Complexes and complex stability Questions in English

(C) Complex salts contain two different metallic elements but give a test for only one of them in solution.
For example,$K_4[Fe(CN)_6]$ dissociates to give $K^+$ and $[Fe(CN)_6]^{4-}$ ions,but it does not give a test for $Fe^{2+}$ ions because the iron is part of the stable coordination entity.

(A) When ammonium hydroxide $(NH_4OH)$ is added to a copper chloride $(CuCl_2)$ solution,initially a pale blue precipitate of copper hydroxide $(Cu(OH)_2)$ is formed.
On adding excess of $NH_4OH$,this precipitate dissolves to form a deep blue complex solution of $[Cu(NH_3)_4]^{2+}$ ion.
The reaction is: $Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+}$.

(C) The solubility of iodine $(I_2)$ in water is low due to its non-polar nature.
When iodide ions $(I^-)$ are added,they react with iodine molecules to form the triiodide ion $(I_3^-)$.
This reaction is represented as: $I_2 + I^- \to I_3^-$.
The formation of the soluble $I_3^-$ complex increases the overall solubility of iodine in the aqueous solution.

(B) The brown ring test is used to detect the presence of nitrate ions $(NO_3^-)$.
When $NO_3^-$ reacts with $FeSO_4$ in the presence of concentrated $H_2SO_4$,it forms a brown-colored complex.
The reaction is: $Fe^{2+} + NO_3^- + 4H^+ \to Fe^{3+} + NO + 2H_2O$.
Then,$Fe^{2+} + NO + 5H_2O \to [Fe(H_2O)_5NO]^{2+}$.
The final complex formed is $[Fe(H_2O)_5NO]SO_4$,which is known as nitroso ferrous sulfate,responsible for the brown ring.

(D) $AgCl$ is sparingly soluble in water but dissolves in aqueous ammonia due to the formation of a soluble complex ion.
The reaction is: $AgCl(s) + 2NH_3(aq) \to [Ag(NH_3)_2]Cl(aq)$.
This complex dissociates in solution as: $[Ag(NH_3)_2]Cl(aq) \rightleftharpoons [Ag(NH_3)_2]^{+}(aq) + Cl^{-}(aq)$.

(C) When a small amount of $NH_4OH$ is added to $CuSO_4$ solution,a pale blue precipitate of $Cu(OH)_2$ is formed.
However,on adding excess $NH_4OH$,this precipitate dissolves to form a deep blue complex,tetraamminecopper$(II)$ sulphate.
The reaction is: $CuSO_4 + 4NH_4OH_{(excess)} \to [Cu(NH_3)_4]SO_4 + 4H_2O$.

(A) The correct answer is $A$.
Anhydrous $CoCl_2$ is blue in colour.
When it comes in contact with moisture or is dissolved in water,it forms a hexahydrate complex $[Co(H_2O)_6]Cl_2$,which is pink in colour.

(A) The reaction proceeds in three steps:
$(i)$ $CuSO_4 + 2 KCN \rightarrow K_2SO_4 + Cu(CN)_2$ (Precipitation of cupric cyanide)
(ii) $2 Cu(CN)_2 \rightarrow 2 CuCN + (CN)_2$ (Decomposition of unstable cupric cyanide)
(iii) $3 KCN + CuCN \rightarrow K_3[Cu(CN)_4]$ (Formation of the stable complex potassium tetracyanocuprate$(I)$)
Thus,the final product is $K_3[Cu(CN)_4]$.

(C) The brown ring test is a common laboratory test used to detect the presence of nitrate ions $(NO_3^-)$ or nitrite ions $(NO_2^-)$ 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 junction of the two layers.
This brown ring is due to the formation of the complex ion $[Fe(H_2O)_5(NO)]^{2+}$,which is known as pentaaquanitrosyliron$(II)$ ion.

(B) When $NaOH$ is added to $Zn^{2+}$ ions,a white precipitate of zinc hydroxide is formed: $Zn^{2+} + 2OH^- \to Zn(OH)_2 \downarrow$ (white precipitate).
On adding excess $NaOH$,the precipitate dissolves to form sodium zincate: $Zn(OH)_2 + 2OH^- \to [Zn(OH)_4]^{2-}$.
In the complex ion $[Zn(OH)_4]^{2-}$,the zinc is present as part of the complex anion.

