General Characteristics Questions in English

(C) $Mn$ has the electronic configuration $[Ar] 3d^5 4s^2$.
$Mn^{2+}$ has the configuration $[Ar] 3d^5$,which is a half-filled $d$-orbital configuration,making it extra stable.
$Mn^{3+}$ has the configuration $[Ar] 3d^4$.
Due to the extra stability of the half-filled $d^5$ subshell,$Mn^{2+}$ is more stable than $Mn^{3+}$.
Therefore,the statement that $Mn^{3+}$ is more stable than $Mn^{2+}$ is incorrect.

(A) The transition metals (with the exception of $Zn$,$Cd$,and $Hg$) are very hard and have low volatility.
Their melting and boiling points are high.
The high melting points of these metals are attributed to the involvement of a greater number of electrons from $(n-1)d$ orbitals in addition to the $ns$ electrons in the interatomic metallic bonding.
Since more electrons participate in metallic bonding,the strength of the metallic bond increases,leading to higher melting points.
Thus,the Assertion is true and the Reason is the correct explanation for the Assertion.

(C) The ionic radii of trivalent lanthanide ions ($La^{3+}$ to $Lu^{3+}$) decrease due to lanthanide contraction.
$La^{3+}$ $(Z=57)$ has the largest radius,while $Lu^{3+}$ $(Z=71)$ has the smallest radius among the lanthanides.
$Y^{3+}$ $(Z=39)$ is a $4d$ transition metal ion and is significantly smaller than the $4f$ lanthanide ions.
Therefore,the order of ionic radii is $Y^{3+} < Lu^{3+} < Eu^{3+} < La^{3+}$.

(D) The general electronic configuration of group $12$ elements is $(n-1)d^{10} ns^2$.
When these elements form $+2$ oxidation state ions,they lose two electrons from the $ns$ orbital.
Therefore,the electronic configuration of $Zn^{2+}$,$Cd^{2+}$,and $Hg^{2+}$ becomes $(n-1)d^{10} ns^0$,which represents a fully filled $d$-subshell.

(B) The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$ and that of $Fe^{3+}$ is $[Ar] 3d^5$.
In $Fe^{3+}$,the $d$-orbital is exactly half-filled $(d^5)$.
According to Hund's rule,half-filled and fully-filled orbitals possess extra stability due to maximum exchange energy and symmetrical distribution of electrons.
Therefore,$Fe^{3+}$ is more stable than $Fe^{2+}$.

(B) The atomic number of $Co$ is $27$.
Electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
When $Co$ forms $Co^{2+}$,it loses two electrons from the $4s$ orbital.
Electronic configuration of $Co^{2+}$ is $[Ar] 3d^7 4s^0$.
The $3d^7$ configuration has $5$ orbitals,where electrons are filled according to Hund's rule: two orbitals are fully filled (paired) and three orbitals are singly occupied (unpaired).
Therefore,the number of unpaired electrons is $3$.

(A) The paramagnetic property of an ion is directly proportional to the number of unpaired electrons present in its $d$-orbitals.
$Cu^{2+} (3d^9)$: $1$ unpaired electron.
$V^{2+} (3d^3)$: $3$ unpaired electrons.
$Cr^{2+} (3d^4)$: $4$ unpaired electrons.
$Mn^{2+} (3d^5)$: $5$ unpaired electrons.
Therefore,the increasing order of paramagnetic property is $Cu^{2+} < V^{2+} < Cr^{2+} < Mn^{2+}$.

(C) The enthalpy of atomization depends on the strength of metallic bonding.
In $s$-block and $p$-block elements,the metallic bond strength generally decreases down a group due to an increase in atomic size,leading to a decrease in the enthalpy of atomization.
However,for $d$-block elements (transition metals),the enthalpy of atomization generally increases down a group because the $d$-orbitals become more involved in metallic bonding as the principal quantum number increases,leading to stronger metallic bonds.
Therefore,the statement that the enthalpy of atomization decreases down a group in $d$-block elements is incorrect.

