Application of Co-ordination compounds Questions in English

(D) In photography,sodium thiosulphate $(Na_2S_2O_3)$,commonly known as 'hypo',is used as a fixing agent.
It dissolves the undecomposed silver bromide $(AgBr)$ from the photographic film by forming a soluble complex,$Na_3[Ag(S_2O_3)_2]$.
The reaction is: $AgBr(s) + 2Na_2S_2O_3(aq) \rightarrow Na_3[Ag(S_2O_3)_2](aq) + NaBr(aq)$.

(D) The reaction of potassium ferrocyanide $(K_4[Fe(CN)_6])$ with concentrated sulfuric acid $(H_2SO_4)$ and water $(H_2O)$ is as follows:
$K_4[Fe(CN)_6] + 6H_2SO_4 + 6H_2O \rightarrow 2K_2SO_4 + FeSO_4 + 3(NH_4)_2SO_4 + 6CO$
As shown in the balanced chemical equation,the gas evolved is carbon monoxide $(CO)$.

(C) In photography,sodium thiosulphate (hypo) is used as a fixing agent.
It reacts with the undecomposed silver bromide $(AgBr)$ on the photographic film to form a soluble complex,sodium dithiosulphatoargentate$(I)$.
The chemical reaction is:
$AgBr + 2Na_2S_2O_3 \to Na_3[Ag(S_2O_3)_2] + NaBr$

(C) In photography,hypo $(Na_2S_2O_3)$ is used as a fixing agent.
It reacts with the undecomposed silver bromide $(AgBr)$ present on the photographic film to form a soluble complex,sodium dithiosulphatoargentate$(I)$.
The reaction is:
$AgBr + 2Na_2S_2O_3 \to Na_3[Ag(S_2O_3)_2] + NaBr$
This process removes the unexposed $AgBr$ from the film,making the image permanent.

(C) In photography,sodium thiosulphate $(Na_2S_2O_3 \cdot 5H_2O)$,commonly known as 'hypo',is used as a fixing agent.
It dissolves the unexposed silver bromide $(AgBr)$ from the photographic film by forming a soluble complex,sodium dithiosulphatoargentate$(I)$,according to the reaction:
$AgBr(s) + 2Na_2S_2O_3(aq) \rightarrow Na_3[Ag(S_2O_3)_2](aq) + NaBr(aq)$.
Therefore,its utility is due to its complex-forming behaviour.

(B) The reaction of copper salts with potassium ferrocyanide in the presence of acetic acid is a characteristic test for $Cu^{2+}$ ions.
$2Cu^{2+} + [Fe(CN)_6]^{4-} \to Cu_2[Fe(CN)_6] \downarrow \text{ (chocolate brown precipitate)}$
The compound formed is copper$(II)$ hexacyanoferrate$(II)$,commonly known as copper ferrocyanide.

(B) The reaction between $FeCl_3$ and $K_4[Fe(CN)_6]$ is a characteristic test for $Fe^{3+}$ ions.
The reaction is: $Fe^{3+} + K_4[Fe(CN)_6] \to KFe[Fe(CN)_6] + 3K^+$.
The product formed is $KFe[Fe(CN)_6]$,which is commonly known as $Fe_4[Fe(CN)_6]_3$ (Prussian blue) or $Fe^{3+}$ ferric-ferrocyanide.

(B) The reaction between ferric ions $(Fe^{3+})$ and potassium ferrocyanide $(K_4[Fe(CN)_6])$ results in the formation of ferric ferrocyanide,which is known as Prussian blue.
The balanced chemical equation is:
$4FeCl_3 + 3K_4[Fe(CN)_6] \to Fe_4[Fe(CN)_6]_3 + 12KCl$
Thus,the correct option is $B$.

(A) The solution must contain $Ni^{2+}$ ions.
When $Ni^{2+}$ reacts with dimethyl glyoxime $(DMG)$ in the presence of an ammoniacal solution,it forms a stable,rose-red colored complex,$[Ni(DMG)_2]$.

(B) Hemoglobin is a protein in red blood cells that carries oxygen. The iron atom at the center of the heme group in hemoglobin must be in the ferrous state $(Fe^{2+})$ to bind oxygen effectively. If the iron is oxidized to the ferric state $(Fe^{3+})$,it forms methemoglobin,which is incapable of binding oxygen.

