JEE Coordination Compounds Questions

We need to evaluate two statements about paramagnetic species and unpaired electrons in coordination compounds.

Statement I: The number of paramagnetic species among $$[CoF_6]^{3-}$$, $$[TiF_6]^{3-}$$, $$V_2O_5$$, and $$[Fe(CN)_6]^{3-}$$ is 3.

- $$[CoF_6]^{3-}$$: Co³⁺ (3d⁶), F⁻ is weak field, so 4 unpaired electrons. Paramagnetic ✓

- $$[TiF_6]^{3-}$$: Ti³⁺ (3d¹), 1 unpaired electron. Paramagnetic ✓

- $$V_2O_5$$: V is in +5 state (3d⁰), no unpaired electrons. Diamagnetic ✗

- $$[Fe(CN)_6]^{3-}$$: Fe³⁺ (3d⁵), CN⁻ is strong field, pairing occurs: t₂g⁵ eg⁰, 1 unpaired electron. Paramagnetic ✓

Number of paramagnetic species = 3. Statement I is TRUE.

Statement II: $$K_4[Fe(CN)_6] < K_3[Fe(CN)_6] < [Fe(H_2O)_6]SO_4.H_2O < [Fe(H_2O)_6]Cl_3$$ is the correct order of unpaired electrons.

- $$K_4[Fe(CN)_6]$$: Fe²⁺ (3d⁶), CN⁻ strong field → t₂g⁶ eg⁰ → 0 unpaired electrons

- $$K_3[Fe(CN)_6]$$: Fe³⁺ (3d⁵), CN⁻ strong field → t₂g⁵ eg⁰ → 1 unpaired electron

- $$[Fe(H_2O)_6]SO_4.H_2O$$: Fe²⁺ (3d⁶), H₂O weak field → t₂g⁴ eg² → 4 unpaired electrons

- $$[Fe(H_2O)_6]Cl_3$$: Fe³⁺ (3d⁵), H₂O weak field → t₂g³ eg² → 5 unpaired electrons

Order: 0 < 1 < 4 < 5. Statement II is TRUE.

Both statements are true, corresponding to Option 1.

The spin-only magnetic moment ($$\mu$$) is calculated using the formula:

$$\mu = \sqrt{n(n+2)}\text{ BM}$$

where $$n$$ is the number of unpaired electrons

A. $$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$$

• Oxidation State: $$\text{Cr}$$ is in the $$+2$$ state ($$\text{Cr}^{2+}$$).
• Electronic Configuration: Ground state $$\text{Cr}$$ is $$[\text{Ar}]\,3d^5 4s^1$$.
  Removing 2 electrons gives $$\text{Cr}^{2+}: [\text{Ar}]\,3d^4$$.
• Crystal Field Splitting: $$\text{H}_2\text{O}$$ is a weak-field ligand, so it forms a high-spin complex ($$t_{2g}^3 \, e_g^1$$).
• Unpaired Electrons ($$n$$): $$4$$
• Magnetic Moment: $$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\text{ BM}$$
• Match: A $$\rightarrow$$ III

B. $$[\text{Co}(\text{H}_2\text{O})_6]^{2+}$$

• Oxidation State: $$\text{Co}$$ is in the $$+2$$ state ($$\text{Co}^{2+}$$).
• Electronic Configuration: Ground state $$\text{Co}$$ is $$[\text{Ar}]\,3d^7 4s^2$$.
  Removing 2 electrons gives $$\text{Co}^{2+}: [\text{Ar}]\,3d^7$$.
• Crystal Field Splitting: $$\text{H}_2\text{O}$$ is a weak-field ligand, leading to a high-spin complex ($$t_{2g}^5 \, e_g^2$$).
• Unpaired Electrons ($$n$$): $$3$$
• Magnetic Moment: $$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\text{ BM}$$
• Match: B $$\rightarrow$$ I

C. $$[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$$

• Oxidation State: $$\text{Cu}$$ is in the $$+2$$ state ($$\text{Cu}^{2+}$$).
• Electronic Configuration: Ground state $$\text{Cu}$$ is $$[\text{Ar}]\,3d^{10} 4s^1$$.
  Removing 2 electrons gives $$\text{Cu}^{2+}: [\text{Ar}]\,3d^9$$.
• Crystal Field Splitting: High-spin arrangement ($$t_{2g}^6 \, e_g^3$$).
• Unpaired Electrons ($$n$$): $$1$$
• Magnetic Moment: $$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\text{ BM}$$
• Match: C $$\rightarrow$$ IV

D. $$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$$

• Oxidation State: $$\text{Mn}$$ is in the $$+2$$ state ($$\text{Mn}^{2+}$$).
• Electronic Configuration: Ground state $$\text{Mn}$$ is $$[\text{Ar}]\,3d^5 4s^2$$.
  Removing 2 electrons gives $$\text{Mn}^{2+}: [\text{Ar}]\,3d^5$$.
• Crystal Field Splitting: $$\text{H}_2\text{O}$$ is a weak-field ligand, leading to a high-spin complex ($$t_{2g}^3 \, e_g^2$$).
• Unpaired Electrons ($$n$$): $$5$$
• Magnetic Moment: $$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\text{ BM}$$
• Match: D $$\rightarrow$$ II
  

Hence, Option D is correct.

Approach: Evaluate each statement about manganate ($$MnO_4^{2-}$$) and permanganate ($$MnO_4^-$$) ions.

Statement A: Both manganate and permanganate ions are tetrahedral.

Both $$MnO_4^{2-}$$ and $$MnO_4^-$$ have Mn bonded to 4 oxygen atoms with no lone pairs on Mn. The geometry is tetrahedral for both. Statement A is correct.

Statement B: Oxidation states of Mn in manganate and permanganate are +7 and +6, respectively.

In $$MnO_4^{2-}$$: $$x + 4(-2) = -2 \implies x = +6$$. In $$MnO_4^-$$: $$x + 4(-2) = -1 \implies x = +7$$. So manganate has Mn in +6 and permanganate has Mn in +7, which is the reverse of what is stated. Statement B is incorrect.

Statement C: Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final product.

Peroxodisulphate ($$S_2O_8^{2-}$$) is a powerful oxidising agent. The oxidation of Mn(II) by peroxodisulphate actually gives permanganate ($$MnO_4^-$$), not manganate. Statement C is incorrect.

Statement D: Manganate ion is paramagnetic and permanganate ion is diamagnetic.

$$MnO_4^{2-}$$ (Mn in +6, $$d^1$$) has 1 unpaired electron, so it is paramagnetic. $$MnO_4^-$$ (Mn in +7, $$d^0$$) has no unpaired electrons, so it is diamagnetic. Statement D is correct.

Statement E: Acidified permanganate reduces oxalate, nitrite, and iodide ions.

Permanganate is an oxidising agent; it oxidises these species, not reduces them. The statement incorrectly says "reduces." Statement E is incorrect.

Correct statements: A and D only, which corresponds to Option 1.

Manganese has atomic number 25 with ground-state configuration $$[Ar]\,3d^{5}\,4s^{2}$$.

In both complexes the overall charge is $$-3$$ and each ligand carries $$-1$$ charge, so for six ligands:$$x+(-6)=-3\;\Longrightarrow\;x=+3$$Hence Mn is in the $$+3$$ oxidation state, that is $$Mn^{3+}$$ whose electronic configuration is $$3d^{4}$$ (because the two $$4s$$ electrons and one $$3d$$ electron are lost).

For an octahedral $$d^{4}$$ ion two spin arrangements are possible:
• high-spin (weak-field ligand)
• low-spin (strong-field ligand)

Case 1: $$[Mn(Br)_6]^{3-}$$
$$Br^-$$ is a weak-field ligand. Thus the complex is high spin:$$t_{2g}^{3}e_{g}^{1}$$The four electrons occupy different orbitals according to Hund’s rule, giving $$n=4$$ unpaired electrons.
Spin-only magnetic moment:
$$\mu_1=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}\;\text{B.M.}\approx4.90\;\text{B.M.}$$

Case 2:$$[Mn(CN)_6]^{3-}$$
$$CN^-$$ is a strong-field ligand. The complex is low spin:$$t_{2g}^{4}e_{g}^{0}$$The occupancy inside $$t_{2g}$$ is $$\uparrow\downarrow\,\uparrow\,\uparrow$$, leaving $$n=2$$ unpaired electrons.
Spin-only magnetic moment:
$$\mu_2=\sqrt{n(n+2)}=\sqrt{2(2+2)}=\sqrt{8}\;\text{B.M.}\approx2.83\;\text{B.M.}$$

Sum of the magnetic moments
$$\mu_{\text{total}}=\mu_1+\mu_2\approx4.90+2.83=7.73\;\text{B.M.}$$

The required sum lies in the range $$7.5\text{-}7.8\;\text{B.M.}$$ (as specified).

The sulphides of the basic-radical cations of the IV group of qualitative analysis precipitate when $$H_2S$$ is passed through the alkaline (NH4Cl + NH4OH) medium.

Colour of fresh sulphide precipitates
  • $$\text{Zn}^{2+} \rightarrow \text{ZnS}$$ : white
  • $$\text{Mn}^{2+} \rightarrow \text{MnS}$$ : flesh-coloured / light pink
  • $$\text{Ni}^{2+} \rightarrow \text{NiS}$$ : black
  • $$\text{Co}^{2+} \rightarrow \text{CoS}$$ : black

The problem statement says that in the subsequent steps a yellow precipitate is obtained. This yellow solid is not the sulphide itself, but the confirmatory test product for cobalt(II):

$$\text{Co}^{2+} + 6\,\text{NO}_2^- + 3\,\text{K}^+ + \text{O} \;\longrightarrow\; \text{K}_3\left[\text{Co}\left(\text{NO}_2\right)_6\right] \downarrow$$

Potassium hexanitritocobaltate(III), $$\text{K}_3[\text{Co}(\text{NO}_2)_6]$$, is a canary-yellow crystalline precipitate and is the standard confirmatory test for $$\text{Co}^{2+}$$ ions.

Therefore the unknown sulphide that finally gives this yellow precipitate must contain cobalt. Among the given options, the only cobalt sulphide is $$\textbf{CoS}$$.

Hence, compound ‘A’ is $$\boxed{\text{CoS}}$$.

Option A which is: CoS

The colour shown by a coordination compound depends on the energy gap $$\Delta_{0}$$ between its $$t_{2g}$$ and $$e_{g}$$ levels.
When white light strikes the complex, a photon of energy exactly equal to $$\Delta_{0}$$ is absorbed and the complementary colour is observed.

Energy of a photon $$E$$ and the wavelength absorbed $$\lambda_{\text{abs}}$$ are related by the Planck relation
$$E = h\nu = \frac{hc}{\lambda_{\text{abs}}}$$.

Therefore, a larger crystal-field splitting $$\Delta_{0}$$ means a larger photon energy and hence a smaller wavelength absorbed, and vice-versa:
$$\boxed{\Delta_{0} \propto \dfrac{1}{\lambda_{\text{abs}}}}$$.

The magnitude of $$\Delta_{0}$$ depends mainly on the nature of the ligands present. The spectrochemical series (from weak field to strong field) is

$$Cl^- \lt H_2O \lt NH_3 \lt \dots \lt CN^-$$.

Now compare the four cobalt(III) complexes (all are low-spin $$d^6$$ species so the metal ion contribution to $$\Delta_{0}$$ is the same):

Case A:

$$[Co(NH_3)_4(H_2O)]^{3+}$$ contains four $$NH_3$$ and one $$H_2O$$.
Average ligand strength is slightly weaker than that in the all-ammine complex.

Case B:

$$[Co(NH_3)_6]^{3+}$$ contains six $$NH_3$$ ligands - stronger than $$H_2O$$ but weaker than $$CN^-$$.

Case C:

$$[Co(CN)_6]^{3-}$$ contains six $$CN^-$$ ligands - the strongest field; hence the largest $$\Delta_{0}$$.

Case D:

$$[CoCl(NH_3)_5]^{2+}$$ contains one $$Cl^-$$, the weakest field ligand of all; therefore it gives the smallest $$\Delta_{0}$$.

Thus the order of crystal-field splitting is
$$[Co(CN)_6]^{3-} \; (C) \;\gt\; [Co(NH_3)_6]^{3+} \; (B) \;\gt\; [Co(NH_3)_4(H_2O)]^{3+} \; (A) \;\gt\; [CoCl(NH_3)_5]^{2+} \; (D).$$

Converting this to wavelength absorbed (inverse relation):

$$\lambda_{\text{abs}}(D) \;\gt\; \lambda_{\text{abs}}(A) \;\gt\; \lambda_{\text{abs}}(B) \;\gt\; \lambda_{\text{abs}}(C).$$

Hence the correct order asked in the question is
$$D \gt A \gt B \gt C.$$

The matching option is Option A.

Electronic configuration of nickel atom is $$[Ar]\,3d^8\,4s^2$$, hence for $$Ni^{2+}$$ it becomes $$[Ar]\,3d^8$$. Thus the d-electron count is 8.

In an octahedral ligand field, the five d-orbitals split into a lower energy set $$t_{2g}$$ and a higher energy set $$e_g$$. The way electrons occupy these levels depends on the relative magnitudes of the octahedral crystal field splitting energy $$\Delta_o$$ and the pairing energy $$P$$.

Case 1: Weak field ligand
Here $$\Delta_o \lt P$$, so electrons occupy orbitals singly before pairing (high spin).
For d⁸: first fill all three $$t_{2g}$$ orbitals with six electrons (each pair in one orbital), then place the remaining two electrons one in each $$e_g$$ orbital. The configuration is $$t_{2g}^6\,e_g^2\,. $$

Case 2: Strong field ligand
Here $$\Delta_o \gt P$$, so electrons pair in the lower set before occupying the higher set (low spin).
For d⁸: all six electrons pair in $$t_{2g}$$, and the last two go into $$e_g$$, again giving $$t_{2g}^6\,e_g^2\,. $$

In both cases the electronic configuration remains $$t_{2g}^6\,e_g^2$$. Hence the distribution of d-electrons in $$Ni^{2+}$$ is not affected by ligand strength.

In the borax bead test, $$Ni^{2+}$$ imparts a violet colour to the non-luminous flame under hot conditions.

Therefore, the correct answer is Option C: $$Ni^{2+}$$.

Crystal‐field splitting energy ($$\Delta$$) and pairing energy ($$P$$) together decide whether a coordination complex will be high spin (maximum unpaired electrons) or low spin (minimum unpaired electrons).

Case 1: Octahedral complexes
In an octahedral field the $$d$$ orbitals split into a lower $$t_{2g}$$ set and an upper $$e_g$$ set with an energy gap of $$\Delta_o$$.

If $$\Delta_o \lt P$$, it costs less energy to promote an electron from $$t_{2g}$$ to $$e_g$$ than to pair it inside $$t_{2g}$$. Hence electrons occupy the $$e_g$$ orbitals before pairing - a high-spin configuration results.
If $$\Delta_o \gt P$$, promotion is costlier than pairing, so electrons pair up within $$t_{2g}$$ first - a low-spin configuration results.
Therefore Statement I is correct.

Case 2: Tetrahedral complexes
For a tetrahedral field the splitting $$\Delta_t$$ is much smaller: $$\Delta_t \approx \tfrac{4}{9}\Delta_o$$. Numerically, $$\Delta_t$$ is almost always less than the usual pairing energy $$P$$.

Because $$\Delta_t \lt P$$ in nearly every tetrahedral complex, promotion of electrons into the higher-energy $$e$$ set is always cheaper than pairing inside the lower $$t_2$$ set. Thus tetrahedral complexes almost invariably adopt high-spin configurations; low-spin tetrahedral complexes are extremely rare.
Hence Statement II is also correct.

Since both statements are correct, the appropriate option is Option D.

Case 1: $$[Fe(CN)_6]^{3-}$$ (octahedral)

Oxidation state of Fe:
$$x + 6(-1) = -3 \;\Rightarrow\; x = +3$$

Electronic configuration of $$Fe^{3+}$$: $$[Ar]\,3d^5$$  ($$d^5$$ system)

$$CN^-$$ is a strong-field ligand, so $$\Delta_o \gt P$$ and a low-spin arrangement occurs.
Filling the $$t_{2g}$$ and $$e_g$$ orbitals: $$t_{2g}^5\,e_g^0$$

Of the five $$t_{2g}$$ electrons, four pair up and one remains single → unpaired electrons = $$1$$.

Case 2: $$[FeF_6]^{3-}$$ (octahedral)

Oxidation state of Fe is again $$+3$$, so $$Fe^{3+}$$ has $$d^5$$.

$$F^-$$ is a weak-field ligand, therefore $$\Delta_o \lt P$$ and a high-spin complex forms.
Filling pattern: $$t_{2g}^3\,e_g^2$$

Each of the five $$d$$ electrons occupies a separate orbital → unpaired electrons = $$5$$.

Case 3: $$[CoF_6]^{3-}$$ (octahedral)

Oxidation state of Co:
$$x + 6(-1) = -3 \;\Rightarrow\; x = +3$$

Electronic configuration of $$Co^{3+}$$: $$[Ar]\,3d^6$$  ($$d^6$$ system)

With weak-field $$F^-$$ ligands (high-spin):
Filling pattern: $$t_{2g}^4\,e_g^2$$

Diagrammatically, two $$t_{2g}$$ orbitals contain one unpaired electron each and both $$e_g$$ orbitals contain one unpaired electron each. Hence unpaired electrons = $$4$$.

Case 4: $$[Mn(CN)_6]^{3-}$$ (octahedral)

Oxidation state of Mn:
$$x + 6(-1) = -3 \;\Rightarrow\; x = +3$$

Electronic configuration of $$Mn^{3+}$$: $$[Ar]\,3d^4$$  ($$d^4$$ system)

$$CN^-$$ is a strong-field ligand; the complex is low-spin:
Filling pattern: $$t_{2g}^4\,e_g^0$$

The four $$t_{2g}$$ electrons produce two paired and two unpaired electrons → unpaired electrons = $$2$$.

Therefore, the numbers of unpaired electrons are
$$[Fe(CN)_6]^{3-}: 1,\quad [FeF_6]^{3-}: 5,\quad [CoF_6]^{3-}: 4,\quad [Mn(CN)_6]^{3-}: 2$$.

This matches Option A: $$1,\,5,\,4,\,2$$.

The reagent $$K_4[Fe(CN)_6]$$ (potassium ferrocyanide) gives characteristic coloured precipitates with only certain metal ions. We shall examine tests (A) to (E) one by one.

$$Cu^{2+} + K_4[Fe(CN)_6] \xrightarrow{\text{acidic}} Cu_2[Fe(CN)_6] \downarrow$$
Copper(II) ferrocyanide is chocolate-brown in colour.
Observation stated in (A) - chocolate brown precipitate - is correct.

$$Fe^{3+} + K_4[Fe(CN)_6] \rightarrow Fe_4[Fe(CN)_6]_3 \downarrow$$
Ferric ferrocyanide is the well-known Prussian blue.
Observation stated in (B) - Prussian blue precipitate - is correct.

$$Zn^{2+} + K_4[Fe(CN)_6] \xrightarrow{\text{neutral/alkaline}} Zn_2[Fe(CN)_6] \downarrow$$
Zinc ferrocyanide separates as a white or faint bluish-white precipitate; neutralisation with $$NH_4OH$$ furnishes the required medium.
Observation stated in (C) - white / bluish-white precipitate - is correct.

Magnesium ions do not give an insoluble ferrocyanide; $$Mg_2[Fe(CN)_6]$$ is soluble. Hence no blue precipitate is obtained.
Observation stated in (D) is incorrect.

