Mix Examples-Co-ordination Chemistry Questions in English

(D) When excess of $AgNO_3$ and $BaCl_2$ are added to solution $X$:
$1.$ Reaction with $AgNO_3$:
$[Co(NH_3)_5Br]Cl_2 + 2AgNO_3 \rightarrow [Co(NH_3)_5Br](NO_3)_2 + 2AgCl(s) (Y)$
Since $1 \, \text{mole}$ of $[Co(NH_3)_5Br]Cl_2$ provides $2 \, \text{moles}$ of $Cl^-$ ions,$0.02 \, \text{mole}$ of the complex yields $0.02 \times 2 = 0.04 \, \text{mole}$ of $AgCl$ precipitate $(Y)$.
$2.$ Reaction with $BaCl_2$:
$[Co(NH_3)_5Cl]SO_4 + BaCl_2 \rightarrow [Co(NH_3)_5Cl]Cl_2 + BaSO_4(s) (Z)$
Since $1 \, \text{mole}$ of $[Co(NH_3)_5Cl]SO_4$ provides $1 \, \text{mole}$ of $SO_4^{2-}$ ions,$0.02 \, \text{mole}$ of the complex yields $0.02 \, \text{mole}$ of $BaSO_4$ precipitate $(Z)$.
Therefore,the number of moles of $Y$ and $Z$ are $0.04$ and $0.02$ respectively.

(D) Statement $(a)$ is incorrect: $Co(III)$ is a $d^6$ ion. With strong field ligands,$\Delta_0 > P$,leading to a low-spin configuration $t_{2g}^6 e_g^0$,which is diamagnetic (magnetic moment = $0$).
Statement $(b)$ is correct: When $\Delta_0 < P$ (weak field),the complex is high-spin,and the $d^6$ configuration is $t_{2g}^4 e_g^2$.
Statement $(c)$ is correct: $en$ is a stronger field ligand than $F^-$,so $\Delta_0$ for $[Co(en)_3]^{3+}$ is larger than for $[CoF_6]^{3-}$. Since $E = hc/\lambda$,higher energy absorption corresponds to a lower wavelength.
Statement $(d)$ is incorrect: The relationship between tetrahedral and octahedral splitting is $\Delta_t = \frac{4}{9} \Delta_0$. Given $\Delta_0 = 18,000 \ cm^{-1}$,$\Delta_t = \frac{4}{9} \times 18,000 = 8,000 \ cm^{-1}$,not $16,000 \ cm^{-1}$.
Therefore,statements $(a)$ and $(d)$ are incorrect.

(C) The chemical formula of the complex is $[Co(NH_{3})_{6}]Cl_{3}$.
Each mole of $[Co(NH_{3})_{6}]Cl_{3}$ contains $3$ moles of ionizable $Cl^{-}$ ions.
The reaction with $AgNO_{3}$ is: $[Co(NH_{3})_{6}]Cl_{3} + 3AgNO_{3} \rightarrow [Co(NH_{3})_{6}](NO_{3})_{3} + 3AgCl$.
Number of moles of $[Co(NH_{3})_{6}]Cl_{3} = \frac{0.3 \; g}{267.46 \; g/mol} \approx 0.0011216 \; mol$.
Since $1$ mole of complex reacts with $3$ moles of $AgNO_{3}$,moles of $AgNO_{3}$ required $= 3 \times 0.0011216 = 0.0033648 \; mol$.
Using the molarity formula $M = \frac{n}{V(L)}$,we have $V(L) = \frac{n}{M} = \frac{0.0033648}{0.125} = 0.0269184 \; L$.
Converting to $mL$,$V = 0.0269184 \times 1000 = 26.92 \; mL$.

(C) In a chromate ion $(CrO_4^{2-})$,the structure consists of one $Cr$ atom bonded to four oxygen atoms. There are two double bonds and two single bonds,totaling $6$ $Cr-O$ bonds $(4\sigma + 2\pi)$.
In a dichromate ion $(Cr_2O_7^{2-})$,there are two $Cr$ atoms linked by one bridging oxygen atom $(Cr-O-Cr)$. Each $Cr$ atom is also bonded to three terminal oxygen atoms. This results in a total of $12$ $Cr-O$ bonds $(8\sigma + 4\pi)$.
The sum of the total number of $Cr-O$ bonds is $6 + 12 = 18$.

