Mix Examples-Co-ordination Chemistry Questions in English

(D) All of the above.
$1.$ $4FeCl_3 + 3K_4[Fe(CN)_6] \to Fe_4[Fe(CN)_6]_3$ (Ferri ferrocyanide,Prussian Blue precipitate).
$2.$ $2FeCl_3 + 3H_2S \to 2FeS + S + 6HCl$ (Formation of sulfur leads to a milky appearance).
$3.$ $FeCl_3 + 3NH_4CNS \to Fe(CNS)_3$ (Blood red color complex) $+ 3NH_4Cl$.

(C) The compound $Fe_4[Fe(CN)_6]_3$ is known as Prussian blue or Turnbull's blue depending on the oxidation states,but specifically,the reaction of $Fe^{3+}$ ions with potassium ferrocyanide $K_4[Fe(CN)_6]$ produces a deep blue precipitate. However,in certain conditions or specific complex formations involving iron and cyanide,brown-coloured complexes can be formed. Among the given options,$Fe_4[Fe(CN)_6]_3$ is the standard ferric ferrocyanide complex. If the question refers to a brown-coloured compound in the context of iron-cyanide chemistry,it is often associated with specific oxidation states or impurities,but $Fe_4[Fe(CN)_6]_3$ is the most chemically significant compound listed.

(B) The reaction between $CuSO_4$ and $K_4[Fe(CN)_6]$ produces a chocolate brown precipitate of $Cu_2[Fe(CN)_6]$,not a blue colouration.
$2CuSO_4 + K_4[Fe(CN)_6] \to Cu_2[Fe(CN)_6] + 2K_2SO_4$ (Chocolate brown).
In contrast,$4NH_4OH + CuSO_4 \to [Cu(NH_3)_4]SO_4 + 4H_2O$ produces a deep blue solution.
Anhydrous $CuSO_4 + 5H_2O \to CuSO_4 \cdot 5H_2O$ produces a blue colour.
$4FeCl_3 + 3Na_4[Fe(CN)_6] \to Fe_4[Fe(CN)_6]_3 + 12NaCl$ produces a Prussian blue precipitate.

(D) Potassium cobaltinitrite,potassium hexanitrocobaltate $(III)$,and Fischer's salt are different names for the same coordination compound,$K_3[Co(NO_2)_6]$.
This complex is known for its characteristic yellow colour.

(D) $1$. $[Cu(NH_3)_6]^{2+}$ is a deep blue coloured complex due to $d-d$ transition.
$2$. $[Zn(H_2O)_6]^{2+}$ is colourless because $Zn^{2+}$ has a $d^{10}$ configuration (no unpaired electrons).
$3$. $[Ni(CN)_4]^{2-}$ has $dsp^2$ hybridization and is square planar,not tetrahedral.
$4$. Nickel dimethyl glyoximate,$[Ni(dmg)_2]$,is a well-known red precipitate used in the gravimetric estimation of nickel.

(A) The mixture $X$ has $0.02 \ mol$ of each complex in $2 \ L$,so $1 \ L$ contains $0.01 \ mol$ of each complex.
Reaction with $AgNO_3$: Only $[Co(NH_3)_5SO_4]Br$ reacts with $AgNO_3$ to form $AgBr$ precipitate.
$[Co(NH_3)_5SO_4]Br + AgNO_3 \to [Co(NH_3)_5SO_4]NO_3 + AgBr(s)$
Since $0.01 \ mol$ of $[Co(NH_3)_5SO_4]Br$ is present in $1 \ L$,moles of $Y (AgBr) = 0.01 \ mol$.
Reaction with $BaCl_2$: Only $[Co(NH_3)_5Br]SO_4$ reacts with $BaCl_2$ to form $BaSO_4$ precipitate.
$[Co(NH_3)_5Br]SO_4 + BaCl_2 \to [Co(NH_3)_5Br]Cl_2 + BaSO_4(s)$
Since $0.01 \ mol$ of $[Co(NH_3)_5Br]SO_4$ is present in $1 \ L$,moles of $Z (BaSO_4) = 0.01 \ mol$.
Therefore,the number of moles of $Y$ and $Z$ are $0.01$ and $0.01$ respectively.

(B) The coordination number of the metal is $6$ and the formula is $MCl_3 \cdot 4H_2O$. Since there is no water of hydration,all $4$ water molecules must be inside the coordination sphere.
Thus,the complex is $[M(H_2O)_4Cl_2]Cl \cdot 0H_2O$.
This means only $1$ chloride ion is ionizable per formula unit.
Number of moles of the complex = $Molarity \times Volume(L) = 0.01 \, M \times 0.200 \, L = 0.002 \, mol$.
Since $1$ mole of complex releases $1$ mole of $Cl^-$,moles of $Cl^-$ = $0.002 \, mol$.
Reaction: $Ag^+ + Cl^- \rightarrow AgCl(s)$.
Moles of $AgNO_3$ required = Moles of $Cl^-$ = $0.002 \, mol$.
Volume of $AgNO_3$ = $\frac{Moles}{Molarity} = \frac{0.002 \, mol}{0.1 \, M} = 0.02 \, L = 20 \, ml$.

