Lanthanoids and Actinoids Questions in English

(B) Lanthanides $(Ln)$ react with halogens $(X_2)$ to form trihalides $(LnX_3)$.
When burned in oxygen $(O_2)$,they form sesquioxides $(Ln_2O_3)$.
When heated with nitrogen $(N_2)$,they form nitrides $(LnN)$.
Therefore,$X = LnX_3$,$Y = Ln_2O_3$,and $Z = LnN$.

(C) The elements following the lanthanoids in the periodic table are influenced by the lanthanoid contraction.
For example, the atomic radius of the metal zirconium, $Zr$, (a period $-5$ transition element) is $159 \ pm$ and that of hafnium, $Hf$, (the corresponding period $-6$ element) is $156 \ pm$.
The radii are very similar even though the number of electrons increases from $40$ to $72$ and the atomic mass increases from $91.22$ to $178.49 \ g/mol$.
The increase in mass and the nearly unchanged radii lead to a steep increase in density from $6.51$ to $13.35 \ g/cm^3$.
Zirconium and hafnium, therefore, have very similar chemical behaviour, having closely similar radii and electron configurations.
Radius-dependent properties such as lattice energies, solvation energies, and stability constants of complexes are also similar.

(A) The $5f$,$6d$,and $7s$ energy levels in actinoids are of comparable energy.
Because of this small energy gap,electrons from all these orbitals can participate in bonding,leading to a larger number of oxidation states.
In contrast,in lanthanoids,the $4f$ orbitals are deeply buried and have a much larger energy difference compared to $5d$ orbitals,restricting the number of oxidation states.

(D) In the lanthanoid series,as we move from $La$ $(Z=57)$ to $Lu$ $(Z=71)$,the ionic radii of the $+3$ ions decrease regularly. This phenomenon is known as lanthanoid contraction.
This decrease occurs because the $4f$ electrons are added,which provide poor shielding for the increasing nuclear charge,leading to a stronger attraction between the nucleus and the outer electrons.
The atomic numbers are: $La$ $(57)$,$Ce$ $(58)$,$Pm$ $(61)$,and $Yb$ $(70)$.
Therefore,the order of ionic radii is $La^{+3} > Ce^{+3} > Pm^{+3} > Yb^{+3}$,which is equivalent to $Yb^{+3} < Pm^{+3} < Ce^{+3} < La^{+3}$.

(B) Across the lanthanide series,the ionic radius of the $Ln^{3+}$ ions decreases due to lanthanide contraction.
As the size of the $Ln^{3+}$ cation decreases,the $Ln-OH$ bond becomes more covalent and less ionic.
Since basicity is directly related to the ease of releasing $OH^-$ ions,the decrease in ionic character leads to a decrease in basicity.
Therefore,the basicity of lanthanide hydroxides decreases from $La(OH)_3$ to $Lu(OH)_3$.

(A) The electronic configuration of $Yb$ $(Z=70)$ is $[Xe] 4f^{14} 6s^2$.
Thus,$Yb^{3+}$ has the configuration $[Xe] 4f^{13}$.
Since it has an unpaired electron,it should be paramagnetic,but $Yb^{3+}$ ions are colorless in aqueous solution because the $f-f$ transitions are not observed in the visible region.
Therefore,the statement that $Yb^{3+}$ is pink is incorrect.

(C) The electronic configuration of $Ce$ $(Z = 58)$ is $[Xe] 4f^1 5d^1 6s^2$.
By losing four electrons,it attains the stable noble gas configuration of $Xe$ $([Xe])$,making the $+4$ oxidation state stable for $Ce$.

(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 occurs due to the poor shielding effect of the $4f$ electrons,which causes the effective nuclear charge to increase,pulling the valence shell closer to the nucleus.

(A) The atomic radii of the given lanthanoids are as follows:
$Eu = 199 \text{ pm}$
$Ce = 183 \text{ pm}$
$Nd = 181 \text{ pm}$
$Ho = 176 \text{ pm}$
Therefore, the correct order of atomic radii is $Eu > Ce > Nd > Ho$.