(N/A) $(i)$ $K_{3}[Co(C_{2}O_{4})_{3}]$: Central metal is $Co$. Coordination number is $6$. Oxidation state: $x + 3(-2) = -3 \implies x = +3$. $Co^{3+}$ $(d^{6})$ is $t_{2g}^{6} e_{g}^{0}$.
$(ii)$ $cis-[Cr(en)_{2}Cl_{2}]Cl$: Central metal is $Cr$. Coordination number is $6$. Oxidation state: $x + 2(0) + 2(-1) = +1 \implies x = +3$. $Cr^{3+}$ $(d^{3})$ is $t_{2g}^{3} e_{g}^{0}$.
$(iii)$ $(NH_{4})_{2}[CoF_{4}]$: Central metal is $Co$. Coordination number is $4$. Oxidation state: $x + 4(-1) = -2 \implies x = +2$. $Co^{2+}$ $(d^{7})$ is $t_{2g}^{5} e_{g}^{2}$.
$(iv)$ $[Mn(H_{2}O)_{6}]SO_{4}$: Central metal is $Mn$. Coordination number is $6$. Oxidation state: $x + 6(0) = +2 \implies x = +2$. $Mn^{2+}$ $(d^{5})$ is $t_{2g}^{3} e_{g}^{2}$.

(N/A) $(i)$ Potassium diaquadioxalatochromate$(III)$ trihydrate.
Oxidation state of $Cr = +3$.
Electronic configuration: $3d^3 (t_{2g}^3)$.
Coordination number $= 6$.
Shape: Octahedral. Stereochemistry: Shows cis-trans isomerism.
Magnetic moment,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$(ii)$ Pentaamminechloridocobalt$(III)$ chloride.
Oxidation state of $Co = +3$.
Electronic configuration: $3d^6 (t_{2g}^6)$.
Coordination number $= 6$.
Shape: Octahedral. Stereochemistry: Shows geometrical isomerism.
Magnetic moment,$\mu = 0 \ BM$ (diamagnetic).
$(iii)$ Trichloridotripyridinechromium$(III)$.
Oxidation state of $Cr = +3$.
Electronic configuration: $3d^3 (t_{2g}^3)$.
Coordination number $= 6$.
Shape: Octahedral. Stereochemistry: Shows facial $(fac)$ and meridional $(mer)$ isomerism.
Magnetic moment,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$(iv)$ Caesium tetrachloroferrate$(III)$.
Oxidation state of $Fe = +3$.
Electronic configuration: $3d^5 (e_g^2 t_{2g}^3)$.
Coordination number $= 4$.
Shape: Tetrahedral. Stereochemistry: Optically inactive.
Magnetic moment,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$(v)$ Potassium hexacyanomanganate$(II)$.
Oxidation state of $Mn = +2$.
Electronic configuration: $3d^5 (t_{2g}^5)$.
Coordination number $= 6$.
Shape: Octahedral. Stereochemistry: Optically inactive.
Magnetic moment,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.

(C) The structure of $CuSO_{4} \cdot 5H_{2}O$ is $[Cu(H_{2}O)_{4}]SO_{4} \cdot H_{2}O$.
In this complex,the $Cu(II)$ ion is coordinated by four water molecules through oxygen atoms,forming $Cu-O$ coordinate bonds. Thus,Statement $I$ is correct.
Regarding Statement $II$,the ligands coordinating with the $Cu(II)$ ion are only water molecules,which are $O$-based ligands. The sulfate ion $(SO_{4}^{2-})$ is not directly coordinated to the $Cu(II)$ ion; it is linked via hydrogen bonding. Therefore,Statement $II$ is incorrect.

(B) The initial moles of $[Co(NH_3)_5SO_4]Br$ are $1\,L \times 0.02\,M = 0.02\,mol$.
The initial moles of $[Co(NH_3)_5Br]SO_4$ are $1\,L \times 0.02\,M = 0.02\,mol$.
The total volume of the mixture is $2\,L$.
When the mixture is divided into two equal parts $(X)$,each part has a volume of $1\,L$.
In $1\,L$ of solution $(X)$,the moles of each complex are halved: $0.01\,mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.01\,mol$ of $[Co(NH_3)_5Br]SO_4$.
The complex $[Co(NH_3)_5SO_4]Br$ dissociates to give $Br^-$ ions,and $[Co(NH_3)_5Br]SO_4$ dissociates to give $SO_4^{2-}$ ions.
Moles of $Br^-$ in $1\,L$ of $(X) = 0.01\,mol$.
Moles of $SO_4^{2-}$ in $1\,L$ of $(X) = 0.01\,mol$.
Reaction with $AgNO_3$: $Br^- + AgNO_3 \rightarrow AgBr(Y) + NO_3^-$. Moles of $Y (AgBr) = 0.01\,mol$.
Reaction with $BaCl_2$: $SO_4^{2-} + BaCl_2 \rightarrow BaSO_4(Z) + 2Cl^-$. Moles of $Z (BaSO_4) = 0.01\,mol$.

