General Characteristics Questions in English

(C) The oxidation states of the given transition elements are as follows:
$Fe$ $(3d^6 4s^2)$: $+2, +3, +4, +5, +6$
$Cr$ $(3d^5 4s^1)$: $+2, +3, +4, +5, +6$
$Mn$ $(3d^5 4s^2)$: $+2, +3, +4, +5, +6, +7$
$Ni$ $(3d^8 4s^2)$: $+2, +3, +4$
Comparing these,$Mn$ exhibits the highest number of oxidation states (from $+2$ to $+7$).

(B) The first ionization enthalpy $(IE_1)$ of the $3d$ transition series elements generally increases from left to right across the period due to the increase in effective nuclear charge.
Among the given elements ($Sc$,$Mn$,$Cu$,$Zn$),$Sc$ $(Z=21)$ is the first element of the $3d$ series.
Since $IE_1$ increases as we move from left to right,the element at the beginning of the series will have the lowest ionization enthalpy.
Therefore,$Sc$ has the lowest $IE_1$ value.

(C) Transition elements are defined as elements that have partially filled $d-$orbitals in their ground state or in any of their common oxidation states.
$Hg$ has an electronic configuration of $[Xe] 4f^{14} 5d^{10} 6s^2$.
Since $Hg$ has a completely filled $d-$orbital $(d^{10})$ in its ground state and does not form ions with partially filled $d-$orbitals,it is not considered a transition element.
Similarly,$Zn$ and $Cd$ are also not considered transition elements for the same reason.

(A) The $5d$ transition series consists of elements from Lanthanum ($La$,atomic number $57$) to Mercury ($Hg$,atomic number $80$).
This series comprises a portion of the $d$-block elements,specifically those where the $5d$ electron shell is being filled progressively.
$Hg$ (Mercury) is the last element of this series,with an atomic number of $80$.
Its electron configuration is $[Xe] 4f^{14} 5d^{10} 6s^2$.
Mercury is well-known for being the only metal that is liquid at room temperature.

(A) Nichrome,an alloy of nickel and chromium in the ratio $80:20$,is specifically used for the construction of gas turbine engines due to its high heat resistance and oxidation resistance.
Titanium alloys are also used in aerospace applications for their ability to withstand stress at high temperatures.

(C) Transition elements are metals and possess metallic properties.
They are generally hard,malleable,and ductile.
They can form alloys with other metals.
Transition elements are good conductors of both heat and electricity.
Therefore,the statement that they do not conduct heat is incorrect.

(C) To determine the number of unpaired electrons,we look at the electronic configuration of each ion:
$Fe^{3+} = [Ar] 3d^5$ ($5$ unpaired electrons)
$Mn^{2+} = [Ar] 3d^5$ ($5$ unpaired electrons)
$Co^{2+} = [Ar] 3d^7$ ($3$ unpaired electrons)
$Sc^{3+} = [Ar] 3d^0$ ($0$ unpaired electrons)
$Zn^{2+} = [Ar] 3d^{10}$ ($0$ unpaired electrons)
$Mn^{3+} = [Ar] 3d^4$ ($4$ unpaired electrons)
Comparing the pairs:
Option $A$: $Co^{2+}$ $(3)$ and $Mn^{3+}$ $(4)$ - Different
Option $B$: $Mn^{2+}$ $(5)$ and $Zn^{2+}$ $(0)$ - Different
Option $C$: $Fe^{3+}$ $(5)$ and $Mn^{2+}$ $(5)$ - Same
Option $D$: $Sc^{3+}$ $(0)$ and $Co^{2+}$ $(3)$ - Different
Therefore,the pair with the same number of unpaired electrons is $Fe^{3+}$ and $Mn^{2+}$.

(D) Scandium ($Sc$,atomic number $21$) has the electronic configuration $[Ar] 3d^1 4s^2$.
It loses three electrons (two from $4s$ and one from $3d$) to achieve the stable noble gas configuration of Argon $([Ar])$.
Therefore,the only common and stable oxidation state exhibited by scandium is $+3$.

