Complexes and complex stability Questions in English

(D) The solubility of silver bromide $(AgBr)$ in sodium thiosulfate $(Na_2S_2O_3)$,commonly known as 'hypo' solution,occurs due to the formation of a soluble complex ion.
The chemical reaction is:
$AgBr_{(s)} + 2 Na_2S_2O_{3(aq)} \rightarrow Na_3[Ag(S_2O_3)_2]_{(aq)} + NaBr_{(aq)}$
In this reaction,the complex ion formed is $[Ag(S_2O_3)_2]^{3-}$.
Therefore,option $D$ is correct.

(A) Molar conductivity is directly proportional to the number of ions produced in the solution.
$1$. $[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$ (Total $5$ ions)
$2$. $[Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^-$ (Total $4$ ions)
$3$. $[Pt(NH_3)_4Cl_2]Cl_2 \rightarrow [Pt(NH_3)_4Cl_2]^{2+} + 2Cl^-$ (Total $3$ ions)
$4$. $[Pt(NH_3)_3Cl_3]Cl \rightarrow [Pt(NH_3)_3Cl_3]^+ + Cl^-$ (Total $2$ ions)
Since $[Pt(NH_3)_6]Cl_4$ produces the maximum number of ions ($5$ ions),it exhibits the highest molar conductivity.

(B) stability constant $(K)$ is an equilibrium constant for the formation of a complex in solution.
It is a measure of the strength of the interaction between the metal ion and the ligand.
$A$ higher value of the stability constant indicates a more stable complex,which implies that the ligand is stronger.
Comparing the given values:
$K(H_2O) = 9.5 \times 10^8$
$K(NH_3) = 4.5 \times 10^{11}$
$K(en) = 3.0 \times 10^{15}$
$K(CN^-) = 2.0 \times 10^{27}$
Since the reaction $Cu^{2+} + 4CN^{-} \rightleftharpoons [Cu(CN)_4]^{2-}$ has the maximum stability constant value $(2.0 \times 10^{27})$,$CN^{-}$ is the strongest ligand.

(C) The total number of chloride ions in $CrCl_3 \cdot 6H_2O$ is $3$.
Since $1/3$ of the total chlorine is precipitated by $AgNO_3$,it means only $1$ chloride ion is present outside the coordination sphere (ionizable).
The formula must be of the form $[CrCl_2(H_2O)_4]Cl \cdot 2H_2O$.

(B) In octahedral complexes,the central metal ion is coordinated with $6$ ligands.
The number of ions produced by the complexes in aqueous solution is determined by their ionization:
$(1) [Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^-$ ($4$ ions)
$(2) [Pt(NH_3)_4Cl_2]Cl_2 \rightarrow [Pt(NH_3)_4Cl_2]^{2+} + 2Cl^-$ ($3$ ions)
$(3) [Pt(NH_3)_3Cl_3]Cl \rightarrow [Pt(NH_3)_3Cl_3]^+ + Cl^-$ ($2$ ions)
$(4) [Pt(NH_3)_2Cl_4] \rightarrow [Pt(NH_3)_2Cl_4]$ (Non-electrolyte,$0$ ions)
Since molar conductance is directly proportional to the number of ions produced,the order is $IV < III < II < I$.

(B) Number of millimoles of the complex $= 50 \, mL \times 0.2 \, M = 10 \, \text{mmol}$.
Number of millimoles of $AgCl$ formed $= \frac{1.435 \, g}{143.5 \, g/mol} = 0.01 \, mol = 10 \, \text{mmol}$.
Since $10 \, \text{mmol}$ of the complex yields $10 \, \text{mmol}$ of $AgCl$,each mole of the complex releases $1$ mole of ionizable $Cl^-$ ions.
This indicates the presence of one $Cl^-$ ion outside the coordination sphere.
Therefore,the formula is $[Co(NH_3)_4Cl_2]Cl$.

