General Characteristics Questions in English

(D) $Mn$ (Manganese) belongs to group $7$ of the periodic table. Its electronic configuration is $[Ar] 3d^5 4s^2$. It can lose all $7$ valence electrons to exhibit a maximum oxidation state of $+7$,as seen in compounds like $KMnO_4$.

(A) Transition elements are defined as elements which have incompletely filled $d$-orbitals in their ground state or in any of their oxidation states. All transition elements are $d$-block elements and they exhibit metallic properties such as high tensile strength,ductility,malleability,and high thermal and electrical conductivity. Therefore,all transition elements are metals.

(A) For ions with the same charge and belonging to the same period,the ionic radius decreases as the atomic number increases due to an increase in effective nuclear charge.
Since the atomic numbers are $Cr$ $(24)$,$Mn$ $(25)$,$Fe$ $(26)$,and $Co$ $(27)$,the ionic radius decreases in the order $Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$.
Therefore,$Cr^{3+}$ has the highest ionic radius.

(C) Among the given elements,$Os$ (Osmium) has the highest density.
Its density is approximately $22.59 \, g/cm^3$,which is the highest among all known naturally occurring elements.

(C) Transition elements exhibit variable valencies because the energy difference between the $(n-1)d$ and $ns$ orbitals is very small.
Therefore,electrons from both $(n-1)d$ and $ns$ orbitals can participate in bond formation,leading to variable oxidation states.

(C) The correct answer is $(C)$.
$d-$block elements show the maximum tendency to form complexes due to the following reasons:
$(I)$ Small atomic/ionic size.
$(II)$ High nuclear charge.
$(III)$ Availability of vacant $d-$orbitals to accept lone pairs of electrons from ligands.

(D) Transitional elements form coloured salts due to the presence of unpaired electrons in the $d$-orbital.

(C) The correct answer is $(C)$ $Cu$.
$Cu$ (Copper) has the atomic number $29$.
Its electronic configuration is $[Ar] 3d^{10} 4s^1$.
Since the last electron enters the $d-$orbital,it is classified as a $d-$block element.

(C) The atomic number $26$ corresponds to the element $Iron$ $(Fe)$.
$Iron$ is a transition metal belonging to the $d$-block of the periodic table.

(C) . Copper,silver and gold; all the three were used for making coins.

(A) The electronic configuration of the element in $+3$ oxidation state is $[Ar]3d^3$.
This means the element has lost $3$ electrons to reach this state.
The total number of electrons in the $+3$ ion is $18 (Ar) + 3 = 21$.
Therefore,the atomic number of the neutral element is $21 + 3 = 24$.

(D) The catalytic activity of transition metals and their compounds is primarily attributed to their ability to adopt multiple oxidation states and their capacity to form complexes. This allows them to provide a large surface area for adsorption and to form intermediate compounds with reactants,thereby lowering the activation energy of the reaction.

(C) The $2^{nd}$ row transition series corresponds to the $4d$ series,which starts after the noble gas Krypton ($Kr$,atomic number $36$).
Its general electronic configuration is represented as $[Kr] 4d^{1-10} 5s^{1-2}$.

(A) The transition elements are placed between the $s$-block and $p$-block elements in the periodic table.
Their properties represent a transition from the highly electropositive $s$-block metals to the less metallic $p$-block elements.
Therefore,the correct option is $(A)$.

(C) Transition elements are defined as elements which have incompletely filled $d-$orbitals in their ground state or in any of their oxidation states. Specifically,they have an incompletely filled $(n-1)d$ subshell and an incompletely filled $ns$ subshell (or outermost shell).

(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.

Ion	Outer Configuration	Unpaired Electrons $(n)$	Magnetic Moment ($B$.$M$.)
$V^{3+}$	$3d^2$	$2$	$\sqrt{2(2+2)} = 2.83$
$Mn^{3+}$	$3d^4$	$4$	$\sqrt{4(4+2)} = 4.90$
$Fe^{3+}$	$3d^5$	$5$	$\sqrt{5(5+2)} = 5.92$
$Cu^{2+}$	$3d^9$	$1$	$\sqrt{1(1+2)} = 1.73$

Comparing the values,$Fe^{3+}$ has the highest number of unpaired electrons $(n=5)$,resulting in the maximum magnetic moment of $5.92 \text{ B.M.}$ Therefore,the correct option is $(C)$.

(C) The statement "All their ions are colourless" is false.
Transition metal ions often exhibit color due to $d-d$ electronic transitions.
Most transition metal ions have partially filled $d-$orbitals,which allow for these transitions,resulting in colored compounds.
Only a few ions,such as $Sc^{3+}$ or $Zn^{2+}$,are colorless due to empty or completely filled $d-$orbitals.

(B) The general electronic configuration of transition elements is $(n-1)d^{1-10} ns^{1-2}$.
This represents elements where the $d-$orbitals are being progressively filled.
Option $B$ is the closest representation of the transition metal configuration pattern among the choices provided.

