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Complexes and complex stability Questions in English
Class 12 Chemistry · Coordination Compounds · Complexes and complex stability
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251
MediumMCQ
Violet colour of $[Ti(H_2O)_6]Cl_3$ on heating changes to $........$
A
Green
B
Colourless
C
White
D
Red
Solution
(B) The complex $[Ti(H_2O)_6]Cl_3$ is a violet coloured coordination compound.
Upon heating,it loses water molecules to form $[Ti(H_2O)_5Cl]Cl_2$,which is colourless.
The reaction is: $[Ti(H_2O)_6]Cl_3 \xrightarrow{\Delta} [Ti(H_2O)_5Cl]Cl_2 H_2O$.
Upon heating,it loses water molecules to form $[Ti(H_2O)_5Cl]Cl_2$,which is colourless.
The reaction is: $[Ti(H_2O)_6]Cl_3 \xrightarrow{\Delta} [Ti(H_2O)_5Cl]Cl_2 H_2O$.
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MediumMCQ
The molar ionic conductances of the octahedral complexes:
$(I)$ $PtCl_4 \cdot 5NH_3$
$(II)$ $PtCl_4 \cdot 4NH_3$
$(III)$ $PtCl_4 \cdot 3NH_3$
$(IV)$ $PtCl_4 \cdot 2NH_3$
Follow the order $-$
$(I)$ $PtCl_4 \cdot 5NH_3$
$(II)$ $PtCl_4 \cdot 4NH_3$
$(III)$ $PtCl_4 \cdot 3NH_3$
$(IV)$ $PtCl_4 \cdot 2NH_3$
Follow the order $-$
A
$I < II < III < IV$
B
$IV < III < II < I$
C
$III < IV < II < I$
D
$IV < III < I < II$
Solution
(B) The molar ionic conductance depends on the number of ions produced in the solution.
$(I)$ $[Pt(NH_3)_5Cl]Cl_3$ dissociates into $4$ ions ($1$ cation + $3$ anions).
$(II)$ $[Pt(NH_3)_4Cl_2]Cl_2$ dissociates into $3$ ions ($1$ cation + $2$ anions).
$(III)$ $[Pt(NH_3)_3Cl_3]Cl$ dissociates into $2$ ions ($1$ cation + $1$ anion).
$(IV)$ $[Pt(NH_3)_2Cl_4]$ is a neutral complex and does not dissociate into ions ($0$ ions).
Since the number of ions follows the order $IV < III < II < I$,the molar ionic conductance follows the same order: $IV < III < II < I$.
$(I)$ $[Pt(NH_3)_5Cl]Cl_3$ dissociates into $4$ ions ($1$ cation + $3$ anions).
$(II)$ $[Pt(NH_3)_4Cl_2]Cl_2$ dissociates into $3$ ions ($1$ cation + $2$ anions).
$(III)$ $[Pt(NH_3)_3Cl_3]Cl$ dissociates into $2$ ions ($1$ cation + $1$ anion).
$(IV)$ $[Pt(NH_3)_2Cl_4]$ is a neutral complex and does not dissociate into ions ($0$ ions).
Since the number of ions follows the order $IV < III < II < I$,the molar ionic conductance follows the same order: $IV < III < II < I$.
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MediumMCQ
Which of the following compounds after mixing cannot produce blue coloration?
A
$K_4[Fe(CN)_6]$ and $FeCl_3$ solution
B
$NH_4OH$ and $CuSO_4$ solution
C
Adding anhydrous $CuSO_4$ to water
D
Excess $KCN$ and $CuSO_4$ solution
Solution
(D) $1$. $K_4[Fe(CN)_6] + FeCl_3$ produces Prussian blue precipitate of $Fe_4[Fe(CN)_6]_3$.
$2$. $NH_4OH$ added to $CuSO_4$ produces a deep blue complex $[Cu(NH_3)_4]SO_4$.
$3$. Anhydrous $CuSO_4$ is white,but it turns blue upon hydration to form $CuSO_4 \cdot 5H_2O$.
$4$. Excess $KCN$ added to $CuSO_4$ forms a colorless complex $[Cu(CN)_4]^{3-}$ because $Cu^{2+}$ is reduced to $Cu^+$ by $CN^-$,and the resulting complex is colorless. Thus,it does not produce a blue color.
$2$. $NH_4OH$ added to $CuSO_4$ produces a deep blue complex $[Cu(NH_3)_4]SO_4$.
$3$. Anhydrous $CuSO_4$ is white,but it turns blue upon hydration to form $CuSO_4 \cdot 5H_2O$.
$4$. Excess $KCN$ added to $CuSO_4$ forms a colorless complex $[Cu(CN)_4]^{3-}$ because $Cu^{2+}$ is reduced to $Cu^+$ by $CN^-$,and the resulting complex is colorless. Thus,it does not produce a blue color.
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View Solution254
MediumMCQ
Which of the following is a strong field ligand?
A
$CN^{-}$
B
$SCN^{-}$
C
$I^{-}$
D
$C_2O_4^{2-}$
Solution
(A) According to the spectrochemical series,ligands are arranged in increasing order of their field strength.
Among the given options,$CN^{-}$ is a strong field ligand because it is capable of causing pairing of electrons in the $d$-orbitals of the central metal ion.
$I^{-}$ is a weak field ligand,while $SCN^{-}$ and $C_2O_4^{2-}$ are intermediate or weak field ligands compared to $CN^{-}$.
Among the given options,$CN^{-}$ is a strong field ligand because it is capable of causing pairing of electrons in the $d$-orbitals of the central metal ion.
$I^{-}$ is a weak field ligand,while $SCN^{-}$ and $C_2O_4^{2-}$ are intermediate or weak field ligands compared to $CN^{-}$.
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EasyMCQ
Which among the following cations forms the most stable complex with the same ligand?
A
$Fe^{2+}$
B
$Co^{2+}$
C
$Mn^{2+}$
D
$Cd^{2+}$
Solution
(B) According to the Irving-Williams series,for the same ligand,the stability of complexes formed by divalent metal ions of the $3d$ series generally increases as the ionic radius decreases from $Mn^{2+}$ to $Zn^{2+}$.
Among the given $3d$ transition metal ions ($Mn^{2+}$,$Fe^{2+}$,$Co^{2+}$),$Co^{2+}$ has the smallest ionic radius and the highest charge density,leading to the strongest metal-ligand interaction.
$Cd^{2+}$ is a $4d$ series element and is generally not compared directly in the $3d$ series stability trend.
Therefore,$Co^{2+}$ forms the most stable complex among the options provided.
Among the given $3d$ transition metal ions ($Mn^{2+}$,$Fe^{2+}$,$Co^{2+}$),$Co^{2+}$ has the smallest ionic radius and the highest charge density,leading to the strongest metal-ligand interaction.
$Cd^{2+}$ is a $4d$ series element and is generally not compared directly in the $3d$ series stability trend.
Therefore,$Co^{2+}$ forms the most stable complex among the options provided.
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View Solution256
EasyMCQ
Identify the highest field strength ligand from the following.
A
$H_2O$
B
$EDTA^{4-}$
C
$en$
D
$S^{2-}$
Solution
(C) According to the spectrochemical series,the order of ligand field strength is:
$S^{2-} < H_2O < EDTA^{4-} < en$.
Therefore,$en$ (ethylenediamine) has the highest field strength among the given options.
$S^{2-} < H_2O < EDTA^{4-} < en$.
Therefore,$en$ (ethylenediamine) has the highest field strength among the given options.
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View Solution257
EasyMCQ
Identify the lowest field strength ligand from the following.
A
$I^{-}$
B
$S^{2-}$
C
$en$
D
$CO$
Solution
(A) According to the spectrochemical series,the order of field strength of ligands is: $I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < N_{3} < F^{-} < OH^{-} < C_{2}O_{4}^{2-} < H_{2}O < NCS^{-} < EDTA^{4-} < NH_{3} < en < NO_{2}^{-} < CN^{-} < CO$.
Comparing the given options ($I^{-}$,$S^{2-}$,$en$,$CO$),the ligand $I^{-}$ is at the beginning of the spectrochemical series,indicating it has the lowest field strength.
Comparing the given options ($I^{-}$,$S^{2-}$,$en$,$CO$),the ligand $I^{-}$ is at the beginning of the spectrochemical series,indicating it has the lowest field strength.
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View Solution258
EasyMCQ
Which among the following cations will form the lowest stability complex if the ligand remains the same?
A
$Cu^{2+}$
B
$Fe^{2+}$
C
$Cd^{2+}$
D
$Ni^{2+}$
Solution
(C) The stability of a metal complex is inversely proportional to the size of the metal cation.
As the size of the cation increases,the electrostatic attraction between the metal ion and the ligand decreases,leading to lower stability.
Among the given cations ($Cu^{2+}$,$Fe^{2+}$,$Cd^{2+}$,$Ni^{2+}$),$Cd^{2+}$ has the largest ionic radius.
Therefore,$Cd^{2+}$ forms the complex with the lowest stability.
