General Characteristics Questions in English

(B) substance is paramagnetic if it contains unpaired electrons $(n > 0)$. If it has no unpaired electrons $(n = 0)$,it is diamagnetic.
$1$. $Cr(ClO_4)_3$: $Cr$ is in $+3$ oxidation state. Electronic configuration of $Cr^{3+}$ is $[Ar]3d^3$. Here,$n = 3$,so it is paramagnetic.
$2$. $KMnO_4$: $Mn$ is in $+7$ oxidation state. Electronic configuration of $Mn^{7+}$ is $[Ar]3d^0$. Here,$n = 0$,so it is diamagnetic.
$3$. $TiCl_3$: $Ti$ is in $+3$ oxidation state. Electronic configuration of $Ti^{3+}$ is $[Ar]3d^1$. Here,$n = 1$,so it is paramagnetic.
$4$. $VOBr_2$: $V$ is in $+4$ oxidation state. Electronic configuration of $V^{4+}$ is $[Ar]3d^1$. Here,$n = 1$,so it is paramagnetic.
Therefore,$KMnO_4$ is not paramagnetic.

(D) The transition elements exhibit variable oxidation states due to the participation of $(n-1)d$ and $ns$ electrons in bonding.
Among the given elements,$Mn \ (Z=25)$ has the electronic configuration $[Ar] \ 3d^5 \ 4s^2$.
It exhibits the largest number of oxidation states ranging from $+2$ to $+7$ $(+2, +3, +4, +6, +7)$ because it has the maximum number of unpaired electrons available for bonding.

(A) The element with the outer electron configuration $3d^5\, 4s^2$ is Manganese $(Mn)$.
This configuration corresponds to the $3d$ transition series.
$Mn$ has $7$ valence electrons ($5$ in $d$-orbital and $2$ in $s$-orbital),allowing it to exhibit a wide range of oxidation states from $+2$ to $+7$ in its compounds.
This is the largest number of oxidation states among the given options.

(C) The electronic configuration of $Gd$ $(Z = 64)$ is $[Xe] 4f^7 5d^1 6s^2$.
For the $Gd^{3+}$ ion,three electrons are removed,resulting in the configuration: $[Xe] 4f^7$.
The number of unpaired electrons $(n)$ is $7$.
The magnetic moment $(\mu)$ is calculated using the spin-only formula: $\mu = \sqrt{n(n + 2)} \ B.M.$
Substituting $n = 7$: $\mu = \sqrt{7(7 + 2)} = \sqrt{7 \times 9} = \sqrt{63} \approx 7.9 \ B.M.$

(D) The magnetic moment (spin only) is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Mn^{2+}$ $(d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$2$. For $Ti^{2+}$ $(d^2)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
$3$. For $V^{2+}$ $(d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$4$. For $Cr^{2+}$ $(d^4)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
The value $4.90 \ BM$ is of the order of $5 \ BM$. Thus,$Cr^{2+}$ is the correct answer.

(B) The electronic configuration of Gadolinium $[Z = 64]$ is $[Xe] 4f^7 5d^1 6s^2$.
In the $4f$ subshell,there are $7$ unpaired electrons.
In the $5d$ subshell,there is $1$ unpaired electron.
Therefore,the total number of unpaired electrons is $7 + 1 = 8$.

(D) The enthalpy of atomisation depends on the strength of metallic bonding,which is determined by the number of unpaired electrons in the $d$-orbital.
$Zn$ has a completely filled $d$-orbital $(3d^{10} 4s^2)$,meaning there are no unpaired electrons available for metallic bonding.
Consequently,$Zn$ exhibits the weakest metallic bonding among the $3d$ transition series,resulting in the lowest enthalpy of atomisation.

(C) The electronic configuration of $Sc$ $(Z=21)$ is $[Ar] 3d^1 4s^2$.
After losing three electrons,it forms $Sc^{3+}$,which has a stable noble gas configuration $([Ar])$.
Due to this stable configuration,$Sc$ does not show variable oxidation states and exhibits only $+3$ oxidation state.

(C) Interstitial compounds are formed when small atoms like $H$,$C$,or $N$ are trapped inside the crystal lattice of transition metals.
These compounds typically exhibit high melting points,high hardness,and retain metallic conductivity.
However,they are generally chemically inert,not reactive.
Therefore,the statement that they are chemically reactive is $INCORRECT$.