(C) To determine the shape,we look at the hybridization and lone pairs of the central atom:
$1$. $XeF_4$: $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar geometry.
$2$. $SF_4$: $sp^3d$ hybridization with $1$ lone pair,resulting in a see-saw geometry.
$3$. $SiF_4$: $sp^3$ hybridization with $0$ lone pairs,resulting in a tetrahedral geometry.
$4$. $BF_4^{-}$: $sp^3$ hybridization with $0$ lone pairs,resulting in a tetrahedral geometry.
$5$. $BrF_4^{-}$: $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar geometry.
$6$. $[Cu(NH_3)_4]^{2+}$: $dsp^2$ hybridization,resulting in a square planar geometry.
$7$. $[FeCl_4]^{2-}$: $sp^3$ hybridization,resulting in a tetrahedral geometry.
$8$. $[PtCl_4]^{2-}$: $dsp^2$ hybridization,resulting in a square planar geometry.
Thus,the species with a square planar shape are $XeF_4$,$BrF_4^{-}$,$[Cu(NH_3)_4]^{2+}$,and $[PtCl_4]^{2-}$.
The total count is $4$.

(D) . Incorrect. Crystal field theory explains the splitting of $d$-orbitals but does not explain the relative strength of ligands (spectrochemical series).
$B$. Correct. Valence bond theory $(VBT)$ provides a qualitative description of bonding but fails to provide a quantitative interpretation of kinetic stability.
$C$. Correct. In $\left[Ni(CN)_4\right]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of electrons,leading to $dsp^2$ hybridization.
$D$. Incorrect. The complex $cis-\left[PtCl_2(en)_2\right]^{2+}$ is a specific isomer. The question asks for the number of possible isomers of this specific configuration,which is just one (the $cis$ form itself).
Therefore,statements $B$ and $C$ are correct.

(A) Statement $I$ is false. When $Ni^{2+}$ reacts with dimethyl glyoxime $(dmg)$ in the presence of $NH_4OH$,it forms a complex $[Ni(dmg)_2]$. In this complex,the chelate rings formed are five-membered,not six-membered.
Statement $II$ is true. The chemical formula for Prussian blue is $Fe_4[Fe(CN)_6]_3$. In this compound,the iron outside the coordination sphere is in the $(+3)$ oxidation state $(Fe^{3+})$,and the iron inside the coordination sphere is in the $(+2)$ oxidation state $([Fe(CN)_6]^{4-})$.

(C) Hybridisation analysis:
$PF_5$: The central atom $P$ has $5$ bond pairs and $0$ lone pairs,resulting in $sp^3 d$ hybridisation.
$BrF_5$: The central atom $Br$ has $5$ bond pairs and $1$ lone pair,resulting in $sp^3 d^2$ hybridisation.
$SF_6$: The central atom $S$ has $6$ bond pairs and $0$ lone pairs,resulting in $sp^3 d^2$ hybridisation.
$[Co(NH_3)_6]^{3+}$: The central metal ion $Co^{3+}$ ($d^6$ configuration) with a strong field ligand $NH_3$ undergoes inner orbital hybridisation,resulting in $d^2 sp^3$ hybridisation.
Conclusion:
Statement $I$ is false because $BrF_5$ is $sp^3 d^2$.
Statement $II$ is false because $[Co(NH_3)_6]^{3+}$ is $d^2 sp^3$.
Therefore,both statements are false.

(C) is $K_2[Ni(CN)_4]$ and $B$ is $K_2[NiCl_4]$.
$1.$ The $IUPAC$ name of $K_2[Ni(CN)_4]$ is Potassium tetracyanonickelate $(II)$ and $K_2[NiCl_4]$ is Potassium tetrachloronickelate $(II)$. Thus,option $(A)$ is correct.
$2.$ In $A$,$CN^-$ is a strong field ligand,so $Ni^{2+}$ $(3d^8)$ undergoes $dsp^2$ hybridization,making it diamagnetic. In $B$,$Cl^-$ is a weak field ligand,so $Ni^{2+}$ $(3d^8)$ undergoes $sp^3$ hybridization,leaving two unpaired electrons,making it paramagnetic. Thus,option $(C)$ is correct.
$3.$ As determined,$A$ has $dsp^2$ hybridization and $B$ has $sp^3$ hybridization. Thus,option $(A)$ is correct.
The sequence of answers is $A, C, A$.