As the size of the cation increases,the electrostatic attraction between the metal ion and the ligand decreases,leading to lower stability.
Among the given cations ($Cu^{2+}$,$Fe^{2+}$,$Cd^{2+}$,$Ni^{2+}$),$Cd^{2+}$ has the largest ionic radius.
Therefore,$Cd^{2+}$ forms the complex with the lowest stability.
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EasyMCQ
Which among the following cations forms a least stable complex if the ligand is same?
A
$Cd^{2+}$
B
$Co^{2+}$
C
$Ni^{2+}$
D
$Fe^{2+}$
Solution
(A) The stability of a complex depends on the charge density and the electronic configuration of the central metal ion.
$Cd^{2+}$ has a larger ionic radius compared to the $3d$ transition metal ions ($Co^{2+}$,$Ni^{2+}$,$Fe^{2+}$).
Due to its larger size and fully filled $d^{10}$ configuration,the electrostatic attraction between the metal ion and the ligand is weaker,resulting in a less stable complex.
$Cd^{2+}$ has a larger ionic radius compared to the $3d$ transition metal ions ($Co^{2+}$,$Ni^{2+}$,$Fe^{2+}$).
Due to its larger size and fully filled $d^{10}$ configuration,the electrostatic attraction between the metal ion and the ligand is weaker,resulting in a less stable complex.
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View Solution260
MediumMCQ
Identify the correct decreasing order of stability of complexes formed by divalent metal ions with the same ligand.
A
$Cu^{2+} > Mn^{2+} > Cd^{2+}$
B
$Cd^{2+} > Mn^{2+} > Cu^{2+}$
C
$Mn^{2+} > Cd^{2+} > Cu^{2+}$
D
$Cu^{2+} > Cd^{2+} > Mn^{2+}$
Solution
(D) The stability of coordination complexes formed by divalent metal ions with the same ligand generally follows the Irving-Williams series.
According to the Irving-Williams series,the stability of complexes increases with the decrease in ionic radius and increase in the charge-to-size ratio.
For the given ions,the order of stability is $Mn^{2+} < Cd^{2+} < Cu^{2+}$.
However,the question asks for the decreasing order of stability.
Comparing the given options,$Cu^{2+}$ is the most stable due to its smaller ionic radius and higher crystal field stabilization energy compared to $Mn^{2+}$ and $Cd^{2+}$.
Thus,the correct decreasing order is $Cu^{2+} > Cd^{2+} > Mn^{2+}$.
According to the Irving-Williams series,the stability of complexes increases with the decrease in ionic radius and increase in the charge-to-size ratio.
For the given ions,the order of stability is $Mn^{2+} < Cd^{2+} < Cu^{2+}$.
However,the question asks for the decreasing order of stability.
Comparing the given options,$Cu^{2+}$ is the most stable due to its smaller ionic radius and higher crystal field stabilization energy compared to $Mn^{2+}$ and $Cd^{2+}$.
Thus,the correct decreasing order is $Cu^{2+} > Cd^{2+} > Mn^{2+}$.
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MediumMCQ
Which of the following cations will form the most stable complex with the same ligand?
A
$Cu^{2+}$
B
$Co^{2+}$
C
$Mn^{2+}$
D
$Fe^{2+}$
Solution
(A) The stability of complexes formed by transition metal ions with the same ligand is governed by the Irving-Williams series.
According to the Irving-Williams series,the stability of complexes for divalent metal ions of the $3d$ series follows the order: $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$.
Among the given options,$Cu^{2+}$ has the highest stability constant for complex formation with a given ligand due to its smaller ionic radius and higher charge density compared to the other listed ions.
Therefore,$Cu^{2+}$ forms the most stable complex.
According to the Irving-Williams series,the stability of complexes for divalent metal ions of the $3d$ series follows the order: $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$.
Among the given options,$Cu^{2+}$ has the highest stability constant for complex formation with a given ligand due to its smaller ionic radius and higher charge density compared to the other listed ions.
Therefore,$Cu^{2+}$ forms the most stable complex.
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View Solution262
DifficultMCQ
Which of the following cations forms the least stable complex with the same ligand?
A
$Co^{2+}$
B
$Fe^{2+}$
C
$Cd^{2+}$
D
$Cu^{2+}$
Solution
(B) The stability of complexes formed by transition metal ions with the same ligand generally follows the Irving-Williams series.
The order of stability for divalent metal ions is: $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$.
Among the given options ($Co^{2+}$,$Fe^{2+}$,$Cd^{2+}$,$Cu^{2+}$),$Fe^{2+}$ has the lowest stability constant because it appears earliest in the series among the transition metals listed.
$Cd^{2+}$ is a $d^{10}$ ion and generally forms less stable complexes compared to the $3d$ transition series ions due to its larger size and lower charge density.
The order of stability for divalent metal ions is: $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$.
Among the given options ($Co^{2+}$,$Fe^{2+}$,$Cd^{2+}$,$Cu^{2+}$),$Fe^{2+}$ has the lowest stability constant because it appears earliest in the series among the transition metals listed.
$Cd^{2+}$ is a $d^{10}$ ion and generally forms less stable complexes compared to the $3d$ transition series ions due to its larger size and lower charge density.
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MediumMCQ
Which among the following is a correct decreasing order of thermodynamic stability of the complexes?
A
$[Co(NH_3)_6]^{3+} > [Ag(CN)_2]^{-} > [Cu(CN)_4]^{2-}$
B
$[Co(NH_3)_6]^{3+} > [Cu(CN)_4]^{2-} > [Ag(CN)_2]^{-}$
C
$[Ag(CN)_2]^{-} > [Co(NH_3)_6]^{3+} > [Cu(CN)_4]^{2-}$
D
$[Cu(CN)_4]^{2-} > [Ag(CN)_2]^{-} > [Co(NH_3)_6]^{3+}$
Solution
(B) The thermodynamic stability of a coordination complex is primarily determined by the charge density of the central metal ion.
Higher charge on the central metal ion leads to stronger electrostatic attraction with the ligands,resulting in higher stability.
The oxidation states of the central metal ions in the given complexes are:
$Co$ in $[Co(NH_3)_6]^{3+}$ is $+3$.
$Cu$ in $[Cu(CN)_4]^{2-}$ is $+2$.
$Ag$ in $[Ag(CN)_2]^{-}$ is $+1$.
Therefore,the order of stability follows the order of oxidation states: $[Co(NH_3)_6]^{3+} > [Cu(CN)_4]^{2-} > [Ag(CN)_2]^{-}$.
Higher charge on the central metal ion leads to stronger electrostatic attraction with the ligands,resulting in higher stability.
The oxidation states of the central metal ions in the given complexes are:
$Co$ in $[Co(NH_3)_6]^{3+}$ is $+3$.
$Cu$ in $[Cu(CN)_4]^{2-}$ is $+2$.
$Ag$ in $[Ag(CN)_2]^{-}$ is $+1$.
Therefore,the order of stability follows the order of oxidation states: $[Co(NH_3)_6]^{3+} > [Cu(CN)_4]^{2-} > [Ag(CN)_2]^{-}$.
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EasyMCQ
Which from the following is a correct stability order of complexes formed by metal ions if the ligand remains the same?
A
$Fe^{2+} > Mn^{2+} > Co^{2+}$
B
$Co^{2+} > Fe^{2+} > Mn^{2+}$
C
$Co^{2+} > Mn^{2+} > Fe^{2+}$
D
$Fe^{2+} > Co^{2+} > Mn^{2+}$
Solution
(B) The stability order of divalent metal ion complexes is governed by the Irving-Williams series.
According to this series,for the same ligand,the stability of complexes of divalent metal ions follows the order: $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$.
This trend is primarily due to the decrease in ionic radius and the increase in crystal field stabilization energy $(CFSE)$ across the series.
Therefore,the correct stability order for the given ions is $Co^{2+} > Fe^{2+} > Mn^{2+}$.
According to this series,for the same ligand,the stability of complexes of divalent metal ions follows the order: $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$.
This trend is primarily due to the decrease in ionic radius and the increase in crystal field stabilization energy $(CFSE)$ across the series.
Therefore,the correct stability order for the given ions is $Co^{2+} > Fe^{2+} > Mn^{2+}$.
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EasyMCQ
When $CuSO_4$ solution in water is treated with concentrated $HCl$,it turns:
A
Violet
B
Yellow
C
Purple
D
Green
Solution
(B) When $CuSO_4$ solution is treated with concentrated $HCl$,the formation of the tetrachlorocuprate$(II)$ complex occurs:
$Cu^{2+} (aq.) + 4Cl^- (aq.) \rightarrow [CuCl_4]^{2-} (aq.)$
This complex ion $[CuCl_4]^{2-}$ is yellow in color.
Therefore,the solution turns yellow.
$Cu^{2+} (aq.) + 4Cl^- (aq.) \rightarrow [CuCl_4]^{2-} (aq.)$
This complex ion $[CuCl_4]^{2-}$ is yellow in color.
Therefore,the solution turns yellow.
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EasyMCQ
The complex having highest electrical conductivity in aqueous solution under similar conditions is . . . . . . .