Ferrocyanides of the alkaline-earth metals Ca, Sr and Ba are readily soluble in water; therefore $$Ba^{2+}$$ does not precipitate with $$K_4[Fe(CN)_6]$$ even after neutralisation with $$NaOH$$.
Observation stated in (E) is incorrect.

Thus only statements (A), (B) and (C) are correct.

The option containing A, B and C only is Option C.

All the four complexes contain iron in the $$+3$$ oxidation state, so the metal ion is $$Fe^{3+}$$ having the electronic configuration $$[Ar]\,3d^5$$.

Case 1: $$[Fe(NH_3)_6]^{3+}$$  and  $$[Fe(H_2O)_6]^{3+}$$    Both $$NH_3$$ and $$H_2O$$ are monodentate ligands of only moderate field strength.    For a $$d^5$$ ion with such ligands, the complex is high-spin:    $$t_{2g}^3\,e_g^2$$  —  CFSE $$=0$$.

Case 2: $$[Fe(Cl)_6]^{3-}$$    $$Cl^-$$ is a very weak-field ligand, so the complex is again high-spin:    $$t_{2g}^3\,e_g^2$$  —  CFSE $$=0$$.

Case 3: $$[Fe(C_2O_4)_3]^{3-}$$    Oxalate $$(C_2O_4^{2-})$$ is a bidentate ligand.    Although its crystal-field strength is similar to $$H_2O$$, the chelate effect gives this complex a much larger overall (stepwise) stability constant.    With a moderate field ligand, $$Fe^{3+}$$ again remains high-spin:    $$t_{2g}^3\,e_g^2$$.

Because of the pronounced chelate effect, $$[Fe(C_2O_4)_3]^{3-}$$ possesses the greatest thermodynamic stability among the four. Hence the most stable complex ion is $$[Fe(C_2O_4)_3]^{3-}$$.

The number of electrons occupying the $$t_{2g}$$ set in this ion is therefore $$X = 3$$.

The oxide of vanadium corresponding to $$V_2O_X$$ with $$X = 3$$ is $$V_2O_3$$. Vanadium(III) oxide behaves like aluminium(III) oxide and chromium(III) oxide: it dissolves both in acids (forming $$V^{3+}$$ salts) and in strong bases (forming vanadates of lower oxidation state after air-oxidation). Such behaviour is termed amphoteric.

Hence, the nature of $$V_2O_3$$ is amphoteric  ⇒  Option D.

We need to find the complex where CFSE ($$\Delta_o$$) is zero.

Key concept: CFSE is zero when the d-orbitals are either empty (d⁰), half-filled (d⁵ in weak field), or completely filled (d¹⁰), because the stabilization from electrons in $$t_{2g}$$ is exactly cancelled by destabilization from electrons in $$e_g$$.

Specifically for an octahedral weak field complex: d⁵ configuration gives CFSE = $$(-0.4 \times 3 + 0.6 \times 2)\Delta_o = 0$$.

Analyze each option:

(A) $$[Fe(en)_3]Cl_3$$: Fe³⁺ is d⁵. en is a strong field ligand, so this is a low-spin complex with CFSE = $$-2.0\Delta_o$$. CFSE is NOT zero.

(B) $$K_4[Fe(CN)_6]$$: Fe²⁺ is d⁶. CN⁻ is a strong field ligand (low spin), CFSE = $$-2.4\Delta_o$$. NOT zero.

(C) $$[Fe(NH_3)_6]Br_2$$: Fe²⁺ is d⁶. NH₃ is a moderate-strong field ligand. CFSE is NOT zero.

(D) $$K_3[Fe(SCN)_6]$$: Fe³⁺ is d⁵. SCN⁻ is a weak field ligand, so this is a high-spin complex. For high-spin d⁵: electrons are distributed as $$t_{2g}^3 e_g^2$$.

CFSE = $$3(-0.4\Delta_o) + 2(0.6\Delta_o) = -1.2\Delta_o + 1.2\Delta_o = 0$$

The correct answer is Option 4: $$K_3[Fe(SCN)_6]$$.

The given molecular composition can produce two octahedral complexes:

$$[Co(NH_3)_5SO_4]Cl$$   and   $$[Co(NH_3)_5Cl]SO_4$$

In the first species, chloride is the counter-ion, whereas sulphate is inside the coordination sphere. In the second species, sulphate is the counter-ion, while chloride is inside the coordination sphere.

Identification of the two isomers by qualitative tests:

Case A: The solution gives a white precipitate with $$AgNO_3$$.
The reagent $$AgNO_3$$ detects free $$Cl^-$$ ions by forming white $$AgCl$$. Therefore the complex that produces this test must have chloride outside the coordination sphere: $$[Co(NH_3)_5SO_4]Cl \;+\; AgNO_3 \;\rightarrow\; AgCl\downarrow$$

Case B: The solution gives a white precipitate with $$BaCl_2$$.
The reagent $$BaCl_2$$ detects free $$SO_4^{2-}$$ ions by forming white $$BaSO_4$$. Hence the complex furnishing this test must have sulphate outside the coordination sphere: $$[Co(NH_3)_5Cl]SO_4 \;+\; BaCl_2 \;\rightarrow\; BaSO_4\downarrow$$

The two compounds differ only in the exchange of $$Cl^-$$ and $$SO_4^{2-}$$ between the coordination sphere and the counter-ion position. Such a pair of complexes, which produce different ions in solution because of this exchange, represents ionisation isomerism.

Therefore the type of isomerism exhibited is Ionisation isomerism  →  Option C.

• (A) $$[\text{FeO}_4]^{2-}$$:

  • Oxidation state of Fe: $$x + 4(-2) = -2 \implies x = +6$$
  • Iron(VI) configuration: Neutral $$\text{Fe}$$ is $$[Ar] 3d^6 4s^2$$, so $$\text{Fe}^{6+}$$ is $$[Ar] 3d^2$$.
  • This is a $$d^2$$ configuration.

• (B) $$[\text{Mn}(\text{CN})_6]^{3-}$$:

  • Oxidation state of Mn: $$x + 6(-1) = -3 \implies x = +3$$
  • Manganese(III) configuration: Neutral $$\text{Mn}$$ is $$[Ar] 3d^5 4s^2$$, so $$\text{Mn}^{3+}$$ is $$[Ar] 3d^4$$.
  • This is a $$d^4$$ configuration.

• (C) $$[\text{Fe}(\text{CN})_6]^{3-}$$:

  • Oxidation state of Fe: $$x + 6(-1) = -3 \implies x = +3$$
  • Iron(III) configuration: Neutral $$\text{Fe}$$ is $$[Ar] 3d^6 4s^2$$, so $$\text{Fe}^{3+}$$ is $$[Ar] 3d^5$$.
  • This is a $$d^5$$ configuration.

• (D) $$\text{Cr}_2(\text{O}-\text{C}(\text{=O})-\text{Me})_4(\text{H}_2\text{O})_2$$ (Chromium(II) acetate dimer):

  • Oxidation state of each Cr: Acetate ($$\text{CH}_3\text{COO}^-$$) carries a $$-1$$ charge, making the complex neutral: $$2x + 4(-1) + 2(0) = 0 \implies x = +2$$
  • Chromium(II) configuration: Neutral $$\text{Cr}$$ is $$[Ar] 3d^5 4s^1$$, so $$\text{Cr}^{2+}$$ is $$[Ar] 3d^4$$.
  • This is a $$d^4$$ configuration.

• (E) $$[\text{NiF}_6]^{2-}$$:

  • Oxidation state of Ni: $$x + 6(-1) = -2 \implies x = +4$$
  • Nickel(IV) configuration: Neutral $$\text{Ni}$$ is $$[Ar] 3d^8 4s^2$$, so $$\text{Ni}^{4+}$$ is $$[Ar] 3d^6$$.
  • This is a $$d^6$$ configuration.

Conclusion:

The complexes containing a central metal ion with a $$d^4$$ configuration are (B) and (D).

Answer: Option C — (B) and (D) only

First write the ground-state electronic configurations of the central metal atoms.

Ni : $$[Ar]\,3d^{8}\,4s^{2}$$
Mn : $$[Ar]\,3d^{5}\,4s^{2}$$

Next, find the oxidation state of the metal in each complex.

$$[Ni(CO)_4]$$ : CO is neutral ⇒ Ni is $$0$$-oxidation.
$$[Ni(CN)_4]^{2-}$$ : $$4(\!-1)$$ from $$CN^-$$ gives $$\;x-4=-2\Rightarrow x=+2\;(Ni^{2+})$$.
$$[NiCl_4]^{2-}$$ : $$4(\!-1)$$ from $$Cl^-$$ gives $$\;x-4=-2\Rightarrow x=+2\;(Ni^{2+})$$.
$$[MnBr_4]^{2-}$$ : $$4(\!-1)$$ from $$Br^-$$ gives $$\;x-4=-2\Rightarrow x=+2\;(Mn^{2+})$$.

Now list the $$d$$-electron counts.

Ni(0) $$3d^{10}$$ (after promotion-hybridisation will pair all 10).
Ni(II) $$3d^{8}$$.
Mn(II) $$3d^{5}$$.

Identify each ligand as strong- or weak-field.

CO and $$CN^-$$ are strong-field (cause pairing).
$$Cl^-$$ and $$Br^-$$ are weak-field (do not cause pairing).

Determine hybridisation, geometry and unpaired electrons one by one.

Case A: $$[Ni(CO)_4]$$

CO is strong field; Ni is in $$0$$ state. The 3d orbitals are fully paired and do not participate in hybridisation. The complex uses $$sp^3$$ hybridisation (4s + 3p) ⇒ tetrahedral. With all electrons paired, $$n=0$$ unpaired ⇒ $$\mu =0\;BM$$. Hence: Tetrahedral, 0 BM (Choice III).

Case B: $$[Ni(CN)_4]^{2-}$$

Ni$$^{2+}$$ is $$3d^{8}$$. $$CN^-$$ is strong field, so the electrons pair up giving configuration $$t_{2g}^{6}e_g^{2}$$ with no unpaired electrons. Square-planar geometry arises from $$dsp^2$$ hybridisation. $$n=0$$ ⇒ $$\mu=0\;BM$$. Hence: Square planar, 0 BM (Choice II).

Case C: $$[NiCl_4]^{2-}$$

Ni$$^{2+}$$ is $$3d^{8}$$ and $$Cl^-$$ is weak field, so no additional pairing occurs. The complex adopts $$sp^3$$ hybridisation ⇒ tetrahedral. For a tetrahedral high-spin $$d^{8}$$ ion there are $$n=2$$ unpaired electrons, so
$$\mu=\sqrt{n(n+2)}=\sqrt{2\times4}\approx2.8\;BM$$. Hence: Tetrahedral, 2.8 BM (Choice I).

Case D: $$[MnBr_4]^{2-}$$

Mn$$^{2+}$$ is $$3d^{5}$$, and the weak-field $$Br^-$$ keeps it high-spin. With $$sp^3$$ hybridisation the geometry is tetrahedral. All five $$d$$ electrons remain unpaired, $$n=5$$, so
$$\mu=\sqrt{5(5+2)}=\sqrt{35}\approx5.9\;BM$$. Hence: Tetrahedral, 5.9 BM (Choice IV).

Collecting the results:
A → III  B → II  C → I  D → IV

Comparing with the given options, this matches Option C.

Co(II) has the electronic configuration $$[Ar]3d^7$$.

The magnetic moment is given by $$ \mu = \sqrt{n(n+2)} \text{ BM} $$, where $$n$$ is the number of unpaired electrons.

Substituting the observed magnetic moment $$\mu = 3.95$$ BM into this formula gives

$$ 3.95 = \sqrt{n(n+2)} $$

and hence

$$ 15.6 \approx n(n+2). $$

Testing integer values of $$n$$ shows that for $$n = 3$$, we have $$3 \times 5 = 15$$ and therefore

$$ \mu = \sqrt{15} = 3.87 \text{ BM}, $$

which is very close to the observed value of 3.95 BM. Thus, there are three unpaired electrons in the complex.

For a $$d^7$$ octahedral complex with three unpaired electrons, the high-spin configuration must be

$$ t_{2g}^5 e_g^2. $$

In this arrangement, the $$t_{2g}$$ orbitals contain five electrons (↑↓ ↑↓ ↑), giving one unpaired electron, and the $$e_g$$ orbitals contain two electrons (↑ ↑), giving two unpaired electrons, for a total of three unpaired electrons.

This matches the given magnetic moment.

The correct answer is Option 3: $$t_{2g}^5 e_g^2$$.

We need to determine the geometry and oxidation state of Ni in $$[NiCl_4]^{2-}$$ (paramagnetic) and $$[Ni(CO)_4]$$ (diamagnetic).

We first analyze $$[NiCl_4]^{2-}$$. Charge balance Ni + 4(−1) = −2 indicates Ni is in the +2 oxidation state and thus has a d⁸ configuration. As Cl⁻ is a weak field ligand with four coordination sites, the complex adopts a tetrahedral geometry. A tetrahedral d⁸ complex has two unpaired electrons, explaining the paramagnetism.

Next, we analyze $$[Ni(CO)_4]$$. CO is a neutral ligand and Ni + 4(0) = 0 shows Ni is in the 0 oxidation state, corresponding to d¹⁰. With all d‐orbitals filled and four coordination sites, the complex remains tetrahedral rather than square planar. The d¹⁰ configuration has no unpaired electrons, which accounts for the observed diamagnetism.

In summary, $$[NiCl_4]^{2-}$$ features Ni(II) in a tetrahedral geometry with two unpaired electrons, while $$[Ni(CO)_4]$$ contains Ni(0) in a tetrahedral geometry with all electrons paired. The correct answer is Option 3.

The standard electrode potentials $$E^\circ(M^{3+}/M^{2+})$$ for the given metals are approximately:
$$Cr^{3+}/Cr^{2+}: -0.41\ \text{V},\; Fe^{3+}/Fe^{2+}: +0.77\ \text{V},\; Mn^{3+}/Mn^{2+}: +1.51\ \text{V},\; Co^{3+}/Co^{2+}: +1.82\ \text{V}$$

A higher $$E^\circ$$ value means the $$M^{3+}$$ ion is a stronger oxidising agent and more readily gains an electron to become $$M^{2+}$$. Hence the highest potential belongs to cobalt, so $$M = Co$$.

The complex is $$[M(CN)_6]^{4-}$$. Let the oxidation state of $$M$$ be $$x$$.

Each $$CN^-$$ ligand carries charge $$-1$$. Therefore:
$$x + 6(-1) = -4 \;\; \Rightarrow \;\; x = +2$$

Thus the metal is $$Co^{2+}$$.

Electronic configuration of neutral cobalt: $$[Ar]\;3d^7\,4s^2$$.
For $$Co^{2+}$$ (lose the two $$4s$$ electrons): $$[Ar]\;3d^7$$.

$$CN^-$$ is a strong-field ligand, so the octahedral complex $$[Co(CN)_6]^{4-}$$ will be low-spin. In an octahedral field the five $$d$$ orbitals split into:
lower set $$t_{2g}$$ (3 orbitals) and upper set $$e_g$$ (2 orbitals).

Low-spin distribution for a $$d^7$$ ion:
$$t_{2g}^6\,e_g^1$$

Hence the number of electrons occupying the $$e_g$$ orbitals is $$1$$.

Final answer: $$1$$ electron.

The reagents $$FeCl_3$$, $$KOH$$ and $$H_2C_2O_4$$ react in two stages.

First, the alkali converts oxalic acid into oxalate ion:
$$H_2C_2O_4 + 2\,KOH \rightarrow K_2C_2O_4 + 2\,H_2O$$

The freshly-generated $$C_2O_4^{2-}$$ (oxalate) then acts as a bidentate ligand toward the $$Fe^{3+}$$ ion obtained from $$FeCl_3$$.
Because $$Fe^{3+}$$ has six vacant coordination sites, three oxalate ions can chelate to it, giving the octahedral complex

$$K_3\,[Fe(C_2O_4)_3] \; \longleftrightarrow \; [Fe(C_2O_4)_3]^{3-} \; = \; \text{(A)}$$

Each $$C_2O_4^{2-}$$ ligand occupies two adjacent sites, so the stoichiometry is of the general type $$[M(AA)_3]$$, where $$(AA)$$ is any symmetrical bidentate ligand.

For an octahedral complex of the form $$[M(AA)_3]$$:

• All six donor atoms are fixed in positions that give only one geometrical arrangement (no cis-trans possibilities).
• However, the three chelate rings can interlock in two nonsuperimposable mirror-image arrangements, conventionally named the $$\Lambda$$ and $$\Delta$$ forms.

Thus such a complex exhibits optical isomerism, producing one pair of enantiomers.

Number of optical isomers $$= 2$$ (one $$\Lambda$$ and one $$\Delta$$).

Therefore, the iron complex (A) shows $$\mathbf{2}$$ optical isomers.

The hybridisation $$sp^3d^2$$ corresponds to an octahedral geometry, so the metal ion must have coordination number $$6$$.

In the given compound $$MCl_4\cdot 3NH_3$$ the six possible ligands are provided by $$3$$ molecules of $$NH_3$$ and some of the chloride ions. Let the number of chloride ions inside the coordination sphere be $$y$$ and the number of chloride ions present as counter-ions be $$(4-y)$$.

When the salt is treated with excess $$AgNO_3$$, only the ionisable (counter) chloride ions precipitate as $$AgCl$$. The number of moles of $$AgCl$$ obtained per mole of the complex is given to be $$x$$.

The problem states that $$x$$ equals the number of lone pairs on the central atom in $$BrF_5$$. In $$BrF_5$$, bromine is surrounded by $$5$$ bond pairs and has $$1$$ lone pair (total steric number $$6$$), so

$$x = 1$$.

Hence, only one chloride ion is outside the coordination sphere:

$$(4-y)=1 \;\;\Longrightarrow\;\; y = 3.$$

Thus the correct structural formula is

$$[\,M(NH_3)_3Cl_3\,]Cl.$$

This is of the general type $$MA_3B_3$$ (where $$A = NH_3$$ and $$B = Cl^-$$) in an octahedral environment.

For an octahedral complex of the form $$MA_3B_3$$, two geometrical (cis-trans) arrangements are possible: Case 1: The three identical ligands occupy one face of the octahedron (fac-isomer).
Case 2: The three identical ligands lie in a meridian passing through the metal (mer-isomer).

Therefore, the number of geometrical isomers exhibited by the complex is $$2$$.

Answer : 2

We need to identify which complexes are yellow among the given low-spin complexes, then sum their spin-only magnetic moments.

The complexes are:

1. $$K_3[Co(NO_2)_6]$$: Co³⁺ (d⁶), low-spin → $$t_{2g}^6 e_g^0$$, 0 unpaired electrons, μ = 0 BM. Yellow in colour. ✓

2. $$K_4[Fe(CN)_6]$$: Fe²⁺ (d⁶), low-spin → $$t_{2g}^6 e_g^0$$, 0 unpaired electrons, μ = 0 BM. Yellow in colour. ✓

3. $$K_3[Fe(CN)_6]$$: Fe³⁺ (d⁵), low-spin → $$t_{2g}^5 e_g^0$$, 1 unpaired electron, μ = 1.73 BM. Red/orange colour, not yellow.

4. $$Cu_2[Fe(CN)_6]$$: Brown/reddish colour.

5. $$Zn_2[Fe(CN)_6]$$: White precipitate.

The yellow complexes are $$K_3[Co(NO_2)_6]$$ and $$K_4[Fe(CN)_6]$$, both with μ = 0 BM.

Sum of magnetic moments = 0 + 0 = 0 BM.

The answer is 0.

We need to find the number of hydrogen atoms in the complex of $$Ni^{2+}$$ with dimethylglyoxime (DMG).