(C) The $5d$ series consists of elements from $Lanthanum$ ($La$,$Z=57$) and $Hafnium$ ($Hf$,$Z=72$) to $Mercury$ ($Hg$,$Z=80$). The statement that it consists of all elements from $La$ to $Hg$ is incorrect because the $4f$ series (lanthanoids) intervenes between $La$ and $Hf$.

(B) The atomic number of copper $(Cu)$ is $29$. The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
In the $+2$ oxidation state, $Cu^{2+}$ loses two electrons, resulting in the configuration $[Ar] 3d^9$.
In the $3d^9$ configuration, there is $n = 1$ unpaired electron.
The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 1$, we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(B) Ferromagnetism is a property shown by substances that are strongly attracted by a magnetic field.
Among the transition metals,$Fe$,$Co$,and $Ni$ are known to exhibit ferromagnetic properties at room temperature.

(C) Bronze is an alloy primarily composed of copper $(Cu)$ and tin $(Sn)$.
It is widely used in the casting of statues and sculptures due to its durability,corrosion resistance,and ability to capture fine details.

(C) The electronic configuration of $Sc^{3+}$ is $[Ar] \ 3d^0$.
Since the $3d$ orbital is completely empty,there are no unpaired electrons present.
Because $d-d$ transitions are not possible in the absence of electrons in the $d$-orbital,the $Sc^{3+}$ ion forms a colourless aqueous solution.

(A) Ferromagnetic substances are those which are strongly attracted by a magnetic field.
Common examples of ferromagnetic elements are $Fe$,$Co$,and $Ni$.
$Cr$ (Chromium) is an antiferromagnetic substance,not a ferromagnetic one.
Therefore,the correct answer is $Cr$.

(C) The electronic configurations of the given lanthanoids are as follows:
$Ce (Z=58) = [Xe] 4f^1 5d^1 6s^2$
$Nd (Z=60) = [Xe] 4f^4 6s^2$
$Sm (Z=62) = [Xe] 4f^6 6s^2$
$Eu (Z=63) = [Xe] 4f^7 6s^2$
$Gd (Z=64) = [Xe] 4f^7 5d^1 6s^2$
$Dy (Z=66) = [Xe] 4f^{10} 6s^2$
$Lu (Z=71) = [Xe] 4f^{14} 5d^1 6s^2$
Among the given options,$Gd (Z=64)$ and $Lu (Z=71)$ both possess exactly one electron in the $5d$-subshell.

(D) The $first$ transition series corresponds to the $3d$ series,which starts with Scandium $(Sc)$.
Scandium has an atomic number of $21$.
The electronic configuration is determined by filling the $1s, 2s, 2p, 3s, 3p, 4s,$ and $3d$ orbitals.
$Sc (Z=21) = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1$.
In terms of the noble gas core,this is written as $[Ar] 3d^1 4s^2$.

(C) The element with atomic number $21$ is Scandium $(Sc)$.
Its electronic configuration is $[Ar] 3d^1 4s^2$.
In the transition metal series,the $3d$ orbital begins to fill after the $4s$ orbital is filled,which is a characteristic feature of transition elements.

(D) The electronic configuration of $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$.
It loses two electrons from the $4s$ orbital to form the $Zn^{2+}$ ion,which has a stable fully-filled $3d^{10}$ configuration.
Therefore,$Zn$ exhibits only a $+2$ oxidation state.

(D) Cupro-nickel alloys (typically containing $70-90\%$ copper and $10-30\%$ nickel) are widely used for marine condenser tubes because they possess excellent resistance to corrosion and biofouling in seawater environments.

(D) The atomic number of $Sc$ is $21$. The electronic configuration of $Sc$ is $[Ar] \ 3d^1 \ 4s^2$. Thus,$Sc^{3+}$ is $[Ar]$,which has $0$ unpaired electrons.
The atomic number of $Zn$ is $30$. The electronic configuration of $Zn$ is $[Ar] \ 3d^{10} \ 4s^2$. Thus,$Zn^{2+}$ is $[Ar] \ 3d^{10}$,which has $0$ unpaired electrons.
Therefore,the number of unpaired electrons in $Sc^{3+}$ and $Zn^{2+}$ are $0$ and $0$ respectively.