(A) . $[Co(NH_3)_4(H_2O)_2]Cl_2$: $Co$ is in $+2$ oxidation state ($d^7$ configuration,paramagnetic). It shows geometrical isomerism (cis and trans forms). Matches: $p, q, s$.
$B$. $[Pt(NH_3)_2Cl_2]$: $Pt$ is in $+2$ oxidation state ($5d^8$ configuration,diamagnetic). It shows geometrical isomerism (cis and trans forms). Matches: $p, r, s$.
$C$. $[Co(H_2O)_5Cl]Cl$: $Co$ is in $+2$ oxidation state ($d^7$ configuration,paramagnetic). It does not show geometrical isomerism. Matches: $q, s$.
$D$. $[Ni(H_2O)_6]Cl_2$: $Ni$ is in $+2$ oxidation state ($3d^8$ configuration,paramagnetic). It does not show geometrical isomerism. Matches: $q, s$.
Therefore,the correct matching is $A$ $\rightarrow p, q, s; B$ $\rightarrow p, r, s; C$ $\rightarrow q, s; D$ $\rightarrow q, s$.

(A) $1.$ When $Cu$ rod $(M)$ is dipped in $AgNO_3$ $(N)$,$Cu$ displaces $Ag^+$ ions: $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$. The formation of $Cu^{2+}$ turns the solution light blue.
$2.$ Addition of $NaCl$ to the solution results in the precipitation of $AgCl$ $(O)$: $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$ (white precipitate).
$3.$ Addition of excess $NH_3$ dissolves the $AgCl$ precipitate by forming a soluble complex: $AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) + Cl^-(aq)$. Simultaneously,$Cu^{2+}$ ions react with $NH_3$ to form an intense blue complex: $Cu^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}(aq)$.
Thus,$M = Cu$,$N = AgNO_3$,and the final solution contains $[Ag(NH_3)_2]^+$ and $[Cu(NH_3)_4]^{2+}$. The correct option is $(A)$.

(A) The complex $[Cr(H_2O)_5Cl]Cl_2$ dissociates in water to release $2$ moles of ionizable $Cl^-$ ions per mole of the complex.
Number of millimoles of $[Cr(H_2O)_5Cl]Cl_2 = 30 \ mL \times 0.01 \ M = 0.3 \ mmol$.
Since each mole of complex provides $2$ moles of $Cl^-$, the total millimoles of $Cl^- = 0.3 \times 2 = 0.6 \ mmol$.
For complete precipitation, $Ag^+$ reacts with $Cl^-$ in a $1:1$ molar ratio $(Ag^+ + Cl^- \rightarrow AgCl(s))$.
Therefore, millimoles of $AgNO_3$ required $= 0.6 \ mmol$.
Using the formula $M = \frac{n}{V(in \ L)}$ or $M = \frac{mmol}{V(in \ mL)}$, we have $0.1 = \frac{0.6}{V}$.
$V = \frac{0.6}{0.1} = 6 \ mL$.

(B) The reaction for the preparation of ammonia is: $(NH_4)_2SO_4 + Ca(OH)_2 \longrightarrow CaSO_4 \cdot 2H_2O + 2NH_3$.
For $1584 \ g$ of $(NH_4)_2SO_4$ $(M = 132 \ g \ mol^{-1})$,the number of moles $n = 1584 / 132 = 12 \ mol$.
This produces $12 \ mol$ of gypsum $(CaSO_4 \cdot 2H_2O)$ and $24 \ mol$ of $NH_3$.
Mass of gypsum $(M = 172 \ g \ mol^{-1})$ $= 12 \times 172 = 2064 \ g$.
The reaction with nickel chloride is: $NiCl_2 \cdot 6H_2O + 6NH_3 \longrightarrow [Ni(NH_3)_6]Cl_2 + 6H_2O$.
For $952 \ g$ of $NiCl_2 \cdot 6H_2O$ $(M = 238 \ g \ mol^{-1})$,the number of moles $n = 952 / 238 = 4 \ mol$.
This reacts with $24 \ mol$ of $NH_3$ to produce $4 \ mol$ of $[Ni(NH_3)_6]Cl_2$.
Mass of $[Ni(NH_3)_6]Cl_2$ $(M = 232 \ g \ mol^{-1})$ $= 4 \times 232 = 928 \ g$.
Total mass $= 2064 \ g + 928 \ g = 2992 \ g$.