(C) When $KCN$ is added in excess to a solution containing $Fe^{2+}$ ions,the $Fe^{2+}$ ions react to form a stable coordination complex.
The reaction is: $Fe^{2+} + 6CN^- \rightarrow [Fe(CN)_6]^{4-}$.
Since the $Fe^{2+}$ ions are now part of the stable complex ion $[Fe(CN)_6]^{4-}$,they are no longer available as free ions in the solution to give the characteristic chemical tests for $Fe^{2+}$.

(D) When an excess of $KCN$ is added to an aqueous solution of copper sulphate $(CuSO_4)$,the initial precipitate of copper$(II)$ cyanide $(Cu(CN)_2)$ is unstable and decomposes to form copper$(I)$ cyanide $(CuCN)$ and cyanogen gas $((CN)_2)$.
Upon adding an excess of $KCN$,the $CuCN$ dissolves to form the stable complex potassium tetracyanocuprate$(I)$,which is $K_3[Cu(CN)_4]$.
The overall balanced chemical equation is:
$2CuSO_4 + 10KCN \rightarrow 2K_3[Cu(CN)_4] + (CN)_2 + 2K_2SO_4$

(D) When $CuSO_4$ reacts with $KCN$,it first forms a precipitate of $Cu(CN)_2$,which is unstable and decomposes to give $CuCN$ and cyanogen gas $(CN)_2$.
$2CuSO_4 + 4KCN \rightarrow 2CuCN + (CN)_2 + 2K_2SO_4$
On adding excess $KCN$,the $CuCN$ precipitate dissolves to form a stable complex,potassium tetracyanocuprate$(I)$,$K_3[Cu(CN)_4]$.
$CuCN + 3KCN \rightarrow K_3[Cu(CN)_4]$

(A) $Lewis$ acid is defined as an electron pair acceptor.
Transition metal ions have vacant $d$-orbitals which can accept electron pairs from ligands to form coordinate bonds.
This ability to accept electron pairs allows them to act as $Lewis$ acids,which is a fundamental requirement for the formation of coordination complexes.

(C) The acidity of metal complexes $[M(H_2O)_6]^{n+}$ depends on the charge density of the metal ion.
Higher charge $(n+)$ and smaller ionic radius lead to higher acidity because the metal ion polarizes the $O-H$ bond of the coordinated water molecules more strongly,facilitating the release of $H^+$.
$1$. Comparing charges: $[M]^{2+}$ ions are less acidic than $[M]^{3+}$ ions.
$2$. Comparing $[Mn(H_2O)_6]^{2+}$ and $[Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ has a smaller ionic radius than $Mn^{2+}$,so $[Ni(H_2O)_6]^{2+}$ is more acidic.
$3$. Comparing $[Fe(H_2O)_6]^{3+}$ and $[Al(H_2O)_6]^{3+}$: $Al^{3+}$ has a smaller ionic radius than $Fe^{3+}$,so $[Al(H_2O)_6]^{3+}$ is more acidic.
Thus,the correct order is $[Mn(H_2O)_6]^{2+} < [Ni(H_2O)_6]^{2+} < [Fe(H_2O)_6]^{3+} < [Al(H_2O)_6]^{3+}$.

(D) In metal carbonyls and metal phosphine complexes,both $CO$ and $PF_3$ act as $\sigma$-donors and $\pi$-acceptors.
$CO$ accepts electron density from the metal into its empty antibonding $\pi^*$ molecular orbital.
$PF_3$ accepts electron density from the metal into its empty $3d$ orbitals (or $\sigma^*$ orbitals depending on the model).
Option $A$ is incorrect because $CO$ is not a $\pi$-donor.
Option $B$ is incorrect as both can be toxic.
Option $C$ is a physical property difference,but the primary chemical distinction in coordination chemistry is the nature of the $\pi$-acceptor orbital.
Option $D$ correctly identifies the difference in the acceptor orbitals involved in back-bonding.