(D) The electronic configuration of the neutral atoms is:
$Ti (Z=22): [Ar] 3d^2 4s^2$
$V (Z=23): [Ar] 3d^3 4s^2$
$Cr (Z=24): [Ar] 3d^5 4s^1$
$Mn (Z=25): [Ar] 3d^5 4s^2$
For the ions in option $(d)$:
$Ti^{2+}: [Ar] 3d^2$
$V^{3+}: [Ar] 3d^2$
$Cr^{4+}: [Ar] 3d^2$
$Mn^{5+}: [Ar] 3d^2$
Since $[Ar]$ corresponds to $1s^2 2s^2 2p^6 3s^2 3p^6$,all these ions possess the $3d^2$ configuration outside the $3p^6$ shell.

(D) $Sc^{3+}$ has the electronic configuration $[Ar] 3d^0 4s^0$.
Since there are no unpaired electrons,it is diamagnetic.

(A) $V^{4+}$ $(3d^1)$,$Ti^{3+}$ $(3d^1)$,and $Ni^{2+}$ $(3d^8)$ have unpaired electrons in their $d$-subshell. Therefore,these ions show $d-d$ transition and exhibit characteristic colours.
$Ti^{4+}$ $(3d^0)$ does not have any unpaired electrons,therefore it is colourless.
Matching the ions with their colours:
$A. V^{4+}$: $IV. Blue$
$B. Ti^{3+}$: $II. Purple$
$C. Ti^{4+}$: $I. Colourless$
$D. Ni^{2+}$: $III. Green$
Thus,the correct matching is $A-IV, B-II, C-I, D-III$.

(D) German silver (nickel silver) is an alloy of copper,zinc,and nickel,often in the proportions $5:2:2$.
It is commonly used in the manufacturing of cheap jewellery,cutlery,and electrical components.

(B) The electronic configuration of vanadium ($V$,$Z=23$) is $[Ar] 3d^3 4s^2$.
For $V^{3+}$,the configuration is $[Ar] 3d^2 4s^0$,which contains $2$ unpaired electrons.
Now,let us check the options:
$Fe^{3+}$ $(Z=26)$: $[Ar] 3d^5$,which has $5$ unpaired electrons.
$Ni^{2+}$ $(Z=28)$: $[Ar] 3d^8$,which has $2$ unpaired electrons.
$Mn^{2+}$ $(Z=25)$: $[Ar] 3d^5$,which has $5$ unpaired electrons.
$Cr^{3+}$ $(Z=24)$: $[Ar] 3d^3$,which has $3$ unpaired electrons.
Thus,$Ni^{2+}$ has the same number of unpaired electrons as $V^{3+}$.

(A) $Cu^{+} = [Ar] 3d^{10}$ (No unpaired electrons,so it is colourless).
$Zn^{2+} = [Ar] 3d^{10}$ (No unpaired electrons,so it is colourless).
Therefore,aqueous solutions of $Cu^{+}$ and $Zn^{2+}$ are colourless.
$Cu^{2+} = [Ar] 3d^9$ (One unpaired electron is present,so it is coloured).
$Fe^{3+}$ is yellow/brown in aqueous solution.
$MnO_4^{-}$ is purple due to charge transfer,despite having no unpaired electrons.
Thus,the statement in option $(A)$ is correct.

(C) Paramagnetic property depends upon the number of unpaired electrons. Higher the number of unpaired electrons,higher the paramagnetic property will be.
$Cu^{2+} = [Ar] 3d^9$,number of unpaired electrons $= 1$
$V^{2+} = [Ar] 3d^3$,number of unpaired electrons $= 3$
$Cr^{2+} = [Ar] 3d^4$,number of unpaired electrons $= 4$
$Mn^{2+} = [Ar] 3d^5$,number of unpaired electrons $= 5$
Hence,the correct order of increasing paramagnetic property is $Cu^{2+} < V^{2+} < Cr^{2+} < Mn^{2+}$.

(C) Spin only magnetic moments depend upon the number of unpaired electrons; the greater the number of unpaired electrons,the greater will be the spin only magnetic moment.
$Mn$ $(Z=25)$: $[Ar] 3d^5 4s^2$
$Mn^{2+}$: $[Ar] 3d^5$,number of unpaired electrons = $5$
$Cr$ $(Z=24)$: $[Ar] 3d^5 4s^1$
$Cr^{2+}$: $[Ar] 3d^4$,number of unpaired electrons = $4$
$V$ $(Z=23)$: $[Ar] 3d^3 4s^2$
$V^{2+}$: $[Ar] 3d^3$,number of unpaired electrons = $3$
Since the magnetic moment $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons,the order of magnetic moment is $Mn^{2+} > Cr^{2+} > V^{2+}$.