(A) $1$. The complex $[Co(en)(NH_3)_3(H_2O)]^{3+}$ has two geometrical isomers,facial $(fac)$ and meridional $(mer)$. Thus,statement $A$ is correct.
$2$. Replacing 'en' with two cyanide ligands $(CN^-)$ gives $[Co(NH_3)_3(H_2O)(CN)_2]^+$. This complex has three geometrical isomers as shown in the provided image. Thus,statement $B$ is correct.
$3$. $Co^{3+}$ has a $d^6$ configuration. Since 'en' and $NH_3$ are strong field ligands,the complex is low-spin and diamagnetic. Thus,statement $C$ is incorrect.
$4$. The crystal field splitting energy $\Delta_0$ is inversely proportional to the wavelength of absorbed light $(\Delta_0 = \frac{hc}{\lambda})$. Since $H_2O$ is a weaker field ligand than $NH_3$,the complex $[Co(en)(NH_3)_3(H_2O)]^{3+}$ has a smaller $\Delta_0$ than $[Co(en)(NH_3)_4]^{3+}$. Therefore,it absorbs light at a longer wavelength. Thus,statement $D$ is correct.

(A) In $[FeCl_4]^-$,$Fe^{3+}$ has the electronic configuration $[Ar] 3d^5$. Since $Cl^-$ is a weak field ligand,it forms a high-spin tetrahedral complex with $sp^3$ hybridization.
$(C)$ $[FeCl_4]^-$ has $5$ unpaired electrons,so its spin-only magnetic moment is $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$. In $[Co(en)(NH_3)_2Cl_2]^+$,$Co^{3+}$ has the configuration $[Ar] 3d^6$. Due to the presence of strong field ligands ($en$ and $NH_3$),the electrons pair up,resulting in $0$ unpaired electrons $(n=0, \mu=0)$. Therefore,$[FeCl_4]^-$ has a higher magnetic moment.
$(B)$ is incorrect because $[Co(en)(NH_3)_2Cl_2]^+$ has $3$ geometrical isomers (trans,cis-cis,and cis-trans).
$(D)$ is incorrect because the hybridization of $Co^{3+}$ in this low-spin octahedral complex is $d^2sp^3$.

(A) $1$. The reaction of $K_4[Fe(CN)_6]$ with $FeSO_4$ in the absence of air produces potassium ferrous ferrocyanide,$K_2Fe[Fe(CN)_6]$,which is a white precipitate $(X)$.
$2$. The brown ring test for nitrates involves the reaction of $Fe^{2+}$ ions with $NO_3^-$ in the presence of $H_2SO_4$ to form the nitrosonium complex $[Fe(H_2O)_5(NO)]SO_4$.
$3$. Therefore,$X$ is $K_2Fe[Fe(CN)_6]$ (Option $C$) and the brown ring complex is $[Fe(H_2O)_5(NO)]SO_4$ (Option $D$).

(A) The reaction is: $2KI + 2K_3[Fe(CN)_6] \longrightarrow I_2 + 2K_4[Fe(CN)_6]$.
$(A)$ This is a redox reaction where $I^-$ is oxidized to $I_2$ and $Fe^{3+}$ is reduced to $Fe^{2+}$. Thus,$(A)$ is true.
$(B)$ The white precipitate formed with $ZnSO_4$ is $K_2Zn_3[Fe(CN)_6]_2$. Thus,$(B)$ is true.
$(C)$ The filtrate contains $I_2$ (or $I_3^-$),which reacts with starch to give a blue colour. Thus,$(C)$ is true.
$(D)$ The white precipitate $K_2Zn_3[Fe(CN)_6]_2$ is soluble in $NaOH$ solution due to the formation of soluble zincate complex $[Zn(OH)_4]^{2-}$. Thus,$(D)$ is true.
All statements $(A), (B), (C),$ and $(D)$ are true. Since the options provided are combinations,the most appropriate choice based on the provided options is $(A), (B), (C),$ and $(D)$ which corresponds to $(ABD)$ or $(BCD)$ depending on the intended set. Given the options,$(ABD)$ is a valid selection.

(A) $(P) [Cr(NH_3)_4Cl_2]Cl$: $Cr^{3+}$ is $d^3$ (paramagnetic). It shows cis-trans isomerism. Matches with $3$.
$(Q) [Ti(H_2O)_5Cl](NO_3)_2$: $Ti^{3+}$ is $d^1$ (paramagnetic). It shows ionisation isomerism. Matches with $1$.
$(R) [Pt(en)(NH_3)_2Cl_2]NO_3$: $Pt^{2+}$ is $d^8$ (diamagnetic,square planar). It shows ionisation isomerism. Matches with $4$.
$(S) [Co(NH_3)_4(NO_3)_2]NO_3$: $Co^{3+}$ is $d^6$ (diamagnetic,strong field ligands). It shows cis-trans isomerism. Matches with $2$.
Therefore,the correct sequence is $P-3, Q-1, R-4, S-2$.

