Mix Examples-d and f-Block Elements Questions in English

(D) Transition metals are widely used as catalysts due to their ability to adopt multiple oxidation states and provide a large surface area for adsorption.
$Nickel$ $(Ni)$ is used in hydrogenation reactions.
$Platinum$ $(Pt)$ is used in the contact process and various oxidation reactions.
$Cobalt$ $(Co)$ is used in the hydroformylation of alkenes.
Therefore,all three metals are used as catalysts.

(C) The hydrated ion $[Fe(H_2O)_6]^{2+}$ is pale green in color.
Similarly,the hydrated ion $[Ni(H_2O)_6]^{2+}$ is also green in color.
Therefore,both $Fe^{2+}$ and $Ni^{2+}$ form green hydrated ions.

(D) Transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and form complexes.
$Ni$ (Nickel) is widely used as a catalyst in hydrogenation reactions (e.g.,Sabatier-Senderens process).
$Co$ (Cobalt) is also used as a catalyst in various industrial processes,such as the hydroformylation of alkenes (oxo process).
Therefore,both $A$ and $B$ are correct.

(D) The king of metals is $Gold$.
$Gold$ forms an amalgam with $Mercury$,which means it is soluble in $Mercury$.
Pure $Gold$ is very soft. Therefore,impurities like $Silver$ and $Copper$ are added to make it harder.
Since all the given statements are correct,the correct option is $D$.

(D) The transition metals have a less tendency to form ions due to the combined effect of high heat of sublimation,high ionisation energy,and low heat of hydration energy.

(D) In $K_2Cr_2O_7$ $(Cr^{6+})$ and $KMnO_4$ $(Mn^{7+})$,the metal ions have a $d^0$ configuration,meaning there are no electrons available for $d-d$ transitions.
However,these compounds are intensely colored due to $Charge \ Transfer \ Spectra$ (specifically,ligand-to-metal charge transfer).
In $K_4[Fe(CN)_6]$,the color is due to the presence of the complex ion,but the question specifically targets compounds that exhibit color without $d-d$ transitions.
Both $K_2Cr_2O_7$ and $KMnO_4$ are classic examples of this phenomenon.
Therefore,the correct answer is $D$ (All of these),as $K_4[Fe(CN)_6]$ also exhibits color due to charge transfer or other electronic transitions in the complex.

(D) The stability of high oxidation states in $d$-block and $p$-block elements generally decreases down a group due to the inert pair effect and the decrease in the strength of the metal-oxygen bond.
$1$. For the oxoanions of $3d$ transition metals,$CrO_4^{2-}$ (Cr is $+6$) is more stable than $MnO_4^{2-}$ (Mn is $+6$) and $FeO_4^{2-}$ (Fe is $+6$).
$2$. Comparing $CrO_4^{2-}$ and $WO_4^{2-}$,the stability of the $+6$ oxidation state increases down the group $(Cr < Mo < W)$,so $WO_4^{2-}$ is more stable than $CrO_4^{2-}$.
$3$. For group $16$ elements,the stability of the $+6$ oxidation state decreases down the group $(Se > Te > Po)$ due to the inert pair effect.
Thus,all the given orders are correct.

(D) The electronic configurations are as follows:
$Cr (Z=24): [Ar] 3d^5 4s^1$. It has $5$ unpaired electrons in the $d$ orbitals.
$Mn (Z=25): [Ar] 3d^5 4s^2$. It has $5$ unpaired electrons in the $d$ orbitals.
$Fe^{3+} (Z=26): [Ar] 3d^5 4s^0$. It has $5$ unpaired electrons in the $d$ orbitals.
Therefore,all $A$,$B$,and $C$ have the same number of unpaired electrons in $d$ orbitals.

(D) Mercury $(Hg)$ has a completely filled $d$-orbital $(5d^{10}6s^2)$.
Due to the relativistic effect,the $6s$ electrons are strongly attracted by the nucleus,making them less available for metallic bonding.
Consequently,the metallic bonds in mercury are very weak compared to other transition metals.
Additionally,mercury has a very high first ionization energy,which prevents the electrons from participating effectively in bonding.
Therefore,both the high ionization energy and the weak metallic bonds contribute to mercury being a liquid at room temperature.