A
$[Co(H_2O)_4Cl_2]Cl$
B
$[Co(H_2O)_3Cl_3]$
C
$[Co(H_2O)_5Cl]Cl_2$
D
$[Co(H_2O)_6]Cl_3$
Solution
(D) Electrical conductivity in an aqueous solution depends on the number of ions produced upon dissociation.
$1$. $[Co(H_2O)_4Cl_2]Cl \rightarrow [Co(H_2O)_4Cl_2]^+ + Cl^-$ (Produces $2$ ions)
$2$. $[Co(H_2O)_3Cl_3]$ (Does not dissociate,$0$ ions)
$3$. $[Co(H_2O)_5Cl]Cl_2 \rightarrow [Co(H_2O)_5Cl]^{2+} + 2Cl^-$ (Produces $3$ ions)
$4$. $[Co(H_2O)_6]Cl_3 \rightarrow [Co(H_2O)_6]^{3+} + 3Cl^-$ (Produces $4$ ions)
Since $[Co(H_2O)_6]Cl_3$ produces the maximum number of ions ($4$ ions),it exhibits the highest electrical conductivity.
$1$. $[Co(H_2O)_4Cl_2]Cl \rightarrow [Co(H_2O)_4Cl_2]^+ + Cl^-$ (Produces $2$ ions)
$2$. $[Co(H_2O)_3Cl_3]$ (Does not dissociate,$0$ ions)
$3$. $[Co(H_2O)_5Cl]Cl_2 \rightarrow [Co(H_2O)_5Cl]^{2+} + 2Cl^-$ (Produces $3$ ions)
$4$. $[Co(H_2O)_6]Cl_3 \rightarrow [Co(H_2O)_6]^{3+} + 3Cl^-$ (Produces $4$ ions)
Since $[Co(H_2O)_6]Cl_3$ produces the maximum number of ions ($4$ ions),it exhibits the highest electrical conductivity.
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EasyMCQ
Which of the following complex ions is the most stable?
A
$[Co(NH_3)_6]^{3+}$
B
$[CoCl_6]^{3-}$
C
$[CoF_6]^{3-}$
D
$[Co(H_2O)_6]^{3+}$
Solution
(A) The stability of a coordination complex depends on the nature of the ligand attached to the central metal ion.
$NH_3$ is a strong field ligand compared to $H_2O$,$F^-$,and $Cl^-$.
According to the spectrochemical series,the order of field strength is $Cl^- < F^- < H_2O < NH_3$.
Stronger ligands form more stable complexes with the central metal ion due to greater crystal field splitting energy.
Therefore,$[Co(NH_3)_6]^{3+}$ is the most stable complex among the given options.
$NH_3$ is a strong field ligand compared to $H_2O$,$F^-$,and $Cl^-$.
According to the spectrochemical series,the order of field strength is $Cl^- < F^- < H_2O < NH_3$.
Stronger ligands form more stable complexes with the central metal ion due to greater crystal field splitting energy.
Therefore,$[Co(NH_3)_6]^{3+}$ is the most stable complex among the given options.
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View Solution268
EasyMCQ
The aqueous solution of which of the following complexes has the least conductivity under identical conditions?
A
$[Cr(H_2O)_6]Cl_3$
B
$[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O$
C
$[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$
D
$[Cr(H_2O)_3Cl_3]$
Solution
(D) The conductivity of an aqueous solution of a coordination complex depends on the number of ions produced upon dissociation in water.
$1$. $[Cr(H_2O)_6]Cl_3 \rightarrow [Cr(H_2O)_6]^{3+} + 3Cl^-$ (Total $4$ ions)
$2$. $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O \rightarrow [Cr(H_2O)_5Cl]^{2+} + 2Cl^-$ (Total $3$ ions)
$3$. $[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O \rightarrow [Cr(H_2O)_4Cl_2]^+ + Cl^-$ (Total $2$ ions)
$4$. $[Cr(H_2O)_3Cl_3]$ does not dissociate into ions in aqueous solution (Total $0$ ions).
Since $[Cr(H_2O)_3Cl_3]$ produces no ions,it has the least conductivity.
$1$. $[Cr(H_2O)_6]Cl_3 \rightarrow [Cr(H_2O)_6]^{3+} + 3Cl^-$ (Total $4$ ions)
$2$. $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O \rightarrow [Cr(H_2O)_5Cl]^{2+} + 2Cl^-$ (Total $3$ ions)
$3$. $[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O \rightarrow [Cr(H_2O)_4Cl_2]^+ + Cl^-$ (Total $2$ ions)
$4$. $[Cr(H_2O)_3Cl_3]$ does not dissociate into ions in aqueous solution (Total $0$ ions).
Since $[Cr(H_2O)_3Cl_3]$ produces no ions,it has the least conductivity.
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View Solution269
EasyMCQ
Which of the following complex ions has the least stability?
A
$[Co(CN)_6]^{3-}$
B
$[Co(NH_3)_6]^{2+}$
C
$[Co(NH_3)_6]^{3+}$
D
$[Co(CO)_6]^{3+}$
Solution
(B) The stability of a coordination complex depends on the oxidation state of the central metal ion and the nature of the ligands.
$1$. Higher oxidation state of the central metal ion generally leads to greater stability due to stronger electrostatic attraction.
$2$. $[Co(NH_3)_6]^{2+}$ contains $Co^{2+}$,whereas $[Co(CN)_6]^{3-}$ and $[Co(NH_3)_6]^{3+}$ contain $Co^{3+}$.
$3$. Among the given options,$[Co(NH_3)_6]^{2+}$ has the lowest oxidation state $(+2)$,making it the least stable complex compared to those with $Co^{3+}$ ions.
Therefore,the correct option is $B$.
$1$. Higher oxidation state of the central metal ion generally leads to greater stability due to stronger electrostatic attraction.
$2$. $[Co(NH_3)_6]^{2+}$ contains $Co^{2+}$,whereas $[Co(CN)_6]^{3-}$ and $[Co(NH_3)_6]^{3+}$ contain $Co^{3+}$.
$3$. Among the given options,$[Co(NH_3)_6]^{2+}$ has the lowest oxidation state $(+2)$,making it the least stable complex compared to those with $Co^{3+}$ ions.
Therefore,the correct option is $B$.
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View Solution270
EasyMCQ
Amongst the following,the most stable complex is . . . . . . .
A
$[Fe(C_2O_4)_3]^{3-}$
B
$[Fe(NH_3)_6]^{3+}$
C
$[Fe(H_2O)_6]^{3+}$
D
$[FeCl_6]^{3-}$
Solution
(A) The stability of a coordination complex is influenced by the chelate effect.
$C_2O_4^{2-}$ (oxalate) is a bidentate ligand,which forms a five-membered chelate ring with the central metal ion.
Complexes containing chelate rings are significantly more stable than those containing only monodentate ligands like $NH_3$,$H_2O$,or $Cl^-$.
Therefore,$[Fe(C_2O_4)_3]^{3-}$ is the most stable complex among the given options due to the chelate effect.
$C_2O_4^{2-}$ (oxalate) is a bidentate ligand,which forms a five-membered chelate ring with the central metal ion.
Complexes containing chelate rings are significantly more stable than those containing only monodentate ligands like $NH_3$,$H_2O$,or $Cl^-$.
Therefore,$[Fe(C_2O_4)_3]^{3-}$ is the most stable complex among the given options due to the chelate effect.
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View Solution271
EasyMCQ
Which among the following is the strongest ligand?
A
$CO$
B
$en$
C
$CN^{-}$
D
$NH_{3}$
Solution
(A)
$CO$ is the strongest ligand among the given options due to the synergic bonding effect.
It acts as a $\sigma$-donor and a $\pi$-acceptor,which strengthens the metal-ligand bond significantly.
$CO$ is the strongest ligand among the given options due to the synergic bonding effect.
It acts as a $\sigma$-donor and a $\pi$-acceptor,which strengthens the metal-ligand bond significantly.
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View Solution272
EasyMCQ
Which complex cannot ionise in solution?
A
$[CoCl_{3}(NH_{3})_{3}]$
B
$K_{4}[Fe(CN)_{6}]$
C
$K_{2}[PtF_{6}]$
D
$[Pt(NH_{3})_{6}]Cl_{4}$
Solution
(A) complex ionises in solution only if it has counter-ions outside the coordination sphere.
In the complex $[CoCl_{3}(NH_{3})_{3}]$,all three $Cl^{-}$ ions are directly bonded to the central $Co^{3+}$ metal ion as ligands (satisfying both primary and secondary valencies).
Since there are no ions outside the square brackets,it does not dissociate into ions in an aqueous solution.
Therefore,it will not be precipitated by the addition of $AgNO_{3}$.
In the complex $[CoCl_{3}(NH_{3})_{3}]$,all three $Cl^{-}$ ions are directly bonded to the central $Co^{3+}$ metal ion as ligands (satisfying both primary and secondary valencies).
Since there are no ions outside the square brackets,it does not dissociate into ions in an aqueous solution.