(A) $1$. $[Cr(CN)_6]^{4-}$: $Cr^{2+}$ is $3d^4$. $CN^-$ is a strong field ligand,so it forms a low spin complex. $t_{2g}^4 e_g^0$. Properties: $P, R, T$.
$2$. $[RuCl_6]^{2-}$: $Ru^{4+}$ is $4d^4$. $4d$ series metals form low spin complexes. $t_{2g}^4 e_g^0$. Properties: $P, R, S, T$.
$3$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $3d^4$. $H_2O$ is a weak field ligand,so it forms a high spin complex. $t_{2g}^3 e_g^1$. Properties: $Q, T$.
$4$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand,so it forms a high spin complex. $t_{2g}^4 e_g^2$. Properties: $P, Q$.
Matching: $I \rightarrow P, R, T$; $II \rightarrow P, R, S, T$; $III \rightarrow Q, T$; $IV \rightarrow P, Q$. The option matching the closest set is $A$.

(A) The relationship between entropy change and cell potential is given by $\Delta S = nF \left( \frac{dE_{cell}}{dT} \right)$.
For option $(A)$: Given $\frac{dE_{cell}}{dT} = \frac{R}{F}$ and for the reaction $M_{(s)} + 2H^{+}_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)}$, $n = 2$. Thus, $\Delta S = 2 \times F \times \frac{R}{F} = 2R$. Since $2R \neq R$, $(A)$ is incorrect.
For option $(B)$: This is a concentration cell. $\Delta H = 0$ for concentration cells. Since the reaction is spontaneous $(E_{cell} > 0)$, $\Delta G < 0$. From $\Delta G = \Delta H - T \Delta S$, we get $\Delta S > 0$. Thus, it is an entropy-driven process. $(B)$ is correct.
For option $(C)$: Racemization involves the conversion of an optically active compound into a racemic mixture. This increases the disorder of the system, so $\Delta S > 0$. $(C)$ is correct.
For option $(D)$: The reaction $[Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O$ involves the replacement of $6$ monodentate ligands with $3$ bidentate ligands, increasing the number of free molecules in the solution ($4$ particles on the left to $7$ particles on the right). This increases entropy, so $\Delta S > 0$. $(D)$ is correct.

(D) The reaction between potassium iodide and potassium ferricyanide is:
$2K_3[Fe(CN)_6] + 2KI \rightleftharpoons 2K_4[Fe(CN)_6] + I_2$
Here,$P$ is $K_4[Fe(CN)_6]$.
$(1)$ From the stoichiometry,$2$ moles of $KI$ are required to produce $2$ moles of $K_4[Fe(CN)_6]$ $(P)$. Thus,the number of moles of $KI$ is $2$.
$(2)$ The reaction of $K_4[Fe(CN)_6]$ with $ZnCl_2$ forms the sparingly soluble complex $K_2Zn_3[Fe(CN)_6]_2$. The number of zinc ions $(Zn^{2+})$ in this formula is $3$.

(A) Paramagnetic species contain at least one unpaired electron. Let us analyze each species:
$1. O_2$: Has $2$ unpaired electrons in $\pi^* 2p$ orbitals. (Paramagnetic)
$2. O_2^{+}$: Has $1$ unpaired electron. (Paramagnetic)
$3. O_2^{-}$: Has $1$ unpaired electron. (Paramagnetic)
$4. NO$: Has $15$ electrons (odd electron species). (Paramagnetic)
$5. NO_2$: Has $23$ electrons (odd electron species). (Paramagnetic)
$6. CO$: Has $14$ electrons,all paired. (Diamagnetic)
$7. K_2[NiCl_4]$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,so it forms a tetrahedral complex with $2$ unpaired electrons. (Paramagnetic)
$8. [Co(NH_3)_6]Cl_3$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,so it forms an octahedral complex with all electrons paired. (Diamagnetic)
$9. K_2[Ni(CN)_4]$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,so it forms a square planar complex with all electrons paired. (Diamagnetic)
The paramagnetic species are $O_2, O_2^{+}, O_2^{-}, NO, NO_2, K_2[NiCl_4]$. The total count is $6$.