(D) The color of lanthanoid ions depends on the presence of unpaired $f$-electrons.
$La^{3+}$ has a configuration of $[Xe] 4f^0$ (no unpaired electrons).
$Gd^{3+}$ has a configuration of $[Xe] 4f^7$ (half-filled,stable,but colorless in many compounds due to high energy transition).
$Lu^{3+}$ has a configuration of $[Xe] 4f^{14}$ (fully filled,no unpaired electrons).
$Sm^{3+}$ has a configuration of $[Xe] 4f^5$,which contains unpaired electrons,allowing for $f-f$ transitions that result in a yellow color.

(D) The actinoids exhibit a range of oxidation states due to the comparable energies of the $5f$,$6d$,and $7s$ orbitals.
Neptunium $(Np)$ and plutonium $(Pu)$ exhibit the maximum number of oxidation states among the actinoids,reaching up to $+7$.
This wide range is possible because the energy levels of the $5f$,$6d$,and $7s$ orbitals are very close,allowing a larger number of electrons to participate in bond formation.

(D) The electronic configuration of $U$ $(Z=92)$ is $[Rn] 5f^3 6d^1 7s^2$. The highest oxidation state shown by $U$ is $+6$.
The electronic configuration of $Pu$ $(Z=94)$ is $[Rn] 5f^6 7s^2$. The highest oxidation state shown by $Pu$ is $+7$.

(B) The given electronic configuration is $[Rn]\, 7s^2\, 5f^3\, 6d^1$.
$1$. The highest principal quantum number $(n)$ is $7$,which indicates that the element belongs to Period $7$.
$2$. The last electron enters the $5f$ orbital,which classifies the element as an $f$-block element (Actinoid series).
$3$. All $f$-block elements belong to Group $3$ (or $III\, B$ in the older $IUPAC$ notation).
Therefore,the element belongs to Period $7$,Group $3$ $(III\, B)$,and $f$-block.

(D) The electronic configuration of Lanthanum $(Z=57)$ is $[Xe] 5d^1 6s^2$.
The filling of the $4f$ orbital begins with Cerium $(Z=58)$,which has the configuration $[Xe] 4f^1 5d^1 6s^2$.
As we proceed through the Lanthanide series,electrons are added to the $4f$ subshell.
Pairing in the $4f$ orbital starts when the $8^{th}$ electron enters the $4f$ subshell,which corresponds to Gadolinium $(Z=64)$.
The configuration of Gadolinium is $[Xe] 4f^7 5d^1 6s^2$,and the next element,Terbium $(Z=65)$,has the configuration $[Xe] 4f^9 6s^2$ (or $[Xe] 4f^8 5d^1 6s^2$),where the $8^{th}$ electron forces pairing in the $4f$ orbital.

(D) The basicity of lanthanoid hydroxides $Ln(OH)_3$ decreases as the ionic radius of the $Ln^{3+}$ ion decreases due to lanthanoid contraction.
As we move from $Pr$ to $Er$ in the lanthanoid series,the ionic radius decreases $(Pr^{3+} > Sm^{3+} > Gd^{3+} > Er^{3+})$.
Therefore,the basicity order is $Pr(OH)_3 > Sm(OH)_3 > Gd(OH)_3 > Er(OH)_3$.
Since acidity is inversely proportional to basicity,the most acidic compound is the one with the smallest ionic radius,which is $Er(OH)_3$.

(A) The electronic configuration $[Xe]\, 4f^{7}\, 5d^{1}\, 6s^{2}$ corresponds to Gadolinium ($Gd$,atomic number $64$).
$Gd$ belongs to the lanthanoid series,which are $f-$block elements,not $d-$block elements.
Therefore,the statement that it is placed in the $d-$block is incorrect.
$Gd$ has $7$ unpaired electrons in the $4f$ orbital and $1$ in the $5d$ orbital,totaling $8$ unpaired electrons,which is the maximum for lanthanoids.
It is a naturally occurring element.

(D) Due to lanthanoid contraction,the ionic radii of $f$-block elements decrease from $La$ to $Lu$ as the atomic number increases.
Since the atomic numbers are $La (57)$,$Ce (58)$,$Pm (61)$,and $Yb (70)$,the ionic radii follow the order: $La^{3+} > Ce^{3+} > Pm^{3+} > Yb^{3+}$.
Therefore,the correct order is $Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$.

(A) The electronic configuration of the element with atomic number $58$ (Cerium,$Ce$) is $[Xe] \ 4f^1 \ 5d^1 \ 6s^2$.
Since the last electron enters the $4f$ orbital,it is a lanthanoid element.
All lanthanoids belong to group $III-B$ (or group $3$) and the $6^{th}$ period of the periodic table.