(D) The complex is $[Co(en)(NH_3)_2Br_2]^+$.
Structure $I$ is the trans-isomer with respect to $Br$ atoms (they are at $180^{\circ}$ to each other).
Structure $II$ is the cis-isomer with respect to $Br$ atoms (they are at $90^{\circ}$ to each other).
Structure $III$ is the mirror image of structure $II$.
Therefore,$II$ and $III$ are enantiomers (optical isomers).
$I$ and $II$ are geometrical isomers (cis-trans).
$I$ and $III$ are also geometrical isomers.
Statement $D$ ($II$ and $III$ are geometrical isomers) is incorrect because they are optical isomers.

(C) The complex $M(ABCDEF)$ represents an octahedral complex where all six ligands are different.
For an octahedral complex of the type $M(abcdef)$,the number of geometrical isomers is $15$.
Since all ligands are different,the complex lacks any plane of symmetry or center of inversion,making it chiral.
Therefore,it exhibits optical isomerism as well.
Thus,the complex exhibits both geometrical and optical isomerism.

(D) The $CuCr_2O_7$ salt dissociates in water into $Cu^{2+}$ and $Cr_2O_7^{2-}$ ions.
$Cu^{2+}$ ions exhibit a blue colour in aqueous solution.
$Cr_2O_7^{2-}$ ions exhibit a yellow colour in aqueous solution.
When these two colours are mixed,they produce a green colour.
Therefore,the correct option is $D$.

(C) The compound $Zn_2[Fe(CN)_6]$ is white in color.
$(NH_4)_3[As(Mo_3O_{10})_4]$ (Ammonium phosphomolybdate/arsenomolybdate derivative) is yellow.
$BaCrO_4$ (Barium chromate) is yellow due to ligand-to-metal charge transfer $(LMCT)$.
$K_3[Co(NO_2)_6]$ (Potassium hexanitrocobaltate$(III)$) is yellow.
$Zn_2[Fe(CN)_6]$ is a white precipitate formed during the test for $Zn^{2+}$ ions.

(D) Option $A$: The reaction $Li_2O + 2KCl \rightarrow 2LiCl + K_2O$ is thermodynamically unfavorable because $Li_2O$ is more stable than $K_2O$.
Option $B$: The reaction $[CoCl(NH_3)_5]^{2+} + 5H^+ \rightarrow Co^{2+} + 5NH_4^+ + Cl^-$ is balanced in terms of atoms and charge ($+2 + 5 = +7$ on left,$+2 + 5 - 1 = +6$ on right; wait,let's re-check: $Co^{2+} + 5NH_4^+ + Cl^-$ gives $+2 + 5 - 1 = +6$. The charge is not balanced).
Option $C$: The reaction $[Mg(H_2O)_6]^{2+} + (EDTA)^{4-} \rightarrow [Mg(EDTA)]^{2-} + 6H_2O$ is balanced.
Option $D$: The reaction $CuSO_4 + 4KCN \rightarrow K_2[Cu(CN)_4] + K_2SO_4$ is balanced: $Cu$ $(1)$,$S$ $(1)$,$O$ $(4)$,$K$ $(4)$,$C$ $(4)$,$N$ $(4)$ on both sides.

(C) . Ziegler-Natta catalyst is a mixture of $Al(C_2H_5)_3$ and $TiCl_4$ $(A-iii)$.
$B$. Brown ring complex is $[Fe(H_2O)_5NO]SO_4$ $(B-ii)$.
$C$. Prussian Blue is $Fe_4[Fe(CN)_6]_3$ $(C-i)$.
$D$. Turnbull Blue is $Fe_3[Fe(CN)_6]_2$ $(D-iv)$.
Therefore,the correct matching is $A-iii, B-ii, C-i, D-iv$.

(D) Excess of aqueous ammonia $(NH_3)$ acts as a ligand and forms soluble complex compounds with various metal salts.
$1$. $AgCl$ reacts with excess $NH_3$ to form a soluble complex: $AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]Cl$.
$2$. $CrCl_3$ reacts with excess $NH_3$ to form a soluble complex: $CrCl_3 + 6NH_3 \rightarrow [Cr(NH_3)_6]Cl_3$.
$3$. $ZnCl_2$ reacts with excess $NH_3$ to form a soluble complex: $ZnCl_2 + 4NH_3 \rightarrow [Zn(NH_3)_4]Cl_2$.
Since all the given compounds form soluble complexes with excess aqueous ammonia,the correct option is $D$.