(B) $AgCl$ is not soluble in $H_2O$ because its lattice energy is greater than its hydration energy. It dissolves in $KCN$ due to the formation of a soluble complex: $AgCl + 2KCN \to K[Ag(CN)_2] + KCl$.
Statement $(II)$ is correct: $4Zn + 10HNO_3(\text{v.v. dil}) \to 4Zn(NO_3)_2 + NH_4NO_3 + 3H_2O$.
Statement $(III)$ is incorrect: $2CuSO_4 + 4KI \to Cu_2I_2 + I_2 + 2K_2SO_4$. The product is $Cu_2I_2$ and $I_2$,not just $Cu_2I_2$ and $K_2SO_4$.
Statement $(IV)$ is correct: Upon heating $CuSO_4 \cdot 5H_2O$ to $1000\,^oC$,it decomposes as: $2CuSO_4 \to 2CuO + 2SO_2 + O_2$.

(D) The melting point of $3d$ transition elements depends on the number of unpaired electrons and the strength of metallic bonding. $Mn$ has a low melting point due to its stable $d^5$ configuration,which leads to weak metallic bonding. $V$ has a high melting point due to a large number of unpaired electrons. The order of melting points for $Co, Ni, Cu, Zn$ is $Co > Ni > Cu > Zn$ because the metallic bond strength decreases as the number of unpaired electrons decreases.
Regarding solubility,$Na_2S$ is an ionic compound and is highly soluble in water. Among metal sulfides,the solubility is determined by the solubility product constant $(K_{sp})$. The $K_{sp}$ of $ZnS$ is higher than that of $CuS$,so $ZnS$ is more soluble than $CuS$. Therefore,the correct order of solubility is $Na_2S > ZnS > CuS$. Option $D$ states $CuS > ZnS > Na_2S$,which is the incorrect order.

(D) $1$. $V_2O_5$ is amphoteric in nature; it reacts with both acids and alkalies to form salts.
$2$. Misch metal is an alloy consisting of a lanthanoid metal (approx $95\%$),iron (approx $5\%$),and traces of $S, C, Ca$ and $Al$.
$3$. The reaction $2Cu^{+} \xrightarrow{aq.} Cu^{2+} + Cu$ represents the disproportionation of $Cu^{+}$ ions in an aqueous medium,which is thermodynamically favorable due to the high hydration energy of $Cu^{2+}$ ions.
$4$. Since all statements are correct,the correct option is $D$.

(D) $[Ar] \, 4s^2 \, 3d^5$ corresponds to atomic number $25$ $(Mn)$,which is a $d$-block element.
For the $III \ B$ group,the magic numbers are $18, 18, 32$.
The first transition series ($3d$-series) contains elements with atomic numbers $21$ to $30$,totaling $10$ elements.
Therefore,all the given statements are correct.

(D) $1$. $Cr^{2+}$ is a stronger reducing agent than $Fe^{2+}$ because $Cr^{2+}$ $(d^4)$ changes to $Cr^{3+}$ ($d^3$,stable $t_{2g}^3$ configuration) while $Fe^{2+}$ $(d^6)$ changes to $Fe^{3+}$ ($d^5$,stable half-filled configuration). However,the $E^o$ value for $Cr^{3+}/Cr^{2+}$ is $-0.41 \ V$ and $Fe^{3+}/Fe^{2+}$ is $+0.77 \ V$,making $Cr^{2+}$ a better reducing agent.
$2$. $Mn^{3+}$ is a stronger oxidising agent than $Cr^{3+}$ because $Mn^{3+}$ $(d^4)$ easily gains an electron to form $Mn^{2+}$ ($d^5$,stable half-filled configuration).
$3$. In group $6$,stability of the highest oxidation state increases down the group. Thus,$WO_3$ is more stable than $CrO_3$.
$4$. Since all statements are correct,the correct option is $D$.

