General Characteristics Questions in English

(A) True: Transition elements have $(n-1)d$ and $ns$ electrons with comparable energies,allowing them to show variable oxidation states.
$(b)$ True: The $E^o$ values for $Mn^{2+}/Mn$,$Zn^{2+}/Zn$,and $Ni^{2+}/Ni$ are more positive than expected due to factors like high hydration enthalpy or stable electronic configurations $(d^5, d^{10})$.
$(c)$ True: The high electronegativity and small size of oxygen and fluorine (in chlorides/fluorides) stabilize the highest oxidation states of transition metals.

(A) True: The magnetic moment of transition metal ions is due to both spin and orbital angular momentum of electrons.
$(b)$ True: For $Mn^{2+}$ ($d^5$ configuration),the number of unpaired electrons $n = 5$. The spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$(c)$ False: Interstitial compounds are non-stoichiometric compounds formed when small atoms like $H, C, N$ are trapped inside the crystal lattice of transition metals. They are typically metallic in nature,not covalent.

(A) In each transition series,as we move from left to right,the nuclear charge increases by one unit at each step.
Due to the poor shielding effect of $d$-electrons,the effective nuclear charge increases,which leads to an increase in the ionization enthalpy.
However,this increase is not regular because of the complex interplay between increasing nuclear charge and increasing shielding effect.

(N/A) Elements of group $3$ to $12$ are known as $d$-block elements.
The general electronic configuration of these elements is $(n-1)d^{1-10}ns^{0-2}$.
These elements are called $d$-block elements because their last electron enters the $d$-orbital.
Characteristics of $d$-block elements:
- They are all metals. They mostly form coloured ions,exhibit variable valency (oxidation states),show paramagnetism,and are often used as catalysts.
- However,$Zn$,$Cd$,and $Hg$,which have the electronic configuration $(n-1)d^{10}ns^{2}$,do not show most of the properties of transition elements.
Transition Elements: Transition metals form a bridge between the chemically active metals of $s$-block elements and the less active elements of groups $13$ and $14$,hence the name "Transition Elements".

(N/A) -block elements are elements belonging to groups $3$ to $12$ in the periodic table,where the last electron enters the $(n-1)d$ orbital. Their general electronic configuration is $(n-1)d^{1-10}ns^{0-2}$.
Transition elements are defined as elements that have partially filled $d$-orbitals in their ground state or in any of their common oxidation states. Thus,$Zn$,$Cd$,and $Hg$ (having $d^{10}$ configuration) are $d$-block elements but not transition elements.
Characteristics of $d$-block elements:
$1$. They are all metals with high melting and boiling points.
$2$. They exhibit variable oxidation states.
$3$. They form coloured ions and compounds.
$4$. They show paramagnetic behaviour due to the presence of unpaired electrons.
$5$. They act as good catalysts due to their ability to adopt multiple oxidation states and provide a large surface area.

(N/A) The two rows of elements at the bottom of the Periodic Table,called the Lanthanoids,$Ce$ $(Z = 58)$ $-$ $Lu$ $(Z = 71)$ and Actinoids,$Th$ $(Z = 90)$ $-$ $Lr$ $(Z = 103)$,are characterized by the outer electronic configuration $(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}$.
The last electron added to each element is filled in an $f$-orbital. These two series of elements are hence called the Inner-Transition Elements ($f$-Block Elements).
They are all metals. Within each series,the properties of the elements are quite similar.
The chemistry of the early actinoids is more complicated than the corresponding lanthanoids,due to the large number of oxidation states possible for these actinoid elements.
Actinoid elements are radioactive. Many of the actinoid elements have been made only in nanogram quantities or even less by nuclear reactions and their chemistry is not fully studied.
The elements after uranium are called Transuranium Elements.

(N/A) transition element is defined as an element which has incompletely filled $d$-orbitals in its ground state or in any one of its oxidation states.
$d$-block elements are those in which the last electron enters the $d$-orbital. Their general electronic configuration is $(n-1)d^{1-10} ns^{0-2}$.
Elements like $Zn$,$Cd$,and $Hg$ have a completely filled $d^{10}$ configuration in their ground state as well as in their common oxidation states $(+2)$.
Since they do not have incompletely filled $d$-orbitals,they do not exhibit the characteristic properties of transition elements.
Therefore,while all transition elements belong to the $d$-block,elements like $Zn$,$Cd$,and $Hg$ are $d$-block elements but not transition elements.