(A) The molarity of the solution is $0.8 \ M$ and the volume is $2.0 \ L$.
Number of moles of $K_4[Fe(CN)_6] = \text{Molarity} \times \text{Volume} = 0.8 \times 2.0 = 1.6 \ \text{mol}$.
One mole of $K_4[Fe(CN)_6]$ dissociates as: $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$.
Total number of ions produced per formula unit = $4 + 1 = 5$ ions.
Total moles of ions = $1.6 \ \text{mol} \times 5 = 8.0 \ \text{mol}$.
Total number of ions = $\text{Moles of ions} \times N_A = 8.0 \times 6.022 \times 10^{23} = 48.176 \times 10^{23} = 4.8176 \times 10^{24}$ ions.
Rounding to the nearest option,the answer is $4.8 \times 10^{24}$.

(B) When a complex salt is passed through a cation exchanger,the metal complex cation is exchanged for $H^+$ ions.
The moles of $NaOH$ used for neutralization = $0.020 \ L \times 0.1 \ mol/L = 0.002 \ mol$.
Since $NaOH$ reacts with $H^+$ in a $1:1$ ratio,the moles of $H^+$ released = $0.002 \ mol$.
Given $0.001 \ mol$ of complex released $0.002 \ mol$ of $H^+$,the charge on the complex cation must be $+2$.
For the complex $[Co(NH_3)_5(NO_3)(SO_4)]$,if the complex is $[Co(NH_3)_5(NO_3)]SO_4$,it dissociates as $[Co(NH_3)_5(NO_3)]^{2+} + SO_4^{2-}$.
The cation $[Co(NH_3)_5(NO_3)]^{2+}$ has a charge of $+2$,which matches the experimental data.
Therefore,the complex is $[Co(NH_3)_5(NO_3)]SO_4$.

(C) The given complexes are of the type $[M(en)_2Cl_2]$.
In structure $I$,the two $Cl$ atoms are at $180^{\circ}$ to each other,representing the $trans$-isomer.
In structures $II$,$III$,and $IV$,the two $Cl$ atoms are at $90^{\circ}$ to each other,representing the $cis$-isomer.
Geometrical isomers are pairs of structures that have the same molecular formula but different spatial arrangements of ligands.
Since $I$ is $trans$ and $II$,$III$,and $IV$ are all $cis$ forms of the same complex,$I$ is the geometrical isomer of $II$,$III$,and $IV$.

(D) The reaction between aqueous copper$(II)$ sulfate and ammonia is: $CuSO_{4(aq)} + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]SO_4(aq)$.
Here,the complex ion $X$ is $[Cu(NH_3)_4]^{2+}$.
$1$. The electronic configuration of $Cu^{2+}$ is $[Ar]3d^9$,which contains one unpaired electron,making it paramagnetic.
$2$. Due to the presence of an unpaired electron in the $d$-orbital,it undergoes $d-d$ transition,making the complex coloured (deep blue).
Therefore,all the given statements are correct.

(D) The aqueous solution of $CuCrO_4$ dissociates into $Cu^{2+}$ and $CrO_4^{2-}$ ions.
$Cu^{2+}$ ions are blue in aqueous solution.
$CrO_4^{2-}$ ions are yellow in aqueous solution.
When blue and yellow colors are mixed,the resulting solution appears green.
Therefore,the green color of the $CuCrO_4$ solution is due to the presence of both blue $Cu^{2+}$ ions and yellow $CrO_4^{2-}$ ions.

(D) $1$. The metal nitrate gives a golden yellow flame,identifying the metal as $Na$. Thus,the metal nitrate is $NaNO_3$.
$2$. $2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$. So,$A$ is $NaNO_2$.
$3$. $NaNO_2$ reacts with thiourea $(NH_2CSNH_2)$ to form $NaSCN$ (Solution $B$). $SCN^-$ is a pseudohalide ion.
$4$. $FeCl_3 + 3NaSCN \rightarrow Fe(SCN)_3 + 3NaCl$. The complex formed is $[Fe(SCN)_3(H_2O)_3]$ or $[Fe(SCN)(H_2O)_5]^{2+}$,which is blood red.
$5$. $SCN^-$ is an ambidentate ligand and can show linkage isomerism (via $S$ or $N$ atom).
$6$. In the complex $[Fe(SCN)(H_2O)_5]^{2+}$,$Fe$ is in $+3$ oxidation state $(3d^5)$. It has $5$ unpaired electrons. Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$
$7$. All statements $A$,$B$,and $C$ are correct. Therefore,the incorrect statement is 'None of these'.

(D) $1$. $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. It undergoes $sp^3$ hybridization,making it tetrahedral and diamagnetic.
$2$. $[HgI_4]^{2-}$: $Hg^{2+}$ has a $d^{10}$ configuration. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with no unpaired electrons,hence it is diamagnetic.
$3$. $[Fe(CO)_4]^{2-}$: $Fe$ is in $-2$ oxidation state $(3d^{10})$. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with no unpaired electrons,hence it is diamagnetic.
Since all the given complexes are diamagnetic and tetrahedral,the correct option is $D$.