(A) The paramagnetic moment depends on the number of unpaired electrons $(n)$.
For $Cu^{2+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^9$. It has $n=1$ unpaired electron.
For $Ti^{3+}$ $(Z=22)$: Electronic configuration is $[Ar] 3d^1$. It has $n=1$ unpaired electron.
Since both $Cu^{2+}$ and $Ti^{3+}$ have the same number of unpaired electrons $(n=1)$,they have the same paramagnetic moment.
Therefore,the correct pair is $Cu^{2+}, Ti^{3+}$.

(B) The colour of transition metal ions is due to the $d-d$ transition of electrons,which requires the presence of unpaired electrons in the $d$-orbitals.
For $Sc^{3+}$: The electronic configuration is $[Ar] 3d^0$. It has no unpaired electrons.
For $Zn^{2+}$: The electronic configuration is $[Ar] 3d^{10}$. It has no unpaired electrons.
Since both $Sc^{3+}$ and $Zn^{2+}$ have no unpaired electrons,they are colourless.

(B) The reaction of ferrous ion $(Fe^{2+})$ with acidified hydrogen peroxide $(H_2O_2)$ results in the oxidation of $Fe^{2+}$ to ferric ion $(Fe^{3+})$:
$2Fe^{2+} + H_2O_2 + 2H^+ \longrightarrow 2Fe^{3+} + 2H_2O$
Thus,the ion $X$ is $Fe^{3+}$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
The number of $d$-electrons is $5$.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Fe^{3+}$,$n = 5$.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.

(A) The atomic number of $Cu$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
For $Cu^{2+}$ ion,two electrons are removed (one from $4s$ and one from $3d$),resulting in the configuration $[Ar] 3d^9$.
In the $3d^9$ configuration,there is $1$ unpaired electron $(n = 1)$.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$.

(D) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Cu^{2+}$ $([Ar] 3d^9)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$2$. For $Ti^{3+}$ $([Ar] 3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$3$. For $Ni^{2+}$ $([Ar] 3d^8)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ BM$.
$4$. For $Mn^{2+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
Since $Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,it exhibits the highest magnetic moment.

(B) The atomic radii of elements in the same group generally increase down the group due to the addition of new shells. However, for elements in the $4d$ and $5d$ series, the atomic radii are often very similar due to the lanthanoid contraction.
$1$. $Zr$ ($4d$ series) and $Hf$ ($5d$ series) have nearly identical atomic radii ($160 \text{ pm}$ and $159 \text{ pm}$ respectively) because of the lanthanoid contraction.
$2$. $Mo$ ($4d$ series) and $W$ ($5d$ series) also exhibit nearly identical atomic radii ($140 \text{ pm}$ and $141 \text{ pm}$ respectively) due to the same phenomenon.
Therefore, pairs $II$ and $III$ have nearly the same atomic radius.

(C) To determine the correct pair,we analyze the electronic configuration and number of electrons for each ion:
$1$. $Lu^{3+}$ $(Z=71)$: $[Xe] 4f^{14}$,electrons = $68$,diamagnetic.
$2$. $Yb^{2+}$ $(Z=70)$: $[Xe] 4f^{14}$,electrons = $68$,diamagnetic.
$3$. $Eu^{3+}$ $(Z=63)$: $[Xe] 4f^{6}$,electrons = $60$,paramagnetic.
$4$. $Pm^{2+}$ $(Z=61)$: $[Xe] 4f^{5}$,electrons = $59$,paramagnetic.
$5$. $Eu^{2+}$ $(Z=63)$: $[Xe] 4f^{7}$,electrons = $61$,paramagnetic.
$6$. $Gd^{3+}$ $(Z=64)$: $[Xe] 4f^{7}$,electrons = $61$,paramagnetic.
$7$. $La^{3+}$ $(Z=57)$: $[Xe]$,electrons = $54$,diamagnetic.
$8$. $Ce^{4+}$ $(Z=58)$: $[Xe]$,electrons = $54$,diamagnetic.
Comparing the options,$Eu^{2+}$ and $Gd^{3+}$ both have $61$ electrons and both possess $7$ unpaired electrons in the $4f$ orbital,making them paramagnetic.