(C) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 273 \ K - 269.28 \ K = 3.72 \ K$.
Using the formula $\Delta T_f = i \times K_f \times m$,where $m = 1 \ m$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$:
$3.72 = i \times 1.86 \times 1
\implies i = \frac{3.72}{1.86} = 2$.
Since the complex undergoes $100\%$ ionization,the van't Hoff factor $i$ represents the total number of ions produced per formula unit.
For $i = 2$,the complex must dissociate into $2$ ions.
Option $B$ is $[Pd(H_2O)_4Cl_2]Cl_2 \cdot 2H_2O$,which dissociates as $[Pd(H_2O)_4Cl_2]Cl_2 \cdot 2H_2O \rightarrow [Pd(H_2O)_4Cl_2]^{2+} + 2Cl^-$. This gives $3$ ions $(i=3)$.
Option $C$ is $[Pd(H_2O)_3Cl_3]Cl \cdot 3H_2O$,which dissociates as $[Pd(H_2O)_3Cl_3]Cl \cdot 3H_2O \rightarrow [Pd(H_2O)_3Cl_3]^+ + Cl^-$. This gives $2$ ions $(i=2)$.
Therefore,the correct formula is $[Pd(H_2O)_3Cl_3]Cl \cdot 3H_2O$.

(B) The aqueous pink solution of cobalt $(II)$ chloride contains the octahedral complex $[Co(H_2O)_6]^{2+}$.
When excess of $HCl$ is added,the following reaction occurs:
$[Co(H_2O)_6]^{2+} (aq) + 4Cl^- (aq) \rightarrow [CoCl_4]^{2-} (aq) + 6H_2O (l)$.
The formation of the tetrahedral complex $[CoCl_4]^{2-}$ results in a deep blue color.
Tetrahedral complexes have smaller crystal field splitting than octahedral complexes,where $\Delta_{t} = \frac{4}{9} \Delta_{0}$.
Therefore,option $(B)$ is correct.

(A) spin-free complex (also known as a high-spin complex) is one where the electrons are arranged to maximize the number of unpaired electrons.
In $[Fe(H_2O)_6]Cl_2$,the central metal ion is $Fe^{2+}$,which has a $d^6$ configuration.
$H_2O$ is a weak field ligand,so it does not cause pairing of electrons in the $d$-orbitals.
Thus,the electrons are arranged as $t_{2g}^4 e_g^2$,resulting in $4$ unpaired electrons,making it a spin-free complex.
In contrast,$CN^-$ and $CO$ are strong field ligands that cause pairing of electrons,leading to low-spin (spin-paired) complexes.
Therefore,$[Fe(H_2O)_6]Cl_2$ is the correct answer.

(D) The electrical conductivity of coordination compounds depends on the number of ions produced in the solution.
$1$. $[Co(NH_3)_6]Cl_3$ dissociates into $4$ ions $([Co(NH_3)_6]^{3+} + 3Cl^-)$.
$2$. $[CoCl(NH_3)_5]Cl_2$ dissociates into $3$ ions $([CoCl(NH_3)_5]^{2+} + 2Cl^-)$.
$3$. $[CoCl_2(NH_3)_4]Cl$ dissociates into $2$ ions $([CoCl_2(NH_3)_4]^+ + Cl^-)$.
$4$. $[CoCl_3(NH_3)_3]$ is a non-electrolyte and does not dissociate into ions.
Therefore,the correct order of electrical conductivity is $[Co(NH_3)_6]Cl_3 > [CoCl(NH_3)_5]Cl_2 > [CoCl_2(NH_3)_4]Cl > [CoCl_3(NH_3)_3]$.

