Complexes and complex stability Questions in English

(D) Silver bromide $(AgBr)$ is insoluble in water but dissolves in a solution of sodium thiosulfate $(Na_2S_2O_3)$,commonly known as hypo.
This dissolution occurs due to the formation of a stable,water-soluble complex ion,$[Ag(S_2O_3)_2]^{3-}$.
The chemical reaction is:
$AgBr(s) + 2Na_2S_2O_3(aq) \rightarrow Na_3[Ag(S_2O_3)_2](aq) + NaBr(aq)$.

(C) $1$. Thermal decomposition of $AgNO_3$: $2AgNO_3 \xrightarrow{\Delta} 2Ag + 2NO_2 + O_2$. Thus,$(W) = Ag$ and $(X) = NO_2$.
$2$. Reaction of $(X)$ with water: $2NO_2 + H_2O \longrightarrow HNO_2 + HNO_3$. This confirms $(X) = NO_2$.
$3$. Reaction of $(W)$ with $HNO_3$: $3Ag + 4HNO_3 \longrightarrow 3AgNO_3 + NO + 2H_2O$. Thus,$(Y) = AgNO_3$.
$4$. Reaction of $(Y)$ with excess $Na_2S_2O_3$: $AgNO_3 + 2Na_2S_2O_3 \longrightarrow Na_3[Ag(S_2O_3)_2] + NaNO_3$. Thus,$(Z) = Na_3[Ag(S_2O_3)_2]$.

(B) The transition metal salt,typically cobalt$(II)$ chloride,exists as $[Co(H_2O)_6]^{2+}$ (pink) in aqueous solution.
When concentrated $HCl$ is added,chloride ions $(Cl^-)$ act as ligands and replace water molecules to form the complex $[CoCl_4]^{2-}$ (blue).
This reaction involves a change in the coordination number of the metal ion from $6$ to $4$ and the formation of a new chloro-complex species in the solution.

(C) When $NaOH(aq)$ is added to $Cr^{3+}(aq)$,a green-blue precipitate of chromium $(III)$ hydroxide is formed:
$Cr^{3+}(aq) + 3OH^-(aq) \rightarrow [Cr(OH)_3(H_2O)_3](s)$ (green-blue precipitate).
On adding excess $NaOH$,the precipitate dissolves to form a soluble complex ion:
$[Cr(OH)_3(H_2O)_3](s) + OH^-(aq) \rightarrow [Cr(OH)_4(H_2O)_2]^-(aq)$ (green solution).
This complex is often represented as $[Cr(OH)_4]^-$. Therefore,the green colour is due to the formation of the $[Cr(OH)_4]^-$ ion.

(B) $HgCl_2$ is a covalent compound that is sparingly soluble in water.
When chloride ions $(Cl^-)$ are added to the solution,$HgCl_2$ reacts with them to form a soluble complex ion,$[HgCl_4]^{2-}$.
The reaction is: $HgCl_2(s) + 2Cl^-(aq) \rightarrow [HgCl_4]^{2-}(aq)$.
This complex formation increases the solubility of $HgCl_2$ in the presence of excess chloride ions.

(C) An aqueous solution of ferrous sulphate $(FeSO_4)$ absorbs nitric oxide $(NO)$ to form a brown-colored hydrated nitrosyl complex,$[Fe(H_2O)_5(NO)]SO_4$.

(C) The brown ring test is a common laboratory test used to detect the presence of nitrate $(NO_3^-)$ or nitrite $(NO_2^-)$ ions in a solution.
When a freshly prepared ferrous sulfate $(FeSO_4)$ solution is added to the sample,followed by the careful addition of concentrated sulfuric acid $(H_2SO_4)$ along the sides of the test tube,a brown ring is formed at the junction of the two layers.
The brown ring is due to the formation of the complex ion $[Fe(H_2O)_5NO]^{2+}$.

