General Characteristics Questions in English

(B) The colour of transition metal ions is due to the presence of unpaired electrons,which allow for $d-d$ transitions.
$Cu^{+}$ $(3d^{10})$: No unpaired electrons,so it is colourless.
$Cu^{2+}$ $(3d^9)$: Has $1$ unpaired electron,so it is coloured.
$Ti^{4+}$ $(3d^0)$: No unpaired electrons,so it is colourless.
$V^{5+}$ $(3d^0)$: No unpaired electrons,so it is colourless.
Therefore,the correct option is $(B)$.

(A) $d-$block elements have higher ionisation potential values compared to $f-$block elements.
This is because the $f-$orbitals have poor shielding effect,which leads to a higher effective nuclear charge experienced by the valence electrons in $f-$block elements.
However,the question asks for the comparison of $d-$block relative to $f-$block,and since $f-$block electrons are more shielded from the nucleus by inner shells,they are held less tightly than $d-$block electrons in corresponding periods.
Therefore,the ionisation potential of $d-$block elements is higher.

(C) Transition elements are characterized by the presence of partially filled $d-$orbitals,which allow them to exhibit variable valency. Therefore,having a $Fixed \ valency$ is not a property of transition elements.

(B) The tendency towards the formation of coloured ions is maximum in $d-$ block elements due to the presence of partially filled $d-$ orbitals,which allow for $d-d$ transitions.

(D) The colour of an aqueous solution of transition metal ions is generally due to the presence of unpaired $d$-electrons,which allow for $d-d$ transitions.
$1$. $Ti^{4+}$ $(3d^0)$: No unpaired electrons,colourless.
$2$. $Cu^{+}$ $(3d^{10})$: No unpaired electrons,colourless.
$3$. $Zn^{2+}$ $(3d^{10})$: No unpaired electrons,colourless.
$4$. $Cr^{3+}$ $(3d^3)$: Contains $3$ unpaired electrons,hence it is coloured.
Therefore,the correct option is $(D)$.

(C) The general electronic configuration of transition elements is $(n-1)d^{1-10} ns^{1-2}$.
For $3d-$ transition elements,the outermost shell is the $4s$ orbital $(n=4)$.
In most of these elements,the $4s$ orbital contains $2$ electrons,so the outermost shell configuration generally remains $ns^2$.

(D) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$.
Since it has $7$ electrons in the outermost shells ($3d$ and $4s$),it can lose all these electrons to show a maximum oxidation state of $+7$.

(D) The correct answer is $(D)$.
An ion forms a coloured solution if it contains unpaired electrons in its $d$-orbitals.
$Cu^{+}$ $(3d^{10})$ has $0$ unpaired electrons.
$Zn^{2+}$ $(3d^{10})$ has $0$ unpaired electrons.
$Ag^{+}$ $(4d^{10})$ has $0$ unpaired electrons.
$Fe^{2+}$ $(3d^{6})$ has $4$ unpaired electrons,which allows for $d-d$ transitions,resulting in a coloured solution.

(D) $Fe$ (Iron) exhibits more than one oxidation state (commonly $+2$ and $+3$) because it is a transition metal belonging to the $d$-block elements.
Transition metals have partially filled $d$-orbitals,allowing them to lose electrons from both $ns$ and $(n-1)d$ orbitals,resulting in variable oxidation states.

(A) The catalytic activity of transition metals is primarily due to their ability to adopt $Variable \ oxidation \ states$ and provide a large surface area with free valencies.
These properties allow them to form unstable intermediate compounds with reactants,providing an alternative reaction path with lower activation energy.
Therefore,the correct option is $(A)$.

(B) The correct option is $(B)$.
Manganese $(Mn)$ is a transition metal with the electronic configuration $[Ar] 3d^5 4s^2$.
It exhibits a wide range of oxidation states due to the involvement of both $3d$ and $4s$ electrons in bonding.
The oxidation states shown by manganese are $+2, +3, +4, +5, +6,$ and $+7$.
Thus,there are $6$ distinct oxidation states.

