General Characteristics Questions in English

(B) The maximum oxidation state of an element in a group is often determined by the number of valence electrons.
For an element showing a maximum oxidation state of $+6$,it implies that there are $6$ valence electrons available for bonding.
This corresponds to the group $VI-B$ (or Group $6$ in modern $IUPAC$ notation).
Therefore,the correct option is $B$.

(D) Transition elements are defined as elements which have incompletely filled $d$-orbitals in their ground state or in any of their oxidation states.
According to the $IUPAC$ definition,the elements of group $3$ to $12$ are considered transition elements.
However,$Zn$,$Cd$,and $Hg$ (group $12$) have completely filled $d$-orbitals $(d^{10})$ in their ground state and common oxidation states,so they are often excluded from the strict definition of transition elements.
Excluding these $3$ elements from the $d$-block elements (which total $40$ elements in the $3d, 4d, 5d,$ and $6d$ series),the number of transition elements is $40 - 3 = 37$.
However,considering the standard classification often taught in textbooks,the total number of transition elements is approximately $40$ (if including group $12$) or $37$ (if excluding group $12$). Given the options provided,$38$ is the closest value often cited in specific contexts or older literature.

(C) Bullet-proof steel alloy is prepared by using $Zr$ (Zirconium). It is used in the manufacturing of high-strength alloys that are resistant to impact and corrosion.

(A) $Niobium$ $(Nb)$ and $tantalum$ $(Ta)$ are transition metals that exhibit exceptional resistance to corrosion due to the formation of a stable,protective oxide layer on their surfaces. This property makes them highly biocompatible and suitable for use in medical implants and surgical instruments.

(D) The first transition series corresponds to the filling of the $3d$ orbitals.
It starts with Scandium ($Sc$,atomic number $Z = 21$) and ends with Zinc ($Zn$,atomic number $Z = 30$).
Therefore,the elements belonging to the first transition series have atomic numbers from $21$ to $30$.

(C) $Ag$ belongs to the second $(4d)$ transition series.
The first transition series consists of elements from $Sc$ $(Z=21)$ to $Zn$ $(Z=30)$.
$Fe$,$V$,and $Cu$ are part of the first transition series ($3d$-series),whereas $Ag$ is part of the second transition series ($4d$-series).

(B) . The electronic configuration of $_{30}Zn$ is $[Ar] 3d^{10} 4s^2$ and $_{80}Hg$ is $[Xe] 4f^{14} 5d^{10} 6s^2$. Since their $d-$orbitals are completely filled,they do not exhibit variable valency.

(D) The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
For the cuprous ion $(Cu^+)$,the configuration is $[Ar] 3d^{10}$. Since the $d-$ orbital is completely filled,there are no unpaired electrons available for $d-d$ transition,making it colourless.
For the cupric ion $(Cu^{2+})$,the configuration is $[Ar] 3d^9$. Since the $d-$ orbital is incompletely filled,it contains one unpaired electron,which allows for $d-d$ transition,making it coloured.

(C) $d-$block elements are known as transition elements.
These elements exhibit variable valency due to the presence of an incomplete $d-$subshell.

(B) The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$.
For $Zn^{2+}$,the configuration is $[Ar] 3d^{10}$.
Since all $10$ electrons in the $3d$ orbital are paired,there are no unpaired electrons.
Therefore,$Zn^{2+}$ is diamagnetic in nature.

(D) Transition metals form a large number of complex compounds due to the following reasons:
$1$. Small size and high ionic charge of the metal ions.
$2$. Availability of vacant $d$-orbitals of appropriate energy to accept lone pairs of electrons from ligands.
Therefore,all the given factors contribute to this property.