Dimethylglyoxime has the molecular formula $$C_4H_8N_2O_2$$. Its structure is:

$$CH_3-C(=NOH)-C(=NOH)-CH_3$$

Each DMG molecule has 8 hydrogen atoms (6 from two $$CH_3$$ groups and 2 from two $$-OH$$ groups).

$$Ni^{2+}$$ forms a square planar complex with two molecules of dimethylglyoxime. Each DMG molecule acts as a bidentate ligand, coordinating through its two nitrogen atoms. The complex is commonly called "nickel dimethylglyoximate" and has the formula $$Ni(DMG)_2$$ or $$[Ni(C_4H_7N_2O_2)_2]$$.

When DMG coordinates with $$Ni^{2+}$$, each DMG molecule loses one proton (from one of its $$-OH$$ groups) to form the anionic ligand $$DMGH^-$$. Since the complex uses two DMG molecules and $$Ni^{2+}$$ has a +2 charge:

- Each DMG loses 1 H: $$8 - 1 = 7$$ H atoms per ligand

- Two ligands contribute: $$7 \times 2 = 14$$ H atoms

The two $$DMGH^-$$ ligands in the complex are held together by intramolecular hydrogen bonds (O-H...O) between the oxime groups, which is a characteristic feature of this complex and gives it additional stability.

The answer is 14.

In an octahedral complex the possible hybridisations are:
- $$d^2sp^3$$ (inner-orbital, produced only when the ligand field is strong enough to force electron pairing in the $$\,(n-1)d\,$$ shell)
- $$sp^3d^2$$ (outer-orbital, produced with weak-field ligands; no pairing in the $$\,(n-1)d\,$$ shell)

Hence the required complex must simultaneously
(i) contain a strong-field ligand set so that $$d^2sp^3$$ hybridisation becomes possible, and
(ii) still possess at least one unpaired electron so that it is paramagnetic.

Step-wise examination of every complex:

Case 1: $$[FeF_6]^{3-}$$
Fe is in the $$+3$$ state ⇒ $$d^5$$.
$$F^-$$ is a weak-field ligand ⇒ high-spin $$sp^3d^2$$ hybridisation.
Hybridisation condition fails ⇒ reject.

Case 2: $$[Fe(CN)_6]^{3-}$$
Fe$$^{3+} : d^5$$.
$$CN^-$$ is a strong-field ligand ⇒ electrons pair in the $$\,(n-1)d\,$ orbitals giving the low-spin configuration $$t_{2g}^5\, (1$$\text{ unpaired}$$)$$.
Hybridisation: $$d^2sp^3$$ (inner-orbital).
Paramagnetic (1 unpaired).
✓ Satisfies both conditions.

Case 3: $$[Mn(CN)_6]^{3-}$$
Mn$$^{3+} : d^4$$.
$$CN^-$$ strong-field ⇒ low-spin $$t_{2g}^4$$ (2 unpaired).
Hybridisation: $$d^2sp^3$$.
Paramagnetic (2 unpaired).
✓ Satisfies both conditions.

Case 4: $$[Co(C_2O_4)_3]^{3-}$$
Co$$^{3+} : d^6$$.
Oxalate $$(C_2O_4)^{2-}$$ is a reasonably strong field for the highly charged Co$$^{3+}$$ ion ⇒ low-spin $$t_{2g}^6$$ (all electrons paired).
Hybridisation: $$d^2sp^3$$, but the complex is diamagnetic.
Fails the paramagnetism requirement ⇒ reject.

Case 5: $$[MnCl_6]^{3-}$$
Mn$$^{3+} : d^4$$ with weak-field $$Cl^-$$ ⇒ high-spin $$sp^3d^2$$.
Hybridisation condition fails ⇒ reject.

Case 6: $$[CoF_6]^{3-}$$
Co$$^{3+} : d^6$$ with weak-field $$F^-$$ ⇒ high-spin $$sp^3d^2$$ (4 unpaired).
Hybridisation condition fails ⇒ reject.

Only two complexes, $$[Fe(CN)_6]^{3-}$$ and $$[Mn(CN)_6]^{3-}$$, satisfy
(i) $$d^2sp^3$$ hybridisation and
(ii) possession of unpaired electrons.

Therefore, the required number of paramagnetic $$d^2sp^3$$ complexes is $$\mathbf{2}$$.

The four complexes given are:
  (i) $$K_2[NiCl_4]$$   (ii) $$[Zn(H_2O)_6]Cl_2$$   (iii) $$K_3[Mn(CN)_6]$$   (iv) $$[Cu(PPh_3)_3I]$$

Step 1 : Identify which of them are diamagnetic.

Case 1: $$[NiCl_4]^{2-}$$ (in $$K_2[NiCl_4]$$)
Ni oxidation state: $$x+4(-1) = -2 \Rightarrow x = +2$$, so $$Ni^{2+}$$ is $$3d^8$$. In a tetrahedral $$Cl^-$$ field (weak field) the ion is high spin with two unpaired electrons ⇒ paramagnetic.

Case 2: $$[Zn(H_2O)_6]^{2+}$$ (in $$[Zn(H_2O)_6]Cl_2$$)
$$Zn^{2+}$$ is $$3d^{10}$$, all electrons paired ⇒ diamagnetic.

Case 3: $$[Mn(CN)_6]^{3-}$$ (in $$K_3[Mn(CN)_6]$$)
$$x+6(-1)=-3 \Rightarrow x = +3$$, so $$Mn^{3+}$$ is $$3d^4$$. $$CN^-$$ is a strong‐field ligand giving a low-spin configuration $$t_{2g}^4e_g^0$$ with two unpaired electrons ⇒ paramagnetic.

Case 4: $$[Cu(PPh_3)_3I]$$ (neutral complex)
Oxidation state: $$Cu + (-1) = 0 \Rightarrow Cu = +1$$, so $$Cu^{+}$$ is $$3d^{10}$$ ⇒ diamagnetic.

Thus the diamagnetic complexes are Case 2 and Case 4. Number of diamagnetic complexes $$= 2$$, hence $$n = 2$$.

Step 2 : Among the elements Ni, Zn, Mn and Cu, compare enthalpies of atomisation. For 3d metals: $$\Delta H_{atom}$$ rises to a maximum near the middle of the series (Cr, Mn, Fe, Co) and then falls, being minimum for Zn.
Approximate values (kJ mol$$^{-1}$$): $$Zn \approx 120$$, $$Cu \approx 330$$, $$Mn \approx 280$$, $$Ni \approx 420$$. Therefore the element with the least enthalpy of atomisation is $$Zn$$.

So the required ion is $$Zn^{n+} = Zn^{2+}$$.

Step 3 : Calculate its spin-only magnetic moment. For an ion with $$n_u$$ unpaired electrons,
$$\mu_{spin} = \sqrt{n_u(n_u+2)}\; \text{BM}$$.

For $$Zn^{2+}$$, electronic configuration $$3d^{10}$$ ⇒ $$n_u = 0$$.
Hence $$\mu_{spin} = \sqrt{0(0+2)} = 0\; \text{BM}$$.

The spin-only magnetic moment, to the nearest integer, is 0.

Evaluate two statements about $$N(CH_3)_3$$ and $$P(CH_3)_3$$ as ligands.

Statement I: "$$N(CH_3)_3$$ and $$P(CH_3)_3$$ can act as ligands to form transition metal complexes."

CORRECT. Both have a lone pair on the central atom (N or P) that can be donated to a transition metal, forming coordinate bonds. They are well-known ligands in coordination chemistry.

Statement II: "The nature of bonding of $$N(CH_3)_3$$ and $$P(CH_3)_3$$ is always same with transition metals."

INCORRECT. $$N(CH_3)_3$$ acts primarily as a $$\sigma$$-donor (donates its lone pair). $$P(CH_3)_3$$, however, can act as both a $$\sigma$$-donor and a $$\pi$$-acceptor — phosphorus has available empty d-orbitals that can accept electron density from filled metal d-orbitals ($$\pi$$-back bonding). This difference in bonding nature is significant.

The correct answer is Option (1): Statement I is correct but Statement II is incorrect.

For this problem we have to verify, for every option (A-D), two separate conditions.
1. Both complexes listed in Set-I must be able to exist as geometrical (cis-trans, fac-mer etc.) isomers.
2. The two complexes listed in Set-II must be ionization isomers of each other (i.e. they must have the same overall formula, but an interchange of an ion inside the coordination sphere with the counter-ion outside).

Condition for geometrical isomerism
• Tetrahedral complexes never show geometrical isomerism.
• Square-planar complexes of the type $$MA_2B_2,\; MA_2BC$$ etc. show cis-trans isomerism.
• Octahedral complexes show geometrical isomerism when at least two different kinds of ligands are present in suitable numbers, e.g. $$MA_4B_2$$ (cis/trans) and $$MA_3B_3$$ (fac/mer).
• Octahedral $$MA_5B$$ possesses only one possible arrangement, hence no geometrical isomerism.

Condition for ionization isomerism
Two complexes are ionization isomers when the total empirical formula (including counter-ions) is the same and an anionic ligand inside the coordination sphere interchanges with an anion that was outside it.

Now we analyse each option.

Option A

Set-I : $$[Ni(CO)_4]$$ is tetrahedral ⇒ no geometrical isomerism. Hence Condition 1 already fails. Option A is rejected.

Option B

Set-I :
• $$[Co(en)(NH_3)_2Cl_2]$$ is octahedral of the type $$M(A\!-\!A)B_2C_2$$ (here en is bidentate), so cis-trans is possible.
• $$[PdCl_2(PPh_3)_2]$$ is square-planar $$MA_2B_2$$, again cis-trans possible. Thus Condition 1 is satisfied.
Set-II : $$[Co(NH_3)_6][Cr(CN)_6]$$ and $$[Cr(NH_3)_6][Co(CN)_6]$$ have completely different ions inside the coordination spheres (the metals themselves are interchanged). They are not obtained by simply exchanging a counter-ion with a ligand. Therefore they are not ionization isomers. Option B is rejected.

Option C

Set-I :
• $$[Co(NH_3)_3(NO_2)_3]$$ is octahedral $$MA_3B_3$$ and exists as fac (facial) and mer (meridional) isomers → geometrical isomerism present.
• $$[Co(en)_2Cl_2]$$ is octahedral $$M(A\!-\!A)_2B_2$$; the two $$Cl^-$$ ligands can be cis (adjacent) or trans (opposite) → geometrical isomerism present. Condition 1 is fulfilled.
Set-II : $$[Co(NH_3)_5Cl]SO_4$$ and $$[Co(NH_3)_5(SO_4)]Cl$$ have identical overall formula $$[Co(NH_3)_5Cl]SO_4$$. One complex contains $$Cl^-$$ inside and $$SO_4^{2-}$$ outside; the other has $$SO_4^{2-}$$ inside and $$Cl^-$$ outside. They are classic ionization isomers. Condition 2 is also fulfilled.

Option D

Set-I : $$[Cr(NH_3)_5Cl]^{2+}$$ is octahedral $$MA_5B$$; all five NH3 positions are equivalent and there is only one possible arrangement. Hence no geometrical isomerism. Condition 1 fails → Option D is rejected (further, its Set-II involves hydrate rather than ionization isomerism).

Only Option C satisfies both required conditions.

Option C which is: Set-I $$[Co(NH_3)_3(NO_2)_3]$$ and $$[Co(en)_2Cl_2]$$; Set-II $$[Co(NH_3)_5Cl]SO_4$$ and $$[Co(NH_3)_5(SO_4)]Cl$$

We need to find the correct IUPAC name of $$[PtBr_2(PMe_3)_2]$$.

To determine this, we recall that ligands are named in alphabetical order ignoring multiplicative prefixes, with simple ligands taking prefixes like di- and tri-, while more complex ligands use bis- or tris- within parentheses; anionic ligands end in “-o” (for example, bromo for $$Br^-$$); and the metal name is followed by its oxidation state in Roman numerals.

First, the bromide anion $$Br^-$$ serves as an anionic ligand named “bromo,” and since there are two bromides, we use “dibromo.” The neutral ligand $$PMe_3$$ (trimethylphosphine) is already composite (it contains the prefix tri-), so we apply “bis” and enclose the ligand name in parentheses to obtain “bis(trimethylphosphine).” Balancing the charges via $$x + 2(-1) + 2(0) = 0$$ shows that $$x = +2$$, corresponding to platinum(II).

Next, arranging the ligand names alphabetically places “bromo” before “trimethylphosphine,” giving the name $$\text{dibromobis(trimethylphosphine)platinum(II)}$$.

The correct answer is Option (3): dibromobis(trimethylphosphine)platinum(II).

We need to determine the geometry of the ion formed when aluminium chloride is dissolved in acidified aqueous solution.

Key Concept: When $$AlCl_3$$ dissolves in acidified aqueous solution, the $$Al^{3+}$$ ion is strongly hydrated by water molecules due to its high charge density.

In aqueous solution, $$AlCl_3$$ dissociates: $$AlCl_3 \rightarrow Al^{3+} + 3Cl^-$$

The $$Al^{3+}$$ ion, being small and highly charged, coordinates with six water molecules to form the hexaaquaaluminium(III) ion: $$Al^{3+} + 6H_2O \rightarrow [Al(H_2O)_6]^{3+}$$

In the complex $$[Al(H_2O)_6]^{3+}$$, the coordination number of aluminium is 6. With 6 ligands around the central metal ion, the geometry is octahedral.

$$Al^{3+}$$ has the electronic configuration $$[Ne]$$ (or $$1s^2 2s^2 2p^6$$), which has no d-electrons. With sp^3d^2 hybridisation, six water molecules arrange themselves in an octahedral geometry around $$Al^{3+}$$.

Answer: Option 1 — Octahedral

We need to identify from the given options the species in which the central atom exhibits $$d^2sp^3$$ hybridization.

Statement (A): $$[Co(NH_3)_6]^{3+}$$ In this complex, the central cobalt atom is in the $$+3$$ oxidation state, giving it a $$3d^6$$ electronic configuration. Ammonia ($$NH_3$$) acts as a strong field ligand, forcing the unpaired electrons in the $$3d$$ orbitals to pair up. This leaves two internal $$3d$$ orbitals vacant, allowing the cobalt atom to undergo $$d^2sp^3$$ (inner orbital octahedral) hybridization. This is TRUE.

Statement (B): $$BrF_5$$ The central bromine atom has 7 valence electrons and forms 5 single bonds with fluorine atoms, leaving 1 lone pair. With a steric number of 6 ($$5 \text{ bond pairs} + 1 \text{ lone pair}$$), the hybridization of bromine is $$sp^3d^2$$ (outer orbital square pyramidal). This is FALSE.

Statement (C): $$[Pt(Cl)_4]^{2-}$$ Platinum(II) belongs to the $$5d$$ transition series and possesses a $$5d^8$$ configuration. Due to the high crystal field splitting energy of $$5d$$ elements, the electrons pair up even with weak ligands like chloride. With a coordination number of 4, it undergoes $$dsp^2$$ hybridization to form a square planar geometry. This is FALSE.

Statement (D): $$SF_6$$ The central sulfur atom has 6 valence electrons, all of which are shared with 6 fluorine atoms to form 6 bond pairs and 0 lone pairs. With a steric number of 6, it uses its outer $$d$$ orbitals, resulting in $$sp^3d^2$$ hybridization with an octahedral geometry. This is FALSE.

Therefore, only the central atom in $$[Co(NH_3)_6]^{3+}$$ shows $$d^2sp^3$$ hybridization.

Answer: Option A — $$[Co(NH_3)_6]^{3+}$$

We need to count how many of the given aqua complexes have an even number of unpaired d-electrons.

Key point: $$H_2O$$ is a weak field ligand, so all these complexes are high-spin (octahedral).

1. $$[V(H_2O)_6]^{3+}$$: V³⁺ has configuration $$[Ar]3d^2$$. High-spin octahedral: $$t_{2g}^2 e_g^0$$. Unpaired electrons = 2 (even).

2. $$[Cr(H_2O)_6]^{2+}$$: Cr²⁺ has configuration $$[Ar]3d^4$$. High-spin octahedral: $$t_{2g}^3 e_g^1$$. Unpaired electrons = 4 (even).

3. $$[Fe(H_2O)_6]^{3+}$$: Fe³⁺ has configuration $$[Ar]3d^5$$. High-spin octahedral: $$t_{2g}^3 e_g^2$$. Unpaired electrons = 5 (odd).

4. $$[Ni(H_2O)_6]^{3+}$$: Ni³⁺ has configuration $$[Ar]3d^7$$. High-spin octahedral: $$t_{2g}^5 e_g^2$$. Unpaired electrons = 3 (odd).

5. $$[Cu(H_2O)_6]^{2+}$$: Cu²⁺ has configuration $$[Ar]3d^9$$. Octahedral: $$t_{2g}^6 e_g^3$$. Unpaired electrons = 1 (odd).

Complexes with even number of unpaired d-electrons: $$[V(H_2O)_6]^{3+}$$ (2) and $$[Cr(H_2O)_6]^{2+}$$ (4). Total = 2.

The correct answer is Option (1): 2.

Count complexes with an even number of electrons in $$t_{2g}$$ orbitals (weak field / high spin octahedral, since all use $$H_2O$$).

Analysis (all are weak-field/high-spin octahedral):

- $$[Fe(H_2O)_6]^{2+}$$: Fe²⁺ = d⁶. High spin: $$t_{2g}^4 e_g^2$$. $$t_{2g}$$ electrons = 4 (even).

- $$[Co(H_2O)_6]^{2+}$$: Co²⁺ = d⁷. High spin: $$t_{2g}^5 e_g^2$$. $$t_{2g}$$ electrons = 5 (odd).

- $$[Co(H_2O)_6]^{3+}$$: Co³⁺ = d⁶. Low spin (Co³⁺ is strong enough): $$t_{2g}^6 e_g^0$$. $$t_{2g}$$ electrons = 6 (even).

- $$[Cu(H_2O)_6]^{2+}$$: Cu²⁺ = d⁹. $$t_{2g}^6 e_g^3$$. $$t_{2g}$$ electrons = 6 (even).

- $$[Cr(H_2O)_6]^{2+}$$: Cr²⁺ = d⁴. High spin: $$t_{2g}^3 e_g^1$$. $$t_{2g}$$ electrons = 3 (odd).

Even $$t_{2g}$$ count: Fe²⁺ (4), Co³⁺ (6), Cu²⁺ (6) → 3 complexes.

The correct answer is Option (2): 3.

We need to determine the magnetic properties of $$[Ni(CO)_4]$$ and $$[NiCl_4]^{2-}$$.

Key Concepts:

(i) The magnetic behaviour of a coordination compound depends on the number of unpaired electrons, which is determined by the metal's oxidation state, its d-electron configuration, and the crystal field splitting caused by the ligands.

(ii) Strong field ligands (like CO) cause large splitting and tend to pair electrons; weak field ligands (like Cl$$^-$$) cause small splitting.

Analysis of [Ni(CO)$$_4$$]:

Step 1: Find the oxidation state of Ni. CO is a neutral ligand, so the complex is neutral. Therefore, Ni is in the 0 oxidation state: Ni(0).

Step 2: Ni(0) has the electron configuration [Ar]3d$$^{8}$$4s$$^2$$, giving 10 valence electrons, which redistribute to 3d$$^{10}$$ in the complex.

Step 3: With a d$$^{10}$$ configuration (all d orbitals are completely filled), there are 0 unpaired electrons. The complex is diamagnetic.

Step 4: The geometry is tetrahedral (sp$$^3$$ hybridization), consistent with 4 ligands around Ni(0).