(D) The reaction between sodium nitroprusside,$Na_2[Fe(CN)_5NO]$,and sulphide ions,$S^{2-}$,is given by:
$Na_2[Fe(CN)_5NO] + Na_2S \rightarrow Na_4[Fe(CN)_5NOS]$
In the reactant,sodium nitroprusside,the oxidation state of $Fe$ is $+2$ (since $CN^-$ is $-1$ and $NO^+$ is $+1$).
In the product,sodium thionitroprusside,$Na_4[Fe(CN)_5NOS]$,the oxidation state of $Fe$ remains $+2$ (since $CN^-$ is $-1$ and $NOS^{3-}$ is $-3$).
Thus,the oxidation state of iron does not change during the reaction.

(B) Correct: Linkage isomerism occurs when an ambidentate ligand like $NCS^-$ or $SCN^-$ is present. Here, $N$ and $S$ are donor atoms, so it is correct.
$(b)$ Correct: In $K_4[Ni(CN)_4]$, $4(+1) + x + 4(-1) = 0$, so $x = 0$.
$(c)$ Incorrect: In $K_3[Cu(CN)_4]$, $Cu$ is in $+1$ oxidation state ($3d^{10}$ configuration). It forms $sp^3$ hybridization, tetrahedral shape, and is diamagnetic $(\mu = 0 \ B.M.)$.
$(d)$ Incorrect: $[Cr(NCS)(NH_3)_5][ZnCl_4]$ exhibits both linkage isomerism and coordination isomerism (due to the presence of two coordination entities).

(C) The correct matching is $A-IV, B-III, C-I, D-II$.
$[Ni(H_2O)_6]^{2+}$ is green $(I)$.
As $H_2O$ ligands are replaced by $en$ (a stronger field ligand), the crystal field splitting energy $(\Delta_o)$ increases.
The wavelength of light absorbed decreases, which shifts the observed color towards the violet end of the spectrum.
The sequence of colors as $en$ replaces $H_2O$ is: $[Ni(H_2O)_6]^{2+}$ (Green) $\rightarrow$ $[Ni(H_2O)_4(en)]^{2+}$ (Pale blue) $\rightarrow$ $[Ni(H_2O)_2(en)_2]^{2+}$ (Blue) $\rightarrow$ $[Ni(en)_3]^{2+}$ (Violet).

(C) Let us analyze each molecule:
$1$. $PCl_5$: The central atom $P$ has $5$ bond pairs and $0$ lone pairs. Hybridisation is $sp^3d$ and the shape is trigonal bipyramidal.
$2$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ has $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. Hybridisation is $dsp^2$ and the shape is square planar.
$3$. $SF_6$: The central atom $S$ has $6$ bond pairs and $0$ lone pairs. Hybridisation is $sp^3d^2$ and the shape is octahedral. This is the correct set.
$4$. $IF_3$: The central atom $I$ has $3$ bond pairs and $2$ lone pairs. Hybridisation is $sp^3d$ and the shape is bent $T$-shaped.

(A) The reaction of $[Ni(H_2O)_6]^{2+}$ with $en$ proceeds in steps by replacing water ligands:
$1$. For $1:1$ ratio: $[Ni(H_2O)_6]^{2+} + en \rightarrow [Ni(H_2O)_4(en)]^{2+} + 2H_2O$. The product is pale blue $(I)$.
$2$. For $1:2$ ratio: $[Ni(H_2O)_6]^{2+} + 2en \rightarrow [Ni(H_2O)_2(en)_2]^{2+} + 4H_2O$. The product is blue/purple $(II)$.
$3$. For $1:3$ ratio: $[Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O$. The product is violet $(III)$.
Thus,the correct match is $A-I, B-II, C-III$.

(C) The reaction of $FeCl_3$ with $NH_3$ produces $Fe(OH)_3$ which is a brown precipitate. Thus,$(A)$ matches with $(III)$.
The reaction of $AgCl$ with $NH_3$ produces $[Ag(NH_3)_2]Cl$ which is a colourless solution. Thus,$(B)$ matches with $(IV)$.
Therefore,the correct sequence is $(A-III, B-IV)$.

(B) The reaction $4FeCl_3 + 3K_4[Fe(CN)_6] \rightarrow Fe_4[Fe(CN)_6]_3 + 12KCl$ produces Prussian blue,which is $Fe_4[Fe(CN)_6]_3$ $(I)$.
Zinc chloride reacts with excess sodium hydroxide to form sodium zincate: $ZnCl_2 + 4NaOH \rightarrow Na_2ZnO_2 + 2NaCl + 2H_2O$ $(V)$.
Ferric chloride is reduced by hydrogen sulfide to ferrous chloride: $2FeCl_3 + H_2S \rightarrow 2FeCl_2 + 2HCl + S$ $(II)$.
Thus,the correct matching is $A-I, B-V, C-II$.