(B) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$1. Sc^{3+} (Z=21): [Ar] 3d^0$. Number of unpaired electrons $(n) = 0$. $\mu = 0 \ B.M.$
$2. Ti^{3+} (Z=22): [Ar] 3d^1$. Number of unpaired electrons $(n) = 1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \ B.M. \approx 1.73 \ B.M.$
$3. Ti^{2+} (Z=22): [Ar] 3d^2$. Number of unpaired electrons $(n) = 2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M. \approx 2.83 \ B.M.$
$4. V^{2+} (Z=23): [Ar] 3d^3$. Number of unpaired electrons $(n) = 3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \ B.M. \approx 3.87 \ B.M.$
Comparing the values: $0 < 1.73 < 2.83 < 3.87$.
Therefore,the correct order is $Sc^{3+} < Ti^{3+} < Ti^{2+} < V^{2+}$.

(D) The first ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge.
For the $3d$ series elements $(Ti, Mn, Ni, Zn)$,the general trend is $Ti < Mn < Ni < Zn$.
$Ti$ $([Ar] 3d^2 4s^2)$ has the lowest ionization enthalpy,while $Zn$ $([Ar] 3d^{10} 4s^2)$ has the highest due to its stable,fully-filled $d$-orbital configuration.

(D) Due to the lanthanoid contraction, the elements of the $5d$ series have atomic radii very similar to those of the corresponding elements of the $4d$ series. $Mo$ ($4d$ series) and $W$ ($5d$ series) exhibit this phenomenon, resulting in similar atomic radii.

(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = \sqrt{15} \ BM$,we have $\sqrt{n(n+2)} = \sqrt{15}$,which implies $n(n+2) = 15$.
Solving for $n$,$n^2 + 2n - 15 = 0$,which factors as $(n+5)(n-3) = 0$. Since $n$ must be positive,$n = 3$.
The atomic number of $M$ is $25$,which corresponds to Manganese $(Mn)$.
The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^{5} 4s^{2}$.
For $M^{x+}$ to have $3$ unpaired electrons,it must have a $3d^{3}$ configuration.
$Mn^{4+}$ has the configuration $[Ar] 3d^{3} 4s^{0}$,which contains $3$ unpaired electrons.
Thus,the number of unpaired electrons is $3$ and the value of $x$ is $4$.

(D) The oxidation state of $+7$ is exhibited by several elements across the periodic table.
$Cl$ (Chlorine) shows $+7$ oxidation state in compounds like $HClO_4$ (perchloric acid).
$Mn$ (Manganese) shows $+7$ oxidation state in $KMnO_4$ (potassium permanganate).
$Np$ (Neptunium) is an actinide that can exhibit a $+7$ oxidation state in compounds like $NpO_2F_3$.
Therefore,all the given elements can show a $+7$ oxidation state.

(C) Ions are colourless if they do not have any unpaired electrons in their $d$-orbitals (i.e.,$d^0$ or $d^{10}$ configuration).
$1$. $Mn^{3+}$ $(3d^4)$,$Co^{3+}$ $(3d^6)$: Both have unpaired electrons,so they are coloured.
$2$. $Fe^{3+}$ $(3d^5)$,$Cr^{3+}$ $(3d^3)$: Both have unpaired electrons,so they are coloured.
$3$. $Zn^{2+}$ $(3d^{10})$: It has a completely filled $d$-subshell with no unpaired electrons,so it is colourless.
$4$. $Sc^{3+}$ $(3d^0)$: It has an empty $d$-subshell with no unpaired electrons,so it is colourless.
$5$. $Ti^{2+}$ $(3d^2)$,$Cu^{2+}$ $(3d^9)$: Both have unpaired electrons,so they are coloured.
Therefore,the pair $Zn^{2+}$ and $Sc^{3+}$ is colourless.

(A) Transition metal ions form coordination complexes because they possess empty valence shell orbitals (specifically $d$-orbitals) that can accept lone pairs of electrons from ligands,which act as Lewis bases.