(A) The density of transition metals generally increases across a period as the atomic mass increases and atomic volume decreases due to the contraction of $d$-orbitals.
Among the given $3d$ transition metals,$Sc$ (Scandium) has the lowest atomic mass and a relatively larger atomic volume compared to the others ($Ti$,$V$,$Cr$).
Therefore,$Sc$ has the lowest density among the options provided.

(C) transition element is defined as an element which has an incompletely filled $d$-orbital in its ground state or in any one of its oxidation states.
Option $C$ represents the configuration $1s^{2}, 2s^{2}2p^{6}, 3s^{2}3p^{6}3d^{2}, 4s^{2}$,which corresponds to Titanium ($Ti$,$Z=22$).
Since it has a partially filled $3d$-orbital $(3d^{2})$,it is a transition element.

(C) Transition metals exhibit high melting and boiling points due to the presence of strong metallic bonding,which involves the overlap of $d$-orbitals.
This strong interatomic interaction leads to the formation of a rigid lattice structure,making these metals $Hard$ and $brittle$ in nature.

(D) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n + 2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 2.83 \ BM$.
$2.83 = \sqrt{n(n + 2)}$
Squaring both sides,we get $8 = n(n + 2)$.
$n^2 + 2n - 8 = 0$.
Solving the quadratic equation: $(n + 4)(n - 2) = 0$.
Since $n$ must be positive,$n = 2$.

(B) The colour of transition metal ions in aqueous solution is primarily due to $d-d$ transitions. Ions with the same number of $d$-electrons often exhibit similar colours.
$1$. $Cr^{+3}$ $(3d^3)$ is violet/green.
$2$. $Fe^{+2}$ $(3d^6)$ is pale green.
$3$. $Ti^{+3}$ $(3d^1)$ is purple.
$4$. $V^{+2}$ $(3d^3)$ is violet.
$5$. $Fe^{+3}$ $(3d^5)$ is yellow/brown.
$6$. $Mn^{+2}$ $(3d^5)$ is pink.
$7$. $Cu^{+}$ $(3d^{10})$ is colourless.
$8$. $Ni^{+2}$ $(3d^8)$ is green.
Comparing the options,$Cr^{+3}$ $(3d^3)$ and $V^{+2}$ $(3d^3)$ both have the same electronic configuration $(d^3)$ and exhibit similar violet/green colours. However,looking at standard chemistry data,$Cr^{+3}$ and $V^{+2}$ are both known to exhibit violet/green shades due to the $d^3$ configuration.

(A) The transition elements (specifically $3d$ elements) exhibit variable oxidation states because the energy difference between the $(n-1)d$ orbitals and the $ns$ orbitals is very small.
Due to this small energy gap,electrons from both the $(n-1)d$ and $ns$ orbitals can participate in bond formation,allowing these elements to show multiple oxidation states.

(D) To find the number of unpaired electrons,we fill the $d$-orbitals according to Hund's rule:
$3d^6$: $5$ orbitals,$4$ unpaired electrons.
$3d^7$: $5$ orbitals,$3$ unpaired electrons.
$3d^8$: $5$ orbitals,$2$ unpaired electrons.
$3d^9$: $5$ orbitals,$1$ unpaired electron.
Thus,$3d^9$ has the smallest number of unpaired electrons $(1)$.

(C) Transition elements are defined as elements which have incompletely filled $d-$orbitals in their ground state or in any one of their oxidation states. Therefore,the characteristic feature of transition elements is the partial filling of $d-$orbitals.

(B) 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.
The electronic configuration of $Mn^{2+}$ becomes $[Ar] 3d^5$.
In the $3d$ subshell,there are $5$ orbitals,and according to Hund's rule,each orbital will be singly occupied.
Therefore,there are $5$ unpaired electrons in $Mn^{2+}$.

