Complexes and complex stability Questions in English

(B) Ammonia $(NH_3)$ acts as a ligand and forms complex compounds with transition metal ions that have vacant $d$-orbitals available for bonding.
$Ag^{+}$,$Cd^{2+}$,and $Fe^{2+}$ are transition metal ions that readily form ammine complexes (e.g.,$[Ag(NH_3)_2]^{+}$,$[Cd(NH_3)_4]^{2+}$,$[Fe(NH_3)_6]^{2+}$).
$Pb^{2+}$ is a $p$-block element ion and does not typically form stable ammine complexes with ammonia in aqueous solution.

(D) When $KCN$ is added to a copper sulfate solution,the following reactions occur:
$1$. $2CuSO_4 + 4KCN \rightarrow 2Cu(CN)_2 + 2K_2SO_4$
$2$. The unstable $Cu(CN)_2$ decomposes: $2Cu(CN)_2 \rightarrow Cu_2(CN)_2 + (CN)_2 \uparrow$
$3$. The $Cu_2(CN)_2$ reacts with excess $KCN$ to form a stable complex: $Cu_2(CN)_2 + 6KCN \rightarrow 2K_3[Cu(CN)_4]$
Thus,the final product is $K_3[Cu(CN)_4]$.

(D) The ionic or atomic radius of $Fe$ depends on its oxidation state and the nature of the ligands attached.
$1$. In $K_4[Fe(CN)_6]$,$Fe$ is in the $+2$ oxidation state.
$2$. In $FeSO_4 \cdot 7H_2O$,$Fe$ is in the $+2$ oxidation state.
$3$. In $FeSO_4(NH_4)_2SO_4 \cdot 6H_2O$ (Mohr's salt),$Fe$ is in the $+2$ oxidation state.
$4$. In $Fe(CO)_5$,$Fe$ is in the $0$ oxidation state.
Since the radius of an atom increases as the oxidation state decreases (due to lower effective nuclear charge),the $Fe$ atom in the $0$ oxidation state $(Fe(CO)_5)$ will have the largest radius compared to $Fe^{2+}$ ions.

(A) When excess $NH_3$ is added to a $CuSO_4$ solution,the pale blue precipitate of $Cu(OH)_2$ initially formed dissolves to form a deep blue complex known as tetraamminecopper$(II)$ sulfate.
The chemical reaction is: $CuSO_4(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]SO_4(aq)$.
The complex ion formed is $[Cu(NH_3)_4]^{2+}$.

(A) In $KMnO_4$,the manganese atom is in the $+7$ oxidation state,which corresponds to a $d^0$ electronic configuration.
Since there are no $d$-electrons available for $d-d$ transition,the color cannot be attributed to $d-d$ transitions.
The intense purple color of $KMnO_4$ arises due to the charge transfer from the oxygen ligand to the metal center $(O^{2-} \rightarrow Mn^{7+})$.
This is known as ligand-to-metal charge transfer $(LMCT)$.

(B) The complex ion $[Co(NH_3)_6]^{3+}$ is a coordination compound where the central metal ion is $Co^{3+}$.
In this octahedral complex,the $NH_3$ ligands cause a specific crystal field splitting.
The aqueous solution of $[Co(NH_3)_6]^{3+}$ exhibits a yellow-orange color due to the absorption of light in the blue-violet region of the visible spectrum.

(C) Transition metal ions form complex compounds due to the following reasons:
$1$. Small size and high nuclear charge of metal ions.
$2$. Availability of vacant $d$-orbitals of appropriate energy to accept lone pairs of electrons from ligands.
$3$. Variable oxidation states.
Coordinate covalent bonds are highly directional in nature,which allows for the formation of specific geometries in complex compounds. Therefore,the statement that coordinate covalent bonds are non-directional is incorrect.

(C) In all the given complexes,the central metal ion $Fe$ is in the $+3$ oxidation state.
$[Fe(C_2O_4)_3]^{3-}$ contains oxalate ions,which are bidentate ligands.
These ligands form a ring structure with the metal ion,known as the chelate effect.
Chelate complexes are significantly more stable than complexes with monodentate ligands due to the increased entropy associated with chelation.