Analysis of [NiCl$$_4$$]$$^{2-}$$:

Step 1: Find the oxidation state. Cl$$^-$$ has charge -1, and the complex has overall charge -2: Ni + 4(-1) = -2, so Ni = +2. Ni$$^{2+}$$ has the configuration [Ar]3d$$^8$$.

Step 2: Cl$$^-$$ is a weak field ligand (low in the spectrochemical series). With 4 ligands and a weak field, the geometry is tetrahedral.

Step 3: In a tetrahedral crystal field with d$$^8$$, the splitting $$\Delta_t$$ is small. The electrons fill as: $$e^4 \, t_2^4$$. Using Hund's rule for the weak field, there are 2 unpaired electrons. The complex is paramagnetic.

The correct answer is Option (4): [Ni(CO)$$_4$$] is diamagnetic, [NiCl$$_4$$]$$^{2-}$$ is paramagnetic.

$$CoCl_3 \cdot nNH_3$$ gives 2 mol AgCl with excess AgNO₃. Find $$x + n$$ where $$x$$ is the oxidation state of Co.

Since 2 mol AgCl is produced, 2 Cl⁻ ions are outside the coordination sphere and 1 Cl is inside as a ligand, giving the formula $$[Co(NH_3)_nCl]Cl_2$$.

The charge on the complex ion is +2, where the inner Cl carries a -1 charge and NH₃ is neutral, so $$x + 0 \times n + (-1) = +2 \implies x = +3$$.

Assuming octahedral geometry with coordination number 6 and ligands consisting of n NH₃ molecules plus 1 Cl, we have $$n + 1 = 6$$, hence $$n = 5$$.

Therefore, $$x + n = 3 + 5 = 8$$.

The correct answer is Option (2): 8.

The color of transition metal complexes depends on d-d transitions, which require unpaired electrons and an energy gap corresponding to visible light. Nickel(II) has a d⁸ electron configuration.

Statement (I): $$[Ni(H_2O)_6]^{2+}$$ is an octahedral complex with water, a weak field ligand. This results in a high-spin configuration with two unpaired electrons. The complex absorbs light in the red region and appears green, which is a known property. Thus, Statement (I) is correct.

Statement (II): $$[Ni(CN)_4]^{2-}$$ is a square planar complex with cyanide, a strong field ligand. The d⁸ configuration leads to all electrons being paired, so no d-d transitions occur. The complex is diamagnetic and appears colorless. Thus, Statement (II) is correct.

Both statements are correct, so the appropriate answer is option B.

A) Since Cr is in +3 oxidation state, the electronic configuration is 4s2 3d3. That means 3 unpaired electrons are present. Hence BM =sqrt(n*(n+2)) = 3.87

B) Since Ni is in +2 oxidation state, the electronic configuration is 4s0 3d8. That means 2 unpaired electrons are present. Hence BM =sqrt(n*(n+2)) = 2.83

C) Since Co is in +3 oxidation state, the electronic configuration is 4s0 3d4. That means 3 unpaired electrons are present. Hence BM =sqrt(n*(n+2)) = 4.90

D) Since Ni is in +2 oxidation state but CN is a strong field ligand, the electronic configuration is 4s0 3d8 with all electrons paired. That means 0 unpaired electrons are present . Hence BM =sqrt(n*(n+2)) = 0

The colour shown by a transition-metal complex depends on the energy gap $$\Delta_{oct}$$ between its two d-electron sets. The light absorbed has frequency $$\nu$$ (or wavenumber $$\tilde{\nu}=1/\lambda$$) given by $$\Delta_{oct}=h\nu=hc\tilde{\nu}$$. Therefore:

larger $$\Delta_{oct}\;\Longrightarrow\;$$ higher energy $$\Longrightarrow\;$$ higher frequency $$\Longrightarrow\;$$ higher wavenumber of light absorbed.

Two main factors decide the magnitude of $$\Delta_{oct}$$:

1. Nature of ligands (spectrochemical series)   Weak-field ligands give small $$\Delta_{oct}$$, strong-field ligands give large $$\Delta_{oct}$$.
$$Cl^- \lt H_2O \lt NH_3 \lt CN^-$$ in ligand strength.

2. Oxidation state of the metal ion   For the same ligands, a higher positive charge on the metal pulls the ligands closer, increasing $$\Delta_{oct}$$.

Let us analyse each complex.

Case A: $$[CoCl(NH_3)_5]^{2+}$$

Co is in $$+3$$ oxidation state (because $$x + (-1) = +2 \Rightarrow x = +3$$). It has five moderate ligands $$NH_3$$ and one weak ligand $$Cl^-$$, so the overall ligand field is weaker than that produced by six $$NH_3$$ molecules.

Case B: $$[Co(CN)_6]^{3-}$$

Co is again $$+3$$. The ligand $$CN^-$$ is a very strong-field ligand. Hence $$\Delta_{oct}$$ is the largest among the four complexes.

Case C: $$[Co(NH_3)_5(H_2O)]^{3+}$$

Co is $$+3$$. There are five $$NH_3$$ (stronger than $$H_2O$$) and one $$H_2O$$. Because $$H_2O$$ is stronger than $$Cl^-$$, the average ligand field here is stronger than in Case A but weaker than in Case B.

Case D: $$[Cu(H_2O)_4]^{2+}$$

Cu is only in the $$+2$$ state, and the ligands are the medium-weak $$H_2O$$. The smaller oxidation state and weaker ligand field give the smallest $$\Delta_{oct}$$. (The geometry is tetragonally distorted octahedral or square-planar, but either way the splitting is smaller than that for the Co$$^{3+}$$ complexes.)

Combining the above comparisons:

$$\Delta_{oct}(D) \lt \Delta_{oct}(A) \lt \Delta_{oct}(C) \lt \Delta_{oct}(B)$$

Since wavenumber of absorbed light is directly proportional to $$\Delta_{oct}$$, the same order applies to $$\tilde{\nu}\,(\text{absorbed})$$:

$$[Cu(H_2O)_4]^{2+} \; \lt \; [CoCl(NH_3)_5]^{2+} \; \lt \; [Co(NH_3)_5(H_2O)]^{3+} \; \lt \; [Co(CN)_6]^{3-}$$

Hence the correct order is D < A < C < B, which corresponds to Option D.

We need to find how many of the given complexes have no electrons in the $$t_2$$ orbitals.

Note: In a tetrahedral field, the d-orbitals split into $$e$$ (lower, $$d_{z^2}$$ and $$d_{x^2-y^2}$$) and $$t_2$$ (higher, $$d_{xy}$$, $$d_{xz}$$, $$d_{yz}$$). For the $$t_2$$ orbitals to have no electrons, the metal must have $$d^0$$ configuration or all electrons must be in the $$e$$ set (at most 4 electrons with 2 in each $$e$$ orbital).

1. $$TiCl_4$$: Ti is in +4 state. $$Ti^{4+}$$: [Ar], $$d^0$$. No d-electrons at all, so no electrons in $$t_2$$. Yes - no electrons in $$t_2$$.

2. $$[MnO_4]^-$$: Mn is in +7 state. $$Mn^{7+}$$: [Ar], $$d^0$$. No d-electrons, so no electrons in $$t_2$$. Yes - no electrons in $$t_2$$.

3. $$[FeO_4]^{2-}$$: Fe is in +6 state. $$Fe^{6+}$$: [Ar], $$d^2$$. In tetrahedral field, 2 electrons go into the lower $$e$$ orbitals. So $$t_2$$ has 0 electrons. Yes - no electrons in $$t_2$$.

4. $$[FeCl_4]^-$$: Fe is in +3 state. $$Fe^{3+}$$: [Ar] $$3d^5$$. In tetrahedral (weak field), electrons fill as $$e^2 t_2^3$$. The $$t_2$$ orbitals have 3 electrons. No - has electrons in $$t_2$$.

5. $$[CoCl_4]^{2-}$$: Co is in +2 state. $$Co^{2+}$$: [Ar] $$3d^7$$. In tetrahedral field, electrons fill as $$e^4 t_2^3$$. The $$t_2$$ orbitals have 3 electrons. No - has electrons in $$t_2$$.

The complexes with no electrons in $$t_2$$ orbitals are: $$TiCl_4$$, $$[MnO_4]^-$$, and $$[FeO_4]^{2-}$$.

The correct answer is Option (3): 3.

We need to evaluate the Assertion (A) and Reason (R) about the complex $$[Co(en)_2Cl_2]^+$$.

Analysis of Assertion (A):

"The total number of geometrical isomers shown by $$[Co(en)_2Cl_2]^+$$ is three."

To determine geometrical isomers, we need to understand the structure:

- Co$$^{3+}$$ is the central metal ion (cobalt in +3 oxidation state: $$d^6$$ configuration)

- $$en$$ (ethylenediamine) is a bidentate ligand that occupies 2 coordination positions

- Two en ligands occupy $$2 \times 2 = 4$$ positions, and two Cl$$^-$$ ligands occupy 2 positions, totaling 6 positions (octahedral)

For an octahedral complex of the type $$[M(AA)_2X_2]$$ (where AA is a bidentate ligand and X is a monodentate ligand), the possible geometrical isomers are:

1. cis isomer: The two Cl ligands are adjacent to each other (at 90 degrees)

2. trans isomer: The two Cl ligands are opposite to each other (at 180 degrees)

There are only 2 geometrical isomers (cis and trans), NOT 3. The assertion that there are three geometrical isomers is incorrect.

Analysis of Reason (R):

"$$[Co(en)_2Cl_2]^+$$ has an octahedral geometry."

Co$$^{3+}$$ has a coordination number of 6 in this complex (2 bidentate en ligands providing 4 donor atoms + 2 Cl atoms = 6 donor atoms). With 6 donor atoms, the geometry is indeed octahedral. This is consistent with the $$d^2sp^3$$ hybridization of Co$$^{3+}$$ (or inner orbital complex with strong field en ligands).

The reason is correct.

Conclusion: Assertion (A) is incorrect (there are only 2 geometrical isomers, not 3), but Reason (R) is correct (the complex does have octahedral geometry).

The correct answer is Option (2): (A) is not correct but (R) is correct.

The complex is MABXL where A, B, X, and L are unidentate ligands and M is the metal atom with $$sp^3$$ hybridization.

Key concept: $$sp^3$$ hybridization gives a tetrahedral geometry.

Tetrahedral complexes do not exhibit geometrical (cis-trans) isomerism. This is because in a tetrahedron, all positions are equivalent with respect to each other — every pair of ligands is adjacent, and there are no "opposite" positions as in square planar or octahedral geometries.

Therefore, the complex MABXL with tetrahedral geometry has 0 geometrical isomers.

(Note: Tetrahedral complexes can show optical isomerism when they have 4 different ligands, but the question specifically asks about geometrical isomers.)

The correct answer is Option (2): 0.

Lead chromate (PbCrO$$_4$$, yellow) dissolves in hot NaOH solution and we are to identify the lead product.

We observe that PbCrO$$_4$$ dissolves in hot NaOH because Pb$$^{2+}$$ is amphoteric:

$$PbCrO_4 + 4NaOH \to Na_2[Pb(OH)_4] + Na_2CrO_4$$

This reaction produces [Pb(OH)$$_4$$]$$^{2-}$$, which is a dianionic complex (charge = $$+2 - 4 = -2$$) with coordination number 4 (four OH$$^-$$ ligands around Pb$$^{2+}$$).

The correct answer is Option D: Dianionic complex with coordination number four.

• (A) $$\text{NH}_4\text{Cl} + \text{NH}_4\text{OH}$$: Group III precipitating reagent, which selectively precipitates $$\text{Al}^{3+}$$ as its hydroxide.
  $$\implies \text{(A)} \rightarrow \text{(III)}$$
• (B) $$\text{NH}_4\text{OH} + \text{Na}_2\text{CO}_3$$: Group V precipitating reagent (or alternative Group V carbonating mixture), which precipitates $$\text{Sr}^{2+}$$ as its carbonate.
  $$\implies \text{(B)} \rightarrow \text{(IV)}$$
• (C) $$\text{NH}_4\text{OH} + \text{NH}_4\text{Cl} + \text{H}_2\text{S}\text{ gas}$$: Group IV precipitating reagent, which precipitates $$\text{Mn}^{2+}$$ as its sulfide.
  $$\implies \text{(C)} \rightarrow \text{(I)}$$
• (D) Dilute $$\text{HCl}$$: Group I precipitating reagent, which precipitates $$\text{Pb}^{2+}$$ as its chloride.
  $$\implies \text{(D)} \rightarrow \text{(II)}$$

Conclusion:

Aligning the systematic inorganic qualitative group separation reagents leads to the correct matching pattern.

Answer: Option A — A-III, B-IV, C-I, D-II

For every complex in List I we will:

(i) find the oxidation state of the central metal ion,
(ii) write its ground-state $$d$$-electron configuration,
(iii) decide whether the ion is high-spin or low-spin (strong-field vs weak-field ligand),
(iv) count the number of unpaired electrons and then match with List II.

Complex A : $$[Cr(H_2O)_6]^{3+}$$

Oxidation state of Cr = +3  ($$0-6(0)+x=+3\Rightarrow x=+3$$).
$$Cr^{3+}$$ ⇒ $$d^3$$ electronic configuration.
$$H_2O$$ is a weak-field ligand, so a high-spin octahedral ion is obtained:
$$t_{2g}^{3}\,e_{g}^{0}$$ - three unpaired electrons.
Hence A corresponds to “3 unpaired electrons” → List II item III.

Complex B : $$[MnF_6]^{3-}$$

Total charge: $$x+6(-1)=-3 \; \Rightarrow \; x=+3$$, so the metal is $$Mn^{3+}\;(d^4)$$.
$$F^-$$ is a weak-field ligand, therefore the complex is high-spin:
$$t_{2g}^{3}\,e_{g}^{1}$$ - four unpaired electrons.
Thus B matches List II item IV (“4 unpaired electrons”).

Complex C : $$[Fe(CN)_6]^{4-}$$

Oxidation state: $$x+6(-1)=-4 \; \Rightarrow \; x=+2$$, so ion is $$Fe^{2+}\;(d^6)$$.
$$CN^-$$ is a strong-field ligand, giving a low-spin octahedral ion:
$$t_{2g}^{6}\,e_{g}^{0}$$ - all electrons paired, zero unpaired electrons.
Therefore C corresponds to List II item I (“0 unpaired electrons”).

Complex D : $$[FeF_6]^{3-}$$

Oxidation state: $$x+6(-1)=-3 \; \Rightarrow \; x=+3$$, so ion is $$Fe^{3+}\;(d^5)$$.
With weak-field $$F^-$$ ligands, the complex is high-spin:
$$t_{2g}^{3}\,e_{g}^{2}$$ - five unpaired electrons.
Hence D matches List II item II (“5 unpaired electrons”).

Putting the four results together:

A → III, B → IV, C → I, D → II.

The combination “A-III, B-IV, C-I, D-II” is Option B.

Final Answer: Option B which is A-III, B-IV, C-I, D-II.

Step 1: Identify the product “A” in the first reaction. In the reaction $$NiS + HNO_3 + HCl \;\longrightarrow\; A + NO + S + H_2O$$ nitrate ion acts as an oxidising agent, converting sulfide to elemental sulfur (S) and itself reduced to NO. The nickel remains in the +2 oxidation state and combines with two chloride ions. Hence $$A \;=\; NiCl_2.$$

Step 2: Write the second reaction with the structure of the diacetyl dioxime ligand. Diacetyl dioxime has the formula $$CH_3C(=N-OH)C(=N-OH)CH_3,$$ and reacts with $$NiCl_2$$ in the presence of ammonium hydroxide as follows: $$NiCl_2 + 2\;NH_4OH + CH_3C(=N-OH)C(=N-OH)CH_3 \;\longrightarrow\; B + NH_4Cl + H_2O.$$ In fact, one molecule of diacetyl dioxime provides two oxime groups that chelate to Ni$$^{2+}$$, giving the neutral complex $$B = Ni[CH_3C(=NOH)C(=NOH)CH_3]_2.$$

Step 3: Draw or visualise the structure of product $$B$$. It is a square-planar complex in which each diacetyl dioxime ligand is bidentate, coordinating through the two nitrogen atoms. Each ligand retains two -OH groups, which form intramolecular hydrogen bonds to the adjacent C=N nitrogens: Intramolecular H-bonds: each $$-OH$$ proton is hydrogen-bonded to the adjacent $$=N$$ atom.

Step 4: Count all the protons in $$B$$. Each ligand $$CH_3C(=NOH)C(=NOH)CH_3$$ has: • Two methyl groups, each with 3 protons → total 6 methyl protons per ligand. • Two oxime -OH protons. Since there are two ligands, total protons in $$B$$ = $$2\times(6\;{\rm methyl}) + 2\times(2\;{\rm OH}) = 12\;{\rm CH_3} + 4\;{\rm OH} = 16\;{\rm H}.$$

Step 5: Determine which protons are involved in hydrogen bonding. The four $$-OH$$ protons are tied up in intramolecular hydrogen bonds to the adjacent $$=N$$ groups. Therefore these 4 protons are involved in hydrogen bonding.

Step 6: Calculate the number of protons not involved in hydrogen bonding. Subtracting the 4 hydrogen-bonded protons from the total 16 gives: $$16 - 4 \;=\; 12.$$

Final Answer: The number of protons in product $$B$$ that do not participate in hydrogen bonding is 12.

We need to determine how many of the given complexes exhibit optical isomerism.

Key Concept: Optical isomers (enantiomers) are non-superimposable mirror images. A complex shows optical isomerism if it lacks an improper axis of symmetry (practically, if it has no plane of symmetry and no centre of symmetry). Generally, cis isomers of octahedral complexes with bidentate ligands are optically active, while trans isomers possess a plane of symmetry and are optically inactive.

Complex 1: $$cis\text{-}[Cr(ox)_2Cl_2]^{3-}$$

This is an octahedral complex with two bidentate oxalate (ox) ligands and two monodentate Cl ligands in the cis arrangement. The cis configuration lacks a plane of symmetry because the two Cl atoms are on the same side, and the bidentate oxalate ligands create an asymmetric arrangement. Its mirror image is non-superimposable.

Result: Optically active.

Complex 2: $$[Co(en)_3]^{3+}$$

This complex has three bidentate ethylenediamine (en) ligands arranged around Co$$^{3+}$$ in an octahedral geometry. With three identical bidentate ligands, the complex adopts a propeller-like arrangement. The mirror image ($$\Delta$$ and $$\Lambda$$ forms) are non-superimposable. This is a classic example of an optically active complex.

Result: Optically active.

Complex 3: $$cis\text{-}[Pt(en)_2Cl_2]^{2+}$$

This octahedral complex has two bidentate en ligands and two Cl ligands in cis positions. Similar to Complex 1, the cis arrangement with bidentate ligands creates an asymmetric structure with no plane of symmetry. The mirror image is non-superimposable.

Result: Optically active.

Complex 4: $$cis\text{-}[Co(en)_2Cl_2]^+$$

This octahedral complex has two bidentate en ligands and two Cl ligands in cis positions. The cis arrangement produces a structure without a plane of symmetry, and its mirror image is non-superimposable. This is another well-known example of an optically active coordination compound.

Result: Optically active.

Complex 5: $$trans\text{-}[Pt(en)_2Cl_2]^{2+}$$

In the trans arrangement, the two Cl ligands are on opposite sides (180 degrees apart). This trans configuration possesses a plane of symmetry that passes through the metal center and the two Cl atoms. The mirror image is superimposable on the original structure.

Result: NOT optically active.

Complex 6: $$trans\text{-}[Cr(ox)_2Cl_2]^{3-}$$

In the trans configuration, the two Cl ligands are positioned opposite to each other. This arrangement introduces a plane of symmetry in the complex, making the mirror image superimposable on the original.