(A, B, C) The reaction of $Ni^{2+}$ with dimethylglyoxime $(DMG)$ in an ammoniacal (basic) medium produces a cherry-red precipitate of nickel$(II)$ dimethylglyoximate,$[Ni(DMG)_2]$.
In this complex,two dimethylglyoximate units are bound to one $Ni^{2+}$ ion.
The two dimethylglyoximate units are held together by strong intramolecular hydrogen bonding.
Each dimethylglyoximate unit forms a five-membered chelate ring with the $Ni^{2+}$ ion through the nitrogen atoms of the oxime groups.
Therefore,statements $A$,$B$,and $C$ are correct,while $D$ is incorrect because the chelate ring formed is five-membered,not six-membered.

(A) The $4 \ L$ solution contains $0.4 \ mol$ of $[Co(NH_{3})_{5}SO_{4}]Br$ and $0.4 \ mol$ of $[Co(NH_{3})_{5}Br]SO_{4}$.
In $2 \ L$ of the mixture,there are $0.2 \ mol$ of each complex.
When $2 \ L$ of the mixture reacts with excess $AgNO_{3}$,only $[Co(NH_{3})_{5}SO_{4}]Br$ reacts to form $AgBr$ precipitate $(Y)$. Thus,$0.2 \ mol$ of $AgBr$ is formed.
When $2 \ L$ of the mixture reacts with excess $BaCl_{2}$,only $[Co(NH_{3})_{5}Br]SO_{4}$ reacts to form $BaSO_{4}$ precipitate $(Z)$. Thus,$0.2 \ mol$ of $BaSO_{4}$ is formed.

(C) In $K_{3}[Co(CO_{3})_{3}]$,$Co$ is in $+3$ oxidation state ($d^{6}$ configuration). $CO_{3}^{2-}$ is a weak field ligand,so it forms an outer orbital complex: $sp^{3}d^{2}$ hybridisation,octahedral geometry,and $4$ unpaired electrons. Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{4(6)} = \sqrt{24} \approx 4.9 \ BM$. Statement-$I$ is true.
For $[Ni(CN)_{4}]^{2-}$,$Ni^{2+}$ is $d^{8}$. $CN^{-}$ is a strong field ligand,so $dsp^{2}$ hybridisation,square planar,$0$ unpaired electrons,$\mu = 0 \ BM$.
For $[MnBr_{4}]^{2-}$,$Mn^{2+}$ is $d^{5}$. $Br^{-}$ is a weak field ligand,so $sp^{3}$ hybridisation,tetrahedral,$5$ unpaired electrons,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.9 \ BM$.
For $[CoF_{6}]^{3-}$,$Co^{3+}$ is $d^{6}$. $F^{-}$ is a weak field ligand,so $sp^{3}d^{2}$ hybridisation,octahedral,$4$ unpaired electrons,$\mu = 4.9 \ BM$.
All values in Statement-$II$ are correct. Thus,both statements are true.

(A) For $[Mn(H_2O)_6]^{2+}$,the configuration is $d^5$ with a weak field ligand $(H_2O)$,resulting in $t_{2g}^3 e_g^2$. The $CFSE$ is $(3 \times -0.4 \Delta_o) + (2 \times 0.6 \Delta_o) = 0 \Delta_o$.
For $[Cr(H_2O)_6]^{2+}$,the configuration is $d^4$ with a weak field ligand $(H_2O)$,resulting in $t_{2g}^3 e_g^1$. The $CFSE$ is $(3 \times -0.4 \Delta_o) + (1 \times 0.6 \Delta_o) = -0.6 \Delta_o$. Since $|-0.6| > 0$,Statement $I$ is true.
For potassium ferricyanide,$K_3[Fe(CN)_6]$,the iron is in $+3$ oxidation state $(d^5)$. $CN^-$ is a strong field ligand,leading to $t_{2g}^5 e_g^0$ with $1$ unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$
For sodium ferrocyanide,$Na_4[Fe(CN)_6]$,the iron is in $+2$ oxidation state $(d^6)$. $CN^-$ is a strong field ligand,leading to $t_{2g}^6 e_g^0$ with $0$ unpaired electrons. $\mu = 0 \ B.M.$
Since $\sqrt{3} > 0$,Statement $II$ is true.