(C) The hydrated $Cu^{2+}$ ion,which exists as $[Cu(H_2O)_6]^{2+}$,exhibits a characteristic blue color due to $d-d$ electronic transitions.

(B) $AgCl$ is sparingly soluble in water due to its very low solubility product $(K_{sp})$.
When $NH_3$ is added,$Ag^{+}$ ions react with $NH_3$ to form a stable soluble complex ion,$[Ag(NH_3)_2]^{+}$.
The reaction is: $AgCl(s) + 2NH_3(aq) \to [Ag(NH_3)_2]^{+}(aq) + Cl^{-}(aq)$.
This complex formation shifts the equilibrium to the right,causing $AgCl$ to dissolve.

(A) When an excess of $NH_3$ solution is added to a $CuSO_4$ solution,the pale blue precipitate of $Cu(OH)_2$ initially formed dissolves to form a deep blue solution.
This deep blue colour is due to the formation of the tetraamminecopper$(II)$ complex,$[Cu(NH_3)_4]^{2+}$.
The reaction is: $Cu^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}(aq)$.

(B) The complex $[Cu(CN)_4]^{2-}$ involves $Cu^{2+}$ ion,which has a $3d^9$ electronic configuration.
However,in this specific complex,the $Cu^{2+}$ ion is reduced by the $CN^-$ ligand to $Cu^+$,which has a $3d^{10}$ configuration.
Since the $3d$ subshell is completely filled $(d^{10})$,there are no unpaired electrons available for $d-d$ transitions.
Consequently,it does not absorb light in the visible region,making it colourless.
It absorbs light in the ultraviolet region due to charge transfer or other electronic transitions.

(A) The brown ring test is used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
$1$. The nitrate ion reacts with concentrated sulphuric acid $(H_2SO_4)$ to produce nitric acid $(HNO_3)$:
$NO_3^- + H_2SO_4 \to HNO_3 + HSO_4^-$
$2$. The nitric acid is then reduced by ferrous sulphate $(FeSO_4)$ to form nitric oxide $(NO)$:
$2HNO_3 + 6FeSO_4 + 3H_2SO_4 \to 3Fe_2(SO_4)_3 + 2NO + 4H_2O$
$3$. The nitric oxide $(NO)$ reacts with the remaining ferrous sulphate to form the brown complex:
$[Fe(H_2O)_6]^{2+} + NO \to [Fe(H_2O)_5NO]^{2+} + H_2O$
The composition of the brown ring is $[Fe(H_2O)_5NO]SO_4$.

(D) The reaction of copper sulphate with potassium cyanide proceeds in steps:
$1$. $CuSO_4 + 2KCN \to Cu(CN)_2 + K_2SO_4$
$2$. The unstable $Cu(CN)_2$ decomposes to give $Cu_2(CN)_2$ and cyanogen gas: $2Cu(CN)_2 \to Cu_2(CN)_2 + (CN)_2$
$3$. $Cu_2(CN)_2$ then reacts with excess $KCN$ to form the stable complex potassium tetracyanocuprate$(I)$: $Cu_2(CN)_2 + 6KCN \to 2K_3[Cu(CN)_4]$
Thus,the final product is $K_3[Cu(CN)_4]$.

(A) When $NH_4OH$ is added to $CuSO_4$ solution,initially a pale blue precipitate of $Cu(OH)_2$ is formed.
Upon adding an excess of $NH_4OH$,this precipitate dissolves to form a deep blue coloured complex,tetraamminecopper$(II)$ sulfate.
The balanced chemical equation is: $CuSO_4 + 4NH_4OH \to [Cu(NH_3)_4]SO_4 + 4H_2O$.
Thus,the correct option is $A$.

(D) $AgCl$ is a white precipitate that is insoluble in water and acids like $HCl$ or $H_2SO_4$.
However,it dissolves in aqueous ammonia $(NH_3)$ due to the formation of a soluble complex,diamminesilver$(I)$ chloride:
$AgCl(s) + 2NH_3(aq) \to [Ag(NH_3)_2]Cl(aq)$.

(A) $NaNO_3$ does not react with $AgCl$ because there is no driving force for the reaction (no formation of a more stable complex,precipitate,or gas).
$Na_2CO_3$ reacts with $AgCl$ to form silver carbonate,which decomposes to silver oxide or metallic silver.
$Na_2S_2O_3$ reacts with $AgCl$ to form a soluble complex: $AgCl + 2Na_2S_2O_3 \to Na_3[Ag(S_2O_3)_2] + NaCl$.
$NH_4OH$ reacts with $AgCl$ to form a soluble diamminesilver$(I)$ chloride complex: $AgCl + 2NH_4OH \to [Ag(NH_3)_2]Cl + 2H_2O$.