Result: NOT optically active.

Counting: Complexes 1, 2, 3, and 4 are optically active. Complexes 5 and 6 are not.

The total number of complexes showing optical isomerism is $$\boxed{4}$$.

$$[Ni(NH_3)_6]^{2+}$$: Ni²⁺ is d⁸. NH₃ is weak field in octahedral... actually NH₃ is moderate/strong field. For octahedral Ni²⁺ with 8 d-electrons: t₂g⁶ eg² gives 2 unpaired electrons.

$$\mu = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} = 2\sqrt{2} \approx 2.83$$ BM.

In units of $$10^{-1}$$ BM: 28.3 ≈ 28.

The answer is $$\boxed{28}$$.

The metallic radius decreases steadily from Sc to Cr in the 3d series because the increasing nuclear charge is not fully screened by the added 3d electrons. After Cr the radius stops decreasing and even rises slightly. Therefore, among the given elements (Sc, Ti, V, Cr, Mn, Zn) the one with the smallest metallic radius is chromium (Cr).

Hence in $$MO_4^{2-}$$ the metal atom is chromium, giving the anion $$CrO_4^{2-}$$ (chromate ion).

Each oxide ion carries a charge of $$-2$$, so for $$CrO_4^{2-}$$ the overall charge balance is
$$x + 4(-2) = -2 \;\;\Longrightarrow\;\; x = +6$$
Thus chromium is in the $$+6$$ oxidation state.

Electronic configuration of a neutral Cr atom: $$[Ar]\,3d^5\,4s^1$$.
To obtain $$Cr^{+6}$$ we remove six electrons (the one 4s electron and five 3d electrons), leaving
$$Cr^{+6}: [Ar]\,3d^{0}$$.

Number of unpaired electrons, $$n = 0$$.

The spin-only magnetic moment is calculated by the formula
$$\mu_{\text{spin}} = \sqrt{n(n+2)}\;\text{BM}$$.

Substituting $$n = 0$$ gives
$$\mu_{\text{spin}} = \sqrt{0(0+2)} = 0\;\text{BM}$$.

Hence the ‘spin only’ magnetic moment of $$MO_4^{2-}$$ is $$\mathbf{0\;BM}$$.

Among the ions $$Ti^{2+}, V^{2+}, Co^{3+}, Cr^{2+}$$, the one that acts as a strong oxidising agent in aqueous solution is $$Co^{3+}$$. A strong oxidising agent easily gets reduced (accepts electrons), and $$Co^{3+}$$ readily accepts an electron to form the more stable $$Co^{2+}$$. The other ions ($$Ti^{2+}, V^{2+}, Cr^{2+}$$) are actually strong reducing agents.

The electronic configuration of cobalt is Co: [Ar] $$3d^7 4s^2$$, so for $$Co^{3+}$$ it becomes [Ar] $$3d^6$$. In aqueous solution, the complex $$[Co(H_2O)_6]^{3+}$$ behaves as a high-spin complex (since water is a weak field ligand), giving the configuration $$t_{2g}^4 e_g^2$$ with 4 unpaired electrons.

Using the spin-only formula for the magnetic moment, we have $$\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} = 4.9 \approx 5 \text{ BM}$$.

Final answer: $$Co^{3+}$$ is the strong oxidising agent and its spin-only magnetic moment is 5 BM.

We need to find the total number of unpaired electrons in $$[Co(NH_3)_6]^{3+}$$ and $$[NiCl_4]^{2-}$$.

For $$[Co(NH_3)_6]^{3+}$$:

Cobalt is in +3 oxidation state: $$Co^{3+}$$ has configuration $$[Ar]3d^6$$.

$$NH_3$$ is a strong field ligand, so this is a low-spin octahedral complex.

In a strong octahedral field, the 6 d-electrons fill the $$t_{2g}$$ orbitals completely: $$t_{2g}^6 e_g^0$$.

Number of unpaired electrons = 0.

For $$[NiCl_4]^{2-}$$:

Nickel is in +2 oxidation state: $$Ni^{2+}$$ has configuration $$[Ar]3d^8$$.

$$Cl^-$$ is a weak field ligand, so this is a tetrahedral complex (common for $$Ni^{2+}$$ with weak field ligands).

In a tetrahedral field with $$d^8$$ configuration, the splitting is: $$e^4 t_2^4$$.

Number of unpaired electrons = 2.

Total unpaired electrons = $$0 + 2 = 2$$.

The answer is $$\boxed{2}$$.

Ambidentate ligands are those that can coordinate through two different donor atoms.

$$NO_2^-$$: Can bind through N (nitro) or O (nitrito) — ambidentate ✓

$$SCN^-$$: Can bind through S (thiocyanato) or N (isothiocyanato) — ambidentate ✓

$$C_2O_4^{2-}$$: Bidentate ligand (binds through two O atoms of the same type) — not ambidentate ✗

$$NH_3$$: Monodentate, binds only through N — not ambidentate ✗

$$CN^-$$: Can bind through C (cyano) or N (isocyano) — ambidentate ✓

$$SO_4^{2-}$$: Not an ambidentate ligand ✗

$$H_2O$$: Monodentate, binds through O — not ambidentate ✗

Number of ambidentate ligands = 3 ($$NO_2^-, SCN^-, CN^-$$).

The answer is $$\boxed{3}$$.

• Identification of Cation ($$\text{M}^{2+}$$): Among group-IV cations ($$\text{Mn}^{2+}, \text{Zn}^{2+}, \text{Co}^{2+}, \text{Ni}^{2+}$$), only $$\text{Co}^{2+}$$ forms a black precipitate ($$\text{A} = \text{CoS}$$) that dissolves in aqua regia to yield $$\text{B} = \text{CoCl}_2$$, which then reacts with $$\text{KNO}_2$$ in acetic acid to give a brown precipitate.

• Complex Formation (C): The brown precipitate is potassium hexanitritocobaltate(III):
  $$\text{C} = \text{K}_3[\text{Co}(\text{NO}_2)_6]$$
• Magnetic Moment Calculation:
  • Oxidation state of Cobalt in the complex is $$+3$$.
  • Electronic configuration: $$\text{Co}^{3+} = [\text{Ar}] 3d^6$$.
  • Since $$\text{NO}_2^-$$ is a strong-field ligand, it causes complete pairing of electrons in the octahedral field ($$t_{2g}^6 e_g^0$$).
  • Number of unpaired electrons ($$n$$) = 0.
  • $$\text{Spin-only magnetic moment } (\mu) = \sqrt{n(n+2)} = 0\text{ BM}$$

Conclusion:

Because all six $$3d$$ electrons are fully paired up by the strong-field nitrito ligands, the complex is completely diamagnetic.

Answer: 0

The Brown Ring Test is used to detect nitrate ions ($$NO_3^-$$). In this test, ferrous sulfate ($$FeSO_4$$) solution is added to the sample, followed by carefully layering concentrated sulfuric acid, resulting in the formation of the brown ring complex $$[Fe(H_2O)_5(NO)]^{2+}$$.

This nitrosyl complex of iron involves NO acting as a neutral ligand: $$NO^+$$ donates to the metal and becomes $$NO$$ in the coordination sphere.

To determine the oxidation state of Fe, we use charge balance. The overall charge on the complex is +2, each water ligand is neutral (charge = 0), and NO is treated as nitrosonium ($$+1$$). Writing the charge equation gives $$\text{Fe} + 5(0) + 1 = +2$$, so $$\text{Fe} = +1$$.

Therefore, the oxidation state of Fe in the brown ring complex is +1. The iron starts as $$Fe^{2+}$$ and is reduced to $$Fe^{+1}$$ upon coordination with NO, an unusual oxidation state stabilized by strong back-bonding with the NO ligand.

The final answer is 1.

$$Pt$$ has atomic number 78. Its electronic configuration is $$[Xe]4f^{14}5d^96s^1$$.

In the complex $$[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$$, the inner complex is $$[Pt(NH_3)_2Cl(NH_2CH_3)]^+$$, so Pt is in +2 oxidation state.

$$Pt^{2+}$$: $$[Xe]4f^{14}5d^8$$

In a square planar complex (which Pt(II) typically forms), the crystal field splits the d-orbitals such that all 8 electrons pair up in the 4 lower energy orbitals.

Number of unpaired electrons = 0.

Spin-only magnetic moment = $$\sqrt{0(0+2)} = 0$$ B.M.

The answer is $$\boxed{0}$$ B.M.

Potassium ferricyanide is $$K_3[Fe(CN)_6]$$. In the anion $$[Fe(CN)_6]^{3-}$$ iron is in the +3 oxidation state and all six ligands are $$CN^-$$.

Step 1 - Formation of the intermediate complex P
Iodide ion is a good entering ligand for the octahedral ferricyanide ion. The substitution that actually occurs is

$$[Fe(CN)_6]^{3-} + I^- \;\rightleftharpoons\; [Fe(CN)_5I]^{3-} + CN^-$$

The reaction is reversible because the liberated $$CN^-$$ can again replace $$I^-$$. However, in a strongly acidic medium the free $$CN^-$$ is protonated to volatile HCN:

$$CN^- + H^+ \rightarrow HCN(g)$$

Removal of $$HCN$$ (by escape of the gas) drives the equilibrium completely to the right, so the only iron-containing species present is

$$P = [Fe(CN)_5I]^{3-}$$

Step 2 - Precipitation of Q with $$ZnCl_2$$
In a slightly acidic solution, the divalent zinc ion combines with the trivalent anion $$[Fe(CN)_5I]^{3-}$$ to give a neutral, sparingly soluble salt. Let the formula of this salt be $$Zn_x[Fe(CN)_5I]_y$$.

For electrical neutrality: $$2x - 3y = 0 \quad\Longrightarrow\quad 2x = 3y$$

The smallest integer solution is $$x = 3,\; y = 2$$. Hence the precipitate is

$$Q = Zn_3[Fe(CN)_5I]_2$$

Step 3 - Number of zinc ions in Q
From the formula of Q, the number of $$Zn^{2+}$$ ions present is $$3$$.

Therefore, the required number is 3.

The complex $$[Pt(NH_3)_2Br_2]$$ is square-planar with the composition $$MA_2B_2$$ (here $$A = NH_3$$ and $$B = Br^-$$).

Square-planar $$MA_2B_2$$ species can exist in two geometrical arrangements:
  • cis : the two $$A$$ ligands are adjacent.
  • trans : the two $$A$$ ligands are opposite.
Hence the “type of isomerism” referred to in the question is cis-trans (geometrical) isomerism.

We must therefore look for complexes in the options that are capable of showing the same cis-trans geometrical isomerism.

Option A : $$[Pt(en)(SCN)_2]$$
  • Coordination number = 4, geometry = square-planar.
  • The bidentate ligand $$en$$ occupies two adjacent positions compulsorily (it must be cis).
  • The two $$SCN^-$$ ligands are forced into the remaining two adjacent positions, so only one spatial arrangement is possible.
  • Therefore no cis-trans isomerism is possible. (The complex can show linkage isomerism through $$S$$ vs $$N$$ binding, but that is a different type.)
  ⇒ Does not match the required isomerism.

Option B : $$[Zn(NH_3)_2Cl_2]$$
  • $$Zn^{2+}$$ with coordination number 4 usually forms a tetrahedral complex.
  • A tetrahedral $$MA_2B_2$$ arrangement is symmetric; cis and trans positions are indistinguishable.
  • Therefore no geometrical isomerism exists.
  ⇒ Does not match the required isomerism.

Option C : $$[Pt(NH_3)_2Cl_4]$$
  • Coordination number = 6, oxidation state $$+4$$, geometry = octahedral.
  • General formula $$MA_2B_4$$ (two $$NH_3$$ and four $$Cl^-$$).
  • In an octahedron, two distinct arrangements exist:
    - cis : the two $$NH_3$$ ligands are 90° apart.
    - trans : the two $$NH_3$$ ligands are 180° apart.
  • Thus the complex does show cis-trans geometrical isomerism.
  ⇒ Matches the required type.

Option D : $$[Cr(en)_2(H_2O)(SO_4)]^{+}$$
  • Coordination number = 6, geometry = octahedral.
  • Two bidentate $$en$$ ligands occupy four sites; $$H_2O$$ and $$SO_4^{2-}$$ occupy the remaining two sites.
  • For the monodentate pair ($$H_2O$$ vs $$SO_4^{2-}$$) we can have:
    - cis : $$H_2O$$ adjacent to $$SO_4^{2-}$$.
    - trans : $$H_2O$$ opposite $$SO_4^{2-}$$.
  • Hence the complex exhibits cis-trans geometrical isomerism; in addition it can be optically active, but that does not affect the present question.
  ⇒ Matches the required type.

Therefore, the complexes capable of showing the same cis-trans geometrical isomerism as $$[Pt(NH_3)_2Br_2]$$ are:

Option C which is: $$[Pt(NH_3)_2Cl_4]$$
Option D which is: $$[Cr(en)_2(H_2O)(SO_4)]^{+}$$

Answer: Option C and Option D.

When $$FeCl_3$$ is treated with $$K_4[Fe(CN)_6]$$ (potassium ferrocyanide), a Prussian blue precipitate is formed. The reaction is:

$$4FeCl_3 + 3K_4[Fe(CN)_6] \rightarrow Fe_4[Fe(CN)_6]_3 + 12KCl$$

The Prussian blue precipitate is ferric ferrocyanide with the formula $$Fe_4[Fe(CN)_6]_3$$. In this compound, $$Fe^{3+}$$ ions (from FeCl$$_3$$) are the external cations, while $$[Fe(CN)_6]^{4-}$$ is the complex anion where Fe is in the +2 oxidation state.

So, the answer is that the Prussian blue precipitate formed is $$Fe_4[Fe(CN)_6]_3$$.

Dimethylglyoxime (DMG) in ammoniacal solution gives a brilliant red precipitate with Ni²⁺ ions. This is a characteristic test for nickel.

The red precipitate is nickel dimethylglyoximate, a chelate complex where Ni²⁺ is coordinated with two DMG molecules through the oxime nitrogen atoms.

A chloride salt solution acidified with dilute HNO$$_3$$ gives a curdy white precipitate [A] on addition of AgNO$$_3$$, and [A] on treatment with NH$$_4$$OH gives a clear solution B.

When AgNO$$_3$$ is added to a chloride solution acidified with dilute HNO$$_3$$, silver chloride precipitates according to the reaction:
$$\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \downarrow \text{ (curdy white precipitate)}$$
This shows [A] is AgCl.

Treatment of AgCl with NH$$_4$$OH leads to formation of the diamminesilver(I) complex by the reaction:
$$\text{AgCl} + 2\text{NH}_3 \rightarrow [\text{Ag}(\text{NH}_3)_2]\text{Cl}$$
This gives a soluble complex and identifies [B] as [Ag(NH$$_3$$)$$_2$$]Cl (diammine silver(I) chloride).

Since [A] is AgCl and [B] is [Ag(NH$$_3$$)$$_2$$]Cl, the correct answer is Option C: AgCl & [Ag(NH$$_3$$)$$_2$$]Cl.

Find the correct order of spin-only magnetic moment for the given complex ions.

Determine the electronic configuration and unpaired electrons.

$$[\text{FeF}_6]^{3-}$$: Fe$$^{3+}$$ has configuration [Ar] 3d$$^5$$. F$$^-$$ is a weak field ligand (high spin). The 5 d-electrons occupy all five orbitals singly: 5 unpaired electrons.

$$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$$ BM

$$[\text{CoF}_6]^{3-}$$: Co$$^{3+}$$ has configuration [Ar] 3d$$^6$$. F$$^-$$ is a weak field ligand (high spin). In octahedral high spin d$$^6$$: t$$_{2g}^4$$ e$$_g^2$$, giving 4 unpaired electrons.

$$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$$ BM

$$[\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$$: Co$$^{3+}$$ has configuration [Ar] 3d$$^6$$. C$$_2$$O$$_4^{2-}$$ (oxalate) is higher than F$$^-$$ in the spectrochemical series, and Co$$^{3+}$$ has a high charge density. This results in a low spin configuration: t$$_{2g}^6$$ e$$_g^0$$, giving 0 unpaired electrons.

$$\mu = 0$$ BM

Order the magnetic moments.

$$[\text{FeF}_6]^{3-}$$ (5.92 BM) $$>$$ $$[\text{CoF}_6]^{3-}$$ (4.90 BM) $$>$$ $$[\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$$ (0 BM)

The correct answer is $$[\text{FeF}_6]^{3-} > [\text{CoF}_6]^{3-} > [\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$$.

We need to identify compounds used to inhibit the growth of tumours.

Analysis of each compound:

(A) EDTA: Used as a chelating agent for removing heavy metals from the body (chelation therapy), not specifically for tumour inhibition.

(B) Coordination Compounds of Pt: Platinum-based coordination compounds are widely used as anticancer drugs. Examples include cisplatin, carboplatin, and oxaliplatin. Used for tumour inhibition. ✓

(C) D-Penicillamine: Used for treatment of Wilson's disease and rheumatoid arthritis (as a chelating agent), not for tumours.

(D) Cis-Platin: Cisplatin (cis-[PtCl$$_2$$(NH$$_3$$)$$_2$$]) is one of the most effective anticancer drugs. It inhibits DNA replication in cancer cells. Used for tumour inhibition. ✓

Note: Cis-platin (D) is itself a coordination compound of Pt (B), so both B and D are correct.

The correct answer is Option 1: B and D Only.

We need to identify the complex that is octahedral, diamagnetic, and most stable.

First, we analyze $$\mathrm{Na}_3[CoCl_6]$$, where cobalt is in the $$3+$$ oxidation state with a $$d^6$$ configuration and chloride is a weak field ligand. In an octahedral weak field, this gives a high spin arrangement of $$t_{2g}^4e_g^2$$ with four unpaired electrons, making the complex paramagnetic and therefore not the answer.

Next, we consider $$[Ni(NH_3)_6]Cl_2$$, in which nickel is $$2+$$ with a $$d^8$$ configuration and ammonia as a ligand. In an octahedral geometry, $$d^8$$ invariably leads to $$t_{2g}^6e_g^2$$ with two unpaired electrons, resulting in paramagnetism and ruling it out.

Then, examining $$K_3[Co(CN)_6]$$, cobalt is again $$3+$$ with $$d^6$$, but cyanide is a strong field ligand that induces a large crystal field splitting. This leads to a low spin configuration of $$t_{2g}^6e_g^0$$ with no unpaired electrons, rendering the complex diamagnetic. Its octahedral geometry combined with high crystal field stabilization energy makes this the most stable candidate that meets all criteria.

Meanwhile, $$[Co(H_2O)_6]Cl_2$$ features cobalt in the $$2+$$ oxidation state with $$d^7$$ and water as a weak field ligand, which yields high spin $$t_{2g}^5e_g^2$$ with three unpaired electrons, again paramagnetic and thus unsuitable.

Therefore, the correct choice is Option 3: $$K_3[Co(CN)_6]$$.

The magnitude of $$d$$-orbital splitting depends primarily on the strength of the ligands according to the Spectrochemical Series. 

For a constant metal ion in the same oxidation state ($$Fe^{3+}$$ in all these cases), the ligand strength determines the extent of splitting.

Ligand Strength Order: 

$$CN^->NH_3>C_2O_4^{2-}>F^-$$

[Fe(CN)$$_6$$]$$^{3-}$$: Cyanide $$(CN^{-})$$ is a strong-field ligand. 

It causes the maximum repulsion with the $$d$$-orbitals of the metal, leading to the largest crystal field splitting energy $$\triangle_o$$.