(D) $1$. $Cr^{+2}$ $(d^4)$ is a strong reducing agent because it changes to $Cr^{+3}$ ($d^3$,stable $t_{2g}^3$ configuration). $Fe^{+2}$ $(d^6)$ is a weaker reducing agent as it changes to $Fe^{+3}$ ($d^5$,stable half-filled configuration). Thus,$Cr^{+2} > Fe^{+2}$ is correct.
$2$. $Mn^{+3}$ $(d^4)$ is a strong oxidising agent because it reduces to $Mn^{+2}$ ($d^5$,stable half-filled configuration). $Cr^{+3}$ $(d^3)$ is stable and does not easily reduce. Thus,$Mn^{+3} > Cr^{+3}$ is correct.
$3$. In group $6$,stability of the highest oxidation state increases down the group. Therefore,$WO_3$ is more stable than $CrO_3$. Thus,$WO_3 > CrO_3$ is correct.

(B) Statement $(I)$ is correct: $ZnCl_2$ is predominantly ionic,while $CdCl_2$ and $HgCl_2$ show significant covalent character due to polarization.
Statement $(II)$ is correct: $Zn$ and $Cd$ have negative standard reduction potentials and react with dilute $HCl$ to release $H_2$ gas. $Hg$ has a positive reduction potential and does not react with non-oxidizing acids.
Statement $(III)$ is incorrect: $Hg^{2+}$ ions react with $OH^-$ to form $HgO$ (yellow/red precipitate),but $Hg_2^{2+}$ (mercurous) forms black $Hg_2O$. The statement is generalized and misleading regarding the specific oxidation states.
Statement $(IV)$ is incorrect: While $Hg$ forms the $Hg_2^{2+}$ ion,$Zn$ and $Cd$ do not form stable $M_2^{2+}$ ions under normal conditions.
Therefore,statements $(III)$ and $(IV)$ are incorrect.

(A) The correct matching is as follows:
$(1)$ General electronic configuration of $d$-block elements is $(n-1)d^{1-10}ns^{1-2}$,which corresponds to $(B)$.
$(2)$ $d$-block elements are also known as transition elements,which corresponds to $(A)$.
$(3)$ Groups $3$ to $12$ are defined by the fact that the last electron enters the $d$-orbital,which corresponds to $(D)$.
$(4)$ Lanthanoid elements range from atomic number $Z = 58$ to $71$,which corresponds to $(C)$.
Therefore,the correct sequence is $(1-B, 2-A, 3-D, 4-C)$.

(C) Copper is a transition metal $(a-ii)$.
Fluorine is a non-metal $(b-i)$.
Silicon is a metalloid $(c-iv)$.
Cerium is a lanthanoid $(d-iii)$.
Therefore,the correct match is $a-ii, b-i, c-iv, d-iii$.

(C) $Cu(II)$ $(d^9)$ complexes have one unpaired electron,so they are paramagnetic. This is correct.
$(B)$ $Cu(I)$ $(d^{10})$ complexes have no unpaired electrons,so they are generally colourless due to the absence of $d-d$ transitions. This is correct.
$(C)$ $Cu(I)$ is unstable in aqueous solution and undergoes disproportionation to $Cu(0)$ and $Cu(II)$,meaning it is easily oxidized to $Cu(II)$. This is correct.
$(D)$ Fehling solution consists of two parts: Fehling $A$ ($CuSO_4$ solution) and Fehling $B$ (alkaline sodium potassium tartrate). The active species is the $Cu(II)$ tartrate complex. This is correct.
Therefore,all $4$ statements are correct.

(B) Statement $I$ is false because neutral $KMnO_{4}$ oxidises $I^{-}$ to $IO_{3}^{-}$ (iodate ion),not $I_{2}$.
Statement $II$ is false because the manganate ion $(MnO_{4}^{2-})$ involves $d\pi-p\pi$ bonding,not $p\pi-p\pi$ bonding,although it is paramagnetic due to one unpaired electron.

(B) In $KMnO_4$,the oxidation state of $Mn$ is $+7$. The electronic configuration is $[Ar] 3d^0$. The number of unpaired electrons $n = 0$. The spin-only magnetic moment $\mu = \sqrt{n(n+2)} = 0 \ BM$.
During the titration of $KMnO_4$ with oxalic acid in an acidic medium,$Mn^{7+}$ is reduced to $Mn^{2+}$.
The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. The number of unpaired electrons $n = 5$.
The spin-only magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
The difference in magnetic moments is $|5.92 - 0| = 5.92 \ BM$.
The nearest integer is $6$.

