$ $(4f^{14})$ are stable due to half-filled and fully-filled $f$-orbitals,respectively. They tend to lose an electron to reach the $+3$ state,thus acting as reducing agents. The statement in the question says they act as oxidising agents,which is incorrect.
$III$. Mischmetal consists of approximately $95 \%$ lanthanoid metal and $5 \%$ iron,along with traces of $S, C, Ca,$ and $Al$. The statement in the question incorrectly reverses the percentages. This statement is incorrect.
$IV$. $La^{3+}$ $([Xe] 4f^0)$ and $Ce^{4+}$ $([Xe] 4f^0)$ have no unpaired electrons,making them diamagnetic. This statement is correct.
Therefore,statements $I$ and $IV$ are correct.

(A) The electronic configurations are:
$Eu (Z=63): [Xe] 4f^7 6s^2$ (Half-filled $f$-orbital)
$Pu (Z=94): [Rn] 5f^6 7s^2$
$Cf (Z=98): [Rn] 5f^{10} 7s^2$
$Sm (Z=62): [Xe] 4f^6 6s^2$
$Gd (Z=64): [Xe] 4f^7 5d^1 6s^2$ (Half-filled $f$-orbital)
$Cm (Z=96): [Rn] 5f^7 6d^1 7s^2$ (Half-filled $f$-orbital)
The elements with half-filled $f$-orbitals are $Eu, Gd,$ and $Cm$.
Total count = $3$.

(D) The ground state electronic configurations are:
$Gd (Z=64) = [Xe] 4f^7 5d^1 6s^2$
$Yb (Z=70) = [Xe] 4f^{14} 6s^2$
For $Gd^{3+}$,three electrons are removed ($1$ from $5d$ and $2$ from $6s$),resulting in $4f^7$. Thus,$x = 7$.
For $Yb^{2+}$,two electrons are removed from $6s$,resulting in $4f^{14}$. Thus,$y = 14$.
The sum $x + y = 7 + 14 = 21$.

(C) Lanthanoids in the $+2$ oxidation state,such as $Eu^{+2}$ and $Yb^{+2}$,act as strong reducing agents.
This is because they tend to lose one electron to acquire the more stable $+3$ oxidation state,which is the most common and stable oxidation state for lanthanoids.

(B) Due to Lanthanide contraction,the ionic radii decrease from $La^{3+}$ to $Lu^{3+}$.
As the size decreases,the polarizing power of the metal ion increases,which increases the covalent character of the $M-OH$ bond. Consequently,the basic strength decreases from $La(OH)_3$ to $Lu(OH)_3$.
Chemical reactivity decreases from $La$ to $Lu$ due to the increase in ionization energy and decrease in size.
$Zr$ and $Hf$ exhibit similar atomic radii due to the Lanthanide contraction,which makes their chemical properties very similar.
Separation of Lanthanides is difficult because their chemical properties are very similar due to the small change in ionic radii.
Therefore,statements $A, B, C,$ and $D$ are correct.

(A) Lanthanoid contraction is the steady decrease in the size of atoms and ions of rare earth elements with an increasing atomic number from lanthanum $(Z = 57)$.
It is caused by the poor shielding effect of $4f$-electrons.
In the $4f$ series,the $4f$-electrons do not shield the outer electrons effectively from the nuclear charge.
This results in a greater effective nuclear charge,which exerts a stronger pull on the electrons towards the nucleus.
Thus,the decrease in atomic and ionic radii is the primary effect of lanthanoid contraction.
Therefore,option $(A)$ is correct.

(C) Transuranium elements are the chemical elements with atomic numbers greater than $92$ (the atomic number of uranium). These elements are synthetic and are produced artificially in nuclear reactors or particle accelerators.

(A) The common oxidation state of $f$-block elements is $+3$.
$Eu$ $(Z=63)$ has the electronic configuration $[Xe] 4f^7 6s^2$. It shows a stable $+2$ oxidation state due to the stable half-filled $4f^7$ configuration.
$Tb$ $(Z=65)$ has the electronic configuration $[Xe] 4f^9 6s^2$. It shows a stable $+4$ oxidation state due to the stable half-filled $4f^7$ configuration after losing four electrons.
Therefore,the other stable oxidation states for $Eu$ and $Tb$ are $+2$ and $+4$ respectively.
Hence,option $A$ is the correct answer.

(C) Due to lanthanide contraction,the ionic radius of lanthanide ions $(Ln^{3+})$ decreases as the atomic number increases from $Ce$ $(Z=58)$ to $Gd$ $(Z=64)$.
As the size of the $Ln^{3+}$ ion decreases,the covalent character of the $Ln-OH$ bond increases according to Fajan's rule.
Consequently,the basic strength of the hydroxides decreases as we move from $Ce(OH)_3$ to $Gd(OH)_3$.
Therefore,the correct increasing order of basic character is $Gd(OH)_3 < Nd(OH)_3 < Ce(OH)_3$.

(C) The actinide series consists of elements with atomic numbers from $89$ to $103$.
$Pa$ (Protactinium,$Z=91$),$Lr$ (Lawrencium,$Z=103$),and $Pu$ (Plutonium,$Z=94$) are all members of the actinide series.
$Nd$ $(Z=60)$ and $Pr$ $(Z=59)$ are lanthanides.
$Lu$ $(Z=71)$ is a lanthanide.
Therefore,the correct set containing only actinides is $Pa, Lr, Pu$.