(D) The $II A$ group precipitate (sulfides) dissolves in dilute $HNO_3$ to form metal nitrates.
When $NH_4OH$ is added to the solution containing $Cu^{2+}$ ions,it forms a deep blue complex,tetraamminecopper$(II)$ ion.
The reaction is: $Cu^{2+} + 4NH_4OH \to [Cu(NH_3)_4]^{2+} + 4H_2O$.
The deep blue color is characteristic of the $[Cu(NH_3)_4]^{2+}$ complex.

(C) The reaction between sodium nitroprusside and sulphide ions in an alkaline medium is given by: $Na_2S + Na_2[Fe(CN)_5(NO)] \to Na_4[Fe(CN)_5(NOS)]$.
The complex formed,$Na_4[Fe(CN)_5(NOS)]$,is known as sodium thionitroprusside,which exhibits a characteristic $violet$ or $purple$ colouration.

(C) $Vitamin \ B_{12}$,also known as cyanocobalamin,is a complex coordination compound.
The central metal ion present in the structure of $Vitamin \ B_{12}$ is $Cobalt$ $(Co^{3+})$.
Therefore,the correct option is $C$.

(A) . In order to make the image permanent,it is necessary to remove the unreduced silver bromide from the surface of the developed film. This operation is called fixing of the image. Fixing is done by dipping the developed film or plate in a sodium thiosulphate (hypo) solution. The hypo solution dissolves the unreduced silver bromide by forming a complex: $AgBr + 2Na_2S_2O_3 \to Na_3[Ag(S_2O_3)_2] + NaBr$. Thus,sodium thiosulphate acts as a complexing agent.

(A) The formation of Prussian blue occurs when $Fe^{3+}$ ions react with potassium ferrocyanide,$K_4[Fe(CN)_6]$.
The chemical reaction is: $4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3$.
This complex,$Fe_4[Fe(CN)_6]_3$,is known as ferric ferrocyanide or Prussian blue.

(D) Blood haemoglobin is a protein in red blood cells that carries oxygen throughout the body.
It contains the metal $Fe$ (iron) as the central atom in its heme group.

(A) The reaction between copper$(II)$ sulfate and potassium ferrocyanide is a double displacement reaction.
The balanced chemical equation is:
$2CuSO_4 + K_4[Fe(CN)_6] \to Cu_2[Fe(CN)_6] + 2K_2SO_4$
Thus,the product formed is copper$(II)$ hexacyanoferrate$(II)$,which is $Cu_2[Fe(CN)_6]$.

(B) Turnbull's blue is $Fe_3[Fe(CN)_6]_2$,which is chemically known as ferrous ferricyanide,$Fe^{II}_3[Fe^{III}(CN)_6]_2$.

(A) The chemical formula for tetrammine copper$(II)$ sulphate is $[Cu(NH_3)_4]SO_4 \cdot H_2O$.
This coordination compound is well known for its deep blue colour,which arises due to $d-d$ transitions in the $Cu^{2+}$ ion.

(D) When potassium ferrocyanide $(K_4[Fe(CN)_6])$ is added to a ferric salt (e.g.,$FeCl_3$),a deep blue precipitate known as Prussian blue is formed.
The chemical reaction is:
$3 K_4[Fe(CN)_6] + 4 FeCl_3 \rightarrow Fe_4[Fe(CN)_6]_3 + 12 KCl$
The compound $Fe_4[Fe(CN)_6]_3$ is ferric ferrocyanide,which is responsible for the Prussian blue colour.

(B) The formation of Prussian blue occurs when $Fe^{3+}$ ions react with potassium ferrocyanide,$K_4[Fe(CN)_6]$.
$Fe_2(SO_4)_3 \to 2Fe^{3+} + 3SO_4^{2-}$
$3K_4[Fe(CN)_6] + 4Fe^{3+} \to Fe_4[Fe(CN)_6]_3 + 12K^+$
$Fe_4[Fe(CN)_6]_3$ is the chemical formula for Prussian blue.

(A) The dissolution of silver sulphide $(Ag_2S)$ in a sodium cyanide $(NaCN)$ solution is a key step in the extraction of silver (MacArthur-Forrest process).
The chemical reaction is:
$Ag_2S + 4NaCN \rightleftharpoons 2Na[Ag(CN)_2] + Na_2S$.
Thus,the complex formed is sodium dicyanoargentate$(I)$,which is $Na[Ag(CN)_2]$.