Therefore,it will not be precipitated by the addition of $AgNO_{3}$.
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EasyMCQ
$CO$ is a stronger ligand than $Cl^{-}$ because:
A
$CO$ is a neutral molecule.
B
$CO$ has $\pi$-bonds.
C
$CO$ is poisonous.
D
$CO$ is more reactive.
Solution
(B) $CO$ is a stronger ligand than $Cl^{-}$ because $CO$ acts as a $\pi$-acid (or $\pi$-acceptor) ligand.
It possesses vacant antibonding molecular orbitals that can accept electron density from the metal $d$-orbitals,forming a synergistic bond (back-bonding),which stabilizes the metal-ligand complex.
It possesses vacant antibonding molecular orbitals that can accept electron density from the metal $d$-orbitals,forming a synergistic bond (back-bonding),which stabilizes the metal-ligand complex.
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View Solution274
MediumMCQ
In which one of the following pairs of ions is there an ion that forms a coordination compound with both aqueous sodium hydroxide and ammonia,and another ion that forms a coordination compound only with aqueous sodium hydroxide?
A
$Zn^{2+}, Al^{3+}$
B
$Al^{3+}, Cu^{2+}$
C
$Pb^{2+}, Cu^{2+}$
D
$Cu^{2+}, Zn^{2+}$
Solution
(A) $Zn^{2+}$ forms a coordination compound with both aqueous sodium hydroxide and ammonia,while $Al^{3+}$ forms a coordination compound only with aqueous sodium hydroxide.
Zinc forms the tetrahydroxozincate$(II)$ complex with $NaOH$: $Zn^{2+} + 4OH^{-} \rightarrow [Zn(OH)_{4}]^{2-}$.
Zinc forms the tetraamminezinc$(II)$ complex with $NH_{3}$: $Zn^{2+} + 4NH_{3(aq)} \rightarrow [Zn(NH_{3})_{4}]^{2+}$.
Aluminum forms the hexahydroxoaluminate$(III)$ complex with aqueous sodium hydroxide: $Al^{3+} + 6OH^{-} \rightarrow [Al(OH)_{6}]^{3-}$.
Aluminum does not form a coordination compound with aqueous ammonia.
Zinc forms the tetrahydroxozincate$(II)$ complex with $NaOH$: $Zn^{2+} + 4OH^{-} \rightarrow [Zn(OH)_{4}]^{2-}$.
Zinc forms the tetraamminezinc$(II)$ complex with $NH_{3}$: $Zn^{2+} + 4NH_{3(aq)} \rightarrow [Zn(NH_{3})_{4}]^{2+}$.
Aluminum forms the hexahydroxoaluminate$(III)$ complex with aqueous sodium hydroxide: $Al^{3+} + 6OH^{-} \rightarrow [Al(OH)_{6}]^{3-}$.
Aluminum does not form a coordination compound with aqueous ammonia.
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View Solution275
MediumMCQ
The stepwise stability constants of a complex are given below. What is its overall reaction stability constant $\beta_4$ ?
$M + L \rightleftharpoons ML ; K_1 = 1.0 \times 10^4$
$ML + L \rightleftharpoons ML_2 ; K_2 = 1.0 \times 10^3$
$ML_2 + L \rightleftharpoons ML_3 ; K_3 = 1.0 \times 10^3$
$ML_3 + L \rightleftharpoons ML_4 ; K_4 = 1.0 \times 10^2$
(Overall reaction : $M + 4L \rightleftharpoons ML_4$)
$M + L \rightleftharpoons ML ; K_1 = 1.0 \times 10^4$
$ML + L \rightleftharpoons ML_2 ; K_2 = 1.0 \times 10^3$
$ML_2 + L \rightleftharpoons ML_3 ; K_3 = 1.0 \times 10^3$
$ML_3 + L \rightleftharpoons ML_4 ; K_4 = 1.0 \times 10^2$
(Overall reaction : $M + 4L \rightleftharpoons ML_4$)
A
$1.0 \times 10^{12}$
B
$12.1 \times 10^3$
C
$1.0 \times 10^6$
D
$1.0 \times 10^8$
Solution
(A) The overall stability constant $\beta_n$ for a complex formation reaction is the product of the stepwise stability constants $K_1, K_2, ..., K_n$.
Given the stepwise constants:
$K_1 = 1.0 \times 10^4$
$K_2 = 1.0 \times 10^3$
$K_3 = 1.0 \times 10^3$
$K_4 = 1.0 \times 10^2$
The overall stability constant $\beta_4$ is calculated as:
$\beta_4 = K_1 \times K_2 \times K_3 \times K_4$
$\beta_4 = (1.0 \times 10^4) \times (1.0 \times 10^3) \times (1.0 \times 10^3) \times (1.0 \times 10^2)$
$\beta_4 = 1.0 \times 10^{(4+3+3+2)} = 1.0 \times 10^{12}$
Given the stepwise constants:
$K_1 = 1.0 \times 10^4$
$K_2 = 1.0 \times 10^3$
$K_3 = 1.0 \times 10^3$
$K_4 = 1.0 \times 10^2$
The overall stability constant $\beta_4$ is calculated as:
$\beta_4 = K_1 \times K_2 \times K_3 \times K_4$
$\beta_4 = (1.0 \times 10^4) \times (1.0 \times 10^3) \times (1.0 \times 10^3) \times (1.0 \times 10^2)$
$\beta_4 = 1.0 \times 10^{(4+3+3+2)} = 1.0 \times 10^{12}$
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View Solution276
EasyMCQ
Which among the following does not form a white precipitate with $AgNO_3$ solution?
A
$[Co(NH_3)_6]Cl_3$
B
$[Co(NH_3)_5Cl]Cl_2$
C
$[Co(NH_3)_4Cl_2]Cl$
D
$[Co(NH_3)_3Cl_3]$
Solution
(D) When a coordination compound is treated with $AgNO_3$ solution,only the chloride ions $(Cl^-)$ present outside the coordination sphere (ionizable chloride) react to form a white precipitate of $AgCl$.
$1$. $[Co(NH_3)_6]Cl_3$ gives $3$ moles of $Cl^-$ ions.
$2$. $[Co(NH_3)_5Cl]Cl_2$ gives $2$ moles of $Cl^-$ ions.
$3$. $[Co(NH_3)_4Cl_2]Cl$ gives $1$ mole of $Cl^-$ ions.
$4$. $[Co(NH_3)_3Cl_3]$ has all three chloride ions inside the coordination sphere. Therefore,it does not provide any ionizable $Cl^-$ ions and does not form a white precipitate with $AgNO_3$ solution.
$1$. $[Co(NH_3)_6]Cl_3$ gives $3$ moles of $Cl^-$ ions.
$2$. $[Co(NH_3)_5Cl]Cl_2$ gives $2$ moles of $Cl^-$ ions.
$3$. $[Co(NH_3)_4Cl_2]Cl$ gives $1$ mole of $Cl^-$ ions.
$4$. $[Co(NH_3)_3Cl_3]$ has all three chloride ions inside the coordination sphere. Therefore,it does not provide any ionizable $Cl^-$ ions and does not form a white precipitate with $AgNO_3$ solution.
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View Solution277
EasyMCQ
Which of the following gives the maximum number of ions in aqueous solution?
A
$Ni(CO)_4$
B
$CoCl_3 \cdot 5 H_2O$
C
$PtCl_4 \cdot 6 NH_3$
D
$[Cr(NH_3)_3(NO_2)_3]$
Solution
(C) $PtCl_4 \cdot 6 NH_3$ is formulated as $[Pt(NH_3)_6]Cl_4$. In aqueous solution,it dissociates as follows:
$[Pt(NH_3)_6]Cl_4 \longrightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$.
This results in a total of $5$ ions ($1$ complex cation and $4$ chloride anions).
Comparing other options:
$1. Ni(CO)_4$ is a neutral complex and does not dissociate into ions.
$2. CoCl_3 \cdot 5 H_2O$ is formulated as $[Co(NH_3)_5Cl]Cl_2$ (assuming ammonia ligands) or $[Co(H_2O)_5Cl]Cl_2$,which dissociates into $3$ ions ($1$ complex cation and $2$ chloride anions).
$3. [Cr(NH_3)_3(NO_2)_3]$ is a neutral coordination complex and does not dissociate into ions.
Therefore,$[Pt(NH_3)_6]Cl_4$ provides the maximum number of ions.
$[Pt(NH_3)_6]Cl_4 \longrightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$.
This results in a total of $5$ ions ($1$ complex cation and $4$ chloride anions).
Comparing other options:
$1. Ni(CO)_4$ is a neutral complex and does not dissociate into ions.
$2. CoCl_3 \cdot 5 H_2O$ is formulated as $[Co(NH_3)_5Cl]Cl_2$ (assuming ammonia ligands) or $[Co(H_2O)_5Cl]Cl_2$,which dissociates into $3$ ions ($1$ complex cation and $2$ chloride anions).
$3. [Cr(NH_3)_3(NO_2)_3]$ is a neutral coordination complex and does not dissociate into ions.