(B) Transition elements are defined as elements which have incompletely filled $d$-orbitals in their ground state or in any of their oxidation states.
The general electronic configuration of transition elements is represented as $(n-1)d^{1-10}\,ns^{1-2}$,where $(n-1)$ represents the inner $d$-orbitals and $n$ represents the outermost shell.

(A) The electronic configurations of the neutral atoms are: $Ti ([Ar] 3d^2 4s^2)$,$V ([Ar] 3d^3 4s^2)$,$Cr ([Ar] 3d^5 4s^1)$,and $Mn ([Ar] 3d^5 4s^2)$.
After the removal of the first electron $(I.P_1)$,the configurations become: $Ti^+ ([Ar] 3d^2 4s^1)$,$V^+ ([Ar] 3d^3 4s^1)$,$Cr^+ ([Ar] 3d^5)$,and $Mn^+ ([Ar] 3d^5 4s^1)$.
The second ionization enthalpy $(I.P_2)$ involves removing an electron from these ions.
$Cr^+$ has a stable half-filled $d^5$ configuration,making it very difficult to remove the second electron,resulting in the highest $I.P_2$.
$Mn^+$ has a $3d^5 4s^1$ configuration,where the $4s$ electron is relatively easier to remove than the $d^5$ electron of $Cr^+$.
Comparing the trends,the order of $I.P_2$ is $Cr > Mn > V > Ti$.

(B) The ionization potential $(I.P.)$ values are determined by electronic configuration and effective nuclear charge.
For the given elements,the values are: $Mn (3d^5 4s^2) = 7.4 \ eV$,$Cr (3d^5 4s^1) = 6.7 \ eV$,$Zn (3d^{10} 4s^2) = 9.4 \ eV$ (approx $9.3 \ eV$),and $Hg (5d^{10} 6s^2) = 10.4 \ eV$ due to lanthanoid contraction.
Matching the values: $I.P.$ of $Mn = 7.4 \ eV$,$I.P.$ of $Cr = 6.7 \ eV$,$I.P.$ of $Hg = 10.4 \ eV$,and $I.P.$ of $Zn = 9.3 \ eV$.
Comparing with options,$I.P.$ of $Mn = 7.4 \ eV$ is correct.

(D) An interstitial compound is formed when small atoms like $H$,$C$,$N$,or $O$ are trapped inside the crystal lattice of transition metals.
Transition metals like $Fe$,$Co$,and $Ni$ have large interstitial spaces in their crystal lattices,allowing them to form interstitial compounds.
Therefore,all of the given metals can form interstitial compounds.

(D) The species $Fe$, $Co$, and $Ni$ belong to the $d$-block elements (transition metals) in the same period ($4^{th}$ period).
In the $d$-block, the atomic radii decrease slightly across a period due to the shielding effect of $d$-electrons, which almost balances the increase in nuclear charge.
Consequently, the atomic radii of $Fe$ $(126 \text{ pm})$, $Co$ $(125 \text{ pm})$, and $Ni$ $(124 \text{ pm})$ are approximately equal.
In contrast, $Na^{+}$, $Mg^{2+}$, and $Al^{3+}$ are isoelectronic but have different nuclear charges, leading to a significant decrease in ionic radii $(Na^{+} > Mg^{2+} > Al^{3+})$.

(A) Transition metals are characterized by variable valency,not fixed valency.
This is due to the participation of both $(n-1)d$ and $ns$ electrons in bonding because their energy levels are very close.
They form coloured compounds due to $d-d$ transitions.
They have high melting and boiling points due to strong metallic bonding involving $d$-electrons.
They have a high tendency to form complexes due to the availability of vacant $d$-orbitals and high charge-to-size ratio.

(C) Transition elements are characterized by the presence of partially filled $d$-orbitals.
Due to the presence of unpaired electrons,they exhibit variable oxidation states.
They have high enthalpies of atomization due to strong interatomic interactions.
They act as good catalysts because of their ability to adopt multiple oxidation states and provide a large surface area for reactions.
Therefore,the statement that transition elements exhibit variable oxidation states is correct.

(C) The element $Pu$ (Plutonium) is named after the planet Pluto.
$Uranus$ $(1781) = Uranium$ $(1789)$
$Neptune$ $(1846) = Neptunium$ $(1940)$
$Pluto$ $(1930) = Plutonium$ $(1940)$
These elements were named after celestial bodies that were considered planets at the time of their discovery or naming.