(B) The complex $cis-[PtCl_2(NH_3)_2]$,commonly known as cisplatin,is widely used as an effective anticancer agent in chemotherapy.
It works by binding to the $DNA$ of cancer cells,which inhibits their replication and leads to cell death.

(D) $Wilkinson's$ catalyst $[RhCl(PPh_3)_3]$ is used for the homogeneous hydrogenation of alkenes.

(D) The reaction between $Ni^{2+}$ and dimethyl glyoxime $(DMG)$ forms a nickel dimethylglyoximate complex,which is a bright red precipitate.
This reaction is typically performed in an ammoniacal medium,which provides an alkaline $pH$ range of $9-11$.
In this $pH$ range,the deprotonation of the $DMG$ ligand occurs,allowing it to coordinate effectively with the $Ni^{2+}$ ion.

(A) The $cis$-isomer of $[Pt(NH_3)_2Cl_2]$,commonly known as cisplatin,is used as an anticancer drug for treating several types of malignant tumours.
When it is injected into the blood stream,the more reactive $Cl$ groups are replaced,allowing the $Pt$ atom to bond to a $N$ atom in guanosine (a part of $DNA$).
This molecule can bond to two different guanosine units,and by bridging between them,it disrupts the normal replication of $DNA$.

(C) Chlorophyll has magnesium $(Mg^{2+})$ as its central metal ion.
It is a porphyrin derivative where the magnesium ion is coordinated to four nitrogen atoms of the porphyrin ring.

(B) The reaction between $FeCl_3$ and $K_4[Fe(CN)_6]$ is a characteristic test for $Fe^{3+}$ ions.
The reaction is: $4FeCl_3 + 3K_4[Fe(CN)_6] \rightarrow Fe_4[Fe(CN)_6]_3 + 12KCl$.
The product $Fe_4[Fe(CN)_6]_3$ is known as Ferric ferrocyanide,which is also called Prussian blue.

(B) Hemoglobin is a red protein responsible for transporting oxygen in the blood of vertebrates.
It consists of a protein called globin and a non-protein part called heme.
The heme group is a porphyrin ring containing a central iron atom in the $Fe^{2+}$ (ferrous) oxidation state.
Therefore,the correct option is $B$.

(B) Hemoglobin is a red pigment present in the red blood cells of vertebrates.
It is a conjugated protein consisting of a protein called globin and a prosthetic group called heme.
The heme group is a complex of $Fe^{2+}$ (ferrous) ion coordinated with a porphyrin ring.
Therefore,the correct ion is $Fe^{2+}$.

(B) When ferric ions $(Fe^{3+})$ react with potassium ferrocyanide $(K_4[Fe(CN)_6])$,they form a deep blue colored precipitate known as Prussian blue.
The chemical reaction is:
$4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3$.
Thus,the compound responsible for the Prussian blue color is $Fe_4[Fe(CN)_6]_3$.

(B) The reaction between a copper salt (e.g.,$CuSO_4$) and potassium ferrocyanide $(K_4[Fe(CN)_6])$ in the presence of acetic acid results in the formation of copper ferrocyanide $(Cu_2[Fe(CN)_6])$.
This compound appears as a characteristic chocolate-brown precipitate.
The chemical equation is: $2Cu^{2+} + [Fe(CN)_6]^{4-} \rightarrow Cu_2[Fe(CN)_6] \downarrow$ (chocolate-brown).

(C) Wilkinson's catalyst contains the metal $Rh$ (Rhodium).
Its chemical formula is $[(Ph_3P)_3RhCl]$.
It is widely used as a homogeneous catalyst in the hydrogenation of alkenes.

(C) The color observed is the complementary color of the light absorbed. According to the color wheel,the complementary color of yellow is blue. Therefore,the complex $[Cu(H_2O)_4]^{2+}$ appears blue.

(A) The aqueous solution of $CuSO_4$ appears blue because it absorbs the complementary color of light.
According to the color wheel,the complementary color of blue is orange-red.
Therefore,$CuSO_4$ absorbs light in the orange-red region of the visible spectrum.

(C) $1$. Vitamin $B_{12}$ (cyanocobalamin) is a coordination complex of Cobalt $(Co)$.
$2$. Hemoglobin is a coordination complex of Iron $(Fe)$.
$3$. Chlorophyll is a coordination complex of Magnesium $(Mg)$,not Calcium $(Ca)$.
$4$. Carbonic anhydrase is a Zinc $(Zn)$ containing enzyme.
Therefore,the statement that chlorophyll contains $Ca$ is incorrect.