Therefore,$[Pt(NH_3)_6]Cl_4$ provides the maximum number of ions.
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View Solution278
EasyMCQ
Which of the following correctly represents copper sulphate pentahydrate?
A
$[Cu(SO_4)(H_2O)_3] \cdot 2H_2O$
B
$[Cu(SO_4)(H_2O)_5]$
C
$[Cu(H_2O)_4]SO_4 \cdot H_2O$
D
None of the above
Solution
(C) The chemical formula of copper sulphate pentahydrate is $[Cu(H_2O)_4]SO_4 \cdot H_2O$.
Copper sulphate pentahydrate contains $5$ molecules of water of crystallisation.
In the crystal structure,four water molecules are coordinated to the $Cu^{2+}$ ion,while the fifth water molecule is hydrogen-bonded to the $SO_4^{2-}$ ion.
Thus,the correct representation is $[Cu(H_2O)_4]SO_4 \cdot H_2O$.
Copper sulphate pentahydrate contains $5$ molecules of water of crystallisation.
In the crystal structure,four water molecules are coordinated to the $Cu^{2+}$ ion,while the fifth water molecule is hydrogen-bonded to the $SO_4^{2-}$ ion.
Thus,the correct representation is $[Cu(H_2O)_4]SO_4 \cdot H_2O$.
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View Solution279
EasyMCQ
The green coloured complex ion of nickel in its aqueous solution is
A
$[Ni(en)_3]^{2+}$
B
$[Ni(H_2O)_2(en)_2]^{2+}$
C
$[Ni(H_2O)_4(en)]^{2+}$
D
$[Ni(H_2O)_6]^{2+}$
Solution
(D) In aqueous solution,nickel ions exist as the hexaaquanickel$(II)$ ion,$[Ni(H_2O)_6]^{2+}$.
This complex has an octahedral geometry with $d^8$ configuration.
Since $H_2O$ is a weak field ligand,no pairing of electrons occurs in the $d$-orbitals.
The presence of two unpaired electrons makes the complex paramagnetic and gives it a characteristic green colour.
This complex has an octahedral geometry with $d^8$ configuration.
Since $H_2O$ is a weak field ligand,no pairing of electrons occurs in the $d$-orbitals.
The presence of two unpaired electrons makes the complex paramagnetic and gives it a characteristic green colour.
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View Solution280
EasyMCQ
Which of the following complex ions is most stable?
A
$[Co(H_2O)_6]^{3+}$
B
$[Co(CN)_6]^{3-}$
C
$[Co(C_2O_4)_3]^{3-}$
D
$[CoF_6]^{3-}$
Solution
(B) The stability of a complex ion is influenced by the nature of the ligand and the chelate effect.
$CN^-$ is a strong field ligand that forms a very stable complex with $Co^{3+}$.
Additionally,the stability of coordination complexes is often determined by the chelate effect,where polydentate ligands like oxalate $(C_2O_4^{2-})$ form more stable complexes than monodentate ligands.
However,among the given options,$[Co(CN)_6]^{3-}$ is exceptionally stable due to the strong field nature of the $CN^-$ ligand,which causes large crystal field splitting energy $(CFSE)$.
Thus,$[Co(CN)_6]^{3-}$ is the most stable complex.
$CN^-$ is a strong field ligand that forms a very stable complex with $Co^{3+}$.
Additionally,the stability of coordination complexes is often determined by the chelate effect,where polydentate ligands like oxalate $(C_2O_4^{2-})$ form more stable complexes than monodentate ligands.
However,among the given options,$[Co(CN)_6]^{3-}$ is exceptionally stable due to the strong field nature of the $CN^-$ ligand,which causes large crystal field splitting energy $(CFSE)$.
Thus,$[Co(CN)_6]^{3-}$ is the most stable complex.
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View Solution281
MediumMCQ
Which one of the following has the highest molar conductivity?
A
Diammine dichloroplatinum $(II)$
B
Tetraamminedichlorocobalt $(III)$ chloride
C
Potassium hexacyano ferrate $(II)$
D
Hexa aquo chromium $(III)$ chloride
Solution
(C) Molar conductivity is directly proportional to the number of ions produced upon dissociation in an aqueous solution.
$1$. $[Pt(NH_3)_2 Cl_2] \rightarrow$ Non-electrolyte ($0$ ions).
$2$. $[Co(NH_3)_4 Cl_2]Cl \rightarrow [Co(NH_3)_4 Cl_2]^+ + Cl^-$ ($2$ ions).
$3$. $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$ ($5$ ions).
$4$. $[Cr(H_2O)_6]Cl_3 \rightarrow [Cr(H_2O)_6]^{3+} + 3Cl^-$ ($4$ ions).
Since $K_4[Fe(CN)_6]$ produces the highest number of ions ($5$ ions),it exhibits the highest molar conductivity.
$1$. $[Pt(NH_3)_2 Cl_2] \rightarrow$ Non-electrolyte ($0$ ions).
$2$. $[Co(NH_3)_4 Cl_2]Cl \rightarrow [Co(NH_3)_4 Cl_2]^+ + Cl^-$ ($2$ ions).
$3$. $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$ ($5$ ions).
$4$. $[Cr(H_2O)_6]Cl_3 \rightarrow [Cr(H_2O)_6]^{3+} + 3Cl^-$ ($4$ ions).
Since $K_4[Fe(CN)_6]$ produces the highest number of ions ($5$ ions),it exhibits the highest molar conductivity.
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View Solution282
MediumMCQ
Cobalt$(III)$ chloride forms a green coloured complex '$X$' with $NH_3$. The number of moles of $AgCl$ formed when an excess of $AgNO_3$ solution is added to $100 \ mL$ of a $1 \ M$ solution of '$X$' is:
A
$0.3$
B
$0.2$
C
$0.1$
D
$1$
Solution
(C) The green coloured complex '$X$' formed by $CoCl_3$ with $NH_3$ is $[Co(NH_3)_4Cl_2]Cl$.
In this complex,only one chloride ion is present outside the coordination sphere as an ionizable chloride.
The dissociation of the complex in water is: $[Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^+ + Cl^-$.
When excess $AgNO_3$ is added,the $Cl^-$ ion reacts to form $AgCl$: $Ag^+ + Cl^- \rightarrow AgCl(s)$.
Thus,$1 \ mole$ of complex '$X$' produces $1 \ mole$ of $AgCl$.
Given: Volume of solution = $100 \ mL = 0.1 \ L$,Molarity = $1 \ M$.
Number of moles of '$X$' = $Molarity \times Volume(L) = 1 \times 0.1 = 0.1 \ mole$.
Since $1 \ mole$ of '$X$' gives $1 \ mole$ of $AgCl$,$0.1 \ mole$ of '$X$' will produce $0.1 \ mole$ of $AgCl$.
In this complex,only one chloride ion is present outside the coordination sphere as an ionizable chloride.
The dissociation of the complex in water is: $[Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^+ + Cl^-$.
When excess $AgNO_3$ is added,the $Cl^-$ ion reacts to form $AgCl$: $Ag^+ + Cl^- \rightarrow AgCl(s)$.
Thus,$1 \ mole$ of complex '$X$' produces $1 \ mole$ of $AgCl$.
Given: Volume of solution = $100 \ mL = 0.1 \ L$,Molarity = $1 \ M$.
Number of moles of '$X$' = $Molarity \times Volume(L) = 1 \times 0.1 = 0.1 \ mole$.
Since $1 \ mole$ of '$X$' gives $1 \ mole$ of $AgCl$,$0.1 \ mole$ of '$X$' will produce $0.1 \ mole$ of $AgCl$.
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View Solution283
MediumMCQ
When $100 \ mL$ of $0.2 \ M$ solution of $CoCl_3 \cdot x NH_3$ is treated with excess of $AgNO_3$ solution,$3.6 \times 10^{22}$ ions are precipitated. The value of $x$ is $\left(N_A = 6 \times 10^{23} \ mol^{-1}\right)$
A
$5$
B
$6$
C
$4$
D
$3$
Solution
(B) The number of moles of $CoCl_3 \cdot x NH_3$ is calculated as: $n = M \times V(L) = 0.2 \ mol/L \times 0.1 \ L = 0.02 \ mol$.
Given that $3.6 \times 10^{22}$ ions are precipitated by $AgNO_3$,the number of moles of $AgCl$ precipitated is $n(AgCl) = \frac{3.6 \times 10^{22}}{6 \times 10^{23}} = 0.06 \ mol$.
Since $0.02 \ mol$ of the complex produces $0.06 \ mol$ of $AgCl$,$1 \ mol$ of the complex produces $\frac{0.06}{0.02} = 3 \ mol$ of $AgCl$.
This implies that there are $3$ chloride ions outside the coordination sphere,meaning the formula is $[Co(NH_3)_x]Cl_3$.
For a coordination number of $6$ for $Co^{3+}$,$x$ must be $6$.
Given that $3.6 \times 10^{22}$ ions are precipitated by $AgNO_3$,the number of moles of $AgCl$ precipitated is $n(AgCl) = \frac{3.6 \times 10^{22}}{6 \times 10^{23}} = 0.06 \ mol$.