(C) $V^{2+}$ acts as a strong reducing agent and liberates hydrogen from a dilute acid. Thus,$(i)$ is correct.
In chemical behavior,the earlier members of the lanthanide series behave more like calcium,not aluminium. Thus,$(ii)$ is incorrect.
The 'silver' $UK$ coins are indeed made of $Cu/Ni$ alloy. Thus,$(iii)$ is correct.
Neptunium $(Np)$ belongs to the actinide series and exhibits a maximum oxidation state of $+7$. Thus,$(iv)$ is correct.
Therefore,statements $(i)$,$(iii)$,and $(iv)$ are correct.

(C) The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons.
For the given $d$-configurations,the number of unpaired electrons are:
$d^1: n=1, \mu = \sqrt{1(3)} = 1.73 \ BM$
$d^3: n=3, \mu = \sqrt{3(5)} = 3.87 \ BM$
$d^4: n=4, \mu = \sqrt{4(6)} = 4.90 \ BM$
$d^5: n=5, \mu = \sqrt{5(7)} = 5.92 \ BM$
Thus,the correct order of increasing magnetic moment is $d^1 < d^3 < d^4 < d^5$.
The sequence given in option $C$ $(d^5, d^3, d^1, d^4)$ is incorrect.

(B) $1$. $a$. Element with highest second ionization enthalpy $(\Delta_{i} H_2)$: $Cu$ $(3d^{10} 4s^1)$ has a stable $d^{10}$ configuration after losing one electron. Removing the second electron from this stable configuration requires very high energy. Thus,$a \rightarrow iii$.
$2$. $b$. Element with highest third ionization enthalpy $(\Delta_{i} H_3)$: $Zn$ $(3d^{10} 4s^2)$ has a stable $d^{10}$ configuration after losing two electrons. Removing the third electron from this stable configuration requires very high energy. Thus,$b \rightarrow iv$.
$3$. $c$. $M$ in $[M(CO)_6]$: According to the $18$-electron rule,$Cr$ $(3d^5 4s^1)$ forms $[Cr(CO)_6]$ where $Cr$ is in $0$ oxidation state,providing $6$ electrons to complete the $18$-electron shell. Thus,$c \rightarrow ii$.
$4$. $d$. Element with highest heat of atomization $(\Delta_{a} H)$: Among the given transition metals,$Ni$ has the highest enthalpy of atomization. Thus,$d \rightarrow v$.
Therefore,the correct match is $a$ $\rightarrow iii, b$ $\rightarrow iv, c$ $\rightarrow ii, d$ $\rightarrow v$.

(A-V, B-II, C-VI, D-IV) Oxidation state of $V$ in $VOCl_2$:
Let oxidation state of $V = x$.
$x + (-2) + 2(-1) = 0$ $\Rightarrow x - 4 = 0$ $\Rightarrow x = +4$.
Thus,$(A)$ matches with $(V)$.
$(B)$ Oxidation state of $Mn$ in $MnO_4^{2-}$:
Let oxidation state of $Mn = x$.
$x + 4(-2) = -2$ $\Rightarrow x - 8 = -2$ $\Rightarrow x = +6$.
Electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1 4s^0$. It has $1$ unpaired electron.
Thus,$(B)$ matches with $(II)$.
$(C)$ Oxidation state of $Ni$ in $[NiCl_4]^{2-}$:
Let oxidation state of $Ni = x$.
$x + 4(-1) = -2 \Rightarrow x = +2$.
Electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8 4s^0$. It has $2$ unpaired electrons.
Thus,$(C)$ matches with $(VI)$.
$(D)$ All lanthanide ions exhibit an oxidation state of $+3$.
Thus,$(D)$ matches with $(IV)$.