(A) $1$. $KMnO_4$ is intensely colored due to $L \rightarrow M$ (Ligand to Metal) charge transfer,where electrons transfer from $O^{2-}$ to $Mn^{7+}$. This is correct.
$2$. $TiF_6^{3-}$ $(d^1)$ is colored due to $d-d$ transition,and $Cu_2Cl_2$ is colorless due to $d^{10}$ configuration. This is incorrect.
$3$. $AgCl$ is white (colorless) and $CrO_2Cl_2$ is colored due to charge transfer,not just polarization. This is incorrect.
$4$. The crystal field splitting energy $\Delta_o$ for $NH_3$ is greater than $H_2O$. Since $E = hc/\lambda$,a higher $\Delta_o$ corresponds to a lower $\lambda$. Thus,$[Co(H_2O)_6]^{2+}$ absorbs at a higher $\lambda$ than $[Co(NH_3)_6]^{2+}$. This is correct.
Note: Both $A$ and $D$ are scientifically correct statements.

(A, B) In metal carbonyls, the back-bonding from metal to $CO$ increases the bond order of $M-C$ and decreases the bond order of $C-O$, resulting in a longer $C-O$ bond length $(d_{C-O})$ compared to free $CO$. Thus, statement $A$ is $TRUE$.
Statement $B$ is $TRUE$ because these complexes differ in the number of water molecules coordinated to the metal ion versus those present as lattice water.
Statement $C$ is $FALSE$ because $dsp^2$ hybridisation involves $d_{x^2-y^2}$ orbital, not $d_{z^2}$.
Statement $D$ is $FALSE$ because facial and meridional isomers are generally optically inactive due to the presence of planes of symmetry.

(B) The correct statement for option $B$ is: In the cyanide process,$2Na[Ag(CN)_2] + Zn \rightarrow Na_2[Zn(CN)_4] + 2Ag$. Here,$Zn$ acts as a reducing agent because it displaces $Ag^+$ to $Ag^0$ and gets oxidized itself.
Option $B$ is correct in its description of the process.
Option $C$ is incorrect because blue vitriol $(CuSO_4 \cdot 5H_2O)$ absorbs the orange-red region of the visible spectrum to appear blue,but the statement as phrased is technically correct regarding its absorption spectrum. However,looking at the options,the chemical formula in option $B$ provided in the original prompt had a typo $(Na_2[Zn(CN)_4]^{+2})$,making it the intended incorrect statement.

(D) $1$. In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
$2$. $CO$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbital,resulting in a $[Ar] 3d^{10}$ configuration. This makes the complex diamagnetic.
$3$. Due to the back-bonding from $Ni$ to the empty $\pi^*$ orbitals of $CO$,the $C-O$ bond order decreases,which leads to an increase in the $C-O$ bond length compared to free $CO$.
$4$. The complex exhibits $sp^3$ hybridization,resulting in a tetrahedral geometry.
$5$. Therefore,all the given statements are correct.

(D) $1$. In complex $(a)$,$Na_2[Fe(CN)_5NO]$,the $NO$ ligand is present as $NO^+$. The oxidation state of $Fe$ is calculated as: $2(+1) + x + 5(-1) + 1 = 0$,which gives $x = +2$. Thus,$Fe$ is in $+2$ state.
$2$. In complex $(b)$,$[Fe(H_2O)_5NO]SO_4$,the $NO$ ligand is present as $NO^+$. The oxidation state of $Fe$ is calculated as: $x + 5(0) + 1 = +2$ (since $SO_4$ is $-2$),which gives $x = +1$. Thus,$Fe$ is in $+1$ state.
$3$. Statement $(d)$ is wrong because the oxidation state of $Fe$ in $(a)$ is $+2$ and in $(b)$ is $+1$.

(C) For an octahedral complex,$CFSE = (-0.4n_{t2g} + 0.6n_{eg}) \Delta_0$.
Case $I$: For $d^7$ configuration with weak field ligands,the distribution is $t_{2g}^5 e_g^2$.
$CFSE = (-0.4 \times 5 + 0.6 \times 2) \Delta_0 = (-2.0 + 1.2) \Delta_0 = -0.8 \Delta_0$.
Number of unpaired electrons $(n)$ = $3$. Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \, BM$.
Case $II$: For $d^2$ configuration,the distribution is $t_{2g}^2 e_g^0$.
$CFSE = (-0.4 \times 2 + 0.6 \times 0) \Delta_0 = -0.8 \Delta_0$.
Number of unpaired electrons $(n)$ = $2$. Magnetic moment $\mu = \sqrt{2(2+2)} = \sqrt{8} \, BM$.
Since both cases satisfy the condition,the correct option is $(C)$.