(B) Representative elements are those belonging to $s$-block and $p$-block (excluding noble gases in some contexts, but generally groups $1, 2$ and $13-17$).
$(A)$ Tellurium $(Te)$ is in group $16$ ($p$-block).
$(B)$ Tantalum $(Ta, Z = 73)$ is a $d$-block element, specifically a transition element.
$(C)$ Thallium $(Tl)$ is in group $13$ ($p$-block).
$(D)$ Astatine $(At)$ is in group $17$ ($p$-block).
Therefore, Tantalum is not a representative element.

(C) The electronic configuration of Tantalum ($Ta$,$Z = 73$) is $[Xe] \ 4f^{14} \ 5d^3 \ 6s^2$.
Since the highest principal quantum number $(n)$ is $6$,the period number is $6$.
The element belongs to the $d$-block because the last electron enters the $d$-orbital. For $d$-block elements,the group number is calculated as $(n-1)d \text{ electrons} + ns \text{ electrons} = 3 + 2 = 5$.
Therefore,Tantalum belongs to period $6$ and group $5$.

Element	Period,Group
$Tc$ $(Z=43)$	$5, 7$
$Tl$ $(Z=81)$	$6, 13$
$Ta$ $(Z=73)$	$6, 5$

(D) The magnetic moment is given by the formula: $\mu = \sqrt{n(n + 2)} \ BM$,where $n$ is the number of unpaired electrons.
As the atomic number $(Z)$ increases across a $d-$block series,the number of unpaired electrons first increases as the $d-$orbitals are filled up to the middle (i.e.,$d^5$ configuration).
After the middle of the series,the number of unpaired electrons begins to decrease as the $d-$orbitals start pairing up.
Therefore,the magnetic moment also follows the same trend: it increases up to the middle of the series and then decreases.
This trend is correctly represented by the graph in option $D$.

(C) The $IUPAC$ name "unununium" corresponds to the atomic number $Z = 111$.
This element is known as Roentgenium $(Rg)$.
It belongs to the $7^{th}$ period and $11^{th}$ group of the periodic table.
Elements with atomic numbers $21-30$,$39-48$,$57, 72-80$,and $89, 104-112$ are classified as transition elements.
Therefore,it is a transition element.

(A) In the $3d$-series,atomic radii generally decrease from $Sc$ to $Cr$ due to the increase in effective nuclear charge. Thus,the order $Sc > Ti > V > Cr$ is correct.
For option $(b)$,the correct order is $Zn > Cu > Co \approx Ni$ because the increase in shielding effect balances the increase in nuclear charge.
For option $(c)$,for isoelectronic species,the ionic radius increases as the negative charge increases and decreases as the atomic number increases. The correct order is $N^{3-} > O^{2-} > F^{-} > S^{2-} > Cl^{-}$.
Therefore,option $(a)$ is the correct statement.

(C) The electronic configuration of $La$ (atomic number $57$) is $[Xe] 5d^1 6s^2$.
Since the last electron enters the $5d$ orbital,$La$ is classified as a $d-$ block element.

(A) The higher oxidation states of transition elements are found in combination with $F$ and $O$ because $F$ and $O$ are the most electronegative elements.
These elements have a high ability to stabilize the higher oxidation states of transition metals.
For example,$Os$ shows the highest oxidation state of $+8$ in $OsO_4$.
Similarly,$Mn$ shows the highest oxidation state of $+7$ in $KMnO_4$.

(A) The density of a metal is determined by its atomic mass and atomic volume.
Osmium $(Os)$ has the highest density among the transition metals,with a value of approximately $22.59 \ g/cm^3$.
Platinum $(Pt)$ has a density of $21.45 \ g/cm^3$,and Gold $(Au)$ has a density of $19.32 \ g/cm^3$.
Chromium $(Cr)$ has a much lower density of $7.19 \ g/cm^3$.
Therefore,Osmium is the most dense metal among the given options.

(A) Ferric salts are more stable than Ferrous salts.
$Fe^{3+}$ ions are more stable compared to $Fe^{2+}$ due to their electronic configuration.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5 4s^0$,which is a half-filled $d$-orbital configuration,making it extra stable.