(B) The electrical conductivity of an aqueous solution of a coordination complex depends on the number of ions produced upon dissociation.
$1$. $[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$. Total ions = $5$.
$2$. $[Cr(NH_3)_6]Cl_3 \rightarrow [Cr(NH_3)_6]^{3+} + 3Cl^-$. Total ions = $4$.
$3$. $[Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^+ + Cl^-$. Total ions = $2$.
$4$. $K_2[PtCl_6] \rightarrow 2K^+ + [PtCl_6]^{2-}$. Total ions = $3$.
Comparing the number of ions: $A (5) > B (4) > D (3) > C (2)$.
Therefore,the order of electrical conductivity is $A > B > D > C$.

(D) $Fe^{+3}$ reacts with $NaOH$ to form $Fe(OH)_3$,which is an insoluble brown-colored precipitate.
$Cr^{+3}$,$Zn^{+2}$,and $Al^{+3}$ form soluble hydroxo-complexes in excess $NaOH$ (e.g.,$[Cr(OH)_4]^-$,$[Zn(OH)_4]^{2-}$,and $[Al(OH)_4]^-$).

(B) One mole of $AgNO_3$ precipitates one mole of chloride ion.
In the given reaction,$0.1 \ mole$ of $CoCl_3(NH_3)_5$ reacts to give $0.2 \ mole$ of $AgCl$,which means $2 \ mole$ of $Cl^{-}$ ions are ionizable.
The complex is $[Co(NH_3)_5Cl]Cl_2$.
The electrolytic solution contains $[Co(NH_3)_5Cl]^{2+}$ and $2Cl^{-}$ as constituent ions.
Thus,it is a $1 : 2$ electrolyte.
$[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+}_{aq} + 2Cl^{-}_{aq}$.
Hence,option $B$ is correct.

(C) The stabilization of coordination compounds due to the formation of rings by polydentate ligands is known as the chelate effect.
Chelating ligands form more stable complexes compared to monodentate ligands due to the increase in entropy upon chelation.
In the given options,$[Fe(C_2O_4)_3]^{-3}$ contains the oxalate ion $(C_2O_4^{2-})$,which is a bidentate chelating ligand.
Since it forms stable five-membered chelate rings with the $Fe^{3+}$ ion,it is the most stable complex among the choices provided.

(A) In coordination compounds,the species present inside the square brackets $[...]$ is the coordination sphere and does not ionize in an aqueous solution.
Only the species present outside the coordination sphere (counter ions) ionize.
For $[Co(NH_3)_3Cl_3]$,there are no counter ions outside the coordination sphere,so it is non-ionizable.
For the other options:
$B$: $[Co(NH_3)_4Cl_2]Cl$ gives $Cl^-$ ions.
$C$: $[Co(NH_3)_5Cl]Cl_2$ gives $2Cl^-$ ions.
$D$: $[Co(NH_3)_6]Cl_3$ gives $3Cl^-$ ions.
Therefore,the correct option is $A$.

(A) In metal carbonyls,the strength of the $CO$ bond is inversely proportional to the extent of back-bonding from the metal to the $CO$ ligand.
Back-bonding increases as the electron density on the metal increases,which in turn weakens the $CO$ bond.
Comparing the given complexes:
$1$. $[V(CO)_6]^-$: The metal center has a negative charge,increasing electron density,leading to maximum back-bonding and the weakest $CO$ bond.
$2$. $[Fe(CO)_5]$: Neutral complex.
$3$. $[Cr(CO)_6]^+$: The metal center has a positive charge,which decreases electron density on the metal,thereby reducing back-bonding.
Since back-bonding is minimal in $[Cr(CO)_6]^+$,the $CO$ bond remains strongest in this complex.

(A) The $Mn$ atom in $KMnO_4$ has a $+7$ oxidation state with an electron configuration of $[Ar] 3d^0 4s^0$.
Since there are no unpaired electrons present,$d-d$ transitions are not possible.
The intense purple colour of $KMnO_4$ arises due to $L \rightarrow M$ (ligand to metal) charge transfer,specifically from the $2p$ orbitals of $O$ to the $3d$ orbitals of $Mn$.

(B) The reaction $Fe(CN)_2 \downarrow + 4KCN \longrightarrow K_4[Fe(CN)_6]$ involves the solid precipitate $Fe(CN)_2$ reacting with potassium cyanide $(KCN)$ to form a soluble complex salt,potassium ferrocyanide $(K_4[Fe(CN)_6])$.
Since the solid precipitate disappears to form a soluble complex,this is classified as a precipitate dissolution reaction.