(A) The compound $ZnFe_2O_4$ is a normal spinel compound.
In a normal spinel structure,the divalent metal ions $(Zn^{2+})$ occupy the tetrahedral voids,and the trivalent metal ions $(Fe^{3+})$ occupy the octahedral voids.

(B) The stability of a metal-ligand bond depends on factors like the charge density of the ligand,the availability of lone pairs for coordinate bond formation,and the electrostatic interaction between the metal and the ligand. Option $(B)$ is incorrect because a larger ligand often leads to steric hindrance,which can destabilize the complex rather than increase its stability.

(A) $CN^{-}$ ion acts as a reducing agent because it can be oxidised to $(CN)_2$.
It also acts as a strong complexing agent due to the presence of a lone pair on the carbon atom,which allows it to form stable coordinate bonds with metal ions.
Therefore,the properties shown by $CN^{-}$ are $(a)$ reducing and $(c)$ complexing.

(D) The coordination compound ${K_4}[Fe(CN)_6]$ dissociates in water as follows:
${K_4}[Fe(CN)_6] \to 4K^{+} + [Fe(CN)_6]^{4-}$
Since the coordination entity $[Fe(CN)_6]^{4-}$ carries a negative charge,it is classified as an anionic complex.

(B) Ammonia acts as a ligand because it donates its lone pair of electrons to $Cu^{2+}$ ions to form the $[Cu(NH_3)_4]^{2+}$ complex.
This reaction occurs in a basic medium.
In an acidic medium,the lone pair of electrons on the nitrogen atom of ammonia is donated to $H^+$ ions (protons) to form ammonium ions $(NH_4^+)$.
Since $NH_4^+$ does not possess a lone pair,it cannot act as a ligand,and thus the complex cannot form.
The reaction is: $Cu^{2+}_{(aq)} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$
In the presence of acid: $NH_3 + H^+ \rightarrow NH_4^+$

(B) The conductivity of an electrolytic solution depends on the number of ions produced upon dissociation.
$1$. $K_2[PtCl_6] \rightarrow 2K^+ + [PtCl_6]^{2-}$ ($3$ ions)
$2$. $[Co(NH_3)_3(NO_2)_3] \rightarrow$ No ions (non-ionizable as it is a neutral complex)
$3$. $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$ ($5$ ions)
$4$. $[Cu(NH_3)_4]SO_4 \rightarrow [Cu(NH_3)_4]^{2+} + SO_4^{2-}$ ($2$ ions)
Since $[Co(NH_3)_3(NO_2)_3]$ does not produce any ions in solution,it is a poor electrolytic conductor.

(A) Molar conductivity is directly proportional to the number of ions produced in the solution upon dissociation.
$A$. $[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^-$ (Total $4$ ions)
$B$. $[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^-$ (Total $3$ ions)
$C$. $[Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^+ + Cl^-$ (Total $2$ ions)
$D$. $[Co(NH_3)_3Cl_3] \rightarrow$ No ions (Non-electrolyte)
Since $[Co(NH_3)_6]Cl_3$ produces the maximum number of ions $(4)$,it exhibits the highest molar conductivity.

(A) The reaction with $AgNO_3$ depends on the presence of ionizable $Cl^-$ ions outside the coordination sphere.
In the complex $[Co(NH_3)_3Cl_3]$,all three $Cl^-$ ions are coordinated to the central metal atom and are not ionizable.
Therefore,it does not release $Cl^-$ ions in the solution and will not form a precipitate of $AgCl$ with $AgNO_3$.

(D) The coordination compound $CoCl_3 \cdot 5NH_3 \cdot H_2O$ is formulated as $[Co(NH_3)_5(H_2O)]Cl_3$.
This complex is known to be pink in colour.

(D) is the correct answer.
In the complex $[Pt(NH_3)_4]Cl_2$,the chloride ions are present outside the coordination sphere as counter ions.
When dissolved in water,it dissociates as: $[Pt(NH_3)_4]Cl_2 \to [Pt(NH_3)_4]^{2+} + 2Cl^-$.
The free $Cl^-$ ions react with $AgNO_3$ to form a white precipitate of $AgCl$ according to the reaction: $Ag^+ + Cl^- \to AgCl \text{ (white precipitate)}$.
Other complexes listed have $Cl^-$ ions inside the coordination sphere,which do not ionize in aqueous solution.