(A) Molar conductivity depends on the number of ions produced by the complex in an aqueous solution.
$1$. $[Co(NH_3)_5Cl]SO_4 \rightarrow [Co(NH_3)_5Cl]^{2+} + SO_4^{2-}$ (Total $2$ ions)
$2$. $[Co(NH_3)_5SO_4]Cl \rightarrow [Co(NH_3)_5SO_4]^+ + Cl^-$ (Total $2$ ions)
$3$. $[Co(NH_3)_3Cl_3]$ is a non-electrolyte and does not ionize in water (Total $0$ ions)
$4$. $[Pt(NH_3)_2Cl_4]$ is a non-electrolyte and does not ionize in water (Total $0$ ions)
Since option $A$ and $B$ both produce $2$ ions,we look at the charge. However,in standard textbook problems of this type,$[Co(NH_3)_5Cl]SO_4$ is often cited for higher conductivity due to the presence of the $SO_4^{2-}$ ion compared to $Cl^-$. Thus,$[Co(NH_3)_5Cl]SO_4$ shows the maximum conductivity.

(D) Synergic bonding occurs in ligands that have vacant $\pi^*$ antibonding orbitals capable of accepting electron density from the filled $d$-orbitals of the central metal atom (back-bonding).
$CO$,$NO$,and $PPh_3$ possess vacant $\pi^*$ orbitals and are known to form metal-ligand $\pi$-back bonds.
$NH_2^-$ is a strong $\sigma$-donor ligand but lacks vacant $\pi^*$ orbitals,therefore it cannot participate in synergic bonding.
Thus,the correct option is $D$.

(B) In metal carbonyls,the extent of back-bonding from the metal to the $CO$ ligand depends on the electron density on the metal atom.
As the negative charge on the complex increases,the electron density on the metal increases,which enhances the back-donation of electrons into the $\pi^*$ antibonding orbitals of $CO$.
This increases the $M-C$ bond order and decreases the $C-O$ bond order.
Therefore,the order of $M-C$ bond order is $[Ni(CO)_4] < [Co(CO)_4]^- < [Fe(CO)_4]^{2-}$.

(D) In coordination chemistry,the conductivity of a complex depends on the number of ions it produces in an aqueous solution.
$CoCl_3 \cdot 6NH_3$ dissociates as $[Co(NH_3)_6]Cl_3$,yielding $4$ ions.
$CoCl_3 \cdot 5NH_3$ dissociates as $[Co(NH_3)_5Cl]Cl_2$,yielding $3$ ions.
$CoCl_3 \cdot 4NH_3$ dissociates as $[Co(NH_3)_4Cl_2]Cl$,yielding $2$ ions.
$CoCl_3 \cdot 3NH_3$ exists as the neutral complex $[Co(NH_3)_3Cl_3]$,which does not dissociate into ions in solution.
Since it produces no ions,$[Co(NH_3)_3Cl_3]$ is non-conducting.

(A) The brown ring test is a common laboratory test for the detection of the nitrate ion $(NO_{3}^{-})$.
When freshly prepared iron$(II)$ sulfate solution is added to a solution containing nitrate ions,followed by the slow addition of concentrated sulfuric acid along the sides of the test tube,a brown ring forms at the interface of the two layers.
The reaction involves the reduction of nitrate to nitric oxide $(NO)$,which then reacts with the hexaaquairon$(II)$ complex to form the brown-colored nitrosyl complex,$[Fe(H_{2}O)_{5}(NO)]^{2+}$.
The chemical equations are:
$NO_{3}^{-} + 3Fe^{2+} + 4H^{+} \rightarrow 3Fe^{3+} + NO + 2H_{2}O$
$[Fe(H_{2}O)_{6}]^{2+} + NO \rightarrow [Fe(H_{2}O)_{5}(NO)]^{2+} + H_{2}O$

(A) The electrical conductance of a complex in aqueous solution depends on the number of ions produced upon dissociation.
$(a) K_2[PtCl_6] \rightarrow 2K^+ + [PtCl_6]^{2-}$. Total ions $= 3$. Conductance $\approx 256 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$(b) [Pt(NH_3)_2Cl_4]$. This is a neutral complex,so it does not dissociate. Total ions $= 0$. Conductance $= 0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$(c) [Pt(NH_3)_3Cl_3]Cl \rightarrow [Pt(NH_3)_3Cl_3]^+ + Cl^-$. Total ions $= 2$. Conductance $\approx 97 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$(d) [Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^-$. Total ions $= 4$. Conductance $\approx 404 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Thus,the order is $256, 0, 97, 404$.