(B) The reaction $Fe(CN)_3 \downarrow + 3KCN \longrightarrow K_3[Fe(CN)_6]$ involves the solid precipitate $Fe(CN)_3$ reacting with $KCN$ to form a soluble complex $K_3[Fe(CN)_6]$.
Therefore,this is a precipitate dissolution reaction.

(B) The reaction $Cd(CN)_2 \downarrow + 2KCN \longrightarrow K_2[Cd(CN)_4]$ involves the dissolution of a solid precipitate $(Cd(CN)_2)$ into a soluble complex salt $(K_2[Cd(CN)_4])$ upon the addition of excess cyanide ions $(CN^-)$.
Therefore,this is a precipitate dissolution reaction.

(B) The reaction involves the dissolution of a solid precipitate of silver iodide $(AgI)$ by sodium thiosulfate $(Na_2S_2O_3)$ to form a soluble complex,sodium dithiosulfatoargentate$(I)$ $(Na_3[Ag(S_2O_3)_2])$.
Since the solid $AgI$ disappears and forms a soluble complex,this is a precipitate dissolution reaction.

(B) The reaction $HgI_2 \downarrow + 2KI \longrightarrow K_2[HgI_4]$ involves the dissolution of the insoluble precipitate $HgI_2$ (mercury$(II)$ iodide) in an excess of potassium iodide $(KI)$ solution to form the soluble complex salt potassium tetraiodomercurate$(II)$,$K_2[HgI_4]$.
Therefore,this is a precipitate dissolution reaction.

(B) The reaction $Cr(OH)_3 \downarrow + NH_3 \text{ (excess)} \longrightarrow [Cr(NH_3)_6]^{3+}$ represents the dissolution of a precipitate into a complex ion.
$Cr(OH)_3$ is a green-coloured precipitate.
Upon adding excess $NH_3$,it forms the hexamminechromium$(III)$ ion,$[Cr(NH_3)_6]^{3+}$,which is a coloured solution (typically violet/purple).
Therefore,this reaction corresponds to the formation of a coloured solution.

(B) When excess $NH_3$ is added to a solution containing $Cu^{2+}$ ions (such as $CuSO_4$),a deep blue coloured complex is formed.
The reaction is: $Cu^{2+}(aq) + 4NH_3(aq) \longrightarrow [Cu(NH_3)_4]^{2+}(aq)$.
This complex,tetraamminecopper$(II)$ ion,is responsible for the characteristic deep blue colour of the solution.
Therefore,the correct classification for this reaction is for a coloured solution.

(B) The reaction between $NiCl_2$ and excess $NH_3$ results in the formation of the complex ion $[Ni(NH_3)_6]^{2+}$.
This complex ion is known to have a characteristic blue-violet color in aqueous solution.
Therefore,the reaction corresponds to the formation of a coloured solution.

(C) The reaction $3KCN + Fe(CN)_3 \downarrow \longrightarrow K_3[Fe(CN)_6]$ involves the dissolution of a precipitate $(Fe(CN)_3)$ into a clear,soluble complex salt $(K_3[Fe(CN)_6])$.
Potassium ferricyanide $(K_3[Fe(CN)_6])$ forms a clear,yellow-coloured solution in water.
However,in the context of classification of reaction products where the formation of a clear solution from a precipitate is observed,it corresponds to category $C$.

(C) The reaction between $CuSO_4$ and excess $KCN$ proceeds as follows:
$1$. Initially,a yellow-green precipitate of $Cu(CN)_2$ is formed,which is unstable.
$2$. It decomposes to form $CuCN$ (white precipitate) and cyanogen gas $(CN)_2$.
$3$. In the presence of excess $KCN$,the $CuCN$ dissolves to form the stable,colourless complex $K_3[Cu(CN)_4]$.
$4$. Therefore,the final observation for the reaction mixture is a clear/colourless solution.
$5$. The correct assignment is $C$.