(N/A) The oxidation states of elements are determined by their electronic configuration,specifically the number of electrons in the outermost shell and the stability of the resulting configuration.
$1$. Non-transition elements: Consider Nitrogen $(Z=7)$ with configuration $[He] 2s^{2} 2p^{3}$. It can show oxidation states like $-3$ (by gaining $3$ electrons to complete the octet) or $+5$ (by losing $5$ valence electrons).
$2$. Transition elements: Consider Manganese $(Z=25)$ with configuration $[Ar] 3d^{5} 4s^{2}$. Due to the availability of both $4s$ and $3d$ electrons,it exhibits variable oxidation states ranging from $+2$ to $+7$. The $+2$ state arises from the loss of $4s^{2}$ electrons,while higher states involve the progressive loss of $3d$ electrons,reaching stability at $+7$ where the $3d$ subshell becomes empty.

(A) Copper $(Cu)$ is widely used as a conductor of electricity in wires due to its high electrical conductivity and ductility.
It is a transition metal that allows the flow of electrons with minimal resistance.

(A) German Silver is an alloy consisting of Copper $(Cu)$,Zinc $(Zn)$,and Nickel $(Ni)$.
It typically contains $50\% \ Cu$,$30\% \ Zn$,and $20\% \ Ni$ by mass.
Despite its name,it does not contain any Silver $(Ag)$.

(A) Interstitial compounds are formed when small atoms like $H$,$C$,or $N$ are trapped inside the crystal lattice of transition metals.
These compounds are very hard and have high melting points.
Their hardness is comparable to that of $Diamond$.

(A) Alloys are formed by atoms with similar atomic sizes. For the formation of substitutional solid solutions (alloys),the difference in the atomic radii of the constituent metals should be less than $15\%$.

(A) $CrO_3$ and $V_2O_5$ exhibit significant covalent character due to the high oxidation state of the transition metal ($Cr^{6+}$ and $V^{5+}$) and the small size of the metal ions.
According to Fajan's rule,high charge density leads to polarization of the oxide ion,resulting in covalent bonding.
Covalent compounds typically have lower melting points compared to purely ionic compounds because the intermolecular forces are weaker than strong electrostatic ionic attractions.

(B) The elements of $Group \ 11$ in the periodic table,which include copper $(Cu)$,silver $(Ag)$,and gold $(Au)$,are historically and commonly known as coinage metals because they have been used to mint coins throughout history.

(N/A) The standard electrode potential $(E^o)$ of a metal is determined by the sum of the enthalpy of atomization,ionization enthalpy,and hydration enthalpy.
$1$. For $Mn^{2+}/Mn$: The $Mn^{2+}$ ion has a stable $d^5$ configuration. The high stability of the $d^5$ configuration makes the reduction of $Mn^{2+}$ to $Mn$ less favorable,resulting in a more negative $E^o$ value.
$2$. For $Zn^{2+}/Zn$: The $Zn^{2+}$ ion has a stable $d^{10}$ configuration. Similar to $Mn^{2+}$,the stability of the $d^{10}$ configuration makes the reduction of $Zn^{2+}$ to $Zn$ less favorable,leading to a more negative $E^o$ value.
$3$. For $Ni^{2+}/Ni$: The $Ni^{2+}$ ion has a very high negative hydration enthalpy,which compensates for the energy required for ionization and atomization,resulting in a more negative $E^o$ value than expected from the general trend.

(A) The first ionisation enthalpy depends on the effective nuclear charge and the electronic configuration of the atom.
$Cr$ has the electronic configuration $[Ar] 3d^5 4s^1$.
$Zn$ has the electronic configuration $[Ar] 3d^{10} 4s^2$.
Although $Cr$ has a stable half-filled $d$-subshell,the effective nuclear charge of $Zn$ $(Z=30)$ is significantly higher than that of $Cr$ $(Z=24)$.
As the atomic number increases across the $3d$ series,the effective nuclear charge increases,which pulls the electrons closer to the nucleus,making them harder to remove.
Therefore,the first ionisation enthalpy of $Zn$ is higher than that of $Cr$ due to its higher effective nuclear charge and the stability of the completely filled $d^{10}$ subshell.

(N/A) Transition elements exhibit high melting points due to the involvement of $(n-1)d$ electrons in addition to $ns$ electrons in the metallic bonding.
This leads to strong interatomic interactions and a greater number of covalent bonds between the metal atoms,which results in a high enthalpy of atomization and consequently,high melting points.