(D) Transition metals are characterized by several unique properties due to the presence of partially filled $d$-orbitals.
$1$. They exhibit variable oxidation states because the energy difference between $(n-1)d$ and $ns$ orbitals is small.
$2$. They have a strong tendency to form coordination complexes due to their small size,high nuclear charge,and availability of vacant $d$-orbitals.
$3$. They form coloured compounds due to $d-d$ transitions of electrons.
Since all the listed properties are characteristic of transition metals,the correct answer is $D$.

(C) For transition elements,as the atomic number increases,electrons enter the empty $d$-orbitals.
Initially,the $d$-orbitals are singly occupied,which increases the number of unpaired electrons and thus increases paramagnetism.
This continues until the $d$-orbitals are half-filled ($d^5$ configuration),which corresponds to the point of maximum paramagnetism.
Beyond this point,electrons start pairing up in the $d$-orbitals,which reduces the number of unpaired electrons and consequently decreases paramagnetism.
Therefore,paramagnetism first increases to a maximum and then decreases.

(B) The transition metals are elements belonging to the $d$-block of the periodic table.
Among the transition metals,$Fe$ (Iron) is the most abundant metal in the earth's crust,ranking fourth among all elements and first among transition metals.

(C) Transition metals 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 $(n - 1) \, d$ subshell and the $ns$ subshell can participate in bond formation.
Therefore,both $ns$ and $(n - 1) \, d$ electrons are involved in exhibiting variable oxidation states.

(B) The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$ and $Cd$ is $[Kr] 4d^{10} 5s^2$.
Because their $d$-subshells are completely filled,the $d$-electrons are not available for bonding.
Consequently,they do not exhibit variable valency like other transition elements.

(C) Paramagnetism is exhibited by species containing unpaired electrons.
In $CuSO_4 \cdot 5H_2O$,$Cu^{2+}$ has $3d^9$ configuration (one unpaired electron).
In $CuCl_2 \cdot 5H_2O$,$Cu^{2+}$ has $3d^9$ configuration (one unpaired electron).
In $NiSO_4 \cdot 6H_2O$,$Ni^{2+}$ has $3d^8$ configuration (two unpaired electrons).
In $CuI$,the copper ion is $Cu^{+}$,which has a $3d^{10}$ electronic configuration.
Since all electrons are paired in the $3d$ subshell,$CuI$ is diamagnetic and does not exhibit paramagnetism.

(C) The stability of an electronic configuration is determined by the exchange energy and symmetry of the orbitals.
Half-filled $(d^5)$ and completely filled $(d^{10})$ $d$-orbitals,along with filled $s$-orbitals,provide extra stability.
In the configuration $[Ar]\, 3d^5\, 4s^2$ (which corresponds to Manganese,$Mn$),the $3d$ subshell is exactly half-filled $(d^5)$ and the $4s$ subshell is completely filled $(s^2)$.
This specific configuration is exceptionally stable,allowing the element to exhibit a wide range of oxidation states,with $+2$ being particularly stable due to the loss of $4s$ electrons leaving a stable $d^5$ core.

(B) In the $d$-block elements,metal-metal bonding is more extensive in the $4d$ and $5d$ series compared to the $3d$ series.
This is primarily because the $4d$ and $5d$ orbitals are larger in size and more diffuse than the $3d$ orbitals.
Due to their larger size,these orbitals can effectively overlap with each other to form stronger metal-metal bonds.
Therefore,the correct reason is the large size of the orbitals which participate in the metal-metal bond formation.

(A) $I$) In $K_2FeO_4$,iron is in the $+ 6$ oxidation state,which is its highest oxidation state.
$II$) The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$. In the $+ 2$ oxidation state $(Fe^{2+})$,it loses two $4s$ electrons,leaving $3d^6$ configuration,which has six electrons in the $3d$ orbitals.
$III$) The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$. This has five unpaired electrons in the $3d$ orbital,making it a common and stable oxidation state due to the half-filled $d$-subshell.
Thus,all three statements are correct.