(C) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Given $\Delta T_f = 3.72$,$m = 1 \ m$,and $K_f = 1.86 \ ^\circ C \ kg \ mol^{-1}$.
Substituting these values: $3.72 = i \times 1.86 \times 1$.
Thus,the van't Hoff factor $i = 2$.
An $i$ value of $2$ indicates that the complex dissociates into $2$ ions in solution.
For the complex $[Pt(H_2O)_4Cl_2]Cl_2 \cdot 2H_2O$,the dissociation is $[Pt(H_2O)_4Cl_2]Cl_2 \cdot 2H_2O \rightarrow [Pt(H_2O)_4Cl_2]^{2+} + 2Cl^-$,which gives $3$ ions $(i=3)$.
For the complex $[Pt(H_2O)_3Cl_3]Cl \cdot 3H_2O$,the dissociation is $[Pt(H_2O)_3Cl_3]Cl \cdot 3H_2O \rightarrow [Pt(H_2O)_3Cl_3]^+ + Cl^-$,which gives $2$ ions $(i=2)$.
Therefore,the correct complex is $[Pt(H_2O)_3Cl_3]Cl \cdot 3H_2O$.

(C) The characteristic test for an ion is given only if the ion is present in the ionization sphere (i.e.,it dissociates in water).
In the complex $K_3[Fe(CN)_6]$,the iron is present as a part of the coordination entity $[Fe(CN)_6]^{3-}$.
Since the coordination sphere does not dissociate to give free $Fe^{3+}$ ions in an aqueous solution,it will not give the characteristic test for $Fe^{3+}$ ions.
Other compounds like $K_2Fe_2(SO_4)_4 \cdot 24H_2O$ (Potash Alum variant),$(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O$ (Mohr's salt),and $Fe_2(SO_4)_3$ are simple salts that dissociate completely in water to provide free iron ions.

(C) When $CuSO_4$ reacts with $KCN$,initially a precipitate of cupric cyanide $Cu(CN)_2$ is formed.
$CuSO_4 + 2KCN \rightarrow Cu(CN)_2 + K_2SO_4$
$Cu(CN)_2$ is unstable and decomposes to give cuprous cyanide $CuCN$ and cyanogen gas $(CN)_2$.
$2Cu(CN)_2 \rightarrow 2CuCN + (CN)_2$
In the presence of excess $KCN$,$CuCN$ dissolves to form the stable complex potassium tetracyanocuprate$(I)$,$K_3[Cu(CN)_4]$.
$CuCN + 3KCN \rightarrow K_3[Cu(CN)_4]$

(C) The reaction of nickel sulfate $(NiSO_4)$ with pyridine $(Py)$ and sodium nitrite $(NaNO_2)$ leads to the formation of the coordination complex $[Ni(Py)_4](NO_2)_2$.
This complex is known for its characteristic deep blue color.
The reaction proceeds as follows: $NiSO_4 + 4Py + 2NaNO_2 \rightarrow [Ni(Py)_4](NO_2)_2 + Na_2SO_4$.

(C) $AgCl$ is sparingly soluble in water. When $NH_4OH$ (aqueous ammonia) is added to $AgCl$,it dissolves due to the formation of a stable,soluble complex ion called diamminesilver$(I)$ ion.
The chemical reaction is:
$AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) + Cl^-(aq)$
Thus,the correct option is $C$.

(A) The coordination number of a metal ion in a complex depends on its size,charge,and the nature of the ligand.
$Ag^{+}$ ion has a $d^{10}$ electronic configuration and typically forms linear complexes with a coordination number of $2$ with strong field ligands like $CN^{-}$.
The reaction is: $Ag^{+} + 2CN^{-} \rightarrow [Ag(CN)_2]^{-}$.
Other ions like $Fe^{2+}$,$Ni^{2+}$,and $Cu^{2+}$ typically form complexes with coordination numbers of $4$ or $6$ (e.g.,$[Fe(CN)_6]^{4-}$,$[Ni(CN)_4]^{2-}$,$[Cu(CN)_4]^{2-}$).