(B) The colour of transition metal ions in aqueous solution is primarily due to $d-d$ transitions,which require the presence of unpaired electrons in the $d$-orbitals.
$1$. Electronic configuration of $Ti$ $(Z=22)$: $[Ar] 3d^2 4s^2$. $Ti^{4+}$ ion: $[Ar] 3d^0 4s^0$. Since there are no unpaired electrons,$Ti^{4+}$ is colourless.
$2$. Electronic configuration of $Ti^{3+}$ $(Z=22)$: $[Ar] 3d^1$. It has $1$ unpaired electron,so it is coloured.
$3$. Electronic configuration of $Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$. It has $4$ unpaired electrons,so it is coloured.
$4$. Electronic configuration of $Fe^{3+}$ $(Z=26)$: $[Ar] 3d^5$. It has $5$ unpaired electrons,so it is coloured.
Therefore,the correct option is $B$.

(B) The first transition series corresponds to the filling of the $3d$ orbitals. This series starts from Scandium $(Z = 21)$ and ends at Zinc $(Z = 30)$. These elements belong to the $4^{th}$ period of the periodic table.

(D) Transition metal.
An element is classified as a transition metal if it has a partially filled $d-$orbital in its ground state or in any of its common oxidation states.
Since the configuration $3d^4 4s^1$ represents a partially filled $d-$subshell,it belongs to the $d-$block elements,which are known as transition metals.

(A) The atomic number of $Cr$ is $24$.
The expected electronic configuration is $[Ar] 3d^4 4s^2$.
However,the actual electronic configuration is $[Ar] 3d^5 4s^1$.
This configuration is more stable because of the symmetrical distribution of electrons in the $d$-orbitals and higher exchange energy associated with the half-filled $d$-subshell.

(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n + 2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
Since $\mu \propto \sqrt{n}$,the ion with the fewest unpaired electrons will have the least magnetic moment.

Ion	Electronic configuration / Unpaired electrons $(n)$
$Cu^{2+}$	$d^9$ / $1$
$Ni^{2+}$	$d^8$ / $2$
$Co^{2+}$	$d^7$ / $3$
$Fe^{2+}$	$d^6$ / $4$

$Cu^{2+}$ has only $1$ unpaired electron,therefore it has the least magnetic moment.

(C) The oxidation state of a transition metal depends on the number of electrons available in the $(n-1)d$ and $ns$ orbitals.
For $3d^5 4s^2$ (Manganese,$Mn$),the total number of valence electrons is $5 + 2 = 7$.
Therefore,it can exhibit a maximum oxidation state of $+7$.
Comparing the given configurations:
$A) 3d^3 4s^2$ (Vanadium,$V$) shows up to $+5$.
$B) 3d^5 4s^1$ (Chromium,$Cr$) shows up to $+6$.
$C) 3d^5 4s^2$ $(Mn)$ shows up to $+7$.
$D) 3d^6 4s^2$ (Iron,$Fe$) shows up to $+6$.
Thus,the configuration $3d^5 4s^2$ shows the highest oxidation state.

(C) The correct answer is $(C)$. Transition metals exhibit a variety of metallic crystal structures,including body-centred cubic $(BCC)$,hexagonal close-packed $(HCP)$,and face-centred cubic $(FCC)$ structures. Therefore,the statement that they crystallize with $BCC$ and $HCP$ structures only is false.

(D) The atomic number of $Fe$ is $26$. The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.
Number of unpaired electrons in $Fe$: $3d^6$ has $4$ unpaired electrons.
Electronic configuration of $Fe^+$ is $[Ar] 3d^6 4s^1$.
Number of unpaired electrons in $Fe^+$: $3d^6$ $(4)$ + $4s^1$ $(1)$ = $5$ unpaired electrons.
Electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
Number of unpaired electrons in $Fe^{2+}$: $3d^6$ has $4$ unpaired electrons.
Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Number of unpaired electrons in $Fe^{3+}$: $3d^5$ has $5$ unpaired electrons.
Both $Fe^+$ and $Fe^{3+}$ have $5$ unpaired electrons,but in standard chemistry problems of this type,$Fe^{3+}$ is the most stable ion with $5$ unpaired electrons. However,based on the options provided,$Fe^{3+}$ is the standard answer for the highest number of unpaired electrons in common oxidation states.

