Lanthanoids and Actinoids Questions in English

(A) In the actinoid series,as the atomic number increases,the ionic radius decreases due to the actinoid contraction,which is analogous to lanthanoid contraction.
This occurs because the $5f$ electrons provide poor shielding for the increasing nuclear charge.
Therefore,the actinoid with the lowest atomic number will have the largest ionic radius in its $+3$ state.
Comparing the atomic numbers: $U$ $(Z=92)$,$Bk$ $(Z=97)$,$Es$ $(Z=99)$,and $Md$ $(Z=101)$.
Since $U$ has the lowest atomic number among the given options,it has the largest size in its $+3$ state.

(D) The lanthanoids are the elements with atomic numbers $57$ to $71$.
$Pm$ (Promethium,$Z=61$),$Er$ (Erbium,$Z=68$),and $Yb$ (Ytterbium,$Z=70$) are lanthanoids.
$Hf$ (Hafnium,$Z=72$) is a $d$-block transition element belonging to the $6th$ period and $4th$ group.

(B) $La$ (Lanthanum) has an atomic number of $57$.
In the $ 3$ oxidation state,its electronic configuration is $[Xe] 4f^0$.
Since there are no unpaired electrons $(n = 0)$,the effective magnetic moment $(\mu_{eff} = \sqrt{n(n 2)} \ BM)$ is $0 \ BM$.

(D) The phenomenon of lanthanide contraction is observed in $4f$ elements due to the increase in effective nuclear charge as we move across the series.
As a result,the atomic and ionic radii decrease continuously from lanthanum $(Z=57)$ to lutetium $(Z=71)$.
Since lutetium $(Lu)$ is the last element in the lanthanide series,it has the highest effective nuclear charge and therefore the smallest ionic radius in the $+3$ state.

(A) The electronic configuration of Gadolinium ($Gd$,$Z=64$) is $[Xe] 4f^7 5d^1 6s^2$.
In this configuration,the $4f$ subshell contains $7$ electrons,which corresponds to a half-filled $4f$ orbital ($f^{14}$ is fully filled,so $f^7$ is half-filled).

(A) $1$. These are bad conductors of heat and electricity: This statement is incorrect. Lanthanoids are metallic in nature and are generally good conductors of heat and electricity,similar to other metals.
$2$. These are soft metals: This is true. Lanthanoids are relatively soft metals,and their hardness increases with increasing atomic number.
$3$. Coordination number is usually greater than six: This is true. Due to their large ionic radii,lanthanoids often exhibit high coordination numbers,typically ranging from $8$ to $12$ in their complexes.
$4$. These are strongly paramagnetic: This is true. Lanthanoids possess unpaired $f$-electrons,which result in paramagnetic behavior. Many lanthanoid ions exhibit strong paramagnetism.

(A) The actinoids are the elements from atomic number $89$ to $103$.
Among the given options,$Cm$ (Curium,atomic number $96$) belongs to the actinoid series.
$Pm$ (Promethium),$Tm$ (Thulium),and $Sm$ (Samarium) are lanthanoids.

(C) The atomic number of $Lu$ is $71$.
Its electronic configuration is $[Xe] 4f^{14} 5d^1 6s^2$.
In this configuration,the $4f$ subshell is completely filled with $14$ electrons,which matches both the expected and observed electronic configurations.

(A) $(1)$ $Sm$ stands for Samarium,which is a rare earth element. Rare earth elements are a group of $17$ elements that include the $15$ lanthanides plus scandium $(Sc)$ and yttrium $(Y)$. Samarium $(Sm)$ is a member of the lanthanide series and is considered a rare earth element.
$(2)$ $Hg$ stands for Mercury,which is a transition metal,not a rare earth element.
$(3)$ $Zn$ stands for Zinc,which is a transition metal,not a rare earth element.
$(4)$ $W$ stands for Tungsten,which is a transition metal,not a rare earth element.

(C) The most stable and common oxidation state of lanthanoids $(Ln)$ is $+3$.
Therefore,when $Ln^{3+}$ reacts with hydroxide ions $(OH^-)$,the resulting compound is $Ln(OH)_3$.

(A) The actinoid series consists of elements with atomic numbers from $89$ to $103$.
- $No$ (Nobelium) has an atomic number of $102$,which places it within the actinoid series.
- $Mo$ (Molybdenum) is a $d$-block element with atomic number $42$.
- $Co$ (Cobalt) is a $d$-block element with atomic number $27$.
- $Ce$ (Cerium) is a lanthanoid element with atomic number $58$.
Therefore,$No$ is the correct actinoid element.