(D) The coordination number $(C.N.)$ of the metal $M$ is $6$ in both complexes.
In the product $[M(gly)_3]$,the metal is bonded to three bidentate glycinate ligands,resulting in $3$ chelate rings,whereas the reactant $[M(gly)_2(NH_3)_2]^+$ has only $2$ chelate rings.
Due to the chelate effect,the formation of more rings increases the stability of the complex.
Both complexes exhibit $d^2sp^3$ hybridization for $C.N. = 6$ (assuming strong field ligands or appropriate metal centers).
Therefore,all the given statements are correct.

(D) $1$. The $\Delta_0$ order depends on the oxidation state of the metal and the strength of the ligand. $[Co(Cl)_6]^{3-}$ has a weak field ligand,$[Fe(H_2O)_6]^{3+}$ has a moderate field ligand,and $[Fe(CN)_6]^{3-}$ has a strong field ligand,so the order is correct.
$2$. The $EAN$ (Effective Atomic Number) for $[Fe(CN)_6]^{4-}$ is $26 - 2 + 12 = 36$ and for $[Fe(CN)_6]^{3-}$ is $26 - 3 + 12 = 35$. Thus,$36 > 35$,so the order is correct.
$3$. According to the spectrochemical series,$CO$ is a stronger $\pi$-acid ligand than $CN^-$,so the order $CN^- < CO$ is correct.
$4$. In the complex $K[PtCl_3(C_2H_4)]$,the metal $Pt$ donates electron density into the $\pi^*$ antibonding orbital of $C_2H_4$ (synergic bonding). This increases the $C-C$ bond length compared to free $C_2H_4$. Therefore,the bond length in the complex is greater than in free $C_2H_4$. The given order $C_2H_4 > K[PtCl_3(C_2H_4)]$ is incorrect.

(B) In solution,coordination compounds like $[Fe(CN)_6]^{3-}$ and $[Fe(CN)_6]^{4-}$ do not dissociate to give free $Fe^{3+}$ or $Fe^{2+}$ ions because the iron is part of the stable complex ion.
$(NH_4)_2SO_4 \cdot FeSO_4 \cdot 6H_2O$ (Mohr's salt) dissociates to give $Fe^{2+}$ ions.
$Fe_2(SO_4)_3$ is an ionic salt that dissociates completely in water as follows:
$Fe_2(SO_4)_3 \rightarrow 2Fe^{3+} + 3SO_4^{2-}$
Thus,$Fe_2(SO_4)_3$ provides $Fe^{3+}$ ions in solution.

(B) $1$. In $[Co(H_2O)_6]^{+2}$,$[Fe(H_2O)_6]^{+3}$,and $[Fe(CN)_6]^{-3}$,the crystal field splitting energy $(\Delta_0)$ increases with higher oxidation state and stronger field ligands. The order is correct.
$2$. In $[Fe(CN)_6]^{-4}$,$Fe$ is in $+2$ state,and in $[Fe(CN)_6]^{-3}$,$Fe$ is in $+3$ state. More back-bonding occurs in $Fe^{+2}$ due to higher electron density,leading to a shorter $Fe-C$ bond. Thus,$[Fe(CN)_6]^{-4} < [Fe(CN)_6]^{-3}$ for bond length. The given order is incorrect.
$3$. $CO$ is a stronger $\pi$-acceptor than $CN^-$. The order is correct.
$4$. In Zeise's salt,$K[PtCl_3(C_2H_4)]$,back-bonding from $Pt$ to $C_2H_4$ increases the $C-C$ bond length compared to free ethene $(C_2H_4)$. Thus,$C_2H_4 < K[PtCl_3(C_2H_4)]$. The given order is incorrect.

(D) To determine if $(n-1)d$ electron pairing occurs,we look at the electronic configuration of the central metal ion in the presence of strong field ligands $(SFL)$.
$1$. For $[Ni(NH_3)_6]^{+2}$: $Ni^{+2}$ is $3d^8$. In an octahedral field,the configuration is $t_{2g}^6 e_g^2$. Since the $e_g$ orbitals are occupied,pairing is not possible in the $d$-orbitals.
$2$. For $[Cr(CN)_6]^{-3}$: $Cr^{+3}$ is $3d^3$. The configuration is $t_{2g}^3 e_g^0$. There are no electrons in the $e_g$ orbitals to pair,and the $t_{2g}$ orbitals are singly occupied.
$3$. For $[Cr(NH_3)_6]^{+3}$: $Cr^{+3}$ is $3d^3$. Similar to the above,the configuration is $t_{2g}^3 e_g^0$,where no pairing of $(n-1)d$ electrons occurs.
Since all these complexes do not involve the pairing of $(n-1)d$ electrons due to their specific $d$-electron configurations,the correct answer is $D$.