(D) Transition elements are characterized by their ability to form complex ions due to the presence of vacant $d$-orbitals,small size,and high charge density. $Ag$,$Au$,and $Cu$ are all transition elements and are well-known for forming various stable complex ions (e.g.,$[Ag(NH_3)_2]^+$,$[Au(CN)_2]^-$,$[Cu(NH_3)_4]^{2+}$). Therefore,the correct answer is $D$.

(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ B.M.}$,where $n$ is the number of unpaired electrons.
For $3d^1$,$n = 1$,$\mu = \sqrt{1(3)} = 1.73 \text{ B.M.}$
For $3d^8$,$n = 2$,$\mu = \sqrt{2(4)} = 2.83 \text{ B.M.}$
For $3d^5$,$n = 5$,$\mu = \sqrt{5(7)} = 5.92 \text{ B.M.}$
For $3d^7$,$n = 3$,$\mu = \sqrt{3(5)} = 3.87 \text{ B.M.}$
Thus,the $3d^5$ configuration has the highest number of unpaired electrons $(n=5)$,resulting in the highest magnetic moment.

(A) The electronic configurations of the given elements are:
$Ti (Z=22): [Ar] 3d^2 4s^2$ (Unpaired electrons = $2$)
$V (Z=23): [Ar] 3d^3 4s^2$ (Unpaired electrons = $3$)
$Cr (Z=24): [Ar] 3d^5 4s^1$ (Unpaired electrons = $6$)
$Fe (Z=26): [Ar] 3d^6 4s^2$ (Unpaired electrons = $4$)
$Sc (Z=21): [Ar] 3d^1 4s^2$ (Unpaired electrons = $1$)
Thus,$Cr$ has the maximum number of unpaired electrons.

(C) Brass is an alloy consisting primarily of copper $(Cu)$ and zinc $(Zn)$.
Typically,it contains $70\%$ $Cu$ and $30\%$ $Zn$.
It is widely used in making utensils,artificial jewelry,and hardware components.

(D) The strength of a metallic bond depends upon the number of unpaired electrons in the $d$-orbitals.
As the number of unpaired electrons increases,the metallic bond strength increases.
$Cr$ (Chromium) has the electronic configuration $[Ar] 3d^5 4s^1$,which contains $6$ unpaired electrons.
Due to the maximum number of unpaired electrons,$Cr$ exhibits the strongest metallic bonding among the given options.

(D) The colour of an ion is due to the presence of unpaired electrons which allow $d-d$ transitions.
Electronic configuration of $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$.
Electronic configuration of $Zn^{2+}$ is $[Ar] 3d^{10}$.
Since all $d$-orbitals are completely filled in $Zn^{2+}$,there are no unpaired electrons.
Therefore,$Zn^{2+}$ is a colourless ion.

(B) $Sc (Z = 21) \to [Ar] 3d^1 4s^2$
The $d-$block elements belonging to groups $3-12$ are known as transition elements because their physical and chemical properties are intermediate between $s-$block and $p-$block elements.
The first transition series (the $3d$ series) starts with Scandium $(Z = 21)$ and ends with Zinc $(Z = 30)$.
Therefore,the first transition element is Scandium.

(A) The atomic number of $Mn$ is $25$. The ground state electronic configuration of neutral $Mn$ is $[Ar] 3d^5 4s^2$.
When $Mn^{2+}$ is formed,two electrons are removed from the outermost $4s$ orbital.
Therefore,the electronic configuration of $Mn^{2+}$ becomes $[Ar] 3d^5 4s^0$ or simply $3d^5$.

(B) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
For $3d^2$,$n = 2$.
For $3d^5$,$n = 5$.
For $3d^7$,$n = 3$.
For $3d^9$,$n = 1$.
Since $3d^5$ has the maximum number of unpaired electrons $(n=5)$,it exhibits the highest magnetic moment.

(D) Transition elements are characterized by their ability to exhibit variable oxidation states due to the participation of both $(n-1)d$ and $ns$ electrons in bonding. Therefore,the statement that they show only one valence state is incorrect.