(A) The lanthanoids are the elements with atomic numbers $57$ to $71$.
$Eu$ (Europium) has an atomic number of $63$,which falls within the lanthanoid series.
$Cm$ (Curium),$Am$ (Americium),and $Np$ (Neptunium) are all actinoid elements (atomic numbers $89$ to $103$).

(D) The actinoids are the elements with atomic numbers $89$ to $103$.
$Gd$ (Gadolinium,$Z=64$),$Lu$ (Lutetium,$Z=71$),and $Pr$ (Praseodymium,$Z=59$) are lanthanoids.
$Pa$ (Protactinium,$Z=91$) is an actinoid element.

(A) The electronic configuration of $Eu$ $(Z=63)$ is $[Xe] 4f^7 6s^2$.
When $Eu$ loses two electrons to form $Eu^{2+}$,its configuration becomes $[Xe] 4f^7$.
This $4f^7$ configuration is half-filled,which provides extra stability to the $Eu^{2+}$ ion.

(B) In the case of lanthanides,as the atomic number $(Z)$ increases,the ionic size decreases due to lanthanide contraction.
As the size of the $Ln^{3+}$ ion decreases,the covalent character of the $Ln-OH$ bond increases,which leads to a decrease in basic strength.
Therefore,the basic character follows the order: $La(OH)_3 > Pr(OH)_3 > Sm(OH)_3 > Lu(OH)_3$.
Thus,$La(OH)_3$ is the strongest base.

(D) In the lanthanoid series,as the atomic number $(Z)$ increases,the ionic radius decreases due to lanthanoid contraction.
Since $Lu$ $(Z = 71)$ has the highest atomic number among the given elements $(Eu: 63, Er: 68, Yb: 70, Lu: 71)$,it has the smallest ionic radius in the $+3$ oxidation state.

(B) Lanthanoids are generally non-radioactive,except for $Pm$ (Promethium),which is radioactive. Therefore,the statement that all lanthanoids are non-radioactive is incorrect.

(C) The electronic configuration of $Cm (Z=96)$ is $[Rn] 5f^7 6d^1 7s^2$.
In the $+3$ oxidation state,it loses three electrons to form $Cm^{3+}$,which has the configuration $[Rn] 5f^7$.
Since the $f$ orbital can hold a maximum of $14$ electrons,$f^7$ represents a half-filled $f$ orbital.

(A) The elements $Np$ (Neptunium),$U$ (Uranium),and $Cf$ (Californium) are all radioactive actinoids.
$Nd$ (Neodymium) is a lanthanoid element and is non-radioactive (stable).

(A) Lanthanoids exhibit a common oxidation state of $+3$.
When reacting with nitrogen $(N^{3-})$,the compound formed is $LnN$.
When reacting with halogens $(X^-)$,the compound formed is $LnX_3$.

(C) Lanthanoids are metallic in nature and are good conductors of heat and electricity.
Most lanthanoid ions are paramagnetic due to the presence of unpaired $4f$ electrons.
Lanthanoids generally exhibit high coordination numbers,typically greater than $6$ (often $8$ or $9$).
However,the statement that 'all lanthanoids are non-radioactive' is incorrect.
Promethium ($Pm$,atomic number $61$) is a well-known radioactive element among the lanthanoids.

(B) The correct statement is that actinoid contraction is greater than lanthanoid contraction.
This is because the $5f$ electrons have poorer shielding effect compared to $4f$ electrons,leading to a greater effective nuclear charge and more significant contraction in the actinoid series.

(D) The reaction of lanthanoids $(Ln)$ with oxygen produces the stable oxide $(Ln_2O_3)$:
$4Ln + 3O_2 \rightarrow 2Ln_2O_3$ (Compound $A$)
When the lanthanoid oxide $(Ln_2O_3)$ reacts with excess $CO_2$,it forms the corresponding carbonate $(Ln_2(CO_3)_3)$:
$Ln_2O_3 + 3CO_2 \rightarrow Ln_2(CO_3)_3$ (Compound $B$)
Therefore,the formula of compound $(B)$ is $Ln_2(CO_3)_3$.

(C) In the lanthanoid series,the atomic size decreases from left to right due to lanthanoid contraction.
As the atomic number increases,the shielding effect of $4f$ electrons is poor,leading to an increase in effective nuclear charge.
Among the given elements ($Ce$,$Pr$,$Pm$,$Sm$),the atomic numbers are $Ce(58)$,$Pr(59)$,$Pm(61)$,and $Sm(62)$.
Since atomic size decreases as atomic number increases,$Sm$ has the smallest atomic size among the options provided.
The order is $Ce > Pr > Pm > Sm$.