Other options: $$NH_{3}$$ is a moderate ligand, while $$C_{2}O_{4}^{2-}$$ and $$F^{-}$$ are weak-field ligands, resulting in smaller splitting.

The complex that will show the largest splitting of $$d$$-orbitals $$\triangle_o$$ is C: $$[Fe(CN)_6]^{3-}$$.

When cobalt chloride (CoCl$$_2$$) is dissolved in water:

The water molecules coordinate with Co$$^{2+}$$ to form the pink-colored hexaaquacobalt(II) complex:

$$ X = [Co(H_2O)_6]^{2+} \quad \text{(Octahedral geometry)} $$

When this solution is treated with concentrated HCl, the water ligands are replaced by chloride ions to form the deep blue tetrachlorocobaltate(II) complex:

$$ [Co(H_2O)_6]^{2+} + 4Cl^- \rightarrow [CoCl_4]^{2-} + 6H_2O $$

$$ Y = [CoCl_4]^{2-} \quad \text{(Tetrahedral geometry)} $$

Therefore: X = $$[Co(H_2O)_6]^{2+}$$, Y = $$[CoCl_4]^{2-}$$, Z = Tetrahedral.

Assertion A: The spin-only magnetic moment for $$[Fe(CN)_6]^{3-}$$ is 1.74 BM, whereas for $$[Fe(H_2O)_6]^{3+}$$ is 5.92 BM.

Fe³⁺ has the electronic configuration [Ar]3d⁵ (5 d-electrons).

In $$[Fe(CN)_6]^{3-}$$: CN⁻ is a strong field ligand, causing pairing. The configuration is $$t_{2g}^5 e_g^0$$ with 1 unpaired electron.

$$\mu = \sqrt{n(n+2)} = \sqrt{1(3)} = 1.73 \approx 1.74$$ BM ✓

In $$[Fe(H_2O)_6]^{3+}$$: H₂O is a weak field ligand, no pairing occurs. The configuration is $$t_{2g}^3 e_g^2$$ with 5 unpaired electrons.

$$\mu = \sqrt{5(7)} = \sqrt{35} = 5.92$$ BM ✓

So Assertion A is true.

Reason R: In both complexes, Fe is in +3 oxidation state. This is true.

However, the same oxidation state does not explain the different magnetic moments. The difference arises because of the different ligand field strengths (strong vs weak field ligands), not the oxidation state. Therefore, R is NOT the correct explanation of A.

The correct answer is Both A and R are true but R is NOT the correct explanation of A.

In $$\text{Ni(CO)}_4$$ and $$\text{Fe(CO)}_5$$, the CO ligands are neutral, so the overall charge on each complex equals the charge on the metal. Since both complexes are uncharged, the metals are in the zero oxidation state. Assertion A is correct.

Reason R states that low oxidation states arise when ligands have $$\pi$$-bonding character beyond $$\sigma$$-bonding. CO forms a synergic bond: it donates a lone pair to the metal ($$\sigma$$-bond) and simultaneously accepts electron density from the metal's filled $$d$$-orbitals into its empty $$\pi^*$$ orbitals ($$\pi$$-back-bonding). This $$\pi$$-interaction stabilises the metal in a low oxidation state. Reason R correctly identifies this $$\pi$$-bonding ability as the cause of the low oxidation state, and it explains Assertion A.

Both A and R are correct, and R is the correct explanation of A. The answer is $$\boxed{\text{Option (D)}}$$.

Ni²⁺ (d⁸, 3d): With weak field ligands (Cl⁻, Br⁻), forms tetrahedral complex (paramagnetic, 2 unpaired electrons, no geometrical isomerism).

Pt²⁺ (d⁸, 5d): Always forms square planar complex (diamagnetic, 0 unpaired electrons, shows cis-trans isomerism).

Changes: A (geometry), B (geometrical isomerism now possible), D (magnetic properties). C (optical isomerism) does not change — neither tetrahedral [MX₂Y₂] nor square planar [MX₂Y₂] shows optical isomerism.

Answer: A, B and D.

The formula for CFSE in an octahedral field is:

• A. $$[Ti(H_2O)_6]^{2+}$$
  • Titanium oxidation state: Ti2+ (3d2 configuration)
  • Electron distribution: t2g2 eg0
  • $$\text{CFSE} = [2 \times (-0.4) + 0 \times 0.6]\Delta_o = \mathbf{-0.8\Delta_o}$$
  A matches with IV


• B. $$[V(H_2O)_6]^{2+}$$
  • Vanadium oxidation state: V2+ (3d3 configuration)
  • Electron distribution: t2g3 eg0
  •$$\text{CFSE} = [3 \times (-0.4) + 0 \times 0.6]\Delta_o = \mathbf{-1.2\Delta_o}$$
B matches with I


• C. $$[Mn(H_2O)_6]^{3+}$$
  • Manganese oxidation state: Mn3+ (3d4 high-spin configuration)
  • Electron distribution: t2g3 eg1
  • $$\text{CFSE} = [3 \times (-0.4) + 1 \times 0.6]\Delta_o = [-1.2 + 0.6]\Delta_o = \mathbf{-0.6\Delta_o}$$
  C matches with II


• D. $$[Fe(H_{2}O)_{6}]^{3+}$$
  • Iron oxidation state: Fe3+ (3d5 high-spin configuration)
  • Electron distribution: t2g3 eg2
  • $$\text{CFSE} = [3 \times (-0.4) + 2 \times 0.6]\Delta_o = [-1.2 + 1.2]\Delta_o = \mathbf{0\Delta_o}$$
  D matches with III

Summary Mapping:

Conclusion:

The correct option order is A-IV, B-I, C-II, D-III.

Correct Option: C

To determine the correct matching between the coordination entities and the wavelengths of light they absorb, we use Crystal Field Theory (CFT).

The wavelength of light absorbed $$\lambda$$ is inversely proportional to the crystal field splitting energy $$\Delta_o$$:

$$E=\frac{hc}{\lambda}\implies \Delta_o\propto\frac{1}{\lambda}$$

Hence, stronger field ligands produce a larger crystal field splitting and absorb light of shorter wavelength.

According to the spectrochemical series,

$$CN^- > NH_3 > H_2O > Cl^-$$

Evaluating the coordination entities

• $$[Co(CN)_6]^{3-}$$
  • $$CN^-$$ is the strongest field ligand among the given ligands.
  • It produces the largest value of $$\Delta_o$$ and therefore absorbs the shortest wavelength.
  • Match: I (310 nm)
• $$[Co(NH_3)_6]^{3+}$$
  • $$NH_3$$ is a strong field ligand but weaker than $$CN^-$$.
  • Hence, it absorbs a longer wavelength than $$[Co(CN)_6]^{3-}$$.
  • Match: II (475 nm)
• $$[CoCl(NH_3)_5]^{2+}$$
  • Replacing one $$NH_3$$ ligand by the weaker field ligand $$Cl^-$$ decreases the crystal field splitting.
  • Therefore, the absorbed wavelength increases compared to $$[Co(NH_3)_6]^{3+}$$.
  • Match: III (535 nm)
• $$[Cu(H_2O)_4]^{2+}$$
  • $$H_2O$$ is a weaker field ligand than $$NH_3$$ and $$CN^-$$.
  • Hence, it produces a smaller value of $$\Delta_o$$ and absorbs the longest wavelength among the given complexes.
  • Match: IV (600 nm)

Therefore, the correct matching is

• A → III
• B → II
• C → I
• D → IV

which corresponds to

$$\boxed{\text{A-III,; B-II,; C-I,; D-IV}}$$

Let us analyze the solubility of each complex in water:

Option 1: (NH$$_4$$)$$_3$$[AsMo$$_3$$O$$_{10}$$]$$_4$$ - Ammonium arsenomolybdate is an insoluble yellow precipitate (used to test for arsenate ions).

Option 2: Fe$$_4$$[Fe(CN)$$_6$$]$$_3$$ - This is Prussian blue (ferric ferrocyanide), which is an insoluble blue pigment.

Option 3: K$$_3$$[Co(NO$$_2$$)$$_6$$] - Potassium cobaltinitrite is an insoluble yellow precipitate (used as a confirmatory test for K$$^+$$ ions).

Option 4: [Fe$$_3$$(OH)$$_2$$(OAc)$$_6$$]Cl - This is a basic iron(III) acetate complex. It is a cationic complex with Cl$$^-$$ as the counter ion, making it water-soluble.

Therefore, the complex that dissolves in water is [Fe$$_3$$(OH)$$_2$$(OAc)$$_6$$]Cl.

For the complex Hexaamminecobalt(III) complex:

• Oxidation state of cobalt $$= +3$$
• Electronic configuration of $$Co^{+3} = 3d^{6}$$
  

Since $$NH_3$$​ acts as a strong field ligand with $$Co^{+3}$$ , the $$3d$$ electrons pair up in the lower energy orbitals.

So the arrangement becomes:

• $$t^6_{2g}$$ $$e^0_g$$
• No unpaired electrons

Therefore:

• Hybridisation = $$d^{2}sp^{3}$$ (inner orbital octahedral complex)
• Magnetic behaviour = Diamagnetic

So the final answer is:

$$d^{2}sp^{3}$$ hybridisation and diamagnetic nature.

Checking the matches:

A. Chlorophyll — Co: INCORRECT. Chlorophyll contains Mg, not Co (Co is in vitamin B12).

B. Water hardness — EDTA: CORRECT. EDTA is used to determine water hardness.

C. Photography — [Ag(CN)₂]⁻: INCORRECT? Actually, in photography, [Ag(S₂O₃)₂]³⁻ is used (hypo/thiosulphate). However, [Ag(CN)₂]⁻ is used in silver extraction. So C is a mismatch.

D. Wilkinson catalyst — [(Ph₃P)₃RhCl]: CORRECT.

E. Chelating ligand — D-Penicillamine: CORRECT (D-Penicillamine is a chelating agent).

Mismatched: A and C only.

The correct answer is Option B: A and C only.

In $$[Co(NH_3)_5Cl]Cl_2$$:

Primary valency (oxidation state of Co) = +3 (since total charge: 5(0) + (-1) + Co = +2 from 2Cl⁻, so Co = +3).

Secondary valency (coordination number) = 6 (5 NH₃ + 1 Cl⁻ in the coordination sphere).

Therefore, primary valency = 3 and secondary valency = 6.

We need to find the primary and secondary valencies of cobalt in $$Co(NH_3)_5ClCl_2$$.

Understand Werner's theory of coordination compounds.

According to Werner's theory:

- Primary valency corresponds to the oxidation state of the metal ion. It is ionisable and is satisfied by negative ions.

- Secondary valency corresponds to the coordination number. It is non-ionisable and represents the number of ligands directly bonded to the metal ion.

Write the coordination compound in proper notation.

$$Co(NH_3)_5ClCl_2$$ can be written as $$[Co(NH_3)_5Cl]Cl_2$$, where:

- The part inside square brackets is the coordination sphere: $$[Co(NH_3)_5Cl]^{2+}$$

- The two $$Cl^-$$ outside are counter ions (satisfy primary valency)

Determine the primary valency (oxidation state).

Let the oxidation state of Co be $$x$$. The overall charge on the complex ion is +2 (to balance two $$Cl^-$$):

$$x + 5(0) + (-1) = +2$$

$$x = +3$$

Primary valency = 3.

Determine the secondary valency (coordination number).

Count the total number of ligands directly attached to Co: 5 $$NH_3$$ molecules + 1 $$Cl^-$$ = 6 ligands.

Secondary valency = 6.

The correct answer is Option (4): 3 and 6.

A double salt is a salt that contains two different cations or anions. It dissociates completely into its constituent ions in aqueous solution and does not retain the identity of its complex ion.

In contrast, a coordination compound retains its complex ion identity in solution.

Let us analyze each compound:

(A) $$FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$$ - This is Mohr's salt, a classic example of a double salt. In solution, it gives $$Fe^{2+}$$, $$NH_4^+$$, and $$SO_4^{2-}$$ ions separately. Double salt.

(B) $$CuSO_4 \cdot 4NH_3 \cdot H_2O$$ - This is $$[Cu(NH_3)_4]SO_4 \cdot H_2O$$, a coordination compound. The $$[Cu(NH_3)_4]^{2+}$$ ion retains its identity in solution.

(C) $$K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$$ - This is potash alum, a double salt. It dissociates completely into $$K^+$$, $$Al^{3+}$$, and $$SO_4^{2-}$$ ions in solution. Double salt.

(D) $$Fe(CN)_2 \cdot 4KCN$$ - This is $$K_4[Fe(CN)_6]$$, potassium ferrocyanide, a coordination compound. The $$[Fe(CN)_6]^{4-}$$ ion retains its identity in solution.

Therefore, the double salts are A and C only.

We need to identify the chiral complex. A complex is chiral if its mirror image is non-superimposable on the original.

Option A: cis-[PtCl₂(en)₂]²⁺

This is an octahedral complex. Ethylenediamine (en) is a bidentate ligand that chelates through two nitrogen atoms. With two en ligands and two Cl⁻ in cis positions, the complex has the form:

The two Cl⁻ are adjacent to each other, and the two en ligands span the remaining positions. This arrangement has no plane of symmetry and no improper rotation axis. The mirror image of this complex cannot be superimposed on the original.

This is analogous to the well-known chiral complex cis-[Co(en)₂Cl₂]⁺, which exists as Λ (lambda) and Δ (delta) enantiomers. This complex is chiral.

Option B: trans-[PtCl₂(en)₂]²⁺

In the trans arrangement, the two Cl⁻ are on opposite sides. This creates a plane of symmetry passing through the two Cl atoms and the metal center. Not chiral.

Option C: cis-[PtCl₂(NH₃)₂]

This is a square planar complex (Pt²⁺ with 4 monodentate ligands). All atoms lie in the same plane (the molecular plane), which is itself a plane of symmetry. Not chiral.

Option D: trans-[Co(NH₃)₄Cl₂]⁺

Octahedral with all monodentate ligands in trans arrangement. Multiple planes of symmetry exist. Not chiral.

The answer is Option A: cis-[PtCl₂(en)₂]²⁺.

A. Mg(NH₄)PO₄ — White precipitate (II)

B. K₃[Co(NO₂)₆] — Potassium cobaltinitrite — Yellow (III)

C. MnO(OH)₂ — Manganese dioxide hydroxide — Brown (I)

D. Fe₄[Fe(CN)₆]₃ — Prussian blue — Blue (IV)

Matching: A-II, B-III, C-I, D-IV

We need to determine the number of unpaired electrons in each complex and arrange them in increasing order.

The number of unpaired electrons depends on the metal's oxidation state, its d-electron count, and whether the ligand creates a strong field (low spin) or weak field (high spin) splitting.

(A) $$[Fe(CN)_6]^{3-}$$:

Iron is in the +3 oxidation state: $$Fe^{3+}$$ has the configuration $$[Ar] 3d^5$$.

$$CN^{-}$$ is a strong field ligand, so we get a low spin octahedral complex.

Low spin $$d^5$$: $$t_{2g}^5 \, e_g^0$$ → 1 unpaired electron.

(B) $$[FeF_6]^{3-}$$:

Iron is in the +3 oxidation state: $$Fe^{3+}$$ has the configuration $$[Ar] 3d^5$$.

$$F^{-}$$ is a weak field ligand, so we get a high spin octahedral complex.

High spin $$d^5$$: $$t_{2g}^3 \, e_g^2$$ → 5 unpaired electrons.

(C) $$[CoF_6]^{3-}$$:

Cobalt is in the +3 oxidation state: $$Co^{3+}$$ has the configuration $$[Ar] 3d^6$$.

$$F^{-}$$ is a weak field ligand, so we get a high spin octahedral complex.

High spin $$d^6$$: $$t_{2g}^4 \, e_g^2$$ → 4 unpaired electrons.

(D) $$[Cr(C_2O_4)_3]^{3-}$$:

Chromium is in the +3 oxidation state: $$Cr^{3+}$$ has the configuration $$[Ar] 3d^3$$.

For $$d^3$$ in an octahedral field, the configuration is always $$t_{2g}^3 \, e_g^0$$ (same for both strong and weak field).

$$d^3$$: $$t_{2g}^3 \, e_g^0$$ → 3 unpaired electrons.

(E) $$[Ni(CO)_4]$$:

Nickel is in the 0 oxidation state: $$Ni^{0}$$ has the configuration $$[Ar] 3d^{8} 4s^2$$.

$$CO$$ is a very strong field ligand. In the tetrahedral $$[Ni(CO)_4]$$ complex, the strong field of $$CO$$ causes all 10 electrons ($$3d^{10}$$) to pair up completely.

$$d^{10}$$ → 0 unpaired electrons.

• E: $$[Ni(CO)_4]$$ → 0 unpaired electrons
• A: $$[Fe(CN)_6]^{3-}$$ → 1 unpaired electron
• D: $$[Cr(C_2O_4)_3]^{3-}$$ → 3 unpaired electrons
• C: $$[CoF_6]^{3-}$$ → 4 unpaired electrons
• B: $$[FeF_6]^{3-}$$ → 5 unpaired electrons

$$ E(0) < A(1) < D(3) < C(4) < B(5) $$

The correct answer is Option A: $$E < A < D < C < B$$.

We need an octahedral, diamagnetic (zero unpaired electrons), low spin complex.

$$[Co(NH_3)_6]^{3+}$$: Co³⁺ has d⁶ configuration. NH₃ is a strong field ligand → low spin. In octahedral low spin d⁶: $$t_{2g}^6 e_g^0$$ → 0 unpaired electrons → diamagnetic. ✓ Octahedral ✓

$$[CoF_6]^{3-}$$: F⁻ is a weak field ligand → high spin d⁶: $$t_{2g}^4 e_g^2$$ → 4 unpaired → paramagnetic. ✗

$$[CoCl_6]^{3-}$$: Cl⁻ is a weak field ligand → high spin → paramagnetic. ✗

$$[NiCl_4]^{2-}$$: This is tetrahedral, not octahedral. ✗

The correct answer is Option 1: $$[Co(NH_3)_6]^{3+}$$.

An ambidentate ligand is a ligand that can coordinate to the central metal atom through two different donor atoms. Examples include NO$$_2^-$$ (through N or O), NCS$$^-$$ (through N or S), and EDTA$$^{4-}$$ (can coordinate through N or O atoms at different sites).

Let us check each set:

Option 1: C$$_2$$O$$_4^{2-}$$, ethylene diamine, H$$_2$$O

- C$$_2$$O$$_4^{2-}$$ (oxalate): bidentate ligand, coordinates through two O atoms (same type) — not ambidentate

- Ethylene diamine: bidentate ligand, coordinates through two N atoms (same type) — not ambidentate

- H$$_2$$O: monodentate ligand, coordinates through O — not ambidentate

None of these are ambidentate.

Option 2: Contains NCS$$^-$$ which is ambidentate (can bind through N or S).

Option 3: Contains NO$$_2^-$$ which is ambidentate (can bind through N or O).

Option 4: Contains both NO$$_2^-$$ and NCS$$^-$$, both ambidentate.

Therefore, the set that does NOT have any ambidentate ligand is Option 1: C$$_2$$O$$_4^{2-}$$, ethylene diamine, H$$_2$$O.

Maximum attraction to a magnetic field means maximum paramagnetic character, which requires the maximum number of unpaired electrons.

Option 1: [Ni(H₂O)₆]²⁺ - Ni²⁺ has d⁸ configuration. H₂O is a weak field ligand. In octahedral field: t₂g⁶ eg² → 2 unpaired electrons.

Option 2: [Co(en)₃]³⁺ - Co³⁺ has d⁶ configuration. en is a strong field ligand. In octahedral field: t₂g⁶ eg⁰ → 0 unpaired electrons (diamagnetic).