(A) The oxidation state of a transition metal depends on the number of unpaired electrons and the availability of $d$-orbitals.
Among the given options,Chromium $(Cr)$ has the electronic configuration $[Ar] 3d^5 4s^1$.
$Cr$ can lose all $6$ valence electrons to exhibit a maximum oxidation state of $+6$ (e.g.,in $CrO_4^{2-}$ or $Cr_2O_7^{2-}$).
$Co$,$Ni$,and $Cu$ generally exhibit lower maximum oxidation states compared to $Cr$ in their common compounds.

(C) The formation of stable complexes by $3d-$ transition metal ions is primarily attributed to two factors:
$1$. Their small size and high ionic charge,which results in a high charge/size ratio (high charge density).
$2$. The presence of vacant $d-$ orbitals of appropriate energy,which can accept lone pairs of electrons from ligands to form coordinate covalent bonds.
Therefore,the correct option is $(c)$.

(D) Paramagnetism in metal ions arises due to the presence of unpaired electrons.
In $3d-$ transition metal ions,the presence of one or more unpaired $d-$ electrons leads to a magnetic moment,making them paramagnetic.
Therefore,the correct option is $(D)$.

(D) $Pt$ is the least reactive metal.
It is classified as a noble metal because it does not react with air,water,or common acids at room temperature.

(A) Ferromagnetism is directly related to the number of unpaired electrons in the $d$-orbitals of the metal atoms.
Comparing the number of unpaired electrons:
$Fe$ $([Ar] 3d^6 4s^2)$: $4$ unpaired electrons.
$Co$ $([Ar] 3d^7 4s^2)$: $3$ unpaired electrons.
$Ni$ $([Ar] 3d^8 4s^2)$: $2$ unpaired electrons.
$Pt$ $([Xe] 4f^{14} 5d^9 6s^1)$: $2$ unpaired electrons.
Since $Fe$ has the maximum number of unpaired electrons,it exhibits the strongest ferromagnetic character.

(D) Transition metals are known to form a large number of interstitial compounds.
In these compounds,small atoms such as $H$,$C$,$N$,and $B$ occupy the interstitial sites (voids) within the crystal lattices of the transition metals.
Since $Fe$,$Co$,and $Ni$ are all transition metals,they all possess the ability to form such compounds.

(A) The electronic configuration of neutral $Zn$ $(Z=30)$ is $[Ar] 3d^{10}4s^2$.
When $Zn$ forms the $Zn^{2+}$ ion,it loses two electrons from the $4s$ orbital.
Therefore,the electronic configuration of $Zn^{2+}$ is $[Ar] 3d^{10}4s^0$.

(C) The electronic configuration of neutral Titanium $(Ti)$ is $[Ar] 3d^2 4s^2$.
To form the $Ti^{4+}$ ion,the atom loses $4$ electrons (two from $4s$ and two from $3d$ orbitals).
Therefore,the electronic configuration of $Ti^{4+}$ is $3d^0 4s^0$.

(D) The density of transition elements in the $3d$ series generally increases across the period from $Sc$ to $Cu$ due to the increase in atomic mass and decrease in atomic radius.
$Sc$ $(Z=21)$ has the lowest density,while $Cu$ $(Z=29)$ has a higher density.
$Zn$ $(Z=30)$ has a slightly lower density than $Cu$ due to the increase in metallic radius caused by the completion of the $3d^{10}$ subshell.
Therefore,the correct order of increasing density is $Sc < Zn < Ni < Cu$.

(C) Transition elements are characterized by the presence of partially filled $d$-orbitals.
While other elements can form paramagnetic or coloured compounds,the ability to form a large number of complex compounds (coordination compounds) is a unique and defining characteristic of transition elements due to their small size,high nuclear charge,and availability of vacant $d$-orbitals for bonding.
Therefore,the correct option is $C$.