(A) The coordination number of $Pt$ is $6$. The complexes can be written as:
$(i) K_2[PtCl_6] \rightarrow 2K^+ + [PtCl_6]^{2-}$ (Total $3$ ions)
$(ii) [Pt(NH_3)_2Cl_4] \rightarrow$ Non-electrolyte (Total $0$ ions)
$(iii) [Pt(NH_3)_3Cl_3]Cl \rightarrow [Pt(NH_3)_3Cl_3]^+ + Cl^-$ (Total $2$ ions)
$(iv) [Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^-$ (Total $4$ ions)
Since molar conductivity is directly proportional to the number of ions produced in solution,the order of conductivity corresponds to the number of ions: $4 > 3 > 2 > 0$.
Matching the values: $(iv) = 404$,$(i) = 256$,$(iii) = 97$,$(ii) = 0$.
Thus,the correct sequence is $256, 0, 97, 404$.

(B) The stability of a coordination complex is directly proportional to its stability constant $(K)$.
$A$ higher value of $K$ indicates that the complex is more stable,which implies that the ligand forming the complex is stronger.
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.

(C) The stability of coordination complexes is significantly influenced by the chelate effect.
Chelating ligands form ring structures with the central metal ion,which increases the stability of the complex compared to complexes with monodentate ligands.
In the given options,$[Fe(C_2O_4)_3]^{3-}$ contains oxalate ions $(C_2O_4^{2-})$,which are bidentate ligands that form stable five-membered chelate rings with the $Fe^{3+}$ ion.
Therefore,$[Fe(C_2O_4)_3]^{3-}$ is the most stable complex among the choices.

(C) Molar conductivity is directly proportional to the number of ions produced in the solution.
$1$. $[Pt(NH_3)_2Cl_2]$ is a non-electrolyte,producing $0$ ions.
$2$. $[Co(NH_3)_4Cl_2]Cl$ dissociates into $2$ ions: $[Co(NH_3)_4Cl_2]^+$ and $Cl^-$.
$3$. $[Cr(H_2O)_6]Br_3$ dissociates into $4$ ions: $[Cr(H_2O)_6]^{3+}$ and $3Br^-$.
$4$. $[Fe(CO)_5]$ is a non-electrolyte,producing $0$ ions.
Since $[Cr(H_2O)_6]Br_3$ produces the maximum number of ions $(4)$,it exhibits the maximum molar conductivity.

(B) $1$. Calculate the moles of the complex: $n(\text{complex}) = \frac{2.675 \ g}{267.5 \ g \ mol^{-1}} = 0.01 \ mol$.
$2$. Calculate the moles of $AgCl$ precipitated: $n(AgCl) = \frac{4.78 \ g}{143.5 \ g \ mol^{-1}} \approx 0.0333 \ mol$.
$3$. Determine the ratio of $AgCl$ to complex: $\frac{n(AgCl)}{n(\text{complex})} = \frac{0.0333}{0.01} \approx 3$.
$4$. Since $3 \ mol$ of $Cl^-$ ions are released per mole of complex,all $3 \ Cl^-$ ions must be outside the coordination sphere.
$5$. Therefore,the formula is $[Co(NH_3)_6]Cl_3$.

(B) The spectrochemical series is an arrangement of ligands in order of increasing field strength. The order of increasing field strength for the given ligands is: $Cl^{-} < NCS^{-} < en < CN^{-}$.
Comparing this with the options,the correct order of decreasing field strength is $CN^{-} > en > NCS^{-} > Cl^{-}$.
Thus,option $B$ is the correct answer.

(A) Molar conductivity depends on the number of ions produced in the solution.
$1$. $[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^-$ (Total $4$ ions)
$2$. $[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^-$ (Total $3$ ions)
$3$. $[Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^+ + Cl^-$ (Total $2$ ions)
$4$. $[Co(NH_3)_4]SO_4 \rightarrow [Co(NH_3)_4]^{2+} + SO_4^{2-}$ (Total $2$ ions)
Since $[Co(NH_3)_6]Cl_3$ produces the maximum number of ions $(4)$,it exhibits the maximum molar conductivity.

(A) The spectrochemical series is an arrangement of ligands in order of their increasing field strength.
According to the spectrochemical series,the order of field strength for the given ligands is:
$Cl^- < NH_3 < en < CO$.
Thus,the correct increasing order is $Cl^- < NH_3 < en < CO$.