(B) $1$. Lanthanoid hydroxides are less basic than actinoid hydroxides due to the higher ionic character of actinoid-oxygen bonds. This statement is correct.
$2$. Lanthanoids generally do not form oxocations (like $UO_2^{2+}$ or $PuO_2^{2+}$),whereas actinoids are well-known for forming stable oxocations. Therefore,the statement that lanthanoids form oxocations is incorrect.
$3$. Among the lanthanoids,only promethium $(Pm)$ is radioactive. This statement is correct.
$4$. Actinoids are known to form oxocations such as $UO_2^{2+}$ and $PuO_2^{2+}$. This statement is correct.

(D) Lawrencium $(Z = 103)$ is the last element of the Actinide series.
The electronic configuration of Lawrencium is $[Rn] \, 5f^{14} \, 6d^1 \, 7s^2$.
Although it was historically debated,experimental and theoretical studies confirm that the $6d$ orbital is occupied in the ground state configuration.

(A) The Lanthanoid contraction is the steady decrease in the atomic and ionic radii of the lanthanoids with increasing atomic number.
$1$. The similarity in size between $Zr$ ($4d$ series) and $Hf$ ($5d$ series) is a direct consequence of the Lanthanoid contraction.
$2$. Because $Zr$ and $Hf$ have similar sizes and chemical properties,they occur together in nature.
$3$. The high density of the sixth period elements is also a result of the Lanthanoid contraction,as the volume of the atoms decreases while the mass increases.
$4$. The $IE_1$ (first ionization energy) of $5d$ series elements is generally higher than that of $3d$ and $4d$ series elements,but this is primarily due to the poor shielding effect of $4f$ electrons and the increased effective nuclear charge,not a direct consequence of the contraction itself in the context of the other options provided.

(C) The actinoid series consists of elements with atomic numbers $89$ to $103$.
$U$ (Uranium,$Z=92$) belongs to the actinoid series.
$Y$ (Yttrium,$Z=39$) is a $d-$block element.
$Ta$ (Tantalum,$Z=73$) is a $d-$block element.
$Pm$ (Promethium,$Z=61$) belongs to the lanthanoid series.

(A) Transuranic elements are those with atomic numbers greater than $92$.
In uranium ores like $U_3O_8$,trace amounts of $Neptunium$ ($Np$,$Z=93$) and $Plutonium$ ($Pu$,$Z=94$) are found due to the neutron capture by $U^{238}$ followed by $\beta$-decay processes.
Therefore,the correct pair is $Np$ and $Pu$.

(D) The element $Am$ (Americium) is an actinoid. Actinoids exhibit a greater range of oxidation states compared to lanthanoids due to the comparable energies of $5f$,$6d$,and $7s$ orbitals.
$Am$ shows $+2, +3, +4, +5, +6,$ and $+7$ oxidation states,which is the maximum among the given options.

(B) The general electronic configuration $(n-2)f^{1-14}(n-1)d^{0-1}ns^2$ represents the $f$-block elements.
For $n=6$,the configuration is $4f^{1-14}5d^{0-1}6s^2$,which corresponds to the Lanthanides.
For $n=7$,the configuration is $5f^{1-14}6d^{0-1}7s^2$,which corresponds to the Actinides.

(A) The energy difference between $5f$ and $6d$ orbitals is smaller than the energy difference between $4f$ and $5d$ orbitals.
Because of this smaller energy gap,the electrons in actinoids can be removed from both $5f$ and $6d$ orbitals.
Consequently,actinoids exhibit a greater variety of oxidation states compared to lanthanoids.

(C) The number of unpaired electrons in a lanthanoid ion is determined by the $4f$ orbital configuration.
For a lanthanoid with $n$ unpaired electrons,the electronic configuration of the $4f$ subshell is $f^n$.
For a lanthanoid with $(14-n)$ unpaired electrons,the electronic configuration is $f^{14-n}$.
According to the principle of electronic configuration in $f$-orbitals,an $f^n$ configuration and an $f^{14-n}$ configuration have the same number of unpaired electrons.
For example,$f^1$ and $f^{13}$ both have $1$ unpaired electron,and $f^2$ and $f^{12}$ both have $2$ unpaired electrons.
Since both have the same number of unpaired electrons,they exhibit the same magnetic behaviour (paramagnetism) and the same colour due to $f-f$ transitions.