(C) The electronic configuration of the element is $[Ar]3d^24s^2$.
Since the last electron enters the $3d$ orbital,the element belongs to the $d-$ block elements.
Therefore,the correct option is $(C)$.

(C) Transition metals are defined as elements that have partially filled $d-$orbitals in their ground state or in any of their oxidation states.
$Chromium$ $(Cr)$,$Titanium$ $(Ti)$,and $Tungsten$ $(W)$ are $d-$block elements and are transition metals.
$Lead$ $(Pb)$ is a $p-$block element belonging to group $14$ and is not a transition metal.

(D) The element with atomic number $22$ is Titanium $(Ti)$.
Its electronic configuration is $[Ar] 3d^2 4s^2$.
The highest oxidation state of a transition element is equal to the total number of electrons in the $(n-1)d$ and $ns$ orbitals.
Therefore,the highest oxidation state $= 2 + 2 = 4$.

(C) -block elements exhibit variable oxidation states and have high charge density,which allows them to form both ionic compounds (e.g.,$FeCl_2$,$MnF_2$) and covalent compounds (e.g.,$CrO_3$,$OsO_4$). They also readily form complex compounds due to the availability of vacant $d$-orbitals.

(C) The atomic number of silver $(Ag)$ is $47$.
Following the Aufbau principle and considering the stability of the half-filled or fully-filled $d$-orbitals,the electron from the $5s$ orbital shifts to the $4d$ orbital to make it fully filled.
Thus,the electronic configuration is $[Kr] 4d^{10} 5s^1$.

(A) The colour of transition metal ions is due to the presence of unpaired electrons which allow $d-d$ transitions.
$Cu^{+}$ has the electronic configuration $[Ar] 3d^{10}$.
Since all $10$ electrons are paired,there are no unpaired electrons available for $d-d$ transitions.
Therefore,$Cu^{+}$ is colourless.

(A) The correct answer is $A$. $Fe - Co - Ni$.
In the $3d$ transition series,as the atomic number increases,the number of $d$-electrons also increases.
This leads to an increase in the screening effect,which effectively counterbalances the increase in nuclear charge.
Consequently,the atomic radii of elements like $Fe$,$Co$,and $Ni$ remain practically the same.

(C) The atomic number of Manganese $(Mn)$ is $25$.
Its electronic configuration is $[Ar] 3d^5 4s^2$.
Since the last electron enters the $3d$ orbital,Manganese belongs to the $d-$ block of the periodic table.

(D) . Cobalt $(CoCl_2)$ is coloured because the $Co^{2+}$ ion has an electronic configuration of $[Ar] 3d^7$,which contains unpaired electrons.
Transition metal ions with unpaired $d$-electrons exhibit $d-d$ transitions,resulting in colour.
In contrast,$Ag^+$,$Hg^{2+}$,and $Zn^{2+}$ have $d^{10}$ configurations (fully filled $d$-orbitals),making their chlorides colourless.

(C) The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n + 2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
$Ti^{3+}$ $(Z=22)$: $3d^1$,$n = 1$.
$Sc^{3+}$ $(Z=21)$: $3d^0$,$n = 0$.
$Mn^{2+}$ $(Z=25)$: $3d^5$,$n = 5$.
$Zn^{2+}$ $(Z=30)$: $3d^{10}$,$n = 0$.
Since $Mn^{2+}$ has the highest number of unpaired electrons $(n = 5)$,it exhibits the highest magnetic moment.