(A) Chlorophyll is a coordination complex of magnesium $(Mg^{2+})$,not calcium $(Ca^{2+})$. Therefore,the statement in option $A$ is incorrect.

(B) The coordination complex cis-$[PtCl_2(NH_3)_2]$,commonly known as cisplatin,is widely used as an effective anti-cancer drug in chemotherapy to treat various types of cancers.

(D) The chelating ligand $D-penicillamine$ is used to treat $Cu$ (copper) poisoning in both plants and animals.
It forms a stable complex with $Cu^{2+}$ ions,which is then excreted from the body,thereby reducing the toxic effects of excess copper.

(D) The color of a coordination compound is determined by the color of the light it transmits or reflects,which is complementary to the color of the light it absorbs.
According to the color wheel,the complementary color of violet is yellow.
Since the complex $[Co(CN)_6]^{3-}$ absorbs violet light,it will appear yellow in its aqueous solution.

(C) Prussian blue is a coordination compound formed by the reaction of $Fe^{3+}$ ions with $[Fe(CN)_6]^{4-}$ ions.
Its chemical formula is $Fe_4[Fe(CN)_6]_3 \cdot xH_2O$.
In this complex,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 $(Fe^{2+})$.

(C) $1$. $Vitamin \ B_{12}$ (Cyanocobalamin) is a coordination compound of cobalt.
$2$. Hemoglobin is a coordination compound of iron $(Fe^{2+})$ and acts as an oxygen carrier in blood.
$3$. Chlorophyll is a coordination compound of magnesium $(Mg^{2+})$,not calcium.
$4$. Carboxypeptidase-$A$ is a zinc-containing enzyme.
Therefore,the statement regarding chlorophyll is incorrect.

(B) Carbon monoxide $(CO)$ binds to hemoglobin $(Hb)$ to form a stable complex known as carboxyhemoglobin $(COHb)$.
This complex is about $200$ times more stable than the oxygen-hemoglobin complex (oxyhemoglobin),which prevents oxygen from being transported in the blood.

(D) $Cu(OH)_2$ is soluble in excess of aqueous ammonia due to the formation of a deep blue soluble complex,tetraamminecopper$(II)$ hydroxide.
The reaction is: $Cu(OH)_2 + 4NH_3 \to [Cu(NH_3)_4](OH)_2$.

(C) Schweitzer's reagent is an aqueous solution of tetraamminediaquacopper$(II)$ hydroxide or,more commonly represented as $[Cu(NH_3)_4]SO_4$ in the context of cellulose dissolution.
It is used to dissolve cellulose to form a viscous solution,which is then extruded into an acid bath to produce rayon (artificial silk).

(B) Cis-platin is a well-known anticancer agent.
The chemical formula of cis-platin is $cis-[PtCl_2(NH_3)_2]$.
In this complex,the $cis$ isomer is specifically effective in inhibiting the growth of cancer cells,making it a vital antitumour agent.

(B) The reagent used for the detection of $Ni^{2+}$ ions is Dimethylglyoxime $(DMG)$.
According to the reaction: $Ni^{2+} + 2DMG \rightarrow [Ni(DMG)_2] + 2H^+$.
Here,$2$ moles of $DMG$ $(A)$ react with $1$ mole of $Ni^{2+}$ to form a complex $[Ni(DMG)_2]$,which is a red-colored precipitate $(C)$.
Therefore,$A = DMG$,$x = 2$,and $C = \text{Red}$.

(A) Carbon monoxide $(CO)$ is a poisonous gas because it binds to hemoglobin in the blood to form carboxyhemoglobin.
This complex is about $200$ to $300$ times more stable than the oxygen-hemoglobin complex.
As a result,the hemoglobin becomes incapable of transporting oxygen to the tissues,leading to suffocation and potentially death.

(A) Carbon monoxide $(CO)$ is absorbed by ammoniacal cuprous chloride $(Cu_2Cl_2)$ solution.
This reaction forms a coordination complex: $Cu_2Cl_2 + 2CO + 4NH_3 \rightarrow 2[Cu(NH_3)_2(CO)]Cl$.
This property is widely used in gas analysis to separate $CO$ from other gases.