(D) In the lanthanoid series,the ionic radius decreases from left to right due to lanthanoid contraction.
As the ionic radius decreases,the covalent character of the $M-OH$ bond increases,which leads to an increase in the acidic nature of the hydroxides.
Therefore,the acidity increases as we move from $Pr$ to $Er$ in the periodic table.
Among the given options,$Er(OH)_3$ is the most acidic.

(C) The atomic number of Curium $(Cm)$ is $96$.
Its electronic configuration is based on the Radon $(Rn)$ core,which is $[Rn] = [Xe]\, 4f^{14}\, 5d^{10}\, 6s^2\, 6p^6$.
The $5f$ subshell is filled after the $6d$ and $7s$ orbitals.
For Curium $(Z=96)$,the configuration is $[Rn]\, 5f^7\, 6d^1\, 7s^2$.

(B) The atomic number of $Yb$ (Ytterbium) is $70$.
The ground state electronic configuration of $Yb$ is $[Xe] \, 4f^{14} \, 6s^2$.
When $Yb$ forms the $Yb^{2+}$ ion,it loses two electrons from the outermost $6s$ orbital.
Therefore,the electronic configuration of $Yb^{2+}$ is $[Xe] \, 4f^{14}$.

(B) The ionic radii of trivalent lanthanides decrease progressively with an increase in atomic number. This phenomenon is known as lanthanide contraction. Therefore,the statement that ionic radii increase with atomic number is false.

(B) $I$. $La(OH)_3$ is the most basic among lanthanide hydroxides because basicity decreases as the ionic radius decreases from $La$ to $Lu$. Thus,statement $I$ is false.
$II$. Due to lanthanide contraction,the ionic radii of $Zr^{4+}$ $(0.79 \ \mathring{A})$ and $Hf^{4+}$ $(0.78 \ \mathring{A})$ are almost identical. Thus,statement $II$ is true.
$III$. $Ce^{4+}$ is a strong oxidizing agent because it readily gains an electron to attain the more stable $Ce^{3+}$ oxidation state. Thus,statement $III$ is true.
Therefore,statements $II$ and $III$ are true.

(A) In lanthanide ions,the $4f$ electrons are well-shielded from the external field by the overlying $5s$ and $5p$ electrons. Thus,the magnetic effect of the orbital motion of the electron is not quenched. Consequently,the magnetic moment must be calculated by taking into account both the spin and orbital contributions.
The magnetic moment is given by $\mu = g\sqrt{J(J+1)}$,where $g = 1 + \frac{S(S+1) - L(L+1) + J(J+1)}{2J(J+1)}$.
For $Dy^{3+}$ $([Xe]4f^9)$,the $4f$ subshell is more than half-filled. In such cases,the spin and orbital angular momenta couple to give a high total angular momentum $J$,resulting in the highest magnetic moment among lanthanoids.
Therefore,both the Assertion and Reason are correct,and the Reason is the correct explanation of the Assertion.

(C) The electronic configuration of $Eu^{2+}$ is $[Xe] 4f^7$,which is a half-filled $f$-orbital configuration,providing extra stability.
The electronic configuration of $Ce^{2+}$ is $[Xe] 4f^1 5d^1$,which is less stable.
Therefore,the Assertion is correct.
Cerium salts are generally not used as catalysts in petroleum cracking; rather,zeolites or other specific catalysts are used. Thus,the Reason is incorrect.

(B) The magnetic moment values of actinides are lower than the theoretically predicted values because the $5f$ electrons are less effectively shielded by the outer shells. This leads to the quenching of the orbital contribution to the magnetic moment,making the observed values smaller than those calculated using the spin-only formula. While actinides are paramagnetic,the reason provided does not explain why the magnetic moments are lower than predicted.

(B) The atomic number of $Eu$ is $63$. The electronic configuration of $Eu$ is $[Xe] 4f^{7} 6s^{2}$.
For $Eu^{2+}$,two electrons are removed from the $6s$ orbital,resulting in $[Xe] 4f^{7}$.
The atomic number of $Ce$ is $58$. The electronic configuration of $Ce$ is $[Xe] 4f^{1} 5d^{1} 6s^{2}$.
For $Ce^{3+}$,three electrons are removed (two from $6s$ and one from $5d$),resulting in $[Xe] 4f^{1}$.