(B) $CuCl_2$ is more stable in an aqueous solution than $Cu_2Cl_2$.
This is because the high hydration enthalpy of $Cu^{2+}(aq)$ compensates for the high second ionization enthalpy required to convert $Cu^+(g)$ to $Cu^{2+}(g)$.
Although the first ionization enthalpy of $Cu$ is lower than the second,the hydration energy released when $Cu^{2+}$ ions are formed in water is significantly higher than that of $Cu^+$ ions,making $Cu^{2+}$ salts more stable in aqueous media.

(N/A) The magnetic moment $(\mu)$ is calculated using the spin-only formula: $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Cr^{3+}$ $(Z=24)$: The electronic configuration is $[Ar] 3d^3$. Here,$n = 3$. Thus,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
For $Co^{2+}$ $(Z=27)$: The electronic configuration is $[Ar] 3d^7$. This results in $n = 3$ unpaired electrons. However,for $Co^{2+}$,there is a significant orbital contribution to the magnetic moment due to the nature of the $d$-orbitals,which increases the observed value to approximately $4.87 \ BM$.

(A) . The difficulty in separating $Zr$ and $Hf$ arises because they possess almost identical atomic and ionic radii.
This phenomenon is a direct consequence of the lanthanoid contraction.
As we move from the $4d$ to the $5d$ series,the filling of $4f$ orbitals occurs before the $5d$ orbitals.
The $4f$ electrons provide poor shielding of the nuclear charge,which leads to an increase in the effective nuclear charge.
This causes the atomic radii of $5d$ elements (like $Hf$) to be very similar to those of the corresponding $4d$ elements (like $Zr$).

(B) According to Fajan's rule,the covalent character of an ionic bond increases with an increase in the charge density of the cation.
As the oxidation state of the transition metal increases,the size of the metal ion decreases and its charge increases.
This leads to a higher charge-to-size ratio,which increases the polarising power of the metal ion.
The metal ion then distorts the electron cloud of the halide ion more effectively,resulting in greater covalent character in the metal-halide bond.

(N/A) The reactivity of transition elements is primarily determined by their ability to lose electrons,which is related to their ionization enthalpy.
As we move from $Sc$ $(Z=21)$ to $Cu$ $(Z=29)$ in the $3d$ series,the effective nuclear charge increases due to poor shielding by $d$-electrons.
This leads to an increase in the first and second ionization enthalpies.
Consequently,the tendency to lose electrons decreases,making the elements less reactive.
Additionally,the increase in the enthalpy of atomization (due to stronger metallic bonding as the number of unpaired electrons increases up to $Cr$) also contributes to the decrease in reactivity.

(A) Osmium $(Os)$ is a transition metal that exhibits an oxidation state of $+8$ in $OsO_4$.
$(B)$ Manganese $(Mn)$ is a $3d$ series element that shows oxidation states ranging from $+2$ to $+7$.
$(C)$ Chromium $(Cr)$ has the highest melting point among the $3d$ series elements due to the involvement of more unpaired electrons in metallic bonding.
Therefore,the correct match is $A-3, B-1, C-2$.

(A-III, B-I, C-V, D-II) $A \rightarrow iii$ ($MnO_2$: $x + 2(-2) = 0 \Rightarrow x = +4$)
$B \rightarrow i$ (The most stable oxidation state of $Mn$ is $+2$)
$C \rightarrow v$ (The maximum stable oxidation state of $Mn$ in oxides is $+7$,as in $Mn_2O_7$)
$D \rightarrow ii$ (The characteristic oxidation state of Lanthanoids is $+3$)

(A-III, B-IV, C-II, D-I) $A \rightarrow iii$ ($Cu$ has a stable $d^{10}$ configuration after losing one electron,making the second ionization energy very high).
$B \rightarrow iv$ ($Zn$ has a stable $d^{10}$ configuration; removing the third electron from the $d$-orbital requires very high energy).
$C \rightarrow ii$ ($Cr(CO)_6$ is a stable metal carbonyl complex).
$D \rightarrow i$ ($Co$ has a high enthalpy of atomization due to strong metallic bonding in the $3d$ series).

(N/A) The elements identified by spectroscopy of their minerals are Rubidium $(Rb)$,Cesium $(Cs)$,Thallium $(Tl)$,Indium $(In)$,Gallium $(Ga)$,and Scandium $(Sc)$.