(A) Generally,paramagnetism is shown by lanthanoid ions. The paramagnetism arises due to the presence of unpaired electrons in the $f$-orbital.
$Lu^{3+}$ has the electronic configuration $[Xe] 4f^{14}$.
Since all $14$ electrons are paired in the $4f$ subshell,$Lu^{3+}$ does not have any unpaired electrons.
Therefore,it is diamagnetic and does not exhibit paramagnetism.

(C) The electronic configuration of $Eu$ $(Z=63)$ is $[Xe] 4f^7 6s^2$. It can lose two electrons to form $Eu^{2+}$ $([Xe] 4f^7)$,which is stable due to the half-filled $f$-orbital. It also exhibits a $+3$ oxidation state.
The electronic configuration of $Gd$ $(Z=64)$ is $[Xe] 4f^7 5d^1 6s^2$. It loses three electrons to form $Gd^{3+}$ $([Xe] 4f^7)$,which is highly stable due to the half-filled $f$-orbital. Thus,the common stable oxidation states are $+2$ for $Eu$ and $+3$ for $Gd$.

(A) The separation of lanthanoids is difficult due to their similar chemical properties. However,$Eu(II)$ shows similarity with alkaline earth metals like $Sr(II)$.
$Eu(II)$ can be precipitated as $EuSO_4$ in the presence of sulfate ions,similar to $SrSO_4$,while other trivalent lanthanoid ions like $Dy(III)$ or $Gd(III)$ remain in the solution.
Therefore,the pair $Eu(II)$ and $Dy(III)$ (or any other trivalent lanthanoid) can be separated by the precipitation method.

(B) The atomic number of $Pr$ (Praseodymium) is $59$.
The electronic configuration of $Pr$ is $[Xe] 4f^3 6s^2$.
When $Pr$ forms a $Pr^{3+}$ ion,it loses three electrons (two from $6s$ and one from $4f$).
Therefore,the electronic configuration of $Pr^{3+}$ is $[Xe] 4f^2$.

(D) The basicity of lanthanide ions depends on their ionic character.
As the atomic number increases from $La$ to $Lu$,the ionic radius decreases due to lanthanide contraction.
As the size of the cation decreases,the covalent character increases and the ionic character decreases.
Therefore,the basicity decreases as the ionic radius decreases.
The correct order of basicity is $La^{3+} > Ce^{3+} > Eu^{3+} > Lu^{3+}$.

(D) The option $(D)$ is false.
In the lanthanide series,the ionic radii of trivalent lanthanides $(Ln^{3+})$ steadily decrease with an increase in atomic number.
This phenomenon is known as lanthanoid contraction,which occurs due to the poor shielding effect of $4f$ electrons.

(C) The atomic number of Samarium $(Sm)$ is $62$.
The ground state electronic configuration of $Sm$ is $[Xe] 4f^6 6s^2$.
When $Sm$ forms the $Sm^{2+}$ ion,it loses two electrons from the $6s$ orbital.
Therefore,the electronic configuration of $Sm^{2+}$ is $[Xe] 4f^6 6s^0$ or simply $[Xe] 4f^6$.

(A) The electronic configuration of uranium $({ }_{92} U)$ is $[Rn] \ 5f^{3} \ 6d^{1} \ 7s^{2}$.
In the $5f$ subshell,there are $3$ unpaired electrons,and in the $6d$ subshell,there is $1$ unpaired electron.
The $7s$ subshell is fully filled.
Therefore,the total number of unpaired electrons is $3 + 1 = 4$.

(A) The atomic number of cerium $(Ce)$ is $58$.
The ground state electronic configuration of $Ce$ is $[Xe] 4f^{1} 5d^{1} 6s^{2}$.
To form the $Ce^{3+}$ ion,three electrons are removed: two from the $6s$ orbital and one from the $5d$ orbital.
Thus,the electronic configuration of $Ce^{3+}$ is $[Xe] 4f^{1}$.

(C) : Incorrect. $Mo(VI)$ and $W(VI)$ are more stable than $Cr(VI)$ because stability increases down the group for group $6$ elements.
$B$: Correct. $Ce^{4+}$ and $Tb^{4+}$ act as strong oxidants because they tend to return to the stable $+3$ oxidation state. $Eu^{2+}$ and $Yb^{2+}$ act as strong reductants because they tend to reach the stable $+3$ oxidation state.
$C$: Incorrect. $Am^{3+}$ has $6$ unpaired electrons $(5f^6)$,while $Cm^{3+}$ has $7$ unpaired electrons $(5f^7)$.
$D$: Correct. The $5f$ orbitals provide poorer shielding than $4f$ orbitals,which leads to a greater effective nuclear charge and thus greater actinoid contraction compared to lanthanoid contraction.
Therefore,statements $B$ and $D$ are correct.

(C) Cerium $(Ce)$ has an atomic number of $58$.
Its electronic configuration is $[Xe] 4f^1 5d^1 6s^2$.
By losing four electrons,it reaches the stable noble gas configuration of Xenon $(4f^0)$,which is why it readily exhibits the $+4$ oxidation state.