(D) The spectrochemical series is an experimental series in which ligands are arranged in order of increasing field strength.
Based on the spectrochemical series,the order of field strength is:
$I^- < Br^- < S^{2-} < SCN^- < Cl^- < F^- < OH^- < H_2O < NCS^- < EDTA^{4-} < NH_3 < en < CN^- < CO$.
Comparing the given options,option $D$ provides a correct subset of this sequence in increasing order of field strength: $Cl^- < F^- < OH^- < H_2O < CN^-$.

(A) The $CN^-$ ion acts as a strong ligand,which facilitates the formation of stable coordination complexes with metal ions,hence it exhibits $(c)$ complexing property.
It also acts as a reducing agent in many chemical reactions involving metal ions,as it can be oxidized to $(CN)_2$,hence it exhibits $(a)$ reducing property.
Therefore,the $CN^-$ ion exhibits both $(a)$ reducing and $(c)$ complexing properties.

(C) The stability of a coordination complex depends on the strength of the ligand field. Stronger ligands cause greater crystal field splitting,leading to higher stability.
The order of ligand strength is: $CN^{-} > NH_3 > H_2O > Cl^{-}$.
Among the given options,$CN^{-}$ is the strongest ligand. Therefore,the complex $[Fe(CN)_6]^{4-}$ is the most stable.

(B) The stability constant of a complex is directly proportional to the strength of the ligand.
Higher stability constant values indicate that the metal-ligand bond is stronger,which implies the ligand is a stronger field ligand.
Comparing the given values: $10^{27} (CN^-) > 10^{15} (H_2O) > 10^{11} (NH_3) > 10^8 (en)$.
Since $CN^-$ has the highest stability constant,it is the strongest ligand among the given options.

(A) The spectrochemical series is an 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 the given ligands: $Cl^{-} < F^{-} < C_2O_4^{2-} < NO_2^{-} < CN^{-}$.
Thus,option $A$ is the correct order.

(A) Conductivity of a coordination complex in solution is directly proportional to the number of ions produced upon dissociation.

Complex	Number of ions
$K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$	$5$
$[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^-$	$4$
$[Cu(NH_3)_4]Cl_2 \rightarrow [Cu(NH_3)_4]^{2+} + 2Cl^-$	$3$
$[Ni(CO)_4] \rightarrow \text{Non-electrolyte}$	$0$

Since $K_4[Fe(CN)_6]$ produces the highest number of ions $(5)$,it exhibits the maximum conductivity.

(A) Molar conductivity of coordination complexes depends on the number of ions produced in the solution.
$i. \ [Pt(NH_3)_3Cl_3]Cl \rightarrow [Pt(NH_3)_3Cl_3]^+ + Cl^-$ ($2$ ions)
$ii. \ [Pt(NH_3)_4Cl_2]Cl_2 \rightarrow [Pt(NH_3)_4Cl_2]^{2+} + 2Cl^-$ ($3$ ions)
$iii. \ [Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^-$ ($4$ ions)
$iv. \ [Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$ ($5$ ions)
As the number of ions increases,molar conductivity increases.
$97$ corresponds to $2$ ions $(i)$,
$229$ corresponds to $3$ ions $(ii)$,
$404$ corresponds to $4$ ions $(iii)$,
$523$ corresponds to $5$ ions $(iv)$.
Therefore,the correct match is $A-ii, B-i, C-iii, D-iv$.

(D) The stability of coordination complexes is influenced by the chelate effect.
Chelating ligands form more stable complexes compared to monodentate ligands due to the increase in entropy.
Among the given options,$EDTA^{4-}$ is a hexadentate ligand that forms a very stable complex with metal ions due to the formation of multiple five-membered chelate rings.
Therefore,$[Ni(EDTA)]^{2-}$ is the most stable complex among the choices provided.

(A) The $CN^-$ ion acts as a reducing agent because the carbon atom can be oxidized to a higher oxidation state.
It also acts as a complexing agent (ligand) because it has a lone pair of electrons on the carbon atom,which it can donate to a central metal ion to form coordination complexes.