(C) The correct matches are:
$(1) \ Cu (Z=29): 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^1 \rightarrow (C)$
$(2) \ Cu^{2+} (Z=29): 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^9 \rightarrow (D)$
$(3) \ Zn^{2+} (Z=30): 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10} \rightarrow (A)$
$(4) \ Cr^{3+} (Z=24): 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^3 \rightarrow (E)$
Thus,the correct sequence is $(1-C, 2-D, 3-A, 4-E)$.

(D) The correct matching is: $(1-D, 2-C, 3-A, 4-B)$.
$(1) \ \text{Cr} \ (Z=24): [\text{Ar}] \ 3d^5 4s^1$ (Matches $D$)
$(2) \ \text{Fe}^{2+} \ (Z=26): [\text{Ar}] \ 3d^6 4s^0$ (Matches $C$)
$(3) \ \text{Ni}^{2+} \ (Z=28): [\text{Ar}] \ 3d^8 4s^0$ (Matches $A$)
$(4) \ \text{Cu} \ (Z=29): [\text{Ar}] \ 3d^{10} 4s^1$ (Matches $B$)

(N/A) The elements named after these scientists are as follows:

Scientist Name	Element Details
$T$. Seaborg	$Z = 106$,Seaborgium $(Sg)$
Mendeleev	$Z = 101$,Mendelevium $(Md)$
Rutherford	$Z = 104$,Rutherfordium $(Rf)$

(B) According to the $IUPAC$ nomenclature for elements with atomic number $Z > 100$,the digits are represented as: $1 = un$,$0 = nil$,$3 = tri$.
Combining these,we get $Un + nil + tri + ium = Unniltrium$.
The symbol is derived from the first letters of the digits: $U + n + t = Unt$.
The official name for the element with atomic number $103$ is Lawrencium,with the symbol $Lr$.

(N/A) $3d$ transition elements: $Sc$ $(Z=21)$ to $Zn$ $(Z=30)$.
$4d$ transition elements: $Y$ $(Z=39)$ to $Cd$ $(Z=48)$.
$5d$ transition elements: $La$ $(Z=57)$ to $Hg$ $(Z=80)$.
$6d$ transition elements: $Ac$ $(Z=89)$ to $Cn$ $(Z=112)$.
$4f$ inner transition elements (Lanthanoids): $Ce$ $(Z=58)$ to $Lu$ $(Z=71)$.
$5f$ inner transition elements (Actinoids): $Th$ $(Z=90)$ to $Lr$ $(Z=103)$.

(N/A) Transition elements are those elements in which the $d$-orbitals are incompletely filled in their ground state or in any of their oxidation states. These are located in groups $3$ to $12$ of the periodic table.
Inner transition elements are those elements in which the $f$-orbitals are incompletely filled in their ground state or in any of their oxidation states. These are known as lanthanoids and actinoids,located in the $f$-block of the periodic table.

(B) The general electronic configuration of $d$-block elements is $(n-1)d^{1-10}ns^{1-2}$.
$(i)$ $1s^2 2s^2 2p^6 3s^2$ is Magnesium $(Mg)$,which is an $s$-block element.
$(ii)$ $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2$ is Zinc $(Zn)$,which is a $d$-block element.
$(iii)$ $1s^2 2s^2 2p^5$ is Fluorine $(F)$,which is a $p$-block element.
$(iv)$ $1s^2 2s^2 2p^6 3s^2 3p^3$ is Phosphorus $(P)$,which is a $p$-block element.
Therefore,only configuration $(ii)$ belongs to the $d$-block elements.

(N/A) $(i)$ Lawrencium $(Lr)$,$Z = 103$
$(ii)$ Seaborgium $(Sg)$,$Z = 106$

(A) Transition elements are less electropositive because their ionization enthalpy values are higher than those of group-$1$ and group-$2$ metals,placing them between $s$-block and $p$-block elements in terms of metallic character.

(N/A) $(i)$ False: Atomic size generally decreases across a period due to increasing effective nuclear charge.
$(ii)$ True: Noble gases have stable closed-shell configurations,making electron gain energetically unfavorable.
$(iii)$ False: The actinoid series starts with $Ac$ $(Z=89)$ or $Th$ $(Z=90)$ depending on definition,but ends with $Lr$ $(Z=103)$,not $Lu$ ($Z=71$,which is a lanthanoid).
$(iv)$ False: While $(n-1)d^{1-10}ns^{1-2}$ is common,there are exceptions like $Pd$ $(4d^{10}5s^0)$ and $Cr$ $(3d^54s^1)$.