(B) $P$. $[Cr(NH_3)_4Cl_2]Cl$: $Cr^{3+}$ $(d^3)$ is paramagnetic. It shows cis-trans isomerism. (Matches $3$)
$Q$. $[Ti(H_2O)_5Cl](NO_3)_2$: $Ti^{3+}$ $(d^1)$ is paramagnetic. It shows ionisation isomerism. (Matches $1$)
$R$. $[Pt(en)(NH_3)Cl]NO_3$: $Pt^{2+}$ $(d^8)$ is diamagnetic. It shows ionisation isomerism. (Matches $4$)
$S$. $[Co(NH_3)_4(NO_3)_2]NO_3$: $Co^{3+}$ ($d^6$,low spin) is diamagnetic. It shows cis-trans isomerism. (Matches $2$)
Thus,the correct matching is $P-3, Q-1, R-4, S-2$.

(C) Let us analyze each statement:
$1$. Hydrated $CuSO_4$ (i.e.,$CuSO_4 \cdot 5H_2O$) contains ionic bonds between $Cu^{2+}$ and $SO_4^{2-}$,covalent bonds within the $SO_4^{2-}$ ion,coordinate bonds between $Cu^{2+}$ and $H_2O$ ligands,and hydrogen bonds between the water molecules and the sulfate ion. Thus,statement $A$ is correct.
$2$. $NO_3^-$ contains covalent bonds between $N$ and $O$ atoms and a coordinate bond from $N$ to one of the $O$ atoms. Thus,statement $B$ is correct.
$3$. $K_4[Fe(CN)_6]$ contains ionic bonds between $K^+$ and $[Fe(CN)_6]^{4-}$,coordinate bonds between $Fe^{2+}$ and $CN^-$ ligands,$AND$ covalent bonds within the $CN^-$ group $(C \equiv N)$. Statement $C$ claims it shows 'only' ionic and coordinate bonds,which is incorrect because it also contains covalent bonds.
$4$. The carbonate ion $(CO_3^{2-})$ exhibits resonance. Thus,statement $D$ is correct.
Therefore,the incorrect statement is $C$.

(C) For an octahedral complex,$CFSE = (-0.4n_{t_{2g}} + 0.6n_{e_g}) \Delta_0$.
Given $CFSE = -0.8 \Delta_0$,we test configurations for $d^n$ ions with weak field ligands.
For a $d^7$ ion $(t_{2g}^5 e_g^2)$: $CFSE = (-0.4 \times 5 + 0.6 \times 2) \Delta_0 = (-2.0 + 1.2) \Delta_0 = -0.8 \Delta_0$. This has $n = 3$ unpaired electrons,so $\mu = \sqrt{3(3+2)} = \sqrt{15} \, BM$.
For a $d^2$ ion $(t_{2g}^2 e_g^0)$: $CFSE = (-0.4 \times 2 + 0.6 \times 0) \Delta_0 = -0.8 \Delta_0$. This has $n = 2$ unpaired electrons,so $\mu = \sqrt{2(2+2)} = \sqrt{8} \, BM$.
Since both configurations are possible with weak field ligands,both $(A)$ and $(B)$ are correct.

(B) $(I)$ The $IUPAC$ name is pentaamminenitrito$-N$-chromium$(III)$ tetrachlorozincate$(II)$. This statement is correct.
$(II)$ The complex does not exhibit geometrical isomerism because the coordination sphere $[Cr(NO_2)(NH_3)_5]^{2+}$ has five identical ligands $(NH_3)$ and one different ligand $(NO_2^-)$,which cannot form cis-trans isomers. This statement is incorrect.
$(III)$ It shows linkage isomerism due to the presence of the ambidentate ligand $NO_2^-$,which can coordinate through $N$ or $O$. This statement is correct.
$(IV)$ It shows coordination isomerism because both the cationic and anionic parts are complex ions,allowing for the exchange of ligands between them. This statement is correct.

(C) The total volume of the mixture is $2 \, L$,containing $0.02 \, mol$ of each complex. Therefore,$1 \, L$ of the mixture contains $0.01 \, mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.01 \, mol$ of $[Co(NH_3)_5Br]SO_4$.
When $1 \, L$ of mixture $X$ reacts with excess $AgNO_3$,only the ionizable $Br^-$ reacts to form $AgBr$ precipitate:
$[Co(NH_3)_5SO_4]Br + AgNO_3 \to [Co(NH_3)_5SO_4]NO_3 + AgBr(s)$
Since there is $0.01 \, mol$ of $[Co(NH_3)_5SO_4]Br$,the moles of $AgBr$ $(Y)$ formed is $0.01 \, mol$.
When $1 \, L$ of mixture $X$ reacts with excess $BaCl_2$,only the ionizable $SO_4^{2-}$ reacts to form $BaSO_4$ precipitate:
$[Co(NH_3)_5Br]SO_4 + BaCl_2 \to [Co(NH_3)_5Br]Cl_2 + BaSO_4(s)$
Since there is $0.01 \, mol$ of $[Co(NH_3)_5Br]SO_4$,the moles of $BaSO_4$ $(Z)$ formed is $0.01 \, mol$.
Thus,the number of moles of $Y$ and $Z$ are $0.01$ and $0.01$ respectively.