(C) The stability of a coordination complex is significantly enhanced by the chelate effect.
Among the given options,$[Fe(C_2O_4)_3]^{3-}$ contains the oxalate ion $(C_2O_4^{2-})$,which is a bidentate ligand.
This ligand forms five-membered chelate rings with the central metal ion,leading to increased thermodynamic stability compared to complexes with monodentate ligands like $H_2O$,$CN^-$,or $Cl^-$.
Therefore,$[Fe(C_2O_4)_3]^{3-}$ is the most stable complex.

(C) The complex is $CrCl_3 \cdot 6H_2O$.
When $AgNO_3$ is added,only the chloride ions $(Cl^-)$ present outside the coordination sphere (ionizable) will react to form $AgCl$ precipitate.
Given that $1/3$ of the total chlorine is precipitated,it means $1$ out of $3$ chlorine atoms is outside the coordination sphere.
Therefore,the formula must be $[CrCl_2(H_2O)_4]Cl \cdot 2H_2O$.
This structure contains $1$ ionizable $Cl^-$ ion.

(D) The stability of coordination compounds is directly proportional to the stability constant $(K)$.
Stability is also directly proportional to the charge on the central metal ion $(CMI)$ and inversely proportional to the size of the $CMI$.
Stronger basic ligands form more stable complexes.
Chelate complexes are generally more stable than non-chelate complexes due to the chelate effect,meaning they have higher stability constants.
Therefore,the statement that chelate complexes have a lower stability constant is incorrect.

(A) Electrical conductivity is directly proportional to the number of ions produced in the solution.
$(I) [Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$ (Total $5$ ions)
$(II) [Cr(NH_3)_6]Cl_3 \rightarrow [Cr(NH_3)_6]^{3+} + 3Cl^-$ (Total $4$ ions)
$(III) [Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^+ + Cl^-$ (Total $2$ ions)
$(IV) K_2[PtCl_6] \rightarrow 2K^+ + [PtCl_6]^{2-}$ (Total $3$ ions)
Comparing the number of ions: $III (2) < IV (3) < II (4) < I (5)$.
Therefore,the correct increasing order of conductivity is $III < IV < II < I$.

(B) $CN^-$ is a strong field ligand according to the spectrochemical series.
Due to its high crystal field splitting energy $(\Delta_o)$,it causes pairing of electrons in the $d$-orbitals.
Therefore,it generally forms low spin complexes.

(B) In an acidic medium,the concentration of $H^+$ ions is high. Ammonia $(NH_3)$ is a Lewis base and reacts with $H^+$ ions to form ammonium ions $(NH_4^+)$ according to the reaction: $NH_3 + H^+ \rightarrow NH_4^+$. Since the lone pair of electrons on nitrogen is now involved in the formation of the $NH_4^+$ ion,it is no longer available to donate to the $Cu^{2+}$ ion to form the coordinate covalent bond required for the complex $[Cu(NH_3)_4]^{2+}$. Therefore,the complex does not form in an acidic medium.

(B) The formation of the complex $[Cu(NH_3)_4]^{2+}$ requires free $NH_3$ molecules to act as ligands.
In an acidic solution,the concentration of $H^+$ ions is high.
These $H^+$ ions react with $NH_3$ (a base) to form ammonium ions: $NH_3 + H^+ \rightarrow NH_4^+$.
Because the $NH_3$ is consumed to form $NH_4^+$,there are no free $NH_3$ molecules available to coordinate with the $Cu^{2+}$ ions to form the complex.

(C) The complex contains $1$ $Co$ atom,$5$ $NH_3$ molecules,$1$ $NO_2$ group,and $2$ $Cl$ atoms.
Since $2$ moles of $AgCl$ are precipitated with $AgNO_3$,it indicates that $2$ $Cl^-$ ions are present outside the coordination sphere (ionizable).
The formula is $[Co(NH_3)_5(NO_2)]Cl_2$.
Dissociation: $[Co(NH_3)_5(NO_2)]Cl_2 \rightarrow [Co(NH_3)_5(NO_2)]^{2+} + 2Cl^-$.
Total ions produced = $1 + 2 = 3$ moles of ions.