(B) Ions exhibit colour in aqueous solution due to the presence of unpaired electrons in their $d$-orbitals,which allow for $d-d$ transitions.
Electronic configurations:
$La^{3+} (Z=57): [Xe] 4f^0 5d^0$ (No unpaired electrons).
$Ti^{3+} (Z=22): [Ar] 3d^1$ (One unpaired electron).
$Lu^{3+} (Z=71): [Xe] 4f^{14}$ (No unpaired electrons).
$Sc^{3+} (Z=21): [Ar] 3d^0$ (No unpaired electrons).
Since $Ti^{3+}$ has one unpaired electron,it will exhibit colour in aqueous solution.

(D) $^{57}La$ and $^{71}Lu$ have a $d^1$ electronic configuration.
Electronic configuration of $^{57}La = [^{54}Xe] 5d^1 6s^2$.
Electronic configuration of $^{71}Lu = [^{54}Xe] 4f^{14} 5d^1 6s^2$.

(D) In the electrochemical series,$Cu$ is placed below hydrogen. Hence,it has a positive value of standard reduction potential $(E^{\circ}_{M^{2+}/M})$.
The standard reduction potentials for the given elements are:
$Ti (Z=21) \Rightarrow -1.63 \ V$
$Cr (Z=24) \Rightarrow -0.90 \ V$
$Ni (Z=28) \Rightarrow -0.257 \ V$
$Cu (Z=29) \Rightarrow +0.337 \ V$
Comparing these values,the highest reduction potential is of $Cu$ because it is the only one with a positive value,while $Ni$,$Cr$,and $Ti$ have negative reduction potential values.

(D) Interstitial compounds are formed when small atoms like $H$,$C$,or $N$ are trapped inside the crystal lattice of transition metals.
$(I)$ They are very hard and rigid.
$(II)$ They have high melting points,which are higher than those of the pure metals.
$(III)$ They retain metallic conductivity.
$(IV)$ They are chemically inert.

(D) The enthalpy of atomisation is the enthalpy change that accompanies the total separation of all atoms in a chemical substance.
The extent of metallic bonding an element undergoes decides the enthalpy of atomisation. The more extensive the metallic bonding of an element,the higher will be its enthalpy of atomisation.
$Cr$ $(3d^5 4s^1)$ has the highest enthalpy of atomisation for the first row of transition elements due to the maximum number of unpaired electrons available for metallic bonding.
$Zn$ $(3d^{10} 4s^2)$ has a completely filled $d$-orbital and has no unpaired electrons to participate in the formation of metallic bonds. Therefore,the metallic bonding in zinc is the weakest,resulting in the lowest enthalpy of atomisation.

(B) The highest oxidation state observed in the first row transition metals is $+7$.
In the first row,$Mn$ (Manganese) with the outer electronic configuration $3d^5, 4s^2$ shows a $+7$ oxidation state.
It has a total of $7$ electrons in its outermost orbitals ($5$ in $3d$ and $2$ in $4s$),which can be involved in bonding,thus allowing it to exhibit a $+7$ oxidation state.

(B) The atomic number of $Cr$ is $24$. Its expected configuration is $3 \ d^4 4 \ s^2$,but due to the extra stability of half-filled $d$-orbitals,it becomes $3 \ d^5 4 \ s^1$.
The atomic number of $Cu$ is $29$. Its expected configuration is $3 \ d^9 4 \ s^2$,but due to the extra stability of completely-filled $d$-orbitals,it becomes $3 \ d^{10} 4 \ s^1$.
Therefore,the correct configuration is $3 \ d^5 4 \ s^1$ and $3 \ d^{10} 4 \ s^1$.

(C) The work function is the minimum energy required to eject an electron from the surface of a metal.
It is generally higher for transition metals compared to alkali metals.
Among the given options,$Na$ and $K$ are alkali metals with low work functions.
Between $Cu$ and $Ag$,$Cu$ has a higher work function (approximately $4.7 \ eV$) compared to $Ag$ (approximately $4.26 \ eV$).
Therefore,$Cu$ has the highest work function value among the choices provided.