(A) The correct answer is $(A)$. The elements $Zn,$ $Cd,$ and $Hg$ have the general electronic configuration $(n-1)d^{10}ns^2$.
Since their $d-$orbitals are completely filled in their ground state as well as in their common oxidation states,they do not exhibit the characteristic properties of transition elements,such as the formation of coloured ions or variable oxidation states.

(B) In the $3d$ series,as we move from $Cr$ to $Cu$,the atomic number increases,which increases the nuclear charge.
However,the electrons are added to the inner $3d$ orbitals.
These $3d$ electrons provide a poor shielding (screening) effect.
As a result,the increase in nuclear charge is almost balanced by the screening effect of the $3d$ electrons,leading to a very negligible change in atomic volume.

(D) The correct answer is $(D)$.
Transition elements are defined as elements that have partially filled $d-$orbitals in their ground state or in any of their oxidation states.
$Fe, Co, Ni$ are $3d$ transition metals.
$Cu, Ag, Au$ are group $11$ transition metals.
$Ti, Zr, Hf$ are group $4$ transition metals.
$Ga, In, Tl$ belong to group $13$ of the $p-$block elements,where the last electron enters the $p-$orbital,not the $d-$orbital.

(B) Most transition metals possess unpaired electrons in their $d$-orbitals ($d^1$ to $d^9$ configuration) in their ground state or common oxidation states. Due to the presence of these unpaired electrons,they exhibit paramagnetic behavior.

(D) All $d-$block elements are metals.
They exhibit variable oxidation states (valency).
They are known to form coloured ions and various complex salts due to the presence of partially filled $d-$orbitals.

(C) The elements $Zn, Cd$,and $Hg$ are considered non-$typical$ transition elements.
This is because they have completely filled $d$-orbitals in their ground state as well as in their common oxidation states,which does not satisfy the general definition of transition elements.

(A) The atomic number of Chromium $(Cr)$ is $24$.
The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
To form $Cr^{3+}$,we remove $3$ electrons: one from the $4s$ orbital and two from the $3d$ orbital.
The electronic configuration of $Cr^{3+}$ becomes $[Ar] 3d^3$.
In the $3d^3$ configuration,there are $3$ unpaired electrons in the $d$-orbitals.
Therefore,the correct option is $A$.

(C) The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$. In its common oxidation state of $+2$,the configuration is $[Ar] 3d^{10}$.
Since the $3d$ subshell is completely filled,there are no unpaired electrons present.
Due to the absence of unpaired electrons,$d-d$ transitions are not possible.
Therefore,$Zn^{2+}$ ions do not form coloured compounds.

(D) Super alloys are high-performance alloys that exhibit excellent mechanical strength and resistance to corrosion at high temperatures. They are typically based on $Fe$,$Ni$,or $Co$.

(A) The transition metal that exhibits oxidation states ranging from $+2$ to $+7$ belongs to group $VII B$ (Group $7$ in modern $IUPAC$ notation).
This is because the metal must have $7$ valence electrons (in the $(n-1)d$ and $ns$ orbitals) to achieve a maximum oxidation state of $+7$,which is characteristic of Manganese $(Mn)$.

(B) The colour of transition metal ions is generally due to $d-d$ transitions,which require the presence of unpaired electrons in the $d$-orbitals.
$1$. $Cr^{3+}$ has an electronic configuration of $[Ar] 3d^3$,which contains $3$ unpaired electrons.
$2$. $Cu^{+}$ has an electronic configuration of $[Ar] 3d^{10}$,which has a completely filled $d$-subshell with no unpaired electrons.
$3$. $Fe^{3+}$ has an electronic configuration of $[Ar] 3d^5$,which contains $5$ unpaired electrons.
$4$. $Cu^{2+}$ has an electronic configuration of $[Ar] 3d^9$,which contains $1$ unpaired electron.
Since $Cu^{+}$ has a $d^{10}$ configuration,it has no unpaired electrons to undergo $d-d$ transitions,making it colourless.
Therefore,the correct option is $(B)$.