(D) weak electrolyte is a substance that does not ionize completely in solution.
Coordination compounds that form complex ions are generally strong electrolytes if they are ionic salts.
However,$[Co(NH_3)_4]SO_4$ is a neutral coordination complex that does not dissociate into ions in aqueous solution,making it a non-electrolyte or a very weak electrolyte compared to the others which are ionic salts that dissociate completely into $K^+$,$[PtCl_6]^{2-}$,$[Co(NH_3)_6]^{3+}$,$[NO_2]^-$,$[Fe(CN)_6]^{4-}$,etc.

(B) Molar conductivity in aqueous solution depends on the number of ions produced upon dissociation.
$(I) K[Co(NH_3)_2(NO_2)_4] \rightarrow K^+ + [Co(NH_3)_2(NO_2)_4]^-$. Total ions = $2$.
$(II) [Cr(NH_3)_3(NO_2)_3] \rightarrow$ Non-electrolyte. Total ions = $0$ (or $1$ molecule).
$(III) [Cr(NH_3)_5NO_2]_3[Co(NO_2)_6]_2 \rightarrow 3[Cr(NH_3)_5NO_2]^{2+} + 2[Co(NO_2)_6]^{3-}$. Total ions = $3 + 2 = 5$.
$(IV) Mg[Cr(NH_3)(NO_2)_5] \rightarrow Mg^{2+} + [Cr(NH_3)(NO_2)_5]^{2-}$. Total ions = $2$.
Comparing the number of ions: $(III) (5) > (I) = (IV) (2) > (II) (0)$.
However,since $(I)$ and $(IV)$ both produce $2$ ions,we look at the charge density. Generally,the order is $(III) > (IV) > (I) > (II)$.
Checking the options,option $(B)$ $II < I < IV < III$ is the most appropriate representation of the increasing order.

(B) The tendency to form complexes is primarily dependent on the charge density of the central metal ion.
Smaller cations have a higher charge density,which allows them to attract ligands more strongly and form stable coordinate bonds.
Therefore,as the size of the cation decreases,the charge density increases,leading to a greater tendency to form complexes.

(D) The stability of a coordination complex is influenced by the chelate effect.
Complexes containing polydentate ligands (chelating agents) are significantly more stable than those with monodentate ligands.
$EDTA^{4-}$ is a hexadentate ligand that forms a very stable complex with metal ions due to the formation of multiple five-membered chelate rings.
Among the given options,$[Ni(EDTA)]^{2-}$ is the most stable because $EDTA$ acts as a hexadentate ligand,creating a high degree of stability through the chelate effect.

(B) The stability constant $(K)$ of a coordination complex is a measure of the strength of the ligand-metal bond.
$A$ higher value of the stability constant indicates a more stable complex,which implies that the ligand is a stronger ligand.
Comparing the given values:
$(a)$ $K = 4.5 \times 10^{11}$
$(b)$ $K = 2.0 \times 10^{27}$
$(c)$ $K = 3.0 \times 10^{15}$
$(d)$ $K = 9.5 \times 10^8$
Since $2.0 \times 10^{27}$ is the largest value,the complex $[Cu(CN)_4]^{2-}$ is the most stable,making $CN^-$ the strongest ligand among the given options.

(D) Ligands are classified based on their ability to cause splitting in the $d$-orbitals.
$CO$ and $CN^-$ are strong field ligands that typically form low-spin complexes.
$NH_3$ is a borderline ligand that can form both high-spin and low-spin complexes depending on the metal ion and its oxidation state.
$NO_2^-$ is also a strong field ligand.
Therefore,$NH_3$ is the correct answer.

(A) The spectrochemical series is an experimental arrangement of ligands in order of increasing field strength.
According to the spectrochemical series,the order of field strength for the given ligands is:
$I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < EDTA^{4-} < NH_3 < en < NO_2^{-} < CN^{-} < CO$.
Comparing this with the given options,the correct order is $Cl^{-} < F^{-} < C_2O_4^{2-} < NO_2^{-} < CN^{-}$.