(D) In the complex $[Co(NH_3)_3Cl_3]$,the three $Cl^-$ ions are bonded to the $Co^{3+}$ metal ion as ligands within the coordination sphere (secondary valency).
Since these $Cl^-$ ions are not present as counter-ions outside the coordination sphere,they do not ionize in water.
Therefore,they do not react with $AgNO_3$ to form a precipitate of $AgCl$.

(A) Molar conductivity depends on the number of ions produced in the solution. The more ions produced,the higher the conductivity.
$1. [Co(NH_3)_3Cl_3] \rightarrow \text{No ions produced (non-electrolyte)}$
$2. [Co(NH_3)_4Cl_2]Cl \rightleftharpoons [Co(NH_3)_4Cl_2]^+ + Cl^- \text{ (2 ions)}$
$3. [Co(NH_3)_5Cl]Cl_2 \rightleftharpoons [Co(NH_3)_5Cl]^{2+} + 2Cl^- \text{ (3 ions)}$
$4. [Co(NH_3)_6]Cl_3 \rightleftharpoons [Co(NH_3)_6]^{3+} + 3Cl^- \text{ (4 ions)}$
Since $[Co(NH_3)_3Cl_3]$ does not ionize in solution,it has the lowest molar conductivity.

(A) Moles of $AgCl$ produced $= \frac{4.78 \ g}{143.5 \ g/mol} = 0.0333 \ mol$.
Moles of complex $= \frac{2.675 \ g}{267.5 \ g/mol} = 0.01 \ mol$.
Number of ionizable $Cl^-$ ions per molecule of complex $= \frac{0.0333 \ mol}{0.01 \ mol} = 3.33 \approx 3$.
Since there are $3$ ionizable chloride ions,the formula is $[Co(NH_3)_6]Cl_3$.

(D) The stability of a coordination complex in solution is determined by the stability constant $(K)$.
Stronger ligands form more stable complexes with metal ions.
According to the spectrochemical series and the nature of the ligand,$CN^-$ is a strong field ligand and forms a very stable complex with $Cd^{2+}$ ions compared to halide ions ($Cl^-$,$Br^-$,$I^-$).
The stability constant for $[Cd(CN)_4]^{2-}$ is significantly higher than that of the halide complexes.
Therefore,$[Cd(CN)_4]^{2-}$ is the most stable complex.

(B) coordination compound acts as a non-electrolyte if it does not dissociate into ions in an aqueous solution.
In the complex $[Co(NH_3)_3(NO_2)_3]$,the coordination sphere is neutral and does not contain any ionizable counter-ions.
Therefore,it does not produce ions in solution and acts as a non-electrolyte.

(D) The correct answer is $(d)$ $Pb^{2+}$.
Ammonia $(NH_3)$ acts as a ligand and forms stable complexes with transition metal ions like $Ag^{+}$,$Cu^{2+}$,and $Cd^{2+}$ due to the availability of vacant $d$-orbitals and their ability to accept lone pairs of electrons.
$Pb^{2+}$ is a post-transition metal ion that does not readily form stable ammine complexes in aqueous solution compared to transition metal ions,as it lacks the specific electronic configuration and $d$-orbital characteristics required for such complexation.

(B) The dissolution of $AgCl$ in aqueous ammonia occurs because of the formation of a soluble complex ion,diamminesilver$(I)$ chloride.
The chemical equation for the reaction is:
$AgCl(s) + 2NH_3(aq) \to [Ag(NH_3)_2]Cl(aq)$
Thus,the correct product formed is $[Ag(NH_3)_2]Cl$.

(B) $AlF_3$ is insoluble in $HF$ because it is an ionic solid with a high lattice energy.
In the presence of $KF$,$AlF_3$ reacts to form a soluble complex salt.
The reaction is: $AlF_3 + 3KF \rightarrow K_3[AlF_6]$.
The coordination number of $Al^{3+}$ is $6$,leading to the formation of the stable octahedral complex $[AlF_6]^{3-}$.