(C) The stability of coordination complexes increases as we move down a group in the periodic table because the $4d$ and $5d$ orbitals are more extended than $3d$ orbitals,leading to better overlap with ligands.
Therefore,the stability order is $[Co(H_2O)_6]^{3+} < [Rh(H_2O)_6]^{3+}$.
Option $C$ is the incorrect observation.

(A) The overall stability constant $(\beta_2)$ is the product of the stepwise formation constants ($K_1$ and $K_2$).
Given:
$K_1 = 4$
$K_2 = 3$
$\beta_2 = K_1 \times K_2$
$\beta_2 = 4 \times 3 = 12$

(D) $\pi$-acid ligands (also known as $\pi$-acceptor ligands) are those that possess empty $\pi^*$ antibonding orbitals capable of accepting electron density from the metal $d$-orbitals. This process is known as back-bonding or synergic bonding. $CO$,$CN^{-}$,and $NO^{\oplus}$ all contain such empty orbitals and are classic examples of $\pi$-acid ligands.

(B) The stability of a coordination complex depends on the nature of the ligand. $CN^-$ is a strong field ligand that forms very stable complexes due to its ability to form strong $\sigma$-bonds and $\pi$-backbonding with the metal center.
Between $[Fe(CN)_6]^{3-}$ and $[Fe(CN)_6]^{4-}$,the complex $[Fe(CN)_6]^{3-}$ has a higher charge density on the metal center $(Fe^{3+})$,which leads to a higher crystal field stabilization energy $(CFSE)$ and greater overall stability compared to the $Fe^{2+}$ complex.

(C) When $CuSO_4$ reacts with $KCN$,initially $Cu(CN)_2$ is formed,which is unstable.
It decomposes to give $CuCN$ and $(CN)_2$.
$2Cu^{2+} + 4CN^- \to 2CuCN + (CN)_2$
In the presence of excess $KCN$,$CuCN$ reacts to form a stable complex:
$CuCN + 3KCN \to K_3[Cu(CN)_4]$
Thus,the final product is $K_3[Cu(CN)_4]$.

(B) The van't Hoff factor $i$ is given by $i = 1 + (y-1) \alpha$,where $y$ is the number of ions produced per formula unit and $\alpha$ is the degree of ionisation.
Given $\alpha = 1$ ($100\%$ ionisation) and $i = 3$,we have $i = y = 3$.
This means the complex must dissociate into $3$ ions.
For $K_2[PtCl_6]$,the dissociation is $K_2[PtCl_6] \rightarrow 2K^{+} + [PtCl_6]^{2-}$.
Total ions $y = 2 + 1 = 3$.
Also,the oxidation state of $Pt$ in $[PtCl_6]^{2-}$ is $x + 6(-1) = -2$,so $x = +4$,which matches $Pt(IV)$.
Thus,$K_2[PtCl_6]$ is the correct complex.

(A) The stability of a coordination complex depends on several factors,including the nature of the ligand and the oxidation state of the central metal ion.
$1$. Strong field ligands like $CN^-$ form more stable complexes compared to weak field ligands like $H_2O$.
$2$. $A$ higher positive oxidation state of the central metal ion generally leads to greater stability due to stronger electrostatic attraction.
$3$. In $[Co(CN)_6]^{3-}$,the cobalt ion is in the $+3$ oxidation state and is coordinated by six strong $CN^-$ ligands,resulting in high crystal field splitting energy and maximum stability.

(C) The stability of a complex is inversely proportional to its dissociation constant $(k_d)$.
Lower the value of $k_d$,higher is the stability of the complex.
Comparing the given values:
$1 \times 10^{-16} > 1 \times 10^{-20} > 1 \times 10^{-37} > 1 \times 10^{-44}$.
The complex $[Fe(CN)_6]^{3-}$ has the minimum dissociation constant $(k_d = 1 \times 10^{-44})$,therefore it is the most stable complex.