Since $0.02 \ mol$ of the complex produces $0.06 \ mol$ of $AgCl$,$1 \ mol$ of the complex produces $\frac{0.06}{0.02} = 3 \ mol$ of $AgCl$.
This implies that there are $3$ chloride ions outside the coordination sphere,meaning the formula is $[Co(NH_3)_x]Cl_3$.
For a coordination number of $6$ for $Co^{3+}$,$x$ must be $6$.
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View Solution284
MediumMCQ
Aqueous ammonia readily dissolves $AgCl$ because
A
$NH_3$ molecules readily solvate $Ag^{+}$ and $Cl^{-}$ ions
B
$NH_3$ molecules abstract chloride from $AgCl$ to form $NH_4Cl$
C
a soluble complex $[Ag(NH_3)_6]^{+}$ is formed
D
a soluble complex $[Ag(NH_3)_2]^{+}$ is formed
Solution
(D) $AgCl$ is sparingly soluble in water but dissolves readily in aqueous ammonia due to the formation of a stable,soluble complex ion,$[Ag(NH_3)_2]^{+}$.
The chemical reaction is as follows:
$AgCl(s) + 2NH_3(aq) \longrightarrow [Ag(NH_3)_2]^{+}(aq) + Cl^{-}(aq)$
The chemical reaction is as follows:
$AgCl(s) + 2NH_3(aq) \longrightarrow [Ag(NH_3)_2]^{+}(aq) + Cl^{-}(aq)$
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View Solution285
MediumMCQ
Which of the following molecules is colourless?
A
$CuSO_4 \cdot 5 H_2 O$ (crystal)
B
$CuSO_4$ (anhydrous)
C
$\left[Cu(NH_3)_4\right]^{2+}_{(aq)}$
D
$\left[CuCl_4\right]^{2-}_{(aq)}$
Solution
(B) Hydrated copper sulphate $CuSO_4 \cdot 5 H_2 O$ is blue in colour due to the presence of water ligands which cause splitting of $d$-orbitals,facilitating $d-d$ transitions.
Anhydrous copper sulphate $CuSO_4$ is colourless because,in the absence of ligands,the $d$-orbitals do not split,making $d-d$ transitions impossible.
Therefore,option $(B)$ is correct.
Anhydrous copper sulphate $CuSO_4$ is colourless because,in the absence of ligands,the $d$-orbitals do not split,making $d-d$ transitions impossible.
Therefore,option $(B)$ is correct.
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View Solution286
MediumMCQ
Which is the correct equation for the reaction of $AgCl$ with $NH_4OH$?
A
$AgCl + NH_4OH \rightarrow AgOH + NH_4^{+} + Cl^{-}$
B
$AgCl + 2NH_4OH \rightarrow [Ag(NH_3)_2]^{+} + Cl^{-} + 2H_2O$
C
$AgCl + 4NH_4OH \rightarrow [Ag(NH_3)_4]^{+} + Cl^{-} + 4H_2O$
D
$2AgCl + NH_4OH \rightarrow Ag_2O + NH_4^{+} + H^{+} + 2Cl^{-}$
Solution
(B) The reaction of $AgCl$ with $NH_4OH$ is a complexation reaction.
$AgCl + 2NH_4OH \rightarrow [Ag(NH_3)_2]^{+} + Cl^{-} + 2H_2O$
$AgCl$ is a white-colored,water-insoluble compound.
In the presence of $NH_3$ (from $NH_4OH$),its solubility increases due to the formation of the soluble complex $[Ag(NH_3)_2]Cl$.
$AgCl + 2NH_4OH \rightarrow [Ag(NH_3)_2]^{+} + Cl^{-} + 2H_2O$
$AgCl$ is a white-colored,water-insoluble compound.
In the presence of $NH_3$ (from $NH_4OH$),its solubility increases due to the formation of the soluble complex $[Ag(NH_3)_2]Cl$.
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View Solution287
EasyMCQ
Which one of the following has the highest molar conductivity?
A
Penta-carbonyl iron$(0)$
B
Hexaqua-chromium$(III)$ bromide
C
Tetraamine-dichloro-cobalt$(III)$ chloride
D
Potassium hexacyanoferrate$(II)$
Solution
(D) The molar conductivity of an electrolyte solution is directly proportional to the number of ions produced per formula unit in the solution.
$(a)$ $[Fe(CO)_5]$ is a neutral complex,it does not dissociate into ions.
$(b)$ $[Cr(H_2O)_5Br]Br_2$ dissociates as $[Cr(H_2O)_5Br]^{2+} + 2Br^-$,providing $3$ ions.
$(c)$ $[Co(NH_3)_4Cl_2]Cl$ dissociates as $[Co(NH_3)_4Cl_2]^+ + Cl^-$,providing $2$ ions.
$(d)$ $K_4[Fe(CN)_6]$ dissociates as $4K^+ + [Fe(CN)_6]^{4-}$,providing $5$ ions.
Since $K_4[Fe(CN)_6]$ produces the maximum number of ions $(5)$,it exhibits the highest molar conductivity.
Thus,$(d)$ is the correct answer.
$(a)$ $[Fe(CO)_5]$ is a neutral complex,it does not dissociate into ions.
$(b)$ $[Cr(H_2O)_5Br]Br_2$ dissociates as $[Cr(H_2O)_5Br]^{2+} + 2Br^-$,providing $3$ ions.
$(c)$ $[Co(NH_3)_4Cl_2]Cl$ dissociates as $[Co(NH_3)_4Cl_2]^+ + Cl^-$,providing $2$ ions.
$(d)$ $K_4[Fe(CN)_6]$ dissociates as $4K^+ + [Fe(CN)_6]^{4-}$,providing $5$ ions.
Since $K_4[Fe(CN)_6]$ produces the maximum number of ions $(5)$,it exhibits the highest molar conductivity.
Thus,$(d)$ is the correct answer.
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View Solution288
MediumMCQ
The formula of the product formed,when sodium thiosulphate solution is added to silver bromide is
A
$Ag_2S_2O_3$
B
$Ag_2S$
C
$Na_3[Ag(S_2O_3)_2]$
D
$Ag_3[Na(S_2O_3)_2]$
Solution
(C) When sodium thiosulphate solution is added to $AgBr$,a complex compound called sodium argentothiosulphate is formed,which is soluble in water.
The chemical reaction is:
$AgBr + 2Na_2S_2O_3 \longrightarrow Na_3[Ag(S_2O_3)_2] + NaBr$
The product formed is $Na_3[Ag(S_2O_3)_2]$.
The chemical reaction is:
$AgBr + 2Na_2S_2O_3 \longrightarrow Na_3[Ag(S_2O_3)_2] + NaBr$
The product formed is $Na_3[Ag(S_2O_3)_2]$.
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View Solution289
MediumMCQ
Identify '$P$' and '$Q$' in the following reaction sequence:
$[Cd(NH_3)_4](NO_3)_2 + KCN + H_2O \rightarrow P + KNO_3 + NH_4OH$
$P \xrightarrow{H_2S} Q$
$[Cd(NH_3)_4](NO_3)_2 + KCN + H_2O \rightarrow P + KNO_3 + NH_4OH$
$P \xrightarrow{H_2S} Q$
A
$P=K_2[Cd(CN)_4], Q=CdS$
B
$P=CdS, Q=K_2[Cd(CN)_4]$
C
$P=Cd(NO_3)_2, Q=CdSO_4$
D
$P=[Cd(OH_2)_4](NO_3)_2, Q=[Cd(NO_3)_4](NO_3)_2$
Solution
(A) The reaction of $[Cd(NH_3)_4](NO_3)_2$ with $KCN$ leads to the formation of the stable cyano-complex $K_2[Cd(CN)_4]$.
Thus,$P = K_2[Cd(CN)_4]$.
When $K_2[Cd(CN)_4]$ reacts with $H_2S$,it forms the yellow precipitate of cadmium sulfide $(CdS)$.
Thus,$Q = CdS$.
The correct option is $A$.
Thus,$P = K_2[Cd(CN)_4]$.
When $K_2[Cd(CN)_4]$ reacts with $H_2S$,it forms the yellow precipitate of cadmium sulfide $(CdS)$.
Thus,$Q = CdS$.
The correct option is $A$.
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View Solution290
MediumMCQ
When $AgCl$ is treated with $KCN$,what happens?
A
$Ag$ is precipitated
B
$A$ complex ion is formed
C
Double decomposition takes place
D
No reaction takes place
Solution
(B) When $AgCl$ reacts with $KCN$,it dissolves to form a soluble complex compound.
The chemical reaction is as follows:
$AgCl(s) + 2KCN(aq) \rightarrow K[Ag(CN)_2](aq) + KCl(aq)$
In this reaction,the silver chloride reacts with potassium cyanide to form the dicyanoargentate$(I)$ complex ion,$[Ag(CN)_2]^-$,which is soluble in water.