(A) Interstitial compounds are formed when small atoms like $H, C, N$ or $O$ are trapped inside the crystal lattice of transition metals.
The characteristics of these compounds are:
$1$. They have high melting points,higher than those of pure metals.
$2$. They are very hard (some borides approach diamond in hardness).
$3$. They retain metallic conductivity.
$4$. They are chemically inert (not more reactive than pure metals).
Therefore,statements $I, II,$ and $III$ are correct.

(A) Technetium $(Tc)$ is a transition metal located in the $5th$ period,directly below manganese $(Mn)$ in the Periodic Table.
Transition metals in the $4d$ and $5d$ series generally exhibit higher melting points,boiling points,and densities compared to their $3d$ counterparts due to stronger metallic bonding resulting from increased effective nuclear charge and more extensive $d$-orbital participation.
Therefore,technetium is expected to have high values for its melting point,boiling point,and density.
Thus,the correct option is $A$.

(A) The electronic configuration of $Zn^{2+}$ is $3d^{10}4s^{0}$.
Since the $d$-subshell is completely filled,there are no vacant orbitals available for $d-d$ transitions.
As a result,$Zn^{2+}$ compounds do not absorb visible light and appear white or colorless.

(D) Transition metals have high enthalpies of atomisation because of the strong interatomic interactions.
These strong interactions are primarily due to the presence of a large number of unpaired electrons in their $(n-1)d$ orbitals.
These unpaired electrons participate in metallic bonding,which leads to stronger metallic bonds compared to other elements.
Therefore,more energy is required to break these bonds to atomise the metal.

(C) Statement $(I)$ is incorrect because $Zn$,$Cd$,and $Hg$ have fully filled $d$-orbitals $(d^{10})$,which leads to weak metallic bonding and consequently low enthalpies of atomisation.
Statement $(II)$ is correct. $Zn$ and $Cd$ exhibit only $+2$ oxidation state,whereas $Hg$ can exhibit $+1$ (as $Hg_2^{2+}$) and $+2$ oxidation states.
Statement $(III)$ is incorrect. Since they have a $d^{10}$ configuration,all electrons are paired,making their compounds diamagnetic,not paramagnetic.
Statement $(IV)$ is correct. Due to weak metallic bonding resulting from the $d^{10}$ configuration,these elements are relatively soft compared to other transition metals.
Therefore,statements $(II)$ and $(IV)$ are correct.

(D) Transition elements are defined as elements which have incompletely filled $d-$orbitals in their ground state or in any of their oxidation states.
$1$. The formation of coloured cations is a characteristic property due to $d-d$ transitions.
$2$. The presence of at least one unpaired electron in a $d-$orbital of a cation is the fundamental definition of transition elements.
$3$. The ability to form complex ions is a characteristic property due to small size,high nuclear charge,and availability of vacant $d-$orbitals.
Since all the given options are characteristic properties of transition elements,the correct answer is $D$.

(B) $1$. Transition metals like $Mn, Fe, Co, Ni,$ and $Cu$ have partially filled $d$-orbitals,which allow for $d-d$ transitions,resulting in the formation of coloured ions in aqueous solution. Thus,statement $I$ is correct.
$2$. These metals are reactive and act as reducing agents. When they react with concentrated nitric acid $(HNO_3)$,they are oxidized,and the nitric acid is reduced to various oxides of nitrogen (such as $NO_2$ or $NO$). Thus,statement $II$ is correct.
$3$. While these metals can form chlorides,they do not all readily form both $MCl_2$ and $MCl_3$. For example,$Cu$ primarily forms $CuCl_2$ $(Cu(II))$ and $CuCl$ $(Cu(I))$,but $CuCl_3$ is not a stable common chloride. Therefore,statement $III$ is not common to all these elements.
Conclusion: Statements $I$ and $II$ are common to all these elements.

(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.
For $Cr^{2+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$.
For $Fe^{2+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.
For $Cu^{2+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^9$. Number of unpaired electrons $(n)$ = $1$.
For $Co^{2+}$ $(Z=27)$: Electronic configuration is $[Ar] 3d^7$. Number of unpaired electrons $(n)$ = $3$.
For $Cu^{+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^{10}$. Number of unpaired electrons $(n)$ = $0$.
Since $Cr^{2+}$ and $Fe^{2+}$ both have $4$ unpaired electrons, they have the same magnetic moment.