(C) The standard electrode potential $(E^{\circ})$ for $M^{3+}/M^{2+}$ follows the order $Mn^{3+}/Mn^{2+} > Fe^{3+}/Fe^{2+} > Cr^{3+}/Cr^{2+}$. Thus,option $A$ is incorrect.
Magnetic moment is calculated as $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons.
For $Mn^{2+}$ $(3d^5)$,$n=5$. For $Cr^{2+}$ $(3d^4)$,$n=4$. For $Fe^{2+}$ $(3d^6)$,$n=4$. The order is $Mn^{2+} > Cr^{2+} \approx Fe^{2+}$. Thus,option $B$ is incorrect.
Oxidizing power depends on the reduction potential and the stability of the lower oxidation state. $MnO_4^-$ $(Mn^{+7})$ is a stronger oxidizing agent than $Cr_2O_7^{2-}$ $(Cr^{+6})$ and $VO_2^+$ $(V^{+5})$. The correct order is $VO_2^+ < Cr_2O_7^{2-} < MnO_4^-$. Thus,option $C$ is correct.
First ionization enthalpy generally increases across the period but shows irregularities due to electronic configuration. The order is $Ti < V < Cr$ is incorrect; the correct order is $Ti < V < Cr$ is not followed due to stability of half-filled $d$-orbitals in $Cr$.

(D) For Statement $(I)$:
The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$. It has $n = 3$ unpaired electrons.
The magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M$.
The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$. It has $n = 3$ unpaired electrons.
Wait,$Cr^{3+}$ is $3d^3$ $(n=3)$,so $\mu = \sqrt{3(5)} = 3.87 \ B.M$. Actually,$Cr^{3+}$ has $3$ unpaired electrons and $Co^{2+}$ has $3$ unpaired electrons. Thus,their magnetic moments are similar. The statement claiming $Co^{2+}$ has a higher magnetic moment is incorrect.
For Statement $(II)$:
The ionisation enthalpies of lanthanoids $(Ce, Pr, Nd)$ are generally higher than those of actinoids $(Th, Pa, U)$ because $4f$ electrons are more effectively shielded than $5f$ electrons,making the valence electrons in actinoids more loosely held.
Therefore,Statement $(I)$ is incorrect and Statement $(II)$ is correct. The correct option is $(D)$.

(C) Statement $I$: Ions with $d^0$ or $d^{10}$ configuration are colourless. $Sc^{3+}$ $(d^0)$,$Ti^{4+}$ $(d^0)$,and $Zn^{2+}$ $(d^{10})$ are colourless.
- $[Sc^{3+}, Ti^{3+}]$: $Sc^{3+}$ is colourless,$Ti^{3+}$ is coloured.
- $[Mn^{2+}, Cr^{2+}]$: Both are coloured ($Mn^{2+}$ is $d^5$,$Cr^{2+}$ is $d^4$).
- $[Cu^{2+}, Zn^{2+}]$: $Cu^{2+}$ is coloured $(d^9)$,$Zn^{2+}$ is colourless.
- $[Ni^{2+}, Ti^{4+}]$: $Ni^{2+}$ is coloured $(d^8)$,$Ti^{4+}$ is colourless.
Only $1$ pair $([Mn^{2+}, Cr^{2+}])$ has both ions coloured. Thus,Statement $I$ is false.
Statement $II$: $Th^{4+}$ is the most stable oxidation state of Thorium and cannot be oxidized further,so it cannot act as a reducing agent. $Eu^{2+}$ is a strong reducing agent as it tends to change to $Eu^{3+}$. Thus,Statement $II$ is false.

(A) Statement $I$: $Ti^{4+}$ $(d^0)$ is colourless due to the absence of unpaired electrons. $V^{2+}$ $(d^3)$,$Mn^{2+}$ $(d^5)$,$Fe^{3+}$ $(d^5)$,and $Cr^{2+}$ $(d^4)$ all have unpaired electrons and are coloured.
The pairs where both ions are coloured are: $[V^{2+}, Mn^{2+}]$,$[Mn^{2+}, Fe^{3+}]$,and $[V^{2+}, Cr^{2+}]$.
Thus,the count is $3$. Statement $I$ is correct.
Statement $II$: $La^{3+}$ $(f^0)$,$Lu^{3+}$ $(f^{14})$,$Ac^{3+}$ $(f^0)$,and $Lr^{3+}$ $(f^{14})$ are diamagnetic. $Yb^{2+}$ $(f^{14})$ and $Ce^{4+}$ $(f^0)$ are also diamagnetic.
The pairs where both ions are diamagnetic are: $[La^{3+}, Yb^{2+}]$,$[Lu^{3+}, Ce^{4+}]$,and $[Ac^{3+}, Lr^{3+}]$.
Thus,the count is $3$. Statement $II$ is correct.