(A) An ion has zero magnetic moment if it has no unpaired electrons (diamagnetic).
$V^{2+} = [Ar] 3d^3$ ($3$ unpaired electrons)
$Zn^{2+} = [Ar] 3d^{10}$ ($0$ unpaired electrons)
$Cu^{2+} = [Ar] 3d^9$ ($1$ unpaired electron)
$Fe^{2+} = [Ar] 3d^6$ ($4$ unpaired electrons)
$Fe^{3+} = [Ar] 3d^5$ ($5$ unpaired electrons)
$Ti^{3+} = [Ar] 3d^1$ ($1$ unpaired electron)
$Sc^{3+} = [Ar] 3d^0$ ($0$ unpaired electrons)
$Ti^{4+} = [Ar] 3d^0$ ($0$ unpaired electrons)
$Ni^{3+} = [Ar] 3d^7$ ($3$ unpaired electrons)
$Co^{3+} = [Ar] 3d^6$ ($4$ unpaired electrons)
$Cu^{+} = [Ar] 3d^{10}$ ($0$ unpaired electrons)
The ions with zero magnetic moment are $Zn^{2+}$,$Sc^{3+}$,$Ti^{4+}$,and $Cu^+$.
There are $4$ such ions.

(B) The magnetic moment $(\mu)$ is related to the number of unpaired electrons '$n$' by the formula: $\mu_{eff} = \sqrt{n(n+2)} \ BM$.
Given $\mu_{eff} = \sqrt{24} \ BM$,we equate $\sqrt{n(n+2)} = \sqrt{24}$,which gives $n(n+2) = 24$.
Solving for $n$,we get $n^2 + 2n - 24 = 0$,which factors to $(n+6)(n-4) = 0$. Since $n$ must be positive,$n = 4$.
Now,let us check the number of unpaired electrons for each ion:
$Mn^{2+}$ $(Z=25)$: $[Ar] 3d^5$,$n = 5$.
$Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$,$n = 4$.
$Fe^{3+}$ $(Z=26)$: $[Ar] 3d^5$,$n = 5$.
$Co^{2+}$ $(Z=27)$: $[Ar] 3d^7$,$n = 3$.
Thus,$Fe^{2+}$ has $4$ unpaired electrons and corresponds to a magnetic moment of $\sqrt{24} \ BM$.

(D) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For the given ions:
$(I) Fe^{2+} ([Ar] 3d^6)$: Number of unpaired electrons $(n)$ = $4$.
$(II) Ti^{2+} ([Ar] 3d^2)$: Number of unpaired electrons $(n)$ = $2$.
$(III) Cu^{2+} ([Ar] 3d^9)$: Number of unpaired electrons $(n)$ = $1$.
$(IV) V^{2+} ([Ar] 3d^3)$: Number of unpaired electrons $(n)$ = $3$.
Comparing the number of unpaired electrons: $1 (III) < 2 (II) < 3 (IV) < 4 (I)$.
Therefore,the increasing order of magnetic moment is $III < II < IV < I$.

(A) The small increase in atomic radii of the $5d$ series elements compared to the $4d$ series is due to the lanthanoid contraction.
This phenomenon occurs because the $4f$ orbitals are filled before the $5d$ orbitals.
The $4f$ electrons provide poor shielding of the nuclear charge,leading to a greater effective nuclear charge,which pulls the valence electrons closer to the nucleus.
Therefore,$X = 4f$ and $Y = 5d$.

(A) $Mo$ (Molybdenum) has an atomic number of $42$.
Its expected configuration is $[Kr] 4d^4 5s^2$,but it adopts the $4d^5 5s^1$ configuration to achieve a half-filled $d$-subshell.
This configuration provides extra stability due to higher exchange energy associated with the half-filled $d$-subshell.
Thus,both the assertion and the reason are true,and the reason correctly explains the assertion.

(C) The atomic number of $Praseodymium$ $(Pr)$ is $59$.
Following the $Aufbau$ principle,we fill the orbitals after the $Xenon$ ($Xe$,atomic number $54$) core.
The remaining $5$ electrons occupy the $4f$ and $6s$ orbitals.
The correct electronic configuration is $[Xe] 4f^3 6s^2$.

(C) The atomic number of $Mn$ is $25$.
The electronic configuration of $Mn$ is $[Ar] 3d^5 4s^2$.
When $Mn$ forms $Mn^{2+}$ ion,it loses two electrons from the $4s$ orbital.
Thus,the electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$.
In the $3d$ subshell,there are $5$ orbitals,and according to Hund's rule,each orbital will be occupied by one electron.
Therefore,there are $5$ unpaired electrons in $Mn^{2+}$ ion.