(D) The maximum oxidation state of a transition metal is determined by the total number of electrons in the $(n-1)d$ and $ns$ orbitals.
For $(n-1)d^3ns^2$,the total number of valence electrons is $3 + 2 = 5$.
For $(n-1)d^5ns^1$,the total number of valence electrons is $5 + 1 = 6$.
For $(n-1)d^8ns^2$,the total number of valence electrons is $8 + 2 = 10$,but only the unpaired electrons or those in the outer shell are typically involved in bonding,limiting the oxidation state.
For $(n-1)d^5ns^2$,the total number of valence electrons is $5 + 2 = 7$.
Therefore,the configuration $(n - 1)d^5ns^2$ can achieve the maximum oxidation state of $+7$ (e.g.,in $Mn$).

(B) The correct option is $(B)$.
To determine the electronic configuration,we look at the atomic numbers of the elements:
$Ti (Z=22): [Ar] 3d^2 4s^2 \implies Ti^{2+} = [Ar] 3d^2$
$V (Z=23): [Ar] 3d^3 4s^2 \implies V^{3+} = [Ar] 3d^2$
$Cr (Z=24): [Ar] 3d^5 4s^1 \implies Cr^{4+} = [Ar] 3d^2$
$Mn (Z=25): [Ar] 3d^5 4s^2 \implies Mn^{5+} = [Ar] 3d^2$
Thus,all ions in option $(B)$ have a $3d^2$ configuration.

(A) $Transition$ metals are known to be efficient catalysts because they possess variable oxidation states and the ability to form unstable intermediate compounds with reactants.
These intermediate compounds provide an alternative pathway with lower activation energy for the reaction,thereby increasing the rate of the reaction.
Thus,$Transition$ metals are the most efficient catalysts.

(B) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
Electronic configurations of the ions are:
$Cr^{2+} (3d^4) \implies n = 4$
$Mn^{2+} (3d^5) \implies n = 5$
$Cu^{2+} (3d^9) \implies n = 1$
$Co^{2+} (3d^7) \implies n = 3$
Since $Mn^{2+}$ has the maximum number of unpaired electrons $(n=5)$,it will show the highest magnetic moment.

(C) The $3d-$ transition series consists of elements with atomic numbers $21$ to $30$ ($Sc$ to $Zn$).
Cobalt $(Co)$ has an atomic number of $27$,which places it in the $3d-$ series.
Its electronic configuration is $[Ar] 3d^7 4s^2$,confirming it belongs to the $3d-$ transition series.

(C) The atomic weights of $Fe$,$Co$,and $Ni$ are approximately $55.85 \ u$,$58.93 \ u$,and $58.69 \ u$ respectively.
Comparing these values: $58.93 \ (Co) > 58.69 \ (Ni) > 55.85 \ (Fe)$.
Therefore,the correct sequence is $Co > Ni > Fe$.

(B) The electronic configuration of the metal $M$ is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$.
Since the last electron enters the $3d$ orbital,the metal belongs to the $d-$ block elements.
Specifically,this configuration corresponds to Copper ($Cu$,atomic number $29$).

(B) Transition elements are defined as elements which have a partially filled $d-$subshell in their ground state or in any of their common oxidation states.
Option $B$ represents an element with the configuration $1s^2, 2s^2 2p^6, 3s^2, 3p^6 3d^2, 4s^2$,which corresponds to Titanium $(Ti)$.
Since the $3d$ subshell is partially filled $(3d^2)$,it is a transition element.

(B) The element with atomic number $105$ is Dubnium.
In $IUPAC$ nomenclature,it is systematically named as Unnilpentium $(Unp)$.

(A) . Elements or ions containing unpaired electrons are paramagnetic.
$_{28}Ni = [Ar] \; 3d^8 4s^2$; $Ni^{2+} = [Ar] \; 3d^8 4s^0$.
In $Ni^{2+}$,the $3d$ subshell has $8$ electrons,which means there are $2$ unpaired electrons.
$_{29}Cu^+ = [Ar] \; 3d^{10} 4s^0$ (diamagnetic).
$_{30}Zn^{2+} = [Ar] \; 3d^{10} 4s^0$ (diamagnetic).
$_{21}Sc^{3+} = [Ar] \; 3d^0 4s^0$ (diamagnetic).
Since $Ni^{2+}$ has unpaired electrons,it is paramagnetic.