(A) According to the spectrochemical series,the strength of ligands is determined by their ability to cause crystal field splitting.
$CN^{-}$ is a strong field ligand because it is a $\pi$-acceptor ligand,which forms strong $\sigma$-bonds and $\pi$-back bonds with the metal center.
$Br^{-}$,$HO^{-}$,and $F^{-}$ are weak field ligands (halides and oxygen donors).
Therefore,$CN^{-}$ is the strongest ligand among the given options.

(D) $CN^{-}$ is a strong field ligand because it is a pseudohalide ion.
Pseudohalide ions are strong coordinating ligands because they have the ability to form a $\sigma$-bond (from the ligand to the metal) and a $\pi$-bond (back-bonding from the metal to the ligand),which stabilizes the metal-ligand complex.

(D) The stability of coordination complexes depends on the nature of the ligand.
$CN^-$ is a strong field ligand and a strong Lewis base,which forms a very stable complex with the $Fe^{3+}$ ion due to strong back-bonding and high crystal field splitting energy.
Among the given options,$[Fe(CN)_6]^{3-}$ is the most stable complex due to the strong field nature of the cyanide ligand.
Therefore,the correct option is $D$.

(D) The stability of a coordination complex depends on the nature of the ligand and the metal ion.
$EDTA^{4-}$ is a hexadentate ligand that forms a very stable chelate complex due to the chelate effect.
Among the given options,${K_2}[Ni(EDTA)]$ contains the $EDTA$ ligand,which forms a highly stable complex with the $Ni^{2+}$ ion.
Therefore,${K_2}[Ni(EDTA)]$ is the most stable complex.

(B) The stability of a metal complex is directly proportional to the charge density of the central metal ion.
$A$ higher charge/size ratio (also known as charge density) of the metal ion leads to stronger electrostatic attraction between the metal ion and the ligands.
Therefore,option $(b)$ is correct.

(A) When $CuSO_4$ reacts with an excess of $KCN$ solution,it initially forms a precipitate of $Cu(CN)_2$,which is unstable and decomposes to form $CuCN$ and cyanogen gas $(CN)_2$.
The $CuCN$ then reacts with excess $KCN$ to form the stable complex $K_3[Cu(CN)_4]$.
The overall balanced chemical equation is:
$2CuSO_4 + 10KCN \to 2K_3[Cu(CN)_4] + 2K_2SO_4 + (CN)_2$.

(B) Dimethyl glyoxime $(DMG)$ is a chelating agent that reacts with $Ni^{2+}$ ions in the presence of ammonium hydroxide $(NH_4OH)$ to form a bright red coloured complex,$Ni(DMG)_2$.

(B) Silver chloride $(AgCl)$ reacts with excess aqueous ammonia $(NH_4OH)$ to form a soluble complex salt,diamminesilver$(I)$ chloride.
The chemical equation is: $AgCl(s) + 2NH_4OH(aq) \to [Ag(NH_3)_2]Cl(aq) + 2H_2O(l)$.
In the aqueous solution,this complex salt dissociates as follows: $[Ag(NH_3)_2]Cl(aq) \to [Ag(NH_3)_2]^+(aq) + Cl^-(aq)$.
Therefore,the cation present in the solution is $[Ag(NH_3)_2]^+$,which is the diamminesilver$(I)$ ion.

(B) Coordination compounds like $[Fe(CN)_6]^{3-}$ and $[Fe(CN)_6]^{4-}$ do not dissociate to give free $Fe^{3+}$ or $Fe^{2+}$ ions in solution because the iron is part of the stable coordination sphere.
$Fe_2(SO_4)_3$ is an ionic salt that dissociates completely in water as follows:
$Fe_2(SO_4)_3 \rightarrow 2Fe^{3+} + 3SO_4^{2-}$.
Therefore,it provides $Fe^{3+}$ ions in the solution.

(B) $Al^{3+}$ does not form an amine complex with excess ammonia because aluminum is a $p-$block element and typically does not form stable coordination complexes with ammonia ligands in the same manner as $d-$block transition metals.
$Cd^{2+}$,$Cu^{2+}$,and $Ag^{+}$ are transition metal ions that readily form stable amine complexes such as $[Cd(NH_3)_4]^{2+}$,$[Cu(NH_3)_4]^{2+}$,and $[Ag(NH_3)_2]^+$,respectively.