(B) The electronic configuration of $Cd$ $(Z=48)$ is $[Kr] 4d^{10} 5s^2$.
Since the $4d$ subshell contains $10$ electrons,it is completely filled.

(C) Interstitial compounds are formed when small atoms like $H$,$C$,or $N$ are trapped inside the crystal lattice of transition metals.
- These compounds retain the chemical properties of the parent transition metals,meaning they are chemically inert.
- They possess very high melting points,which are higher than those of the pure parent metals.
- They are extremely hard,with some borides being comparable to diamond in hardness.
- They maintain the electrical and thermal conductivity of the parent metal.
- Therefore,the statement that their chemical properties are different from the parent metal is incorrect.

(A) The electronic configuration of Manganese $(Mn)$ is $[Ar] 3d^5 4s^2$.
Due to the presence of $7$ valence electrons ($5$ in the $3d$ subshell and $2$ in the $4s$ subshell),it can exhibit a wide range of oxidation states from $+2$ to $+7$.

(B) In $Zn$ $(3d^{10} 4s^2)$,all electrons in the $d$-orbitals are paired.
Due to the absence of unpaired electrons,the metallic bonds in $Zn$ are weak.
Consequently,$Zn$ is a soft metal compared to other transition elements.

(C) The colour of transition metal ions depends on the presence of unpaired electrons in the $d$-orbitals,which allow for $d-d$ transitions.
$Cu^{+}$ has an electronic configuration of $[Ar] 3d^{10}$.
Since all $10$ electrons are paired in the $3d$ subshell,there are no unpaired electrons.
Therefore,$Cu^{+}$ does not exhibit $d-d$ transitions and does not form coloured compounds.

(A) The electronic configuration of $Sc$ is $[Ar] 3d^1 4s^2$.
In the $+3$ oxidation state,$Sc^{3+}$ has the configuration $[Ar] 3d^0$.
Since $Sc^{3+}$ has no unpaired electrons in its $d$-orbitals,it cannot undergo $d-d$ transitions,making its compounds colourless.

(A) In the $3d-$ series elements,the density and atomic mass increase from $Sc$ to $Zn$ across the period.
Since $Sc$ $(Z = 21)$ is the first element in the $3d-$ series,it has the lowest atomic mass and the lowest density among the given options.

(B) Most transition elements are very hard due to strong metallic bonding involving $d$-electrons.
However,elements like $Zn, Cd,$ and $Hg$ have completely filled $d$-orbitals ($d^{10}$ configuration),which results in weak metallic bonding.
Therefore,$Zn$ is a soft element compared to $Co, W,$ and $Mo$.

(C) The highest oxidation state exhibited by any transition element is $+8$.
This is observed in elements like Ruthenium $(Ru)$ and Osmium $(Os)$ in their tetroxides,such as $RuO_4$ and $OsO_4$.

(C) $(C)$
Across the $3d-$ series, the density of the elements generally increases with an increase in atomic number due to the increase in atomic mass and the decrease in atomic radius.
Therefore, the correct decreasing order of density is $Ni > Fe > Cr > V$.

(C) The electronic configuration of titanium $(Z=22)$ is $[Ar] 3d^2 4s^2$. In its $Ti^{4+}$ state,it loses all $4$ valence electrons,resulting in a $3d^0$ configuration,which makes it colourless due to the absence of $d-d$ transitions.
The electronic configuration of copper $(Z=29)$ is $[Ar] 3d^{10} 4s^1$. In its $Cu^+$ state,it loses the $4s$ electron,resulting in a $3d^{10}$ configuration. Since the $d$-subshell is completely filled,no $d-d$ transition is possible,making $Cu^+$ compounds colourless.

(D) The electronic configuration of $Sc$ $(Z = 21)$ is $[Ar] 3d^1 4s^2$.
In its common oxidation state of $+3$,it forms $Sc^{3+}$ ion with the configuration $[Ar] 3d^0$.
Since there are no unpaired electrons in the $d$-orbital,$d-d$ transition is not possible,which makes its compounds colourless.