(B) The atomic number of chromium $(Cr)$ is $24$.
According to the Aufbau principle,the expected configuration is $[Ar] \, 3d^4 4s^2$.
However,due to the extra stability associated with half-filled $d$-orbitals $(d^5)$,one electron from the $4s$ orbital shifts to the $3d$ orbital.
Therefore,the actual electronic configuration is $[Ar] \, 3d^5 4s^1$.

(A) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
When $Ni$ forms $Ni^{2+}$ ion,it loses two electrons from the $4s$ orbital.
The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
In the $3d$ subshell,there are $5$ orbitals. According to Hund's rule,filling $8$ electrons results in $3$ paired electrons and $2$ unpaired electrons.
Therefore,the number of unpaired electrons in $Ni^{2+}$ is $2$.

(D) The electronic configuration of Manganese $(Z = 25)$ is $[Ar] 3d^5 4s^2$.
After the removal of two electrons,the configuration of $Mn^{2+}$ becomes $[Ar] 3d^5$.
This $3d^5$ configuration is exceptionally stable due to the half-filled $d$-subshell.
Therefore,removing the third electron requires a significantly higher amount of energy,making the third ionization enthalpy of Manganese the highest among the given elements.

(A) An ion is colourless if it does not have any unpaired electrons in its $d$-orbitals.
$1. _{21}Sc^{3+}: [Ar] 3d^0$. It has no unpaired electrons,so it is colourless.
$2. _{26}Fe^{2+}: [Ar] 3d^6$. It has $4$ unpaired electrons,so it is coloured.
$3. _{22}Ti^{3+}: [Ar] 3d^1$. It has $1$ unpaired electron,so it is coloured.
$4. _{25}Mn^{2+}: [Ar] 3d^5$. It has $5$ unpaired electrons,so it is coloured.
Therefore,the correct option is $A$.

(A) In the case of $Tl$ (Thallium),the $Tl^{+}$ ion is more stable than the $Tl^{3+}$ ion due to the inert pair effect,where the $6s^{2}$ electrons are reluctant to participate in bonding.
Consequently,$Tl^{3+}$ acts as a strong oxidizing agent and gets reduced to $Tl^{+}$:
$Tl^{3+} + 2e^{-} \to Tl^{+}$
In contrast,for $Cu$,$Cr$,and $V$,the higher oxidation states ($Cu^{2+}$,$Cr^{3+}$,$VO^{2+}$) are generally more stable in aqueous solution due to higher hydration energy or electronic configuration stability.

(C) Due to lanthanide contraction,the atomic radii of $Zr$ and $Hf$ are nearly identical.
Lanthanide contraction is explained by the poor shielding effect of $4f$ electrons.
In multi-electron atoms,inner electrons shield outer electrons from the nuclear charge.
The shielding efficiency follows the order $s > p > d > f$.
Because the $4f$ subshell has a very poor shielding effect,the outer electrons experience a higher effective nuclear charge,leading to a decrease in atomic size.
This effect causes elements of the $5d$ series (like $Hf$) to have radii very similar to their counterparts in the $4d$ series (like $Zr$).

(A) $Fe^{3+} \to [Ar] 3d^{5} 4s^{0}$,number of unpaired electrons $= 5$
$Fe^{2+} \to [Ar] 3d^{6} 4s^{0}$,number of unpaired electrons $= 4$
$Co^{2+} \to [Ar] 3d^{7} 4s^{0}$,number of unpaired electrons $= 3$
$Co^{3+} \to [Ar] 3d^{6} 4s^{0}$,number of unpaired electrons $= 4$
Thus,$Fe^{3+}$ has the maximum number of unpaired electrons.

(D) Paramagnetic character is due to the presence of unpaired electrons in the $d$-orbitals of transition metal atoms or ions.