(A) The lanthanoid series elements typically exhibit a $+3$ oxidation state. However,Cerium $(Z=58)$ is well known to exhibit a $+4$ oxidation state because it achieves a stable noble gas configuration $([Xe]4f^0)$ upon losing four electrons.

(N/A) In actinoids,$5f$ orbitals are filled. These $5f$ orbitals have a poorer shielding effect than $4f$ orbitals (in lanthanoids).
Thus,the effective nuclear charge experienced by electrons in valence shells in the case of actinoids is much greater than that experienced by lanthanoids.
Hence,the size contraction in actinoids is greater as compared to that in lanthanoids.

(N/A) As we move along the lanthanoid series,the atomic number increases gradually by $1$. This means that the number of electrons and protons present in an atom also increases by $1$. As electrons are being added to the same shell,the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of a proton is more pronounced than the increase in the interelectronic repulsions due to the addition of an electron. Also,with the increase in atomic number,the number of electrons in the $4f$ orbital also increases. The $4f$ electrons have poor shielding effect. Therefore,the effective nuclear charge experienced by the outer electrons increases. Consequently,the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.
Consequences of lanthanoid contraction:
$i$. There is similarity in the properties of the second and third transition series.
$ii$. Separation of lanthanoids is possible due to lanthanoid contraction.
$iii$. It is due to lanthanoid contraction that there is a variation in the basic strength of lanthanoid hydroxides. (Basic strength decreases from $La(OH)_3$ to $Lu(OH)_3$).

(N/A) In the lanthanoid series,the $+3$ oxidation state is the most common,meaning $Ln(III)$ compounds are predominant. However,$+2$ and $+4$ oxidation states are also observed in certain solutions or solid compounds.

(N/A) $(i)$ Electronic configuration: The general electronic configuration for lanthanoids is $[Xe] \ 4f^{0-14} \ 5d^{0-1} \ 6s^2$ and that for actinoids is $[Rn] \ 5f^{1-14} \ 6d^{0-1} \ 7s^2$. Unlike $4f$ orbitals,$5f$ orbitals are not deeply buried and participate in bonding to a greater extent.
$(ii)$ Atomic and ionic sizes: Similar to lanthanoid contraction,actinoids exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater in actinoids due to the poorer shielding effect of $5f$ orbitals compared to $4f$ orbitals.
$(iii)$ Oxidation states: The principal oxidation state of lanthanoids is $+3$. However,$+2$ and $+4$ states are also observed due to the stability of empty,half-filled,or fully-filled $f$-orbitals. Actinoids exhibit a greater range of oxidation states (e.g.,$+3, +4, +5, +6, +7$) because the $5f, 6d,$ and $7s$ energy levels are of comparable energies.
$(iv)$ Chemical reactivity: Lanthanoids are highly reactive,similar to calcium,but reactivity decreases with increasing atomic number. Actinoids are highly reactive metals,especially when finely divided. They react with boiling water to form oxides and hydrides and combine with most non-metals at moderate temperatures. They are generally unaffected by alkalies,and their reaction with nitric acid is limited due to the formation of a protective oxide layer.

(N/A) An alloy is a solid solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys usually possess different physical properties than those of the component elements.
An important alloy of lanthanoids is Mischmetal. It contains lanthanoids $(94-95 \%)$,iron $(5 \%)$,and traces of $S$,$C$,$Si$,$Ca$,and $Al$.
Uses:
$(1)$ Mischmetal is used in cigarette and gas lighters.
$(2)$ It is used in flame-throwing tanks.
$(3)$ It is used in tracer bullets and shells.

Inner transition elements are those elements in which the last electron enters the $f$-orbital. These elements are also known as $f$-block elements,where the $4f$ and $5f$ orbitals are progressively filled.
Lanthanoids range from atomic numbers $58$ to $71$ ($4f$ series).
Actinoids range from atomic numbers $90$ to $103$ ($5f$ series).
Comparing the given atomic numbers:
$29$ is $Cu$ ($d$-block),
$59$ is $Pr$ (Lanthanoid),
$74$ is $W$ ($d$-block),
$95$ is $Am$ (Actinoid),
$102$ is $No$ (Actinoid),
$104$ is $Rf$ ($d$-block).
Therefore,the atomic numbers of the inner transition elements are $59, 95,$ and $102$.