(C) When $CuSO_4$ reacts with $KCN$,it first forms a precipitate of cupric cyanide $(Cu(CN)_2)$,which is unstable and decomposes to form cuprous cyanide $(Cu_2(CN)_2)$ and cyanogen gas $((CN)_2)$.
$2CuSO_4 + 4KCN \rightarrow 2Cu(CN)_2 + 2K_2SO_4$
$2Cu(CN)_2 \rightarrow Cu_2(CN)_2 + (CN)_2$
The cuprous cyanide $(Cu_2(CN)_2)$ then dissolves in excess $KCN$ to form the stable soluble complex potassium tetracyanocuprate$(I)$,$K_3[Cu(CN)_4]$.
$Cu_2(CN)_2 + 6KCN \rightarrow 2K_3[Cu(CN)_4]$
Overall reaction:
$2CuSO_4 + 10KCN \rightarrow 2K_3[Cu(CN)_4] + 2K_2SO_4 + (CN)_2$

(B) The trans-effect is defined as the effect of a coordinated ligand upon the rate of substitution of ligands attached to the trans position. The experimental order of the trans-effect for the given species is $NH_3 < Br^{-} < C_6H_5^{-} < CN^{-}$.
Thus,the correct increasing order is $NH_3 < Br^{-} < C_6H_5^{-} < CN^{-}$.

(C) The octahedral complex requires a coordination number of $6$.
In $CoCl_3 \cdot 3NH_3$,the complex is formulated as $[Co(NH_3)_3Cl_3]$.
Since all $3$ chloride ions are inside the coordination sphere,they are not ionizable.
Therefore,it will not produce a precipitate of $AgCl$ when treated with silver nitrate $(AgNO_3)$.

(A) Number of moles of $CoCl_3 \cdot 6 NH_3 = \frac{2.675}{267.5} = 0.01 \ mol$.
Number of moles of $AgCl = \frac{4.78}{143.5} = 0.03 \ mol$.
Since $0.01 \ mol$ of the complex $CoCl_3 \cdot 6 NH_3$ yields $0.03 \ mol$ of $AgCl$ upon treatment with excess $AgNO_3$,it indicates that $3$ chloride ions are present in the ionisable sphere.
Therefore,the formula of the complex is $[Co(NH_3)_6]Cl_3$.

(D) The number of moles of $CoCl_3 \cdot 6H_2O$ is $0.1 \ M \times 0.1 \ L = 0.01 \ mol$.
The number of moles of $AgCl$ precipitated is $\frac{1.2 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.02 \ mol$.
Since $0.01 \ mol$ of complex produces $0.02 \ mol$ of $AgCl$,$1 \ mol$ of complex produces $2 \ mol$ of $AgCl$.
This indicates that there are $2$ chloride ions outside the coordination sphere.
Therefore,the complex is $[Co(H_2O)_5 Cl]Cl_2 \cdot H_2O$.

(C) The $C-O$ bond length in metal carbonyl complexes depends on the extent of back-bonding (synergic bonding) from the metal to the $CO$ ligand.
Greater back-bonding increases the electron density in the antibonding $\pi^*$ orbital of $CO$,which decreases the $C-O$ bond order and increases the $C-O$ bond length.
$PMe_3$ is a better $\sigma$-donor ligand compared to $PF_3$ and $CO$.
Therefore,in $[Ni(CO)_3(PMe_3)]$,the electron density on the $Ni$ atom is higher,which leads to greater back-donation into the $CO$ ligands.
Consequently,$[Ni(CO)_3(PMe_3)]$ exhibits the maximum $C-O$ bond length.

(A) The brown ring test is a common chemical test for the detection of nitrate ions $(NO_3^-)$.
When a freshly prepared ferrous sulfate $(FeSO_4)$ solution is added to a solution containing nitrate ions,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 reaction involves the reduction of nitrate to nitric oxide $(NO)$,which then reacts with the hexaaquairon$(II)$ complex to form the brown-colored complex known as pentaaquanitrosyliron$(II)$ sulfate,represented as $[Fe(H_2O)_5NO]SO_4$.

(C) The stability of a coordination complex is determined by its stability constant $(K)$.
$A$ higher value of the stability constant indicates a more stable complex.
$CN^{-}$ is a strong field ligand (a $\pi$-acceptor ligand) compared to $NH_3$,which leads to a stronger metal-ligand bond and a higher stability constant for the $[Cu(CN)_4]^{2-}$ complex.
Therefore,both the higher stability constant and the stronger ligand nature of $CN^{-}$ contribute to the greater stability of $[Cu(CN)_4]^{2-}$.