(A) The brown ring test is used to detect the presence of nitrate $(NO_3^-)$ or nitrite $(NO_2^-)$ ions in a solution.
When a freshly prepared ferrous sulfate $(FeSO_4)$ solution is added to the nitrate solution,followed by the careful addition of concentrated sulfuric acid $(H_2SO_4)$ along the sides of the test tube,a brown ring is formed at the interface.
The brown ring is due to the formation of the complex $[Fe(H_2O)_5 NO]^{2+}$.
In this complex,the iron is in the $+1$ oxidation state.

(D) The Bronsted-Lowry definition defines acids as proton $(H^+)$ donors and bases as proton acceptors.
In options $A$,$B$,and $C$,there is a clear transfer of a proton $(H^+)$ from the acid to the base.
In the reaction $[Cu(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+} + 4H_2O$,there is no proton transfer.
Instead,$NH_3$ acts as a Lewis base by donating a lone pair of electrons to the $Cu^{2+}$ ion (Lewis acid) to form a coordinate covalent bond.
Therefore,this reaction is an acid-base reaction according to the Lewis definition but not the Bronsted-Lowry definition.

(A) The acidity of metal complexes depends on the charge-to-size ratio (ionic potential) of the central metal ion.
$Al^{3+}$ has a higher charge $(+3)$ and a smaller ionic radius compared to $Mg^{2+}$ $(+2)$.
This results in a higher charge density for $Al^{3+}$,which polarizes the $O-H$ bond in the coordinated water molecules more effectively,facilitating the release of $H^+$ ions.
Therefore,$[Al(H_2O)_6]^{3+}$ is a stronger acid than $[Mg(H_2O)_6]^{2+}$.
Both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(C) $NF_3$ is a weaker ligand than $N(CH_3)_3$ because fluorine is highly electronegative,which withdraws electron density from the nitrogen atom,making the lone pair less available for donation.
In contrast,$N(CH_3)_3$ is a stronger ligand because the methyl groups are electron-releasing,increasing the electron density on the nitrogen atom.
$NF_3$ does not ionize to give $F^{-}$ ions in aqueous solution; it is a covalent molecule.
Therefore,the Assertion is correct,but the Reason is incorrect.

(C) In the $CrO_4^{2-}$ ion,chromium is in the $+VI$ oxidation state,meaning it has a $d^0$ electronic configuration.
Since there are no unpaired $d-$ electrons,the color cannot be attributed to $d-d$ transitions.
The intense yellow color arises due to the $Cr \to O$ charge transfer,where an electron is excited from an oxygen-centered orbital to a chromium-centered orbital.

(D) Nickel dimethylglyoxime,represented as $[Ni(DMG)_2]$,is a square planar coordination compound.
It is a neutral,non-ionic complex that is insoluble in water and is used for the gravimetric estimation of $Ni^{2+}$ ions.

(B) The color of a complex is complementary to the color of the light it absorbs.
Since red light is absorbed,the complex will appear blue-green.
Among the given options,$[Cu(NH_3)_4]^{2+}$ is a well-known deep blue-colored complex,which corresponds to the absorption of red light.

(A) The complex $[Cu(H_2O)_6]^{2+}$ is blue in aqueous solution.
When $HCl$ is added,the $Cl^-$ ligand replaces one $H_2O$ molecule to form $[CuCl(H_2O)_5]^+$.
This substitution leads to a change in the crystal field splitting energy,resulting in a color change from blue to green/yellowish-green depending on the concentration of $HCl$.

(C) $AgCl$ is insoluble in water,conc. $HCl$,and $CCl_4$.
It dissolves in $NH_4OH$ solution due to the formation of a soluble complex salt.
The reaction is: $AgCl + 2NH_4OH \to [Ag(NH_3)_2]Cl + 2H_2O$.
The product formed is Diamminesilver$(I)$ chloride.

(C) In the $[Fe(CN)_6]^{3-}$ complex,the central metal ion is $Fe^{3+}$,which has a $d^5$ electronic configuration.
$CN^-$ is a strong field ligand,causing pairing of electrons,resulting in one unpaired electron $(t_{2g}^5 e_g^0)$.
Since $CN^-$ is a strong field ligand and $Fe^{3+}$ has a high charge density,the crystal field splitting energy is high,making the complex very stable.