(A) The aqueous solution of $CoCl_2$ contains the octahedral complex $[Co(H_2O)_6]^{2+}$,which is reddish-pink in colour.
When concentrated $HCl$ is added,the chloride ions $(Cl^-)$ act as ligands and replace the water molecules to form the tetrahedral complex $[CoCl_4]^{2-}$.
The formation of the tetrahedral $[CoCl_4]^{2-}$ complex is responsible for the deep blue colour observed in the solution.
Therefore,the correct complex ion is $[CoCl_4]^{2-}$.

(D) In the metal carbonyl complex $[Fe(CO)_5]$,the $Fe-C$ bond possesses both $\sigma$ and $\pi$ character.
The $\sigma$ bond is formed by the donation of electron pairs from the $CO$ ligand into the vacant $d$-orbitals of the $Fe$ atom.
The $\pi$ bond is formed by the back-donation of electrons from the filled $d$-orbitals of the $Fe$ atom into the vacant antibonding $\pi^*$ orbitals of the $CO$ ligand.
This phenomenon is known as synergic bonding.

(A) The dissociation constant $(K_d)$ is the reciprocal of the overall stability or association constant $(\beta_n)$.
$K_d = \frac{1}{\beta_4}$
Given $\beta_4 = 2.1 \times 10^{13}$.
$K_d = \frac{1}{2.1 \times 10^{13}}$
$K_d \approx 0.476 \times 10^{-13} = 4.76 \times 10^{-14}$.

(D) The reaction of coordination compounds with $AgNO_3$ depends on the number of chloride ions present outside the coordination sphere (ionizable chlorides).
$1$. $[Co(NH_3)_6]Cl_3$ gives $3$ moles of $AgCl$ per mole of compound.
$2$. $[Co(NH_3)_5Cl]Cl_2$ gives $2$ moles of $AgCl$ per mole of compound.
$3$. $[Co(NH_3)_4Cl_2]Cl$ gives $1$ mole of $AgCl$ per mole of compound.
Since all these compounds contain at least one ionizable $Cl^-$ ion,they all produce at least $1$ mole of $AgCl$ precipitate upon reaction with excess $AgNO_3$.

(D) The stability of coordination complexes is influenced by the chelate effect.
$[Co(en)_3]^{3+}$ contains the bidentate ligand ethylenediamine $(en)$,which forms stable five-membered rings with the central metal ion.
Complexes containing polydentate ligands (chelating agents) are significantly more stable than those containing only monodentate ligands like $NH_3$ due to the increase in entropy upon chelation.
Therefore,$[Co(en)_3]^{3+}$ is the most stable complex among the given options.

(A) The reducing agent is a species that can easily lose an electron to get oxidized.
In the complex $[Co(CN)_6]^{4-}$,the oxidation state of $Co$ is $+2$ ($x + 6(-1) = -4$,so $x = +2$).
$Co^{2+}$ has a $d^7$ electronic configuration.
In the presence of strong field ligands like $CN^-$,the configuration becomes $t_{2g}^6 e_g^1$.
The single electron in the $e_g$ orbital is very loosely held and can be easily lost to form the stable $t_{2g}^6$ configuration (which corresponds to $Co^{3+}$ in $[Co(CN)_6]^{3-}$).
Therefore,$[Co(CN)_6]^{4-}$ is a very strong reducing agent.

(D) $Ni^{2+}$ reacts with dimethyl glyoxime $(DMG)$ to form a complex which appears as a cherry-red precipitate.
This complex,$[Ni(DMG)_2]$,is stabilized by intramolecular hydrogen bonding between the two $DMG$ ligands.
$Ni^{2+} + 2DMG \xrightarrow{+2OH^{-}, -2H_2O} \underbrace{[Ni(DMG)_2]}_{\text{Cherry red precipitate}}$

(D) In an alkaline solution,the equilibrium is: $Cu^{2+}_{(aq)} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$.
In an acidic solution,the $H^+$ ions react with the lone pair of $NH_3$ to form $NH_4^+$ ions: $NH_3 + H^+ \rightarrow NH_4^+$.
Since $NH_4^+$ does not have a lone pair of electrons,it cannot act as a ligand to form a coordinate bond with the $Cu^{2+}$ ion. Therefore,no complex is formed in an acidic solution.