(D) When $CuSO_4$ reacts with an excess of $KCN$ solution,it first forms a precipitate of $Cu(CN)_2$,which is unstable and decomposes to form $CuCN$ and $(CN)_2$.
$2CuSO_4 + 4KCN \to 2CuCN + (CN)_2 + 2K_2SO_4$
$CuCN$ then reacts with excess $KCN$ to form the stable complex $K_3[Cu(CN)_4]$.
$CuCN + 3KCN \to K_3[Cu(CN)_4]$
The overall reaction is:
$2CuSO_4 + 10KCN \to 2K_3[Cu(CN)_4] + (CN)_2 + 2K_2SO_4$
Thus,the final product is $K_3[Cu(CN)_4]$.

(C) The aqueous solution of $CoCl_2$ contains the octahedral complex $[Co(H_2O)_6]^{2+}$,which is pink in color.
Upon adding excess concentrated $HCl$,the chloride ions $(Cl^-)$ act as ligands and displace the water molecules.
This leads to the formation of the tetrahedral complex $[CoCl_4]^{2-}$,which is blue in color.
The reaction is: $[Co(H_2O)_6]^{2+} + 4Cl^- \rightarrow [CoCl_4]^{2-} + 6H_2O$.

(D) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,which has the chemical formula $K_2[HgI_4]$.
In this reagent,the active species responsible for the detection of ammonia is the tetraiodomercurate$(II)$ ion,represented as $[HgI_4]^{2-}$.
When it reacts with ammonia,it forms a brown precipitate known as the iodide of Millon's base.

(B) The brown ring test for nitrate ions $(NO_3^-)$ involves adding freshly prepared ferrous sulfate $(FeSO_4)$ solution to the nitrate solution,followed by the careful addition of concentrated sulfuric acid $(H_2SO_4)$ along the sides of the test tube.
The reaction produces a brown-colored complex,nitrosoferrous sulfate,which is represented as $[Fe(H_2O)_5(NO)]SO_4$ or simply $[Fe(H_2O)_5(NO)]^{2+}$.
This complex is responsible for the characteristic brown ring formed at the junction of the two layers.

(A) Cobalt chloride $(CoCl_2)$ in aqueous solution exists as the hexahydrate complex $[Co(H_2O)_6]^{2+}$.
This complex ion imparts a characteristic pink color to the solution.

(A) The overall reaction for the formation of $[Ag(NH_3)_2]^+$ is the sum of the two given steps:
$Ag^{+} + 2NH_3 \rightleftharpoons [Ag(NH_3)_2]^+$
The overall equilibrium constant $(K_{overall})$ or formation constant $(K_f)$ is the product of the equilibrium constants of the individual steps:
$K_f = K_1 \times K_2$
$K_f = (1.6 \times 10^3) \times (6.8 \times 10^3)$
$K_f = 10.88 \times 10^6 = 1.088 \times 10^7$

(A) Ionic conductivity depends on the number of ions produced in the solution.
$K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$ (Total $5$ ions)
$[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^-$ (Total $4$ ions)
$[Cu(NH_3)_4]Cl_2 \rightarrow [Cu(NH_3)_4]^{2+} + 2Cl^-$ (Total $3$ ions)
$[Ni(CO)_4]$ is a neutral complex and does not dissociate into ions.
Since $K_4[Fe(CN)_6]$ produces the highest number of ions ($5$ ions),it exhibits the maximum ionic conductivity.

(C) The $CN^-$ ion acts as a ligand that forms a complex with metal ions by donating its electron pair to the empty d-orbitals of the metal.
Thus,it acts as a complexing agent.
Additionally,$CN^-$ is a strong reducing agent because it can donate electrons to metal ions,often reducing them to lower oxidation states (e.g.,in the formation of $[Au(CN)_2]^-$).
Therefore,it exhibits both $(a)$ and $(c)$ properties.

(A) The stability of complex compounds formed by $3d$ transition metal ions generally increases with the increase in the effective nuclear charge and decrease in ionic radius across the period. The Irving-Williams series for the stability of complexes is $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$. Among the given options,$Cu^{2+}$ forms the most stable complex.