The chemical reaction is as follows:
$AgCl(s) + 2KCN(aq) \rightarrow K[Ag(CN)_2](aq) + KCl(aq)$
In this reaction,the silver chloride reacts with potassium cyanide to form the dicyanoargentate$(I)$ complex ion,$[Ag(CN)_2]^-$,which is soluble in water.
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View Solution291
DifficultMCQ
The conductivity measurement of a coordination compound of cobalt$(III)$ shows that it dissociates into $3$ ions in solution. The compound is
A
hexaamminecobalt$(III)$ chloride
B
pentaamminesulphatocobalt$(III)$ chloride
C
pentaamminechloridocobalt$(III)$ sulphate
D
pentaamminechloridocobalt$(III)$ chloride
Solution
(D) coordination compound of $Co(III)$ that dissociates into $3$ ions in solution must have two ionizable ions outside the coordination sphere,given the oxidation state of cobalt is $+3$.
$(A)$ $[Co(NH_{3})_{6}]Cl_{3} \rightarrow [Co(NH_{3})_{6}]^{3+} + 3Cl^{-}$ ($4$ ions)
$(B)$ $[Co(NH_{3})_{5}(SO_{4})]Cl \rightarrow [Co(NH_{3})_{5}(SO_{4})]^{+} + Cl^{-}$ ($2$ ions)
$(C)$ $[Co(NH_{3})_{5}Cl]SO_{4} \rightarrow [Co(NH_{3})_{5}Cl]^{2+} + SO_{4}^{2-}$ ($2$ ions)
$(D)$ $[Co(NH_{3})_{5}Cl]Cl_{2} \rightarrow [Co(NH_{3})_{5}Cl]^{2+} + 2Cl^{-}$ ($3$ ions)
Thus,the compound that dissociates into $3$ ions is $[Co(NH_{3})_{5}Cl]Cl_{2}$.
Hence,option $(D)$ is the correct answer.
$(A)$ $[Co(NH_{3})_{6}]Cl_{3} \rightarrow [Co(NH_{3})_{6}]^{3+} + 3Cl^{-}$ ($4$ ions)
$(B)$ $[Co(NH_{3})_{5}(SO_{4})]Cl \rightarrow [Co(NH_{3})_{5}(SO_{4})]^{+} + Cl^{-}$ ($2$ ions)
$(C)$ $[Co(NH_{3})_{5}Cl]SO_{4} \rightarrow [Co(NH_{3})_{5}Cl]^{2+} + SO_{4}^{2-}$ ($2$ ions)
$(D)$ $[Co(NH_{3})_{5}Cl]Cl_{2} \rightarrow [Co(NH_{3})_{5}Cl]^{2+} + 2Cl^{-}$ ($3$ ions)
Thus,the compound that dissociates into $3$ ions is $[Co(NH_{3})_{5}Cl]Cl_{2}$.
Hence,option $(D)$ is the correct answer.
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MediumMCQ
Cold ferrous sulphate solution on absorption of $NO$ develops brown colour due to the formation of
A
paramagnetic $[Fe(H_{2}O)_{5}(NO)]SO_{4}$
B
diamagnetic $[Fe(H_{2}O)_{5}(N_{3})]SO_{4}$
C
paramagnetic $[Fe(H_{2}O)_{5}(NO_{3})][SO_{4}]_{2}$
D
diamagnetic $[Fe(H_{2}O)_{4}(SO_{4})]NO_{3}$
Solution
(A) The brown ring test involves the reaction of cold $FeSO_{4}$ solution with $NO$ gas to form the brown-colored complex $[Fe(H_{2}O)_{5}(NO)]SO_{4}$.
$FeSO_{4} + NO + 5H_{2}O \rightarrow [Fe(H_{2}O)_{5}(NO)]SO_{4}$
In this complex,iron is in the $+1$ oxidation state ($d^7$ configuration). The $NO$ ligand acts as $NO^+$,making the iron center $Fe^+$. The complex exhibits a magnetic moment of $3.89 \ BM$,which corresponds to $3$ unpaired electrons,confirming it is paramagnetic.
$FeSO_{4} + NO + 5H_{2}O \rightarrow [Fe(H_{2}O)_{5}(NO)]SO_{4}$
In this complex,iron is in the $+1$ oxidation state ($d^7$ configuration). The $NO$ ligand acts as $NO^+$,making the iron center $Fe^+$. The complex exhibits a magnetic moment of $3.89 \ BM$,which corresponds to $3$ unpaired electrons,confirming it is paramagnetic.
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DifficultMCQ
Silver chloride dissolves in excess of ammonium hydroxide solution. The cation present in the resulting solution is
A
$[Ag(NH_3)_6]^+$
B
$[Ag(NH_3)_4]^+$
C
$Ag^+$
D
$[Ag(NH_3)_2]^+$
Solution
(D) $AgCl$ dissolves in excess $NH_4OH$ to form a soluble complex,diamminesilver$(I)$ chloride.
The chemical reaction is:
$AgCl(s) + 2NH_4OH(aq) \rightarrow [Ag(NH_3)_2]Cl(aq) + 2H_2O(l)$
The complex dissociates in solution as:
$[Ag(NH_3)_2]Cl \rightarrow [Ag(NH_3)_2]^+ + Cl^-$
Thus,the cation present in the resulting solution is $[Ag(NH_3)_2]^+$.
The chemical reaction is:
$AgCl(s) + 2NH_4OH(aq) \rightarrow [Ag(NH_3)_2]Cl(aq) + 2H_2O(l)$
The complex dissociates in solution as:
$[Ag(NH_3)_2]Cl \rightarrow [Ag(NH_3)_2]^+ + Cl^-$
Thus,the cation present in the resulting solution is $[Ag(NH_3)_2]^+$.
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MediumMCQ
Which of the following species exhibits both $LMCT$ and paramagnetism?
A
$MnO_4^{2-}$
B
$MnO_4^{-}$
C
$Cr_2O_7^{2-}$
D
$CrO_4^{2-}$
Solution
(A) $LMCT$ (Ligand to Metal Charge Transfer) causes intense color in oxoanions of transition metals.
$MnO_4^{2-}$ (manganate ion) has $Mn$ in the $+6$ oxidation state,which corresponds to a $3d^1$ electronic configuration.
Since it has one unpaired electron,it is paramagnetic.
Additionally,it exhibits color due to $LMCT$ from $O^{2-}$ to $Mn^{+6}$.
In contrast,$MnO_4^{-}$,$Cr_2O_7^{2-}$,and $CrO_4^{2-}$ have $Mn^{+7}$ $(3d^0)$ and $Cr^{+6}$ $(3d^0)$ respectively,which are diamagnetic.
$MnO_4^{2-}$ (manganate ion) has $Mn$ in the $+6$ oxidation state,which corresponds to a $3d^1$ electronic configuration.
Since it has one unpaired electron,it is paramagnetic.
Additionally,it exhibits color due to $LMCT$ from $O^{2-}$ to $Mn^{+6}$.
In contrast,$MnO_4^{-}$,$Cr_2O_7^{2-}$,and $CrO_4^{2-}$ have $Mn^{+7}$ $(3d^0)$ and $Cr^{+6}$ $(3d^0)$ respectively,which are diamagnetic.
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MediumMCQ
The reaction of nitroprusside anion with sulphide ion gives purple colouration due to the formation of
A
the tetranionic complex of iron $(II)$ coordinating to one $NOS^{4-}$ ion
B
the dianionic complex of iron $(II)$ coordinating to one $NCS^{-}$ ion
C
the trianionic complex of iron $(III)$ coordinating to one $NOS^{-}$ ion
D
the tetranionic complex of iron $(III)$ coordinating to one $NCS^{-}$ ion
Solution
(A) The reaction between sodium nitroprusside $(Na_2[Fe(CN)_5NO])$ and sulphide ion $(S^{2-})$ is a standard test for the detection of sulphur.
The reaction is as follows:
$Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NOS]$
This complex dissociates in solution as:
$Na_4[Fe(CN)_5NOS] \rightleftharpoons 4Na^+ + [Fe(CN)_5NOS]^{4-}$
The resulting complex $[Fe(CN)_5NOS]^{4-}$ is a tetranionic complex of iron $(II)$ where the $NO^+$ ligand is replaced by $NOS^{3-}$ (or effectively coordinates as $NOS^{3-}$ to the $Fe^{2+}$ center),resulting in a deep purple colouration.
The reaction is as follows:
$Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NOS]$
This complex dissociates in solution as:
$Na_4[Fe(CN)_5NOS] \rightleftharpoons 4Na^+ + [Fe(CN)_5NOS]^{4-}$
The resulting complex $[Fe(CN)_5NOS]^{4-}$ is a tetranionic complex of iron $(II)$ where the $NO^+$ ligand is replaced by $NOS^{3-}$ (or effectively coordinates as $NOS^{3-}$ to the $Fe^{2+}$ center),resulting in a deep purple colouration.