Option 3: [Zn(H₂O)₆]²⁺ - Zn²⁺ has d¹⁰ configuration → 0 unpaired electrons (diamagnetic).

Option 4: [Co(H₂O)₆]²⁺ - Co²⁺ has d⁷ configuration. H₂O is a weak field ligand. In octahedral field: t₂g⁵ eg² → 3 unpaired electrons.

[Co(H₂O)₆]²⁺ has the maximum number of unpaired electrons (3), so it shows maximum paramagnetic character and maximum attraction to a magnetic field.

This matches option 4.

Assertion A: [CoCl(NH₃)₅]²⁺ absorbs at lower wavelength than [Co(NH₃)₅(H₂O)]³⁺.

Both have Co³⁺ (d⁶). Cl⁻ is a weaker field ligand than H₂O. Weaker field → smaller Δ → longer wavelength absorbed. So [CoCl(NH₃)₅]²⁺ absorbs at LONGER wavelength. A is False.

Reason R: Wavelength of light depends on oxidation state. This is True — higher oxidation state generally means larger crystal field splitting.

A is false but R is true.

For octahedral complexes, Crystal Field Stabilization Energy (CFSE) is calculated using:

• Each electron in $$t_{2g}$$​ contributes $$−0.4Δ_{0​}$$
• Each electron in $$e_g$$​ contributes $$+0.6Δ_{0}$$​

A. $$[Cu(NH_{3})_{6}]^{2+}]$$

Since, $$Cu=+2$$ = $$3d^9$$

arrangement $$t^{6}_{2g}e^{3}_{g}$$

CFSE:

$$\left(6\times\ -0.4\right)+\left(3\times\ +0.6\right)$$

$$=-2.4+1.8$$

$$=-0.6\triangle_o$$

So, A$$\longrightarrow\ $$ I

B. $$[Ti(H_{2}O)_{6}]^{3+}]$$

Since, $$Ti=+3$$ = $$3d^1$$

arrangement $$t^{1}_{2g}e^{0}_{g}$$

CFSE:

$$1\times\ \left(-0.4\right)$$

$$=-0.4\triangle_o$$

So, B$$\longrightarrow\ $$ IV

C. $$[Fe(CN)_{6}]^{3-}]$$

Since, $$Fe=+3$$ = $$3d^5$$

CN is strong Field ligand (it means their will be pairing)

arrangement $$t^{5}_{2g}e^{0}_{g}$$

CFSE:

$$5\times\ \left(-0.4\right)$$

$$=-2.0\triangle_o$$

So, C$$\longrightarrow\ $$ II

D. $$[NiF_{6}]^{4-}]$$

Since, $$Ni=+2$$ = $$3d^8$$

arrangement $$t^{6}_{2g}e^{2}_{g}$$

CFSE:

$$\left(6\times\ -0.4\right)+\left(2\times\ +0.6\right)$$

$$=-2.4+1.2$$

$$=-1.2\triangle_o$$

So, D$$\longrightarrow\ $$ III

Therefore, Option B is correct.

The complex [Cr(ox)₂ClBr]³⁻ has chromium with two bidentate oxalate ligands and two monodentate ligands (Cl and Br).

This is an octahedral complex of the type [M(AA)₂ab] where AA is a bidentate ligand and a, b are different monodentate ligands.

For [M(AA)₂ab] type complexes, there are 3 possible stereoisomers:

• cis isomer (Cl and Br on same side): This has no plane of symmetry, so it exists as a pair of optical isomers (Δ and Λ). That gives 2 isomers.
• trans isomer (Cl and Br on opposite sides): This has a plane of symmetry, so only 1 isomer.

Total stereoisomers = 2 (optical) + 1 (trans) = 3.

This matches option 1: 3.

A meridional (mer) isomer is possible in octahedral complexes of the type $$[Ma_3b_3]$$, where three identical ligands occupy positions that include one on the axis and two in the equatorial plane (forming a T-shape or meridian).

Analyzing each option:

$$[Co(NH_3)_3(NO_2)_3]$$: This is an octahedral complex of the type $$[Ma_3b_3]$$ with three NH$$_3$$ and three NO$$_2^-$$ ligands. This can exist as both facial (fac) and meridional (mer) isomers.

$$[Pt(NH_3)_2Cl_2]$$: Square planar complex — shows cis/trans isomerism, not fac/mer.

$$[Co(en)_2Cl_2]$$: Contains bidentate en ligands — shows cis/trans isomerism.

$$[Co(en)_3]$$: Contains only one type of bidentate ligand — shows optical isomerism (D/L), not fac/mer.

Therefore, $$[Co(NH_3)_3(NO_2)_3]$$ can exist as a meridional isomer.

Fe³⁺ has d⁵ configuration.

[Fe(H₂O)₆]³⁺: H₂O is weak field. All 5 electrons unpaired (high spin). $$\mu = \sqrt{5 \times 7} = \sqrt{35} = 5.92$$ BM.

[Fe(CN)₆]³⁻: CN⁻ is strong field. t₂g⁵ configuration → 1 unpaired electron. $$\mu = \sqrt{1 \times 3} = \sqrt{3} = 1.732$$ BM.

Nessler's reagent is an alkaline solution of potassium tetraiodomercurate(II). 

Its chemical formula is:

$$\ K_2\left[HgI\right]_4$$

Based on this formula, the elements present are:

• K: Potassium
• Hg: Mercury
• I: Iodine
  

Oxygen is not a constituent element of the reagent itself.

The correct answer is Option D. Oxygen

We need to match each coordination complex with its number of unpaired electrons.

A. $$[Cr(CN)_6]^{3-}$$

Chromium in $$[Cr(CN)_6]^{3-}$$: Oxidation state of Cr = +3 (since $$x + 6(-1) = -3$$, so $$x = +3$$).

$$Cr^{3+}$$: Electronic configuration = $$[Ar] 3d^3$$

$$CN^-$$ is a strong field ligand, so electrons pair up in $$t_{2g}$$ orbitals first.

Configuration: $$t_{2g}^3 e_g^0$$ → 3 unpaired electrons → matches II

B. $$[Fe(H_2O)_6]^{2+}$$

Iron in $$[Fe(H_2O)_6]^{2+}$$: Oxidation state of Fe = +2.

$$Fe^{2+}$$: Electronic configuration = $$[Ar] 3d^6$$

$$H_2O$$ is a weak field ligand, so no pairing occurs beyond what is necessary.

Configuration: $$t_{2g}^4 e_g^2$$ (high spin) → 4 unpaired electrons → matches IV

C. $$[Co(NH_3)_6]^{3+}$$

Cobalt in $$[Co(NH_3)_6]^{3+}$$: Oxidation state of Co = +3.

$$Co^{3+}$$: Electronic configuration = $$[Ar] 3d^6$$

$$NH_3$$ is a strong field ligand for $$Co^{3+}$$, so all electrons pair in $$t_{2g}$$.

Configuration: $$t_{2g}^6 e_g^0$$ (low spin) → 0 unpaired electrons → matches I

D. $$[Ni(NH_3)_6]^{2+}$$

Nickel in $$[Ni(NH_3)_6]^{2+}$$: Oxidation state of Ni = +2.

$$Ni^{2+}$$: Electronic configuration = $$[Ar] 3d^8$$

For octahedral $$Ni^{2+}$$ ($$d^8$$), regardless of ligand field strength, the configuration is $$t_{2g}^6 e_g^2$$.

This always gives 2 unpaired electrons → matches III

Therefore, the correct matching is:

A → II, B → IV, C → I, D → III

The correct answer is Option A.

We need to identify the complex cation that has two isomers.

Option A: $$[\text{Co}(\text{H}_2\text{O})_6]^{3+}$$

This complex has only one type of ligand (H$$_2$$O). It has the form [MA$$_6$$], which has only one possible structure. No isomers.

Option B: $$[\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+}$$

This is of the type [MA$$_5$$B]. In an octahedral complex, there is only one way to arrange 5 identical ligands and 1 different ligand. No geometric isomers. Also no linkage isomers since Cl coordinates only through Cl.

Option C: $$[\text{Co}(\text{NH}_3)_5\text{NO}_2]^{2+}$$

This is of the type [MA$$_5$$B]. While there are no geometric isomers, the NO$$_2^-$$ ligand is an ambidentate ligand — it can coordinate through nitrogen (nitro, -NO$$_2$$) or through oxygen (nitrito, -ONO). This gives two linkage isomers:

1. $$[\text{Co}(\text{NH}_3)_5(\text{NO}_2)]^{2+}$$ (nitro — N-bonded)

2. $$[\text{Co}(\text{NH}_3)_5(\text{ONO})]^{2+}$$ (nitrito — O-bonded)

Option D: $$[\text{Co}(\text{NH}_3)_5\text{Cl}]^{+}$$

Same type as Option B. Only one geometric arrangement and Cl is not ambidentate. No isomers.

Only $$[\text{Co}(\text{NH}_3)_5\text{NO}_2]^{2+}$$ has exactly two isomers (linkage isomers).

The correct answer is Option C: $$[\text{Co}(\text{NH}_3)_5\text{NO}_2]^{2+}$$.

First find the oxidation state and hence the number of $$d$$-electrons present in every metal centre.

$$[Fe(H_2O)_6]^{2+}:$$
$$Fe = +2 \;\Rightarrow\; d^{6}$$

$$[Mn(H_2O)_6]^{2+}:$$
$$Mn = +2 \;\Rightarrow\; d^{5}$$

$$[Co(NH_3)_6]^{3+}:$$
$$Co = +3 \;\Rightarrow\; d^{6}$$

$$[FeCl_4]^{-}:$$
Total charge $$= -1$$, each $$Cl^{-}$$ contributes $$-1$$, therefore
$$x - 4 = -1 \;\Rightarrow\; x = +3$$
$$Fe = +3 \;\Rightarrow\; d^{5}$$

Next decide the geometry and the ligand field strength.

• $$[Fe(H_2O)_6]^{2+}$$ and $$[Mn(H_2O)_6]^{2+}$$ are octahedral with the weak-field ligand $$H_2O$$ ⇒ high-spin.
• $$[Co(NH_3)_6]^{3+}$$ is octahedral, but $$NH_3$$ is a stronger field ligand and, with the highly charged $$Co^{3+}$$, gives a low-spin complex.
• $$[FeCl_4]^{-}$$ is tetrahedral because the large ligand $$Cl^{-}$$ prefers tetrahedral geometry; $$Cl^{-}$$ is weak field, so the complex is always high-spin.

Write the electronic arrangements:

Case 1 - Octahedral high-spin $$d^{6}$$
$$t_{2g}^{4}\,e_g^{2}$$

Case 2 - Octahedral high-spin $$d^{5}$$
$$t_{2g}^{3}\,e_g^{2}$$

Case 3 - Octahedral low-spin $$d^{6}$$
$$t_{2g}^{6}\,e_g^{0}$$

Case 4 - Tetrahedral high-spin $$d^{5}$$
Level order reverses, so the lower set is $$e$$ and the higher set is $$t_2$$:
$$e^{2}\,t_{2}^{3}$$

Now match each complex with the patterns listed in List-I:

P : $$t_{2g}^{6} e_g^{0}$$ ⇒ octahedral low-spin $$d^{6}$$ ⇒ $$[Co(NH_3)_6]^{3+}\;(3)$$

Q : $$t_{2g}^{3} e_g^{2}$$ ⇒ octahedral high-spin $$d^{5}$$ ⇒ $$[Mn(H_2O)_6]^{2+}\;(2)$$

R : $$e^{2} t_2^{3}$$ ⇒ tetrahedral high-spin $$d^{5}$$ ⇒ $$[FeCl_4]^{-}\;(4)$$

S : $$t_{2g}^{4} e_g^{2}$$ ⇒ octahedral high-spin $$d^{6}$$ ⇒ $$[Fe(H_2O)_6]^{2+}\;(1)$$

Therefore
$$P \to 3,\; Q \to 2,\; R \to 4,\; S \to 1$$

The option showing this combination is
Option D which is: P → 3; Q → 2; R → 4; S → 1.

We need to find the number of unpaired electrons in a metal ion whose calculated magnetic moment is 4.90 BM.

The spin-only magnetic moment (in Bohr Magnetons, BM) is given by:

where $$n$$ is the number of unpaired electrons. This formula arises from quantum mechanics, where the spin magnetic moment depends on the total spin quantum number $$S = n/2$$, and $$\mu = \sqrt{4S(S+1)} = \sqrt{n(n+2)}$$ BM.

We need a positive integer $$n$$ such that $$n^2 + 2n \approx 24$$. Testing values:

$$n = 3$$: $$3^2 + 2(3) = 9 + 6 = 15$$ (too small)

$$n = 4$$: $$4^2 + 2(4) = 16 + 8 = 24$$ (matches!)

$$n = 5$$: $$5^2 + 2(5) = 25 + 10 = 35$$ (too large)

With $$n = 4$$: $$\mu = \sqrt{4 \times 6} = \sqrt{24} = 4.899 \approx 4.90$$ BM. This matches the given value.

The metal ion has 4 unpaired electrons.

We are given that the magnetic moment of $$[Mn(NCS)_6]^{x-}$$ is 6.06 BM.

Using the spin-only magnetic moment formula $$\mu = \sqrt{n(n+2)}$$ BM, where $$n$$ is the number of unpaired electrons:

$$6.06 = \sqrt{n(n+2)}$$, so $$36.72 = n(n+2)$$.

For $$n = 5$$: $$5(5+2) = 35$$, giving $$\mu = \sqrt{35} = 5.92$$ BM. The observed value of 6.06 BM is slightly higher than the spin-only value due to orbital contribution, which is characteristic of high-spin $$d^5$$ systems. So we have $$n = 5$$ unpaired electrons.

Now, $$Mn^{2+}$$ has the electronic configuration $$[Ar] 3d^5$$, which gives 5 unpaired electrons. Since $$NCS^-$$ is a weak field ligand, no pairing occurs in the high-spin octahedral complex, giving configuration $$t_{2g}^3 e_g^2$$ with 5 unpaired electrons. This matches.

Each $$NCS^-$$ ligand has a charge of $$-1$$. With 6 ligands, total ligand charge = $$-6$$. The overall charge on the complex is $$x-$$:

$$+2 + 6(-1) = -x$$, so $$-4 = -x$$, giving $$x = 4$$.

Hence, the answer is $$x = 4$$.

Seven coordination compounds are given; find the number of paramagnetic species.

[Ni(CN)$$_4$$]$$^{2-}$$: Ni$$^{2+}$$ has configuration [Ar] 3d$$^8$$. CN$$^-$$ is a strong field ligand causing pairing. With 4 ligands, this complex is square planar (dsp$$^2$$ hybridisation). All 8 d-electrons are paired. Diamagnetic.

[Ni(CO)$$_4$$]: Ni is in the 0 oxidation state with configuration [Ar] 3d$$^{10}$$. With 10 d-electrons, all are paired regardless of ligand field. Diamagnetic.

[NiCl$$_4$$]$$^{2-}$$: Ni$$^{2+}$$ (d$$^8$$). Cl$$^-$$ is a weak field ligand, so the complex is tetrahedral (sp$$^3$$ hybridisation). In tetrahedral geometry, d$$^8$$ has 2 unpaired electrons. Paramagnetic.

[Fe(CN)$$_6$$]$$^{4-}$$: Fe$$^{2+}$$ (d$$^6$$). CN$$^-$$ is a strong field ligand causing complete pairing: t$$_{2g}^6$$ e$$_g^0$$. Zero unpaired electrons. Diamagnetic.

[Cu(NH$$_3$$)$$_4$$]$$^{2+}$$: Cu$$^{2+}$$ (d$$^9$$). With 9 d-electrons, there is always 1 unpaired electron regardless of geometry. Paramagnetic.

[Fe(CN)$$_6$$]$$^{3-}$$: Fe$$^{3+}$$ (d$$^5$$). CN$$^-$$ is a strong field ligand: t$$_{2g}^5$$ e$$_g^0$$. One unpaired electron. Paramagnetic.

[Fe(H$$_2$$O)$$_6$$]$$^{2+}$$: Fe$$^{2+}$$ (d$$^6$$). H$$_2$$O is a weak field ligand: t$$_{2g}^4$$ e$$_g^2$$. Four unpaired electrons. Paramagnetic.

The paramagnetic species among these complexes are [NiCl$$_4$$]$$^{2-}$$, [Cu(NH$$_3$$)$$_4$$]$$^{2+}$$, [Fe(CN)$$_6$$]$$^{3-}$$, and [Fe(H$$_2$$O)$$_6$$]$$^{2+}$$, giving a total of 4 paramagnetic complexes.

Answer: 4

$$CoCl_4^{2-}$$: Co²⁺ has d⁷ configuration. In tetrahedral field (always high spin):

$$e^4 t_2^3$$ (4 electrons in e, 3 in t₂). $$m = 4$$.

Unpaired electrons: e has 4 electrons (2 pairs, 0 unpaired), t₂ has 3 (all unpaired). Total unpaired = 3.

Sum = $$m$$ + unpaired = 4 + 3 = 7.

We need to find the d-electronic configuration of $$CoCl_4^{2-}$$ in a tetrahedral crystal field. The oxidation state of Co in $$CoCl_4^{2-}$$ can be determined from $$x + 4(-1) = -2$$, giving $$x = +2$$. Therefore, $$Co^{2+}$$ has the electronic configuration: $$[Ar]3d^7$$.

In a tetrahedral field, the d-orbitals split into two sets: $$e$$ (lower energy), consisting of $$d_{z^2}, d_{x^2-y^2}$$ (2 orbitals), and $$t_2$$ (higher energy), consisting of $$d_{xy}, d_{xz}, d_{yz}$$ (3 orbitals).

Because $$CoCl_4^{2-}$$ contains weak field ligands ($$Cl^-$$), it is a high spin complex. Distributing the seven d-electrons in this high spin tetrahedral configuration gives: $$e$$ can accommodate up to 4 electrons (2 orbitals × 2), and $$t_2$$ can accommodate up to 6 electrons (3 orbitals × 2). Filling accordingly yields $$e^4 t_2^3$$, so $$m = 4$$ and $$n = 3$$.

The number of unpaired electrons is determined from the occupancy: in $$e^4$$ both orbitals are fully paired, contributing 0 unpaired electrons, while in $$t_2^3$$ each of the three orbitals has one electron, contributing 3 unpaired electrons. Thus the complex has 3 unpaired electrons.

The sum of $$m$$ and the number of unpaired electrons is $$4 + 3 = 7$$.

The answer is 7.

Both complexes contain Cr$$^{3+}$$, which has the electronic configuration [Ar]3d$$^3$$.

[Cr(CN)$$_6$$]$$^{3-}$$: CN$$^-$$ is a strong field ligand. The d$$^3$$ electrons occupy the three t$$_{2g}$$ orbitals with one electron each.

Number of unpaired electrons = 3

[Cr(H$$_2$$O)$$_6$$]$$^{3+}$$: H$$_2$$O is a weak field ligand. However, for d$$^3$$ configuration, even with weak field ligands, the three electrons go into the three t$$_{2g}$$ orbitals one each (since there are only 3 electrons, no pairing occurs regardless of crystal field strength).

Number of unpaired electrons = 3

The spin-only magnetic moment is: $$\mu = \sqrt{n(n+2)}$$ BM

Since both complexes have the same number of unpaired electrons (n = 3):

$$\mu_{eff}[Cr(CN)_6]^{3-} = \mu_{eff}[Cr(H_2O)_6]^{3+} = \sqrt{3(3+2)} = \sqrt{15}$$ BM

$$\frac{\mu_{eff}[Cr(CN)_6]^{3-}}{\mu_{eff}[Cr(H_2O)_6]^{3+}} = 1$$

Manganese has atomic number 25, so its electronic configuration is $$[\text{Ar}]\,3d^5\,4s^2$$. In the complex $$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$$, Mn is in the +2 oxidation state, giving a $$d^5$$ configuration. Since water is a weak-field ligand, no pairing occurs and all five $$d$$-electrons remain unpaired.