(C) Generally,$d$-block elements are called transition elements because they contain an inner partially filled $d$-subshell.
Thus,their general electronic configuration is represented as $(n - 1)d^{1 - 10}ns^{1 - 2}$.

(C) The colour of transition metal ions is primarily due to the presence of unpaired $d-$ electrons.
These electrons undergo $d-d$ transitions by absorbing light from the visible region,which results in the complementary colour being observed.

(B) To find the number of unpaired $d-$ electrons,we look at the electronic configuration of each ion:
$Zn$ $(Z=30)$: $[Ar] 3d^{10} 4s^2$ ($0$ unpaired electrons).
$Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6 4s^0$. In $3d^6$,the electrons are filled as: $\uparrow\downarrow, \uparrow, \uparrow, \uparrow, \uparrow$. This results in $4$ unpaired electrons.
$Ni^{3+}$ $(Z=28)$: $[Ar] 3d^7 4s^0$. In $3d^7$,the electrons are filled as: $\uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow, \uparrow$. This results in $3$ unpaired electrons.
$Cu^{+}$ $(Z=29)$: $[Ar] 3d^{10} 4s^0$ ($0$ unpaired electrons).
Therefore,$Fe^{2+}$ has the maximum number of unpaired $d-$ electrons.

(A) An amalgam is an alloy of mercury with another metal.
Mercury readily forms amalgams with most metals,such as $Ag$,$Zn$,$Au$,and $Na$.
However,mercury does not form amalgams with iron $(Fe)$,platinum $(Pt)$,tungsten $(W)$,or tantalum $(Ta)$.

(C) Paramagnetism in transition metals arises due to the presence of one or more unpaired electrons in their $(n-1)d$ orbitals.
These unpaired electrons generate a magnetic moment,which causes the substance to be attracted by an external magnetic field.

(B) Transition elements exhibit multiple oxidation states due to the availability of vacant $d-$orbitals.
They form coloured ions due to $d-d$ electronic transitions.

(A) To determine the number of unpaired electrons,we look at the electronic configuration of the ions:
$1$. $Mn^{2+}$ $(Z=25)$: The configuration is $[Ar] 3d^5$. It has $5$ unpaired electrons.
$2$. $Fe^{2+}$ $(Z=26)$: The configuration is $[Ar] 3d^6$. It has $4$ unpaired electrons.
$3$. $Co^{2+}$ $(Z=27)$: The configuration is $[Ar] 3d^7$. It has $3$ unpaired electrons.
$4$. $Ni^{2+}$ $(Z=28)$: The configuration is $[Ar] 3d^8$. It has $2$ unpaired electrons.
Therefore,$Mn^{2+}$ has the maximum number of unpaired electrons.

(D) Transition metals exhibit variable oxidation states because both the $ns$ and $(n-1)d$ electrons can participate in bonding.
Therefore,the maximum number of oxidation states is determined by the total number of electrons in the $ns$ and $(n-1)d$ orbitals.

(D) Iron ($Fe$,atomic number $26$) has the electronic configuration $[Ar] 3d^6 4s^2$.
It belongs to the $d$-block,specifically the first transition series ($3d$ series).
In the long form of the periodic table,it is placed in group $8$.
Historically,in the Mendeleev periodic table and some older classifications,it was categorized under group $VIII$ (or $VIII B$ in some conventions).
Therefore,option $(d)$ is the correct statement among the choices provided.

(B) The electronic configuration of $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$.
Since the $3d$ subshell is completely filled,there are no unpaired electrons available for bonding,and the energy required to remove an electron from the stable $d^{10}$ configuration is very high.
Therefore,it does not exhibit variable valency like other transition elements.
Thus,the correct option is $(B)$.

(C) transition element is defined as an element which has an incompletely filled $d$-orbital in its ground state or in any of its oxidation states.
The general electronic configuration of transition elements is $(n-1)d^{1-10} ns^{1-2}$.
$Ni$ (Nickel) has the atomic number $28$ and its electronic configuration is $[Ar] 3d^8 4s^2$. Since it has a partially filled $d$-subshell,it is a transition element.
$Al$ (Aluminum) is a $p$-block element.
$As$ (Arsenic) is a $p$-block element.
$Rb$ (Rubidium) is an $s$-block element.
Therefore,the correct option is $C$.

(D) Transition elements readily form coordinate compounds due to the following reasons:
$(i)$ High nuclear charge.
$(ii)$ Small atomic and ionic size.
$(iii)$ Availability of vacant $d$-orbitals for bonding.

(C) $Transition$ elements exhibit variable valency because the energy difference between $(n-1)d$ and $ns$ orbitals is very small,allowing electrons from both to participate in bond formation.