(D) low-spin complex is formed when the crystal field splitting energy $(\Delta_o)$ is greater than the pairing energy $(P)$. This typically occurs with strong field ligands like $NH_3$ and $CN^-$.
In $[Fe(NH_3)_6]^{3+}$,the central metal ion is $Fe^{3+}$ ($d^5$ configuration). $NH_3$ is a strong field ligand,which causes the electrons to pair up in the $t_{2g}$ orbitals,resulting in a low-spin configuration $(t_{2g}^5 e_g^0)$.
In contrast,$[Ni(NH_3)_6]^{2+}$ $(d^8)$ and $[Cr(NH_3)_6]^{3+}$ $(d^3)$ do not exhibit low-spin/high-spin variations due to their specific electronic configurations. $[Ti(H_2O)_6]^{3+}$ $(d^1)$ also does not exhibit spin states.
Therefore,$[Fe(NH_3)_6]^{3+}$ is the correct answer.

(B) The stability increases with the chelate effect as the denticity of the ligand increases ($NH_3$ is monodentate,$en$ is bidentate,$trien$ is tetradentate). This order is correct.
$(b)$ The stability of a complex depends on the field strength of the ligand. $NO_2^-$ is a stronger field ligand than $NH_3$. Therefore,the correct order should be $[Fe(H_2O)_6]^{3+} < [Fe(NH_3)_6]^{3+} < [Fe(NO_2)_6]^{3-}$. The given order is incorrect.
$(c)$ For the same group,stability increases down the group due to the increase in crystal field splitting energy $(\Delta_o)$ as $Z_{eff}$ increases from $Co^{3+}$ to $Ir^{3+}$. This order is correct.
$(d)$ Stability increases with the increase in the oxidation state of the central metal ion due to stronger electrostatic attraction. This order is correct.

(C) The stability of coordination complexes is influenced by the nature of the ligand and the crystal field splitting energy.
Among the given complexes,$[Fe(CN)_6]^{4-}$ is the most stable.
$CN^-$ is a strong field ligand that causes large crystal field splitting,leading to a low-spin $d^6$ configuration for $Fe^{2+}$.
This results in a high crystal field stabilization energy $(CFSE)$.
While $[Fe(CN)_6]^{3-}$ is also stable,$[Fe(CN)_6]^{4-}$ is generally considered more stable due to the higher charge density and stronger back-bonding associated with the $Fe^{2+}$ state in this specific geometry.

(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: $Cl^{-} < F^{-} < C_2O_4^{2-} < NO_2^{-} < CN^{-}$.
Thus,option $A$ is the correct order.

(C) $\pi$-acid ligand (or $\pi$-acceptor ligand) is a ligand that can accept electron density from the metal atom into its vacant $\pi$ or $\pi^*$ orbitals.
$N_2$,$CO$,and $C_2H_4$ possess vacant $\pi^*$ antibonding molecular orbitals,allowing them to act as $\pi$-acid ligands.
However,the peroxide ion $O_2^{2-}$ has a filled $\pi^*$ antibonding molecular orbital and lacks vacant orbitals of appropriate energy to accept electron density from the metal.
Therefore,$O_2^{2-}$ does not act as a $\pi$-acid ligand.