(C) The given complexes are dinuclear platinum complexes with bridging chlorine atoms.
In the first structure,the $PPh_3$ groups are on the same side of the $Pt-Pt$ axis,which corresponds to the $cis$-isomer.
In the second structure,the $PPh_3$ groups are on opposite sides of the $Pt-Pt$ axis,which corresponds to the $trans$-isomer.
Since these isomers differ in the spatial arrangement of ligands around the metal centers,they exhibit geometrical isomerism.

(B) The complex $[CoBr_2(NH_3)_2(en)]^{\oplus}$ has the general formula $[M(AA)a_2b_2]$.
In structure $I$,the two $Br$ atoms are $cis$ to each other,and the two $NH_3$ ligands are $trans$ to each other.
In structure $II$,the two $Br$ atoms are $trans$ to each other,and the two $NH_3$ ligands are $cis$ to each other.
In structure $III$,the two $Br$ atoms are $cis$ to each other,and the two $NH_3$ ligands are $cis$ to each other.
Structure $II$ is a $trans$ isomer (with respect to $Br$ atoms),while structures $I$ and $III$ are $cis$ isomers.
Structure $I$ and $III$ are optically active because they lack a plane of symmetry.
Structure $II$ is optically inactive due to the presence of a plane of symmetry.
Therefore,the statement '$II$ and $III$ are optically active isomers' is wrong because $II$ is optically inactive.

(A) $1$. Bis(glycinato)Zinc$(II)$ is $[Zn(gly)_2]$. Since $Zn^{2+}$ is $d^{10}$,it forms a tetrahedral complex which is achiral and optically inactive. Thus,option $A$ is incorrect.
$2$. $[NiCl_4]^{2-}$ is tetrahedral $(sp^3)$,while $[PtCl_4]^{2-}$ is square planar $(dsp^2)$. This statement is correct.
$3$. $[Ni(CN)_4]^{4-}$ involves $Ni(0)$,which is $d^{10}$. It forms a tetrahedral complex $(sp^3)$. Thus,option $C$ is incorrect.
$4$. $[Ni(CN)_4]^{2-}$ is $d^8$ square planar (diamagnetic,$\mu = 0$) and $[Ni(CO)_4]$ is $d^{10}$ tetrahedral (diamagnetic,$\mu = 0$). This statement is correct.
Note: Both $A$ and $C$ are technically incorrect statements based on standard coordination chemistry.

(B, C) The complex $[Ni(DMG)_2]$ is a square planar complex.
$1$. The $IUPAC$ name is Bis(dimethylglyoximato)nickel$(II)$,which is correct.
$2$. The structure of the complex is stabilized by strong intramolecular hydrogen bonding between the oxime groups of the two dimethylglyoximate ligands.
$3$. The ligand dimethylglyoximate $(DMG^-)$ is a bidentate ligand that coordinates through two nitrogen atoms,not two different donor sites.
$4$. The $E.A.N.$ of $Ni^{2+}$ in this complex is $28 - 2 + (2 \times 4) = 34$,but this does not make it an oxidizing agent; it is a stable complex used in the gravimetric estimation of nickel.
Therefore,statements $B$ and $C$ are correct.

(B) $(I)$ The reaction is $Na_2[Fe(CN)_5(NO)] + Na_2S \to Na_4[Fe(CN)_5(NOS)]$. In this reaction,the oxidation state of iron remains $+2$. Thus,the statement is $F$.
$(II)$ $Pt(IV)$ is more stable than $Ni(IV)$ because $Pt$ is a $5d$ element and $Ni$ is a $3d$ element; higher oxidation states are more stable for heavier elements in a group. Thus,the statement is $T$.
$(III)$ Magnesium is highly reactive with oxygen and nitrogen at high temperatures,so welding is performed in an inert atmosphere like Helium. Thus,the statement is $T$.
$(IV)$ $LiAlH_4$ reacts with water to release $H_2$ gas: $LiAlH_4 + 4H_2O \to LiOH + Al(OH)_3 + 4H_2$. Thus,the statement is $T$.
The correct order is $FTTT$.

(A) The reaction of $CuSO_4$ with $H_2S$ produces a black precipitate of copper$(II)$ sulfide $(CuS)$:
$CuSO_{4(aq)} + H_2S_{(g)} \rightarrow CuS_{(s)} (M) + H_2SO_{4(aq)}$
When $CuS$ is treated with excess $KCN$,it undergoes a redox reaction where $Cu(II)$ is reduced to $Cu(I)$ and $CN^-$ is oxidized to cyanogen gas $(CN)_2$:
$2CuS_{(s)} + 8KCN_{(aq)} \rightarrow 2K_3[Cu(CN)_4]_{(aq)} (N) + (CN)_{2(g)} (O) + 2K_2S_{(aq)}$
Thus,the final products $N$ and $O$ are $[Cu(CN)_4]^{3-}$ and $(CN)_2$ respectively.