(C) In the lanthanoid series,as the atomic number increases from $Ce$ $(Z=58)$ to $Lu$ $(Z=71)$,the ionic radii of the trivalent ions $(Ln^{3+})$ decrease steadily. This phenomenon is known as lanthanoid contraction.
Therefore,the decreasing order of ionic radii for the given ions is: $Ce^{3+} > Pm^{3+} > Sm^{3+} > Gd^{3+}$.

(A) The general oxidation state for lanthanoids is $+3$.
$Ce$ (Cerium) has the electronic configuration $[Xe] 4f^1 5d^1 6s^2$.
Due to the stability of the empty $f$-orbital after losing four electrons,$Ce$ exhibits a $+4$ oxidation state in addition to $+3$.

(B) The correct answer is $B$.
Lanthanoids generally exhibit a $+3$ oxidation state.
However,some elements show $+2$ or $+4$ oxidation states due to the stability of $f^0$,$f^7$,or $f^{14}$ configurations.
Gadolinium $(Gd)$ has the electronic configuration $[Xe] \ 4f^7 \ 5d^1 \ 6s^2$.
Upon losing three electrons,it forms $Gd^{3+}$ with the configuration $[Xe] \ 4f^7$,which is a half-filled stable $f$-subshell configuration.
Due to this extra stability,$Gd$ predominantly exhibits only the $+3$ oxidation state.

(A) The correct option is $A$.
$Lr$ $(Z=103)$ has the electronic configuration $[Rn] 5f^{14} 6d^{1} 7s^{2}$.
Due to the completely filled $5f$ orbital,$Lr$ exhibits only the $+3$ oxidation state.

(C) The first inner transition series is known as the $Lanthanoid$ series,which includes elements from atomic number $58$ $(Ce)$ to $71$ $(Lu)$.
$Pr$ $(Praseodymium)$ has an atomic number of $59$,which falls within the range of the $Lanthanoid$ series.
$Bk$ $(Berkelium)$,$Pu$ $(Plutonium)$,and $Fm$ $(Fermium)$ belong to the second inner transition series,known as the $Actinoid$ series.
Therefore,the correct option is $C$.

(A) The actinoids exhibit a range of oxidation states due to the comparable energies of $5f$,$6d$,and $7s$ orbitals.
While the most common oxidation state is $+3$,the elements like $Np$ and $Pu$ exhibit oxidation states up to $+7$.
Therefore,the highest oxidation state exhibited by actinoids is $+7$.

(D) Promethium $(Pm)$ is the only radioactive lanthanide.

(D) Due to lanthanide contraction,the ionic radius of $M^{3+}$ ions decreases as the atomic number increases from $Ce$ to $Lu$.
As the size of the $M^{3+}$ ion decreases,the covalent character of the $M-OH$ bond increases,which leads to a decrease in the basic strength of the hydroxides.
The order of ionic radii is $Ce^{3+} > Eu^{3+} > Yb^{3+} > Lu^{3+}$.
Therefore,the order of basic strength of the hydroxides is $Ce(OH)_3 > Eu(OH)_3 > Yb(OH)_3 > Lu(OH)_3$.
Thus,$Ce(OH)_3$ is the most basic hydroxide.

(D) Lanthanides and actinides are both considered inner transition elements and commonly exhibit a $+3$ oxidation state.
However,a key point of dissimilarity is their radioactivity.
All actinides are radioactive in nature,whereas among lanthanides,only promethium $(Pm)$ is radioactive,and the rest are non-radioactive.

(A) The lanthanoid series elements typically exhibit a $+3$ oxidation state.
However,$Ce$ (Cerium,atomic number $58$) is well known to exhibit a $+4$ oxidation state because it attains a stable noble gas configuration $([Xe]4f^0)$ upon losing four electrons.
Therefore,the correct option is $A$.

(B) The correct answer is $B$.
$Pm$ (Promethium) is a lanthanoid,not an actinoid.
$Pm$ is the only radioactive element among the lanthanoids.
All actinoids are radioactive.
Therefore,the statement that $Pm$ is a radioactive element among actinoids is incorrect.

(B) The correct answer is $B$.
Pyrophoric Misch metal is an alloy of lanthanoid metals (about $95\%$) and iron (about $5\%$) with traces of $S, C, Ca,$ and $Al$.
It is highly pyrophoric,meaning it produces sparks when struck or scratched,which makes it ideal for use in gas lighters and bullet tips.