(D) The stability of coordination complexes is influenced by several factors:
$1$. The oxidation state of the central metal ion: Higher oxidation states lead to greater stability due to stronger electrostatic attraction. Thus,$[Co(NH_3)_6]^{3+}$ is more stable than $[Co(NH_3)_6]^{2+}$.
$2$. Nature of the ligand: Strong field ligands like $CN^{-}$ form more stable complexes than weak field ligands like halides.
$3$. Chelate effect: Complexes with chelating ligands (like $en$) are significantly more stable than those with monodentate ligands (like $NH_3$) due to the increase in entropy. Therefore,$[Cu(en)_2]^{2+}$ is more stable than $[Cu(NH_3)_4]^{2+}$.
$4$. The statement in option $D$ is incorrect because the chelate effect makes the ethylenediamine complex more stable.

(C) In complexes of phosphines $(PR_3)$,the vacant $d-$orbitals on the phosphorus atom participate in $M \xrightarrow{\pi} P$ back-bonding (synergic bonding).
This is a characteristic feature of $\pi-$acid ligands that possess vacant $d-$orbitals.

(B) In metal carbonyls,the extent of back-bonding from the metal $d$-orbitals to the $\pi^*$ antibonding orbitals of $CO$ depends on the electron density on the metal.
$1$. As the negative charge on the metal increases (e.g.,$[V(CO)_6]^-$),the metal becomes more electron-rich,leading to greater back-donation into the $CO$ $\pi^*$ orbitals.
$2$. Increased back-donation weakens the $C-O$ bond (decreasing its bond order and $IR$ stretching frequency) and strengthens the $M-C$ bond.
$3$. Conversely,in the cation $[Mn(CO)_6]^+$,the metal is electron-deficient,resulting in minimal back-donation. Thus,the $C-O$ bond remains closer to that of free $CO$ (stronger),and the $M-C$ bond is weaker.
$4$. Comparing the frequencies: $[V(CO)_6]^-$ $(1860 \, cm^{-1})$ < $[Cr(CO)_6]$ $(2000 \, cm^{-1})$ < $[Mn(CO)_6]^+$ $(2090 \, cm^{-1})$. The $C-O$ bond is strongest in the cation and weakest in the anion.

(A) In $PR_3$ (phosphine ligands),the phosphorus atom possesses vacant $d-$orbitals.
These vacant $d-$orbitals can accept electron density from the metal $d-$orbitals through back-bonding,which is a key component of synergic bonding.
Other ligands like $CO$,$NO^{+}$,and $C_6H_6$ primarily use their $\pi^*$ antibonding orbitals for back-bonding,not their $d-$orbitals.

(B) The complex $[Ni(DMG)_2]$ is a square planar complex formed by $Ni^{2+}$ and two dimethylglyoximate $(DMG^-)$ ligands.
$1$. The $DMG^-$ ligand is a bidentate ligand that coordinates through one nitrogen atom and one oxygen atom,but in the complex,the two $DMG^-$ ligands are held together by strong intramolecular hydrogen bonds between the $O-H$ and $O$ groups.
$2$. This hydrogen bonding provides extra stability to the complex.
$3$. The $IUPAC$ name is bis(dimethylglyoximato)nickel$(II)$,not nickelate$(II)$ as it is a neutral complex.
$4$. The $EAN$ of $Ni$ in $[Ni(DMG)_2]$ is $28 - 2 + 2(4) = 34$,but this does not make it an oxidizing agent.
$5$. Therefore,the most characteristic and correct statement is that it is extra-stabilized by hydrogen bonding.

(C) Ozone $(O_3)$ acts as a strong oxidizing agent.
It oxidizes potassium ferrocyanide $(K_4[Fe(CN)_6])$ to potassium ferricyanide $(K_3[Fe(CN)_6])$.
The chemical reaction is:
$2K_4[Fe(CN)_6] + O_3 + H_2O \rightarrow 2K_3[Fe(CN)_6] + 2KOH + O_2$

(B) When ammonia $(NH_3)$ is added to a cupric salt solution,it acts as a strong ligand and replaces the water molecules coordinated to the $Cu^{2+}$ ion.
The reaction results in the formation of the tetraamminecopper$(II)$ complex,which is deep blue in color.
The chemical equation is: $Cu^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}(aq)$.