(C) The molarity of the solution is $1.6 \ M$ and the volume is $2 \ L$.
Number of moles of $K_4[Fe(CN)_6] = \text{Molarity} \times \text{Volume} = 1.6 \times 2 = 3.2 \ mol$.
Each formula unit of $K_4[Fe(CN)_6]$ dissociates as: $K_4[Fe(CN)_6] \rightarrow 4K^{+} + [Fe(CN)_6]^{4-}$.
Total number of ions produced per formula unit $= 4 + 1 = 5$ ions.
Total number of ions $= \text{Number of moles} \times N_A \times 5$.
Total number of ions $= 3.2 \times 6.022 \times 10^{23} \times 5 = 16 \times 6.022 \times 10^{23} \approx 9.635 \times 10^{24}$.

(D) Zeise's salt,$K[PtCl_3(\eta^2-C_2H_4)]$,has a $dsp^2$ hybridised $Pt$ center and is square planar.
In ferrocene,$Fe(C_5H_5)_2$,the $Fe$ atom is bonded to two cyclopentadienyl rings,each acting as a pentahapto ligand,so the coordination number of $Fe$ is $10$ (or $6$ depending on definition,but definitely not $2$).
In $Mn_2(CO)_{10}$,the $EAN$ of $Mn$ is calculated as: $Atomic \ number - Oxidation \ state + 2 \times (Number \ of \ CO \ ligands) + 1 \text{ (for Mn-Mn bond)} = 25 - 0 + 10 + 1 = 36$.
Since all statements $A$,$B$,and $C$ are incorrect,the correct answer is $D$.

(A) $(A). [Ag(CN)_2]^-$: $Ag^+$ is $d^{10}$,coordination number $2$,geometry is linear,magnetic moment is $0 \, B.M.$ (Matches $2$).
$(B). [Cu(CN)_4]^{3-}$: $Cu^+$ is $d^{10}$,coordination number $4$,geometry is tetrahedral,magnetic moment is $0 \, B.M.$ (Matches $4$).
$(C). [Cu(CN)_6]^{4-}$: $Cu^{2+}$ is $d^9$,coordination number $6$,geometry is octahedral,magnetic moment is $1.73 \, B.M.$ (Matches $5$).
$(D). [Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ is $d^9$,coordination number $4$,geometry is square planar,magnetic moment is $1.73 \, B.M.$ (Matches $1$).
$(E). [Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $d^6$,strong field ligand $CN^-$,low spin octahedral,magnetic moment is $0 \, B.M.$ (Matches $3$).
Therefore,the correct sequence is $A-2, B-4, C-5, D-1, E-3$.

(D) low spin complex is formed when the crystal field splitting energy $(\Delta_o)$ is greater than the pairing energy $(P)$,which typically occurs with strong field ligands like $CN^-$ or $NH_3$.
$1$. In $[Fe(CN)_6]^{3-}$,$Fe^{3+}$ is a $d^5$ system. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in a low spin complex.
$2$. In $[Co(NH_3)_6]^{3+}$,$Co^{3+}$ is a $d^6$ system. $NH_3$ is a strong field ligand,causing pairing of electrons,resulting in a low spin complex.
$3$. In $[Mn(CN)_6]^{3-}$,$Mn^{3+}$ is a $d^4$ system. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in a low spin complex.
Since all the given complexes involve strong field ligands that force electron pairing,all of them are low spin complexes.

(A) The total volume of the solution is $2 \, L$. Since we are taking $1 \, L$ of the mixture,the amount of each complex in the $1 \, L$ sample is $0.01 \, mol$.
$1$. Reaction with $AgNO_3$:
$[Co(NH_3)_5SO_4]Br$ dissociates to give $Br^-$ ions,which react with $Ag^+$ to form $AgBr$ precipitate.
$[Co(NH_3)_5Br]SO_4$ does not give $Br^-$ ions in the solution.
Thus,$Y = 0.01 \, mol$ of $AgBr$.
$2$. Reaction with $BaCl_2$:
$[Co(NH_3)_5SO_4]Br$ does not give $SO_4^{2-}$ ions in the solution.
$[Co(NH_3)_5Br]SO_4$ dissociates to give $SO_4^{2-}$ ions,which react with $Ba^{2+}$ to form $BaSO_4$ precipitate.
Thus,$Z = 0.01 \, mol$ of $BaSO_4$.
Therefore,$Y = 0.01$ and $Z = 0.01$.

(D) In the complex $K_4[Fe(CN)_6]$,the oxidation states are: $Fe = +2$,$C = +2$,and $N = -3$.
In the products $Fe^{3+}$,$CO_2$,and $NO_3^-$,the oxidation states are: $Fe = +3$,$C = +4$,and $N = +5$.
Comparing these:
$1$. $Fe$ changes from $+2$ to $+3$ (Oxidation).
$2$. $C$ changes from $+2$ to $+4$ (Oxidation).
$3$. $N$ changes from $-3$ to $+5$ (Oxidation).
Therefore,the statement that 'Carbon is not oxidized' is incorrect.