(A) For $d$-block elements,the maximum oxidation state is determined by the sum of $ns$ and $(n-1)d$ electrons.
$(a)$ $Mn (25) = [Ar] 3d^{5} 4s^{2}$. Maximum $O$.$S$. $= 5 + 2 = +7$.
$(b)$ $Cr (24) = [Ar] 3d^{5} 4s^{1}$. Maximum $O$.$S$. $= 5 + 1 = +6$.
$(c)$ $Cu (29) = [Ar] 3d^{10} 4s^{1}$. Common $O$.$S$. $= +1, +2$.
$(d)$ $Fe (26) = [Ar] 3d^{6} 4s^{2}$. Common $O$.$S$. $= +2, +3$.
Thus,only $Mn$ exhibits a $+7$ oxidation state among the given elements.

(C) The electronic configuration of $Ce$ $(Z=58)$ is: $[Xe] 4f^{1} 5d^{1} 6s^{2}$.
In the $+4$ oxidation state,$Ce^{4+}$ loses four electrons to achieve the stable noble gas configuration of $Xe$ $(Z=54)$.
The configuration of $Ce^{4+}$ is $[Xe] 4f^{0} 5d^{0} 6s^{0}$.
Since all valence orbitals $(4f, 5d, 6s)$ are empty,$Ce^{4+}$ attains the stable configuration of the nearest inert gas,making it highly stable.

(A) The electronic configurations of the given elements are as follows:
$Fe$ $(Z=26)$: $[Ar] 3d^6 4s^2$. It has $4$ unpaired electrons.
$Cu$ $(Z=29)$: $[Ar] 3d^{10} 4s^1$. It has $1$ unpaired electron.
$Co$ $(Z=27)$: $[Ar] 3d^7 4s^2$. It has $3$ unpaired electrons.
$Ni$ $(Z=28)$: $[Ar] 3d^8 4s^2$. It has $2$ unpaired electrons.
Thus,the maximum number of unpaired electrons is present in $Fe$.

(D) $Sc^{3+}$ has an electronic configuration of $[Ar] 3d^0$.
Since there are no electrons in the $d$-orbital,$d-d$ transitions are not possible,making the ion colourless.

(D) Transition elements are defined as elements which have incompletely filled $d$-orbitals in their ground state or in any of their oxidation states.
Generally,the $d$-block elements are referred to as transition elements.
The general electronic configuration of these elements is represented as $(n-1) d^{1-10}, n s^{1-2}$.

(D) The first ionisation enthalpy $(IE_1)$ of lanthanoids generally increases with an increase in atomic number due to the lanthanoid contraction,which leads to a decrease in atomic size and stronger attraction of the nucleus for the valence electrons.
Among the given elements,$Ce$ $(Z=58)$,$La$ $(Z=57)$,$Gd$ $(Z=64)$,and $Yb$ $(Z=70)$,the element $Yb$ has the highest atomic number.
$Yb$ has a stable electronic configuration of $[Xe] 4f^{14} 6s^2$. Due to the completely filled $4f$ subshell and the highest effective nuclear charge among the options,it requires the most energy to remove the first electron.
Therefore,$Yb$ has the highest $IE_1$.

(D) The atomic number of lutetium $(Lu)$ is $71$.
The electronic configuration of neutral $Lu$ is $[Xe] 4f^{14} 5d^1 6s^2$.
In its $+3$ oxidation state,$Lu$ loses three electrons (two from $6s$ and one from $5d$),resulting in the configuration $[Xe] 4f^{14}$.
Since the $4f$ subshell is completely filled with $14$ electrons,there are no unpaired electrons in the $f$ orbitals.
Therefore,the number of unpaired electrons is $0$.

(A) In the lanthanoid series,the basic strength of hydroxides decreases as the atomic number increases from $La$ to $Lu$. This is due to the lanthanoid contraction,which leads to a decrease in ionic radius and an increase in covalent character of the $M-OH$ bond. Since $La^{3+}$ has the largest ionic radius among the given options,the $La-OH$ bond is the most ionic,making $La(OH)_3$ the strongest base.