(D) The colour of transition metal ions is due to the presence of unpaired electrons,which allow for $d-d$ transitions.
$Cr^{3+}$ has $3$ unpaired electrons $([Ar] 3d^3)$.
$Co^{2+}$ has $3$ unpaired electrons $([Ar] 3d^7)$.
$Cr^{2+}$ has $4$ unpaired electrons $([Ar] 3d^4)$.
$Cu^{+}$ has the electronic configuration $[Ar] 3d^{10}$,meaning it has no unpaired electrons.
Therefore,$Cu^{+}$ is colourless.

(C) The elements $Zn$,$Cd$,and $Hg$ belong to group $12$ of the periodic table.
These elements have a completely filled $d$-orbital configuration ($n-1)d^{10}ns^2$ in their ground state as well as in their common oxidation states.
Due to this,they do not exhibit the characteristic properties of transition elements,such as variable oxidation states,formation of colored ions,or catalytic activity.
Therefore,they are often referred to as 'non-typical' or 'normal' transition elements,but in the context of the $d$-block classification,they are grouped with transition elements despite lacking typical transition metal characteristics.

(C) Iron $(Fe)$ has the atomic number $26$ and its electronic configuration is $[Ar] 3d^6 4s^2$.
Since the last electron enters the $d$-orbital,it is classified as a transition element (or $d$-block element).

(B) They are inert towards many common reagents.
Noble metals are resistant to corrosion and oxidation in moist air.
They do not react with most common chemical reagents,which makes them highly stable and chemically inert.

(D) Transition elements show variable oxidation states due to the participation of $(n-1)d$ orbital electrons along with $ns$ orbital electrons in bond formation.

(A) The atomic number of chromium $(Cr)$ is $24$.
The ground state electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
When $Cr$ forms a $Cr^{2+}$ ion,it loses two electrons,first from the $4s$ orbital and then from the $3d$ orbital.
Therefore,the electronic configuration of $Cr^{2+}$ is $[Ar] 3d^4 4s^0$ or simply $3d^4 4s^0$.

(A) . The hardness of $Cr$ is due to the presence of covalent bonds formed by unpaired $d$-electrons.
Metallic lustre is due to the presence of free electrons in the metallic bond.

(A) . $Cr$ has the highest $M.P.$ and $B.P.$ in the first transition series due to the maximum number of unpaired electrons,which leads to stronger metallic bonding.

(B) The first ionization energy of $3d$ transition elements generally increases across the period due to an increase in effective nuclear charge.
Among the given elements ($Sc$,$Ti$,$V$,$Mn$),$Sc$ (atomic number $21$) is the first element in the $3d$ series.
As we move from left to right across the $3d$ series,the effective nuclear charge increases,which leads to a stronger attraction between the nucleus and the valence electrons,thereby increasing the ionization energy.
Therefore,$Sc$ has the lowest first ionization energy among the given options.

(B) The correct answer is $(B)$ $Zn$.
The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$.
The first ionisation potential removes one $4s$ electron,leaving $[Ar] 3d^{10} 4s^1$.
The second ionisation potential involves removing the remaining $4s$ electron.
Since the $3d^{10}$ subshell is completely filled,it provides a strong shielding effect,which reduces the effective nuclear charge felt by the $4s$ electron,making it easier to remove than expected compared to other transition metals.

(A) The electronic configuration of the ion $X^{+3}$ is $[Ar]3d^4$.
The total number of electrons in $X^{+3}$ is $18 + 4 = 22$.
Since the element $X$ is in the $+3$ oxidation state,it has lost $3$ electrons to form the $X^{+3}$ ion.
Therefore,the atomic number of the neutral element $X$ is $22 + 3 = 25$.

(D) The general electronic configuration of transition elements (d-block elements) is represented as $(n - 1)d^{1-10}ns^{1-2}$.
This can be expanded to include the inner shells as $(n - 1)s^2p^6d^{1-10}ns^{1-2}$.
Thus,the configuration is $(n - 1)s^2p^6d^{1-10}ns^1 \text{ or } ns^2$.

(A) The electronic configurations are as follows:
$Zn^{2+} (Z=30): [Ar] 3d^{10} 4s^0$. It has no unpaired electrons,so it is colourless.
$Ni^{2+} (Z=28): [Ar] 3d^8 4s^0$. It has two unpaired electrons,so it is coloured.
$Cr^{3+} (Z=24): [Ar] 3d^3 4s^0$. It has three unpaired electrons,so it is coloured.
Therefore,only $Zn^{2+}$ is colourless,while $Ni^{2+}$ and $Cr^{3+}$ are coloured.

(C) The atomic number of scandium $(Sc)$ is $21$.
Its electronic configuration is $[Ar] 3d^1 4s^2$.
To achieve a stable noble gas configuration $([Ar])$,scandium loses three electrons (two from $4s$ and one from $3d$).
Therefore,the most common and stable oxidation state of scandium is $+ 3$.

(D) The correct option is $(D)$. Transition metals exhibit variable valency due to the presence of vacant $d$-orbitals and the small energy difference between $(n-1)d$ and $ns$ orbitals.