(D) The stability of a particular oxidation state in Lanthanoid elements is determined by the combined effect of the energy required to remove electrons (ionization energy) and the energy released when the resulting ions are hydrated (hydration energy).$ \Delta H_{total} = \Delta H_{ionization} + \Delta H_{hydration} $. Therefore,the correct option is $D$.

(B) The most common and stable oxidation state for all lanthanoid elements is $+3$.
Although some lanthanoids exhibit $+2$ or $+4$ oxidation states in specific compounds,the $+3$ state is characteristic of the entire series due to the stability provided by the electronic configuration.

(D) The actinide series consists of elements from atomic number $89$ to $103$.
These elements involve the filling of the $5f$ orbitals.
The general electronic configuration for the actinide series is given by $[Rn] 5f^{0-14} 6d^{0-2} 7s^2$.
Here,$[Rn]$ represents the noble gas core of Radon $(Z=86)$.

(D) $Ce$ has the electronic configuration $[Xe] 4f^1 5d^1 6s^2$.
Upon losing four electrons,it forms $Ce^{+4}$ with the configuration $[Xe] 4f^0 5d^0 6s^0$.
Since $Ce^{+4}$ achieves a stable noble gas configuration,it is well known to exhibit the $+4$ oxidation state.

(B) Misch metal is an alloy that consists of a lanthanoid metal (approximately $95 \%$) and iron (approximately $5 \%$).
It also contains traces of sulfur $(S)$,carbon $(C)$,calcium $(Ca)$,and aluminum $(Al)$.

(D) The incorrect statement regarding lanthanoids is given in option $(D)$.
Lanthanoids exhibit colour in solution due to $f-f$ transitions,not $d-d$ transitions,because the $4f$ orbitals are partially filled in their ions.

(B) Among the lanthanoids (atomic numbers $57-71$),$Promethium$ ($Pm$,atomic number $61$) is the only element that is radioactive.

(A) The elements in which the $4f$ orbitals are progressively filled are known as Lanthanoids.
These elements range from atomic number $58$ $(Ce)$ to $71$ $(Lu)$.
They are also known as $4f$-block elements or rare earth elements.
Therefore,the correct option is $A$.

(C) $Ce^{4+}$ acts as a strong oxidizing agent because it has a strong tendency to gain an electron to attain the stable $+3$ oxidation state $(Ce^{4+} + e^- \rightarrow Ce^{3+})$. Therefore,the statement that $Ce^{4+}$ is a reducing agent is incorrect.

(B) Lanthanoids exhibit a $+3$ oxidation state as the most stable configuration is acquired by losing $3 \ e^{-}$ ions.

(B) The most stable oxidation state for lanthanoids is $+3$.
$Ce^{4+}$ is a strong oxidizing agent in aqueous solution because it easily gains an electron to form the more stable $Ce^{3+}$ ion.
Therefore,$Ce^{4+}$ is not stable in aqueous solution and is not known to exist as a free ion in such conditions.
Thus,the statement '$Ce^{4+}$ in aqueous solution is not known' is considered correct in the context of its stability.

(A) $Ln(III)$ ions are often colored due to $f-f$ transitions,which occur because of the presence of partly filled $4f$ orbitals,allowing them to absorb certain wavelengths from the visible region of the spectrum.
Therefore,the statement that "$Ln(III)$ compounds are generally colourless" is incorrect.

(C) $Al$ atom is smaller than $Mg$ atom because $Al$ lies to the right of $Mg$ in the same period and has a greater effective nuclear charge.
$A$ cation is smaller than its parent atom,so $Al^{3+}$ is smaller than $Al$. Therefore,$Al^{3+}$ is the smallest in size among the given species.
Thus,$Statement-I$ is correct.
Among lanthanides,there is a gradual decrease in atomic and ionic radii as we move from left to right due to lanthanoid contraction.
The atomic radii of lanthanides generally decrease across the series. $Eu$ does not show an exceptionally high atomic radius compared to its neighbors in the context of the general trend of lanthanoid contraction.
Thus,$Statement-II$ is false.

$(A)$ In the $4f$-series, as the atomic number increases from $La$ to $Lu$, the atomic and ionic radii decrease gradually.
This decrease in atomic and ionic size from lanthanum to lutetium is known as lanthanide contraction.
Since the atomic number of $Pr$ $(59)$, $Gd$ $(64)$, and $Tm$ $(69)$ increases in that order, their ionic radii decrease accordingly.
Therefore, the correct order of ionic radii is $Pr^{3+} (101 \text{ pm}) > Gd^{3+} (94 \text{ pm}) > Tm^{3+} (87 \text{ pm})$.