(C) The electronic configurations are as follows:
$(1)$ $Ti$ $(Z=22)$ is $[Ar] 3d^2 4s^2$. $Ti^+$ is $[Ar] 3d^2 4s^1$ or more specifically $(n-1)d^2 ns^1$. However,looking at the options,$(1)$ matches $(C)$ $(n-1)d^2 ns^0$ (assuming $4s$ electron is lost).
$(2)$ $_{30}Zn$ is $[Ar] 3d^{10} 4s^2$,which is $(n-1)d^{10} ns^2$ $(F)$.
$(3)$ $_{24}Cr$ is $[Ar] 3d^5 4s^1$,which is $(n-1)d^5 ns^1$ $(D)$.
$(4)$ $_{29}Cu$ is $[Ar] 3d^{10} 4s^1$,which is $(n-1)d^{10} ns^1$ $(B)$.
Therefore,the correct match is $(1-C), (2-F), (3-D), (4-B)$.

(N/A) Down the group,with the increase in atomic number,the atomic radii increase because of the addition of new shells. Hence,the atomic radii of elements of the second transition series are greater than the corresponding elements of the first transition series.
The atomic radii of elements of the second and third transition series are almost the same because of the intervention of the $4f$-orbitals,which must be filled before the $5d$ series of elements begins. The filling of $4f$ before the $5d$ orbital results in a regular decrease in atomic radii,called lanthanoid contraction. The lanthanoid contraction counterbalances the increase in atomic size with the increasing atomic number.
Example: $Zr$ and $Hf$ have very similar chemical properties and have nearly the same atomic radii because of lanthanoid contraction.
The lanthanoid contraction is due to the imperfect shielding of one electron by another in the same set of orbitals. However,the shielding of one $4f$ electron by another is less than that of one $d$ electron by another,and as the nuclear charge increases along the series,there is a fairly regular decrease in the size of the entire $4f^n$ orbitals.

(N/A) Complex compounds are those in which the metal ions bind a number of anions or neutral molecules,giving complex species with characteristic properties. Examples include: $[Fe(CN)_{6}]^{3-}$,$[Fe(CN)_{6}]^{4-}$,$[Cu(NH_{3})_{4}]^{2+}$,and $[PtCl_{4}]^{2-}$.
The transition elements have a great tendency to form complex compounds because of:
$(i)$ Small size of metal ions.
$(ii)$ High nuclear charge.
$(iii)$ Availability of vacant $d$-orbitals of suitable energy to accommodate lone pairs of electrons donated by the ligands.

(N/A) Interstitial compounds are those which are formed when small atoms like $B$,$H$,$C$,or $N$ are trapped inside the crystal lattice of metals.
The transition elements form interstitial compounds with these elements. The $B$,$H$,$C$,and $N$ occupy the vacant spaces (interstitial sites) in the crystal lattice,and as a result,the compounds formed are extremely hard.
They are classified as borides,hydrides,carbides,or nitrides depending on the nature of atoms trapped in the vacant sites. They are usually non-stoichiometric,neither typically covalent nor ionic. Examples include $TiC$,$Mn_4N$,$Fe_3H$,$VH_{0.58}$,and $TiH_{1.7}$.
The important physical and chemical characteristics of these compounds are:
$(i)$ High melting points,even higher than those of pure metals.
$(ii)$ They are extremely hard. Some borides approach diamond in hardness.
$(iii)$ They retain metallic conductivity.
$(iv)$ They are chemically inert.

(N/A) According to $Fajan's$ rule,as the oxidation state of the metal ion increases,its size decreases and its charge density increases. This leads to an increase in the polarizing power of the metal ion. Consequently,the metal ion polarizes the electron cloud of the halide ion more effectively,resulting in an increase in the covalent character of the bond.

(N/A) The properties of elements of the second and third transition series are similar due to the lanthanoid contraction.
As a result of the lanthanoid contraction,the elements of the third transition series have atomic radii that are almost identical to those of the corresponding elements in the second transition series.
This similarity in atomic radii leads to a significant similarity in their chemical and physical properties.

(B) The atomic number of Gadolinium $(Gd)$ is $Z = 64$.
Its ground state electronic configuration is $[Xe] 4f^7 5d^1 6s^2$.
For the $Gd^{3+}$ ion,three electrons are removed (one from $5d$ and two from $6s$):
$Gd^{3+} = [Xe] 4f^7$.
Here,there are $7$ unpaired electrons $(n = 7)$ in the $4f$ orbital.
The formula for the spin-only magnetic moment $(\mu)$ is $\mu = \sqrt{n(n+2)} \ B.M.$
$\mu = \sqrt{7(7+2)} = \sqrt{7 \times 9} = \sqrt{63} \approx 7.9 \ B.M.$
Therefore,the correct option is $B$.