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EasyMCQ
$PbCl_{2}$ is insoluble in cold water. Addition of $HCl$ increases its solubility due to
A
formation of soluble complex anions like $[PbCl_{3}]^{-}$
B
oxidation of $Pb(II)$ to $Pb(IV)$
C
formation of $[Pb(H_{2}O)_{6}]^{2+}$
D
formation of polymeric lead complexes
Solution
(A) $PbCl_{2}$ is sparingly soluble in cold water. Upon adding $HCl$,the excess $Cl^{-}$ ions react with $PbCl_{2}$ to form soluble complex anions.
The reaction is as follows:
$PbCl_{2(s)} + Cl^{-} \rightarrow [PbCl_{3}]^{-}_{(aq)}$
$PbCl_{2(s)} + 2Cl^{-} \rightarrow [PbCl_{4}]^{2-}_{(aq)}$
These complex ions are soluble in water,which increases the overall solubility of $PbCl_{2}$ in the presence of $HCl$.
The reaction is as follows:
$PbCl_{2(s)} + Cl^{-} \rightarrow [PbCl_{3}]^{-}_{(aq)}$
$PbCl_{2(s)} + 2Cl^{-} \rightarrow [PbCl_{4}]^{2-}_{(aq)}$
These complex ions are soluble in water,which increases the overall solubility of $PbCl_{2}$ in the presence of $HCl$.
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DifficultMCQ
$A$ first row transition metal $(M)$ does not liberate $H_{2}$ gas from dilute $HCl$. $1 \ mol$ of aqueous solution of $MSO_{4}$ is treated with excess of aqueous $KCN$ and then $H_{2}S_{(g)}$ is passed through the solution. The amount of $MS$ (metal sulphide) formed from the above reaction is . . . . . . $mol$.
A
$2$
B
$1$
C
$3$
D
$0$
Solution
(D) The metal $M$ that does not liberate $H_{2}$ gas from dilute $HCl$ is $Cu$ (Copper),as it has a positive standard reduction potential $(E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V)$.
When $CuSO_{4}$ is treated with excess $KCN$,it forms a very stable complex $[Cu(CN)_{4}]^{3-}$.
The reaction is: $2CuSO_{4} + 10KCN \rightarrow 2K_{3}[Cu(CN)_{4}] + (CN)_{2} + 2K_{2}SO_{4}$.
Because the complex $[Cu(CN)_{4}]^{3-}$ is extremely stable (a 'perfect' complex),the concentration of free $Cu^{2+}$ ions in the solution is negligible.
Therefore,when $H_{2}S_{(g)}$ is passed through this solution,the ionic product does not exceed the solubility product $(K_{sp})$ of $CuS$,and no precipitate of $CuS$ is formed.
Thus,the amount of $MS$ formed is $0 \ mol$.
When $CuSO_{4}$ is treated with excess $KCN$,it forms a very stable complex $[Cu(CN)_{4}]^{3-}$.
The reaction is: $2CuSO_{4} + 10KCN \rightarrow 2K_{3}[Cu(CN)_{4}] + (CN)_{2} + 2K_{2}SO_{4}$.
Because the complex $[Cu(CN)_{4}]^{3-}$ is extremely stable (a 'perfect' complex),the concentration of free $Cu^{2+}$ ions in the solution is negligible.
Therefore,when $H_{2}S_{(g)}$ is passed through this solution,the ionic product does not exceed the solubility product $(K_{sp})$ of $CuS$,and no precipitate of $CuS$ is formed.
Thus,the amount of $MS$ formed is $0 \ mol$.
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MediumMCQ
Identify the correct statements from the following:
$A$. $[Fe(C_2O_4)_3]^{3-}$ is the most stable complex among $[Fe(OH)_6]^{3-}$,$[Fe(C_2O_4)_3]^{3-}$ and $[Fe(SCN)_6]^{3-}$.
$B$. The stability of $[Cu(NH_3)_4]^{2+}$ is greater than that of $[Cu(en)_2]^{2+}$.
$C$. The hybridization of $Fe$ in $K_4[Fe(CN)_6]$ is $d^2sp^3$.
$D$. $[Fe(NO_2)_3Cl_3]^{3-}$ exhibits linkage isomerism.
$E$. $NO_2^-$ and $SCN^-$ ligands are $NOT$ ambidentate ligands.
Choose the correct answer from the options given below:
$A$. $[Fe(C_2O_4)_3]^{3-}$ is the most stable complex among $[Fe(OH)_6]^{3-}$,$[Fe(C_2O_4)_3]^{3-}$ and $[Fe(SCN)_6]^{3-}$.
$B$. The stability of $[Cu(NH_3)_4]^{2+}$ is greater than that of $[Cu(en)_2]^{2+}$.
$C$. The hybridization of $Fe$ in $K_4[Fe(CN)_6]$ is $d^2sp^3$.
$D$. $[Fe(NO_2)_3Cl_3]^{3-}$ exhibits linkage isomerism.
$E$. $NO_2^-$ and $SCN^-$ ligands are $NOT$ ambidentate ligands.
Choose the correct answer from the options given below:
A
$A, B, C, D$ and $E$
B
$B, C$ and $D$ only
C
$A, C$ and $D$ only
D
$A, C$ and $E$ only
Solution
(C) is correct: The chelate effect makes $[Fe(C_2O_4)_3]^{3-}$ more stable than complexes with monodentate ligands like $OH^-$ or $SCN^-$.
$B$ is incorrect: $[Cu(en)_2]^{2+}$ is more stable than $[Cu(NH_3)_4]^{2+}$ because $en$ (ethylenediamine) is a bidentate ligand that forms stable $5-$membered chelate rings.
$C$ is correct: In $K_4[Fe(CN)_6]$,$Fe$ is in the $+2$ oxidation state ($d^6$ configuration). $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $d^2sp^3$ hybridization.
$D$ is correct: $NO_2^-$ is an ambidentate ligand,so $[Fe(NO_2)_3Cl_3]^{3-}$ can exhibit linkage isomerism.
$E$ is incorrect: $NO_2^-$ and $SCN^-$ are classic examples of ambidentate ligands.
Therefore,statements $A, C,$ and $D$ are correct.
$B$ is incorrect: $[Cu(en)_2]^{2+}$ is more stable than $[Cu(NH_3)_4]^{2+}$ because $en$ (ethylenediamine) is a bidentate ligand that forms stable $5-$membered chelate rings.
$C$ is correct: In $K_4[Fe(CN)_6]$,$Fe$ is in the $+2$ oxidation state ($d^6$ configuration). $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $d^2sp^3$ hybridization.
$D$ is correct: $NO_2^-$ is an ambidentate ligand,so $[Fe(NO_2)_3Cl_3]^{3-}$ can exhibit linkage isomerism.
$E$ is incorrect: $NO_2^-$ and $SCN^-$ are classic examples of ambidentate ligands.
Therefore,statements $A, C,$ and $D$ are correct.
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DifficultMCQ
$5.33 \text{ g}$ of $CrCl_3 \cdot 6H_2O$,which is a $1:3$ electrolyte,is dissolved in water and passed through a cation exchanger. The chloride ions in the eluted solution,on treatment with $AgNO_3$,result in $8.61 \text{ g}$ of $AgCl$. The ratio of moles of complex reacted and moles of $AgCl$ formed is . . . . . . $\times 10^{-2}$. (Nearest integer) [Molar mass in $\text{g mol}^{-1}$: $Cr = 52, Ag = 108, Cl = 35.5, H = 1, O = 16$]
A
$20$
B
$50$
C
$80$
D
$33$
Solution
(D) $1$. Calculate the molar mass of $CrCl_3 \cdot 6H_2O$: $52 + (35.5 \times 3) + (6 \times 18) = 52 + 106.5 + 108 = 266.5 \text{ g/mol}$.
$2$. Calculate the moles of the complex: $\text{Moles} = \frac{5.33 \text{ g}}{266.5 \text{ g/mol}} = 0.02 \text{ mol}$.
$3$. Calculate the molar mass of $AgCl$: $108 + 35.5 = 143.5 \text{ g/mol}$.
$4$. Calculate the moles of $AgCl$ formed: $\text{Moles} = \frac{8.61 \text{ g}}{143.5 \text{ g/mol}} = 0.06 \text{ mol}$.
$5$. The ratio of moles of complex reacted to moles of $AgCl$ formed is $\frac{0.02}{0.06} = \frac{1}{3} \approx 0.3333$.
$6$. Expressing this as $\times 10^{-2}$,we get $33.33 \times 10^{-2}$. The nearest integer is $33$.
$2$. Calculate the moles of the complex: $\text{Moles} = \frac{5.33 \text{ g}}{266.5 \text{ g/mol}} = 0.02 \text{ mol}$.
$3$. Calculate the molar mass of $AgCl$: $108 + 35.5 = 143.5 \text{ g/mol}$.
$4$. Calculate the moles of $AgCl$ formed: $\text{Moles} = \frac{8.61 \text{ g}}{143.5 \text{ g/mol}} = 0.06 \text{ mol}$.
$5$. The ratio of moles of complex reacted to moles of $AgCl$ formed is $\frac{0.02}{0.06} = \frac{1}{3} \approx 0.3333$.
$6$. Expressing this as $\times 10^{-2}$,we get $33.33 \times 10^{-2}$. The nearest integer is $33$.
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