(B) The ground state electronic configurations are as follows:
$Yb (Z=70): [Xe] 4f^{14} 6s^2$
$Eu (Z=63): [Xe] 4f^7 6s^2$
$Cd (Z=48): [Kr] 4d^{10} 5s^2$
$W (Z=74): [Xe] 4f^{14} 5d^4 6s^2$
Among the given options,$Yb$ has a completely filled $4f$ orbital $(4f^{14})$.

(B) The electronic configuration of $La (Z=57)$ is $[Xe] 5d^1 6s^2$.
It does not contain any electrons in the $4f$ subshell.
For the other elements:
$Nd (Z=60)$ is $[Xe] 4f^3 6s^2$.
$Eu (Z=63)$ is $[Xe] 4f^7 6s^2$.
$Lu (Z=71)$ is $[Xe] 4f^{14} 5d^1 6s^2$.
Thus,$La$ is the correct answer.

(B) Chemical twins are pairs of elements that exhibit very similar chemical properties due to the lanthanoid contraction.
$Nb-Ta$, $Mo-W$, and $Tc-Re$ are examples of chemical twins.
$Zr-Hf$ (Hafnium) are chemical twins, but $Zr-Rf$ (Rutherfordium) are not, as $Rf$ belongs to the actinoid series and has different chemical characteristics.

(D) $Zr$ $(Z=40)$ belongs to the $5^{th}$ period,while $Hf$ $(Z=72)$ belongs to the $6^{th}$ period.
Due to lanthanoid contraction,$Zr$ and $Hf$ exhibit nearly identical atomic and ionic radii,making them chemical twins.
Since they belong to different periods ($5^{th}$ and $6^{th}$),the statement that they belong to the same period is incorrect.

(C) The $Gd^{3+}$ ion has an electronic configuration of $[Xe] 4f^7$. Although it contains $7$ unpaired electrons,it is colourless because the $f-f$ transitions are forbidden by the Laporte selection rule and the energy gap is too large for visible light absorption.
$La^{3+}$ $(4f^0)$ and $Lu^{3+}$ $(4f^{14})$ are colourless due to the absence of unpaired electrons.
$Eu^{3+}$ $(4f^6)$ is coloured due to $f-f$ transitions.
Therefore,$Gd^{3+}$ is the correct answer.

(C) Due to the lanthanoid contraction, the atomic radii of elements of the second transition series are very similar to those of the corresponding elements of the third transition series.
This phenomenon occurs because the $4f$-electrons do not screen the nuclear charge effectively.
As a result, pairs like $Zr-Hf$, $Nb-Ta$, $Mo-W$, and $Tc-Re$ are known as 'chemical twins' due to their nearly identical chemical properties and atomic sizes.

(C) Ruby is a mineral of aluminium,i.e.,$Al_{2}O_{3}$. It does not contain silicon.

(D) $Cu^{+}$ has a $3d^{10}$ electronic configuration.
Due to the absence of unpaired electrons and a completely filled $d$-subshell,$d-d$ transitions are not possible,resulting in a colorless aqueous solution.
In contrast,$Fe^{3+}$ $(3d^5)$,$Fe^{2+}$ $(3d^6)$,and $Cu^{2+}$ $(3d^9)$ have unpaired electrons and exhibit color.

(D) The third row transition elements belong to the $5d$ series.
Among these,Ruthenium $(Ru)$ and Osmium $(Os)$ exhibit the highest oxidation state of $+8$.
This is observed in compounds like $RuO_4$ and $OsO_4$.

(A) Substances like iron $(Fe)$,cobalt $(Co)$,nickel $(Ni)$,gadolinium $(Gd)$,and chromium dioxide $(CrO_2)$ are attracted very strongly by an external magnetic field.
Such substances are called ferromagnetic substances.
These materials can be permanently magnetized even after the removal of the external magnetic field.
Among the given options,$Ni$ is a ferromagnetic element.