(B) Transition elements are defined as elements which have incompletely filled $d$-orbitals in their ground state or in any of their oxidation states.
These include elements with atomic numbers:
$21-30$ (first transition series),
$39-48$ (second transition series),
$57$ and $72-80$ (third transition series),
$89$ and $104-112$ (fourth transition series).
In the given set $21, 25, 42, 72$:
$21$ $(Sc)$,$25$ $(Mn)$,$42$ $(Mo)$,and $72$ $(Hf)$ are all transition elements.
Therefore,the correct set is $21, 25, 42, 72$.

(C) The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
For $Cr^{2+}$ ion,two electrons are removed,resulting in the configuration $[Ar] 3d^4$.
This means there are $n = 4$ unpaired electrons.
The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$
Substituting $n = 4$:
$\mu = \sqrt{4(4+2)} \ B.M. = \sqrt{4 \times 6} \ B.M. = \sqrt{24} \ B.M. \approx 4.90 \ B.M.$

(C) In the $3d$-series,$Manganese$ $(Mn)$ exhibits the maximum number of oxidation states,ranging from $+2$ to $+7$.
$Zinc$ $(Zn)$ has a completely filled $d$-orbital $(3d^{10})$ in its ground state and its common oxidation state $(+2)$,therefore it is not considered a transition element.
$Scandium$ $(Sc)$ exhibits only one oxidation state,which is $+3$.
$Copper$ $(Cu^{+})$ undergoes a disproportionation reaction in an aqueous solution: $2Cu^{+}_{(aq)} \longrightarrow Cu^{2+}_{(aq)} + Cu_{(s)}$.
Thus,the correct matching is: $a-iii, b-iv, c-i, d-ii$.

(C) The spin-only magnetic moment $(\mu)$ depends on the number of unpaired electrons $(n)$ and is given by the formula: $\mu = \sqrt{n(n+2)} \ BM$.
The electronic configurations and number of unpaired electrons for the ions are:
$1$. For $Mn^{2+}$ $(Z=25)$: $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \ BM$.
$2$. For $Cr^{2+}$ $(Z=24)$: $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \ BM$.
$3$. For $Ti^{2+}$ $(Z=22)$: $[Ar] 3d^2$. Number of unpaired electrons $(n)$ = $2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ BM$.
Comparing the values,the correct order of spin-only magnetic moment is $Mn^{2+} > Cr^{2+} > Ti^{2+}$.

(D) The covalent radii for the given transition elements are as follows:
$Ni = 125 \text{ pm}$
$Cu = 128 \text{ pm}$
$Zn = 137 \text{ pm}$
$Mn = 137 \text{ pm}$
Comparing these values, $Ni$ has the smallest covalent radius among the given options.

(D) The colour of transition metal ions is due to the presence of unpaired electrons which allow $d-d$ transitions.
Electronic configurations:
$Zn^{2+} (Z=30): [Ar] 3d^{10} 4s^0$ (All electrons are paired,so it is colourless).
$Ni^{2+} (Z=28): [Ar] 3d^8 4s^0$ (Has $2$ unpaired electrons,so it is coloured).
$Cr^{3+} (Z=24): [Ar] 3d^3 4s^0$ (Has $3$ unpaired electrons,so it is coloured).
Therefore,only $Zn^{2+}$ is colourless,while $Ni^{2+}$ and $Cr^{3+}$ are coloured.

(B) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 1.73 \ BM$,we have $1.73 = \sqrt{n(n+2)}$.
Squaring both sides,$3 = n(n+2)$,which gives $n^2 + 2n - 3 = 0$.
Solving for $n$,we get $(n+3)(n-1) = 0$,so $n = 1$ (since $n$ cannot be negative).
Vanadium $(V)$ has the atomic number $23$ and electronic configuration $[Ar] 3d^3 4s^2$.
For $n=1$,the vanadium ion must have one unpaired electron.
In $VCl_4$,vanadium is in the $+4$ oxidation state $(V^{4+})$.
The configuration of $V^{4+}$ is $[Ar] 3d^1$,which contains $1$ unpaired electron.
Thus,the formula is $VCl_4$.