(N/A) Lanthanoids primarily show a stable $+3$ oxidation state. They display a limited number of oxidation states because the energy gap between $4f$,$5d$,and $6s$ orbitals is quite large.
On the other hand,the energy difference between $5f$,$6d$,and $7s$ orbitals in actinoids is very small.
Consequently,actinoids exhibit a wide range of oxidation states.
For example,uranium $(U)$ and plutonium $(Pu)$ display $+3, +4, +5,$ and $+6$ oxidation states,while neptunium $(Np)$ displays $+3, +4, +5,$ and $+7$ oxidation states.

(N/A) The last element in the actinoid series is lawrencium,$Lr$. Its atomic number is $103$ and its electronic configuration is $[Rn] \, 5f^{14} \, 6d^{1} \, 7s^{2}$. The most common oxidation state displayed by it is $+3$; because after losing $3$ electrons it attains stable $f^{14}$ configuration.

(D) The atomic number of $Ce$ is $58$. The electronic configuration of $Ce$ is $[Xe] 4f^{1} 5d^{1} 6s^{2}$.
To form $Ce^{3+}$,three electrons are removed (two from $6s$ and one from $5d$ or $4f$ depending on the energy levels; for $Ce^{3+}$,the configuration is $[Xe] 4f^{1}$).
Thus,the electronic configuration of $Ce^{3+}$ is $[Xe] 4f^{1}$.
There is $n = 1$ unpaired electron in the $4f$ orbital.
The 'spin-only' magnetic moment formula is $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 1$:
$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.732 \ BM$.

(N/A) The lanthanoids that exhibit $+2$ and $+4$ oxidation states are listed in the table below:

$+2$ Oxidation State	$+4$ Oxidation State
$Nd \ (60), Sm \ (62), Eu \ (63), Tm \ (69), Yb \ (70)$	$Ce \ (58), Pr \ (59), Nd \ (60), Tb \ (65), Dy \ (66)$

Correlation with electronic configuration:
$1$. $Ce^{4+}$ $([Xe] \ 4f^0)$ attains a stable noble gas configuration.
$2$. $Tb^{4+}$ $([Xe] \ 4f^7)$ attains a stable half-filled $f$-orbital configuration.
$3$. $Eu^{2+}$ $([Xe] \ 4f^7)$ attains a stable half-filled $f$-orbital configuration.
$4$. $Yb^{2+}$ $([Xe] \ 4f^{14})$ attains a stable fully-filled $f$-orbital configuration.

(A) Electronic configuration:
The general electronic configuration for lanthanoids is $[Xe]^{54} 4f^{0-14} 5d^{0-1} 6s^{2}$ and that for actinoids is $[Rn]^{86} 5f^{1-14} 6d^{0-1} 7s^{2}$. Unlike $4f$ orbitals,$5f$ orbitals are not deeply buried and participate in bonding to a greater extent.
Oxidation states:
The principal oxidation state of lanthanoids is $+3$. However,sometimes we also encounter oxidation states of $+2$ and $+4$ due to the extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states because the $5f, 6d,$ and $7s$ levels are of comparable energies. Again,$+3$ is the principal oxidation state for actinoids.
Chemical reactivity:
In the lanthanoid series,the earlier members are more reactive,with reactivity comparable to $Ca$. With an increase in atomic number,they behave similar to $Al$. Actinoids are highly reactive metals,especially when finely divided. When added to boiling water,they give a mixture of oxide and hydride. Actinoids combine with most non-metals at moderate temperatures. Alkalies have no action on them,and they are only slightly affected by nitric acid due to the formation of a protective oxide layer.

(N/A) The electronic configurations are determined based on the Aufbau principle and the noble gas core notation:

Atomic Number	Electronic Configuration
$61$	$[Xe] \, 4f^{5} \, 6s^{2}$
$91$	$[Rn] \, 5f^{2} \, 6d^{1} \, 7s^{2}$
$101$	$[Rn] \, 5f^{13} \, 7s^{2}$
$109$	$[Rn] \, 5f^{14} \, 6d^{7} \, 7s^{2}$