The spin-only magnetic moment is given by $$\mu = \sqrt{n(n+2)}$$ B.M., where $$n$$ is the number of unpaired electrons. Here $$n = 5$$, so $$\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92$$ B.M., which rounds to $$\boxed{6}$$ B.M.

The complex [Cr(H₂O)₅Cl]Cl₂ has:

• 1 Cl in the coordination sphere (not ionizable)
• 2 Cl⁻ as counter ions (ionizable)

Moles of complex = 0.01 × 20/1000 = 0.0002 mol

Moles of ionizable Cl⁻ = 2 × 0.0002 = 0.0004 mol

For precipitation: Ag⁺ + Cl⁻ → AgCl

Moles of AgNO₃ needed = 0.0004 mol

Volume of 0.1 M AgNO₃ = 0.0004/0.1 = 0.004 L = 4 mL

The volume is $$\mathbf{4}$$ mL.

We need to find the total number of moles of AgCl precipitated on addition of excess AgNO$$_3$$ to one mole each of the given complexes.

Key Principle: Only ionic (free) chloride ions outside the coordination sphere react with AgNO$$_3$$ to form AgCl precipitate. Chloride ions inside the coordination sphere (bonded to the metal) do not precipitate.

Complex 1: [Co(NH$$_3$$)$$_4$$Cl$$_2$$]Cl

Coordination sphere: [Co(NH$$_3$$)$$_4$$Cl$$_2$$]$$^+$$ (2 Cl inside).

Free chloride: 1 Cl$$^-$$ outside.

Moles of AgCl = 1

Complex 2: [Ni(H$$_2$$O)$$_6$$]Cl$$_2$$

Coordination sphere: [Ni(H$$_2$$O)$$_6$$]$$^{2+}$$ (no Cl inside).

Free chloride: 2 Cl$$^-$$ outside.

Moles of AgCl = 2

Complex 3: [Pt(NH$$_3$$)$$_2$$Cl$$_2$$]

This is a neutral complex with all chlorides inside the coordination sphere.

Free chloride: 0.

Moles of AgCl = 0

Complex 4: [Pd(NH$$_3$$)$$_4$$]Cl$$_2$$

Coordination sphere: [Pd(NH$$_3$$)$$_4$$]$$^{2+}$$ (no Cl inside).

Free chloride: 2 Cl$$^-$$ outside.

Moles of AgCl = 2

Total moles of AgCl = 1 + 2 + 0 + 2 = $$\boxed{5}$$

Potassium ferrocyanide is K₄[Fe(CN)₆].

Fe is in +2 oxidation state: Fe²⁺ has electron configuration [Ar]3d⁶.

CN⁻ is a strong field ligand, causing pairing (low spin complex).

In octahedral low spin d⁶: $$t_{2g}^6 e_g^0$$

The $$t_{2g}$$ set has 6 electrons arranged in 3 orbitals, meaning 3 pairs.

The number of pairs of electrons in $$t_{2g}$$ = 3.

An ambidentate ligand is a ligand that can coordinate to a metal through two different donor atoms.

In the complex $$M(en)(SCN)_4$$:

en (ethylenediamine): H₂N-CH₂-CH₂-NH₂. It coordinates through two nitrogen atoms. Both donor atoms are the same element (N), so it is a bidentate ligand but NOT ambidentate.

SCN⁻ (thiocyanate): This is an ambidentate ligand as it can coordinate through either:

- Sulfur atom (thiocyanato, S-bonded: M-SCN)

- Nitrogen atom (isothiocyanato, N-bonded: M-NCS)

There are 4 SCN⁻ ligands in the complex, and each one is ambidentate.

The number of ambidentate ligands is 4.

Find the denticity of the ligand present in Fehling's reagent.

Identify the components of Fehling's reagent.

Fehling's reagent is a mixture of:

Fehling's solution A: Aqueous CuSO$$_4$$

Fehling's solution B: Alkaline sodium potassium tartrate (Rochelle salt, NaKC$$_4$$H$$_4$$O$$_6$$)

The ligand in Fehling's reagent is the tartrate ion ($$\text{C}_4\text{H}_4\text{O}_6^{2-}$$), which is 2,3-dihydroxybutanedioate.

Determine the denticity.

The tartrate ion coordinates to Cu$$^{2+}$$ through its donor atoms. In Fehling's reagent, tartrate acts as a bidentate ligand, coordinating through two oxygen atoms (one from a hydroxyl group and one from a carboxylate group).

Denticity = 2.

The correct answer is 2.

The homoleptic octahedral complex of Co²⁺ with H₂O is [Co(H₂O)₆]²⁺.

Co²⁺ has electronic configuration: [Ar] 3d⁷.

H₂O is a weak field ligand, so the complex is high spin.

In a high-spin octahedral d⁷ complex:

$$t_{2g}^5 \, e_g^2$$

The t₂g orbitals (dxy, dxz, dyz) have 5 electrons distributed as: ↑↓, ↑↓, ↑

Number of unpaired electrons in t₂g = 1 (two orbitals are fully paired, one has 1 unpaired).

The number of unpaired electrons in t₂g is $$\mathbf{1}$$.

We need to find the sum of bridging carbonyls in W(CO)₆ and Mn₂(CO)₁₀.

W(CO)₆: Tungsten hexacarbonyl is a mononuclear complex with octahedral geometry. All 6 CO ligands are terminal (bonded only to W). There are 0 bridging COs.

Mn₂(CO)₁₀: Dimanganese decacarbonyl has two Mn atoms connected by a direct Mn-Mn bond. Each Mn has 5 terminal CO ligands in a staggered arrangement. There are 0 bridging COs. (This is different from Co₂(CO)₈ which has 2 bridging COs in one of its structures.)

Sum of bridging carbonyls = $$0 + 0 = 0$$.

The answer is $$\boxed{0}$$.

This question is about the MacArthur-Forrest cyanide process for gold extraction.

Leaching reaction (formation of complex A): Gold reacts with dilute NaCN solution in the presence of oxygen:

$$4Au(s) + 8NaCN(aq) + O_2(g) + 2H_2O(l) \rightarrow 4Na[Au(CN)_2](aq) + 4NaOH(aq)$$

The gold complex formed is $$[Au(CN)_2]^-$$ (dicyanoaurate(I) ion). So complex A = $$[Au(CN)_2]^-$$.

Recovery with zinc (formation of complex B): Zinc displaces gold from the cyanide complex:

$$2[Au(CN)_2]^-(aq) + Zn(s) \rightarrow 2Au(s) + [Zn(CN)_4]^{2-}(aq)$$

The zinc complex formed is $$[Zn(CN)_4]^{2-}$$ (tetracyanozincate(II) ion). So complex B = $$[Zn(CN)_4]^{2-}$$.

Match with options: Option A: $$[Au(CN)_2]^-$$ and $$[Zn(CN)_4]^{2-}$$ — Matches both A and B. Correct.

Option B: $$[Au(CN)_4]^-$$ and $$[Zn(CN)_2(OH)_2]^{2+}$$ — Incorrect coordination numbers and charges.

Option C: $$[Au(CN)_2]^-$$ and $$[Zn(OH)_4]^{2-}$$ — Complex B is wrong.

Option D: $$[Au(CN)_4]^{2-}$$ and $$[Zn(CN)_6]^{4-}$$ — Both are incorrect.

The correct answer is Option A: $$[Au(CN)_2]^-$$ and $$[Zn(CN)_4]^{2-}$$.

We need to identify which ligand property stabilizes low oxidation states of metals in complexes.

In low oxidation state complexes (e.g., metal carbonyls like $$Ni(CO)_4$$, $$Fe(CO)_5$$), the metal has zero or low positive oxidation state. This means there is excess electron density on the metal.

For the complex to be stable in a low oxidation state, the ligands must be able to accept electron density from the metal through back-bonding (also called $$\pi$$-back donation). This involves:

- The metal donates electron density from its filled $$d$$-orbitals into the empty $$\pi^*$$ (antibonding) orbitals of the ligand.

- This stabilizes the complex by delocalizing the excess electron density away from the metal center.

Common $$\pi$$-acceptor ligands include $$CO$$, $$CN^-$$, $$NO^+$$, phosphines, etc. These ligands have empty orbitals of suitable symmetry ($$\pi^*$$ or $$d$$-orbitals) to accept electrons from the metal.

Therefore, low oxidation states are stabilized when ligands have good $$\pi$$-accepting character.

Hence, the correct answer is Option A.

We need to identify the compound responsible for the Prussian blue precipitate formed when $$Fe^{3+}$$ reacts with potassium ferrocyanide.

- $$Fe^{3+}$$ cation (from a ferric salt)

- Potassium ferrocyanide: $$K_4[Fe(CN)_6]$$, which provides the $$[Fe(CN)_6]^{4-}$$ ion

When $$Fe^{3+}$$ reacts with $$[Fe(CN)_6]^{4-}$$:

$$4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3 \downarrow$$

- Charge from $$Fe^{3+}$$: $$4 \times (+3) = +12$$

- Charge from $$[Fe(CN)_6]^{4-}$$: $$3 \times (-4) = -12$$

- Net charge = 0 $$\checkmark$$

The compound $$Fe_4[Fe(CN)_6]_3$$ is known as Prussian blue (or ferric ferrocyanide). It is a deep blue insoluble precipitate and is a classic qualitative test for $$Fe^{3+}$$ ions.

Hence, the correct answer is Option D.

We need to evaluate the two statements about CuSO₄·5H₂O.

In the crystal structure of CuSO₄·5H₂O (blue vitriol), the Cu²⁺ ion is surrounded by 6 ligands in an octahedral arrangement: 4 water molecules in the equatorial plane (coordinated through their oxygen atoms) and 2 oxygen atoms from two different SO₄²⁻ ions in the axial positions. The 5th water molecule is not directly coordinated to Cu²⁺ but is held in the crystal lattice by hydrogen bonds between the coordinated water molecules and sulphate ions. The structure can be represented as: $$[Cu(H₂O)₄]SO₄·H₂O$$ where the sulphate oxygen atoms also coordinate to Cu²⁺.

Statement I says "In CuSO₄·5H₂O, Cu-O bonds are present." Cu²⁺ forms bonds with the oxygen atoms of the water molecules and the oxygen atoms of sulphate ions, so Cu-O bonds are indeed present. Therefore, Statement I is correct.

Statement II says "In CuSO₄·5H₂O, ligands coordinating with Cu(II) ion are O-and S-based ligands." The ligands coordinating with Cu²⁺ are water molecules (H₂O) — these coordinate through oxygen atoms (O-based ligands) — and sulphate ions (SO₄²⁻) — these also coordinate through oxygen atoms, not through sulphur atoms. All the ligands coordinate through oxygen atoms only; there are no S-based ligands since sulphur does not directly bond to Cu²⁺. Thus, Statement II is incorrect.

Statement I is correct but Statement II is incorrect. The answer is Option C: Statement I is correct but Statement II is incorrect.

We need to evaluate both statements about nickel complexes.

Background: Nickel electronic configuration

$$Ni$$ has atomic number 28: $$[Ar] 3d^8 4s^2$$

$$Ni^{2+}$$: $$[Ar] 3d^8$$ (2 unpaired electrons)

$$Ni^{0}$$: $$[Ar] 3d^8 4s^2$$ (in complexes, effectively $$3d^{10}$$ with strong field ligands)

Analysis of Statement I:

$$[Ni(CN)_4]^{2-}$$:

- $$Ni^{2+}$$ has $$3d^8$$ configuration

- $$CN^-$$ is a strong field ligand, so it forces pairing of electrons

- The two unpaired electrons in $$3d$$ pair up, vacating one 3d orbital

- Hybridization: $$dsp^2$$ → square planar geometry

- All electrons are paired → diamagnetic

This part of Statement I is correct.

$$[Ni(CO)_4]$$:

- $$Ni$$ is in zero oxidation state ($$Ni^0$$): $$[Ar] 3d^{8} 4s^2$$

- $$CO$$ is a very strong field ligand (causes pairing)

- In $$Ni(CO)_4$$, the effective configuration becomes $$3d^{10}$$ as the 4s electrons move to 3d orbitals

- With all 3d orbitals filled, the hybridization is $$sp^3$$ → tetrahedral

- All electrons are paired → $$[Ni(CO)_4]$$ is diamagnetic, NOT paramagnetic

Statement I says $$[Ni(CO)_4]$$ is paramagnetic, which is wrong.

Therefore, Statement I is FALSE.

Analysis of Statement II:

$$[NiCl_4]^{2-}$$:

- $$Ni^{2+}$$: $$3d^8$$ (2 unpaired electrons)

- $$Cl^-$$ is a weak field ligand → no pairing of electrons

- Hybridization: $$sp^3$$ → tetrahedral geometry

- Has 2 unpaired electrons → paramagnetic

$$[Ni(CO)_4]$$:

- As analyzed above: $$3d^{10}$$ configuration

- $$sp^3$$ → tetrahedral geometry

- All electrons paired → diamagnetic

Statement II claims both have the same d-electron configuration, same geometry, and are both paramagnetic.

- d-electron configurations are different: $$3d^8$$ ($$Ni^{2+}$$) vs $$3d^{10}$$ ($$Ni^0$$)

- Both are tetrahedral (same geometry) — this is correct

- $$[Ni(CO)_4]$$ is diamagnetic, not paramagnetic — this is wrong

Therefore, Statement II is also FALSE.

The correct answer is Option D: Both Statement I and Statement II are false.

We need to match each complex with its hybridization by analyzing the electronic configuration and geometry of the central metal ion in each complex.

A. $$Ni(CO)_4$$: Nickel in $$Ni(CO)_4$$ is in the 0 oxidation state (since CO is a neutral ligand). The electronic configuration of Ni (Z = 28) is $$[Ar] 3d^8 4s^2$$. In $$Ni(CO)_4$$, CO is a strong field ligand, so all electrons pair up in the 3d orbitals, giving a $$3d^{10}$$ configuration with all d-orbitals filled. Since no d-orbitals are available, the metal uses one 4s and three 4p orbitals for bonding, resulting in $$sp^3$$ hybridization with a tetrahedral geometry. So A matches with I ($$sp^3$$).

B. $$[Ni(CN)_4]^{2-}$$: Nickel is in the +2 oxidation state, giving a $$3d^8$$ configuration. CN$$^-$$ is a strong field ligand, which forces the electrons to pair up. The 8 d-electrons pair into four of the five 3d orbitals, leaving one 3d orbital empty. This empty $$3d$$ orbital, along with one $$4s$$ and two $$4p$$ orbitals, undergoes $$dsp^2$$ hybridization, forming a square planar geometry. So B matches with IV ($$dsp^2$$).

C. $$[Co(CN)_6]^{3-}$$: Cobalt is in the +3 oxidation state, giving a $$3d^6$$ configuration. CN$$^-$$ is a strong field ligand, so all six d-electrons pair into three orbitals, leaving two 3d orbitals empty. These two empty 3d orbitals combine with one 4s and three 4p orbitals to give $$d^2sp^3$$ hybridization (inner orbital complex) with an octahedral geometry. So C matches with III ($$d^2sp^3$$).

D. $$[CoF_6]^{3-}$$: Cobalt is again in the +3 oxidation state with a $$3d^6$$ configuration. However, $$F^-$$ is a weak field ligand, so no pairing of electrons occurs beyond what is already present. The 3d orbitals are not available for bonding (they are occupied). The metal uses the outer orbitals — two 4d, one 4s, and three 4p orbitals — giving $$sp^3d^2$$ hybridization (outer orbital complex) with an octahedral geometry. So D matches with II ($$sp^3d^2$$).

The correct matching is: A-I, B-IV, C-III, D-II.

Hence, the correct answer is Option B.

Crystal field stabilization energy (CFSE) determines the stabilization of a complex. We need to find which complex has the maximum CFSE.

Option A: $$[Ti(H_2O)_6]^{3+}$$

$$Ti^{3+}$$ has configuration $$3d^1$$. $$H_2O$$ is a weak field ligand, so octahedral splitting gives:

$$t_{2g}^1 e_g^0$$

CFSE = $$-0.4\Delta_o$$

Option B: $$[Co(H_2O)_6]^{2+}$$

$$Co^{2+}$$ has configuration $$3d^7$$. $$H_2O$$ is a weak field ligand (high spin):

$$t_{2g}^5 e_g^2$$

CFSE = $$5(-0.4) + 2(0.6) = -0.8\Delta_o$$

Option C: $$[Co(CN)_6]^{3-}$$

$$Co^{3+}$$ has configuration $$3d^6$$. $$CN^-$$ is a strong field ligand (low spin):

$$t_{2g}^6 e_g^0$$

CFSE = $$6(-0.4)\Delta_o = -2.4\Delta_o$$

Additionally, $$Co^{3+}$$ has a higher charge than $$Co^{2+}$$ or $$Ti^{3+}$$, and $$CN^-$$ causes a much larger crystal field splitting ($$\Delta_o$$) compared to $$H_2O$$ or $$NH_3$$. The actual CFSE in absolute terms is very large.

Option D: $$[Cu(NH_3)_4]^{2+}$$

$$Cu^{2+}$$ has configuration $$3d^9$$. This is a square planar complex:

CFSE for $$d^9$$ in square planar geometry is relatively moderate.

Comparing all options, $$[Co(CN)_6]^{3-}$$ has the maximum CFSE because:

1. $$Co^{3+}$$ ($$d^6$$) with strong field $$CN^-$$ ligand forms a low spin complex with all 6 electrons in $$t_{2g}$$.

2. The coefficient is $$-2.4\Delta_o$$, the highest among the options.

3. $$CN^-$$ produces a very large $$\Delta_o$$.

Hence, the correct answer is Option C.

We need to arrange the coordination compounds in increasing order of magnetic moments.

Magnetic moment $$\mu = \sqrt{n(n+2)}$$ BM, where $$n$$ is the number of unpaired electrons.

Determine the number of unpaired electrons in each complex.

A. $$[FeF_6]^{3-}$$: Fe is in +3 oxidation state, so $$Fe^{3+}$$ has configuration $$3d^5$$. $$F^-$$ is a weak field ligand, so no pairing occurs (high spin). Unpaired electrons = 5.

$$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92$$ BM

B. $$[Fe(CN)_6]^{3-}$$: Fe is in +3 oxidation state, so $$Fe^{3+}$$ has $$3d^5$$. $$CN^-$$ is a strong field ligand, so maximum pairing occurs (low spin): $$t_{2g}^5 e_g^0$$. Unpaired electrons = 1.

$$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73$$ BM

C. $$[MnCl_6]^{3-}$$ (high spin): Mn is in +3 oxidation state, so $$Mn^{3+}$$ has $$3d^4$$. $$Cl^-$$ is a weak field ligand and it is given as high spin: $$t_{2g}^3 e_g^1$$. Unpaired electrons = 4.

$$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$$ BM

D. $$[Mn(CN)_6]^{3-}$$: Mn is in +3 oxidation state, so $$Mn^{3+}$$ has $$3d^4$$. $$CN^-$$ is a strong field ligand (low spin): $$t_{2g}^4 e_g^0$$. Unpaired electrons = 2.

$$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83$$ BM

Arrange in increasing order of magnetic moment.

$$B (1.73) < D (2.83) < C (4.90) < A (5.92)$$

The increasing order is: $$B < D < C < A$$

The correct answer is Option B.