JEE Coordination Compounds Questions

In the sodium nitroprusside test the reagent used is sodium nitroprusside, $$Na_2[Fe(CN)_5NO]$$. In the anion complex $$[Fe(CN)_5NO]^{2-}$$ the six ligands present are five cyanide ions, $$CN^-$$, and one nitrosyl ligand, $$NO$$.

When sulphide ion $$S^{2-}$$ is added to an aqueous solution of sodium nitroprusside, the deep-purple coloured species $$[Fe(CN)_5NOS]^{4-}$$ is produced:

$$[Fe(CN)_5NO]^{2-} + S^{2-} \;\longrightarrow\; [Fe(CN)_5NOS]^{4-}$$

Comparing the coordination spheres before and after the reaction, we see that one of the original ligands, $$NO$$, is replaced by the thionitrosyl (thiocyanato-N-oxide) ligand $$NOS^-$$. The five $$CN^-$$ ligands remain unchanged.

Thus, during the test one coordinated ligand is converted to $$NOS^-$$.

Option A which is: NOS$$^{-}$$

For octahedral $$d^6$$ cobalt(III) complexes the chief electronic transition seen in the UV-visible spectrum is the spin-allowed $$t_{2g}^6e_g^0 \rightarrow t_{2g}^5e_g^1$$ (a $$d-d$$ transition).
The energy of this transition equals the crystal-field splitting energy $$\Delta_0$$; hence

$$E = h\nu = \frac{hc}{\lambda_{\max}} = \Delta_0 \; \Longrightarrow \; \lambda_{\max} \propto \frac{1}{\Delta_0}$$

So, the larger the splitting $$\Delta_0$$, the smaller (bluer) is the absorption wavelength, and vice-versa.

Because the metal ion is the same (Co(III)) and the geometry is the same (octahedral), the magnitude of $$\Delta_0$$ depends almost entirely on the ligand field strength. The relevant order from the spectrochemical series is

$$CN^- \gt NH_3 \gt H_2O \gt Cl^-$$

Therefore, the descending order of $$\Delta_0$$ (strongest field to weakest field) for the given complexes is:

$$[Co(CN)_6]^{3-} \; \gt \; [Co(NH_3)_6]^{3+} \; \gt \; [Co(NH_3)_5(H_2O)]^{3+} \; \gt \; [Co(NH_3)_5(Cl)]^{2+}$$

Converting this into the ascending order of the absorption wavelength $$\lambda_{\max}$$ (since $$\lambda_{\max} \propto 1/\Delta_0$$) gives:

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

Thus the correct sequence matches Option A.

Option A which is: [Co(CN)$$_6$$]$$^{3-}$$ < [Co(NH$$_3$$)$$_6$$]$$^{3+}$$ < [Co(NH$$_3$$)$$_5$$(H$$_2$$O)]$$^{3+}$$ < [Co(NH$$_3$$)$$_5$$(Cl)]$$^{2+}$$

Facial-Meridional (fac-mer) isomerism occurs in octahedral complexes of the type $$[MA_3B_3]$$, where three identical ligands can either occupy the same face of the octahedron (facial) or be arranged along a meridian (meridional).

Let us analyze each option:

Option 1: $$[Co(en)_2Cl_2]^+$$ - This is of the type $$[M(AA)_2B_2]$$, which shows cis-trans isomerism, not fac-mer isomerism.

Option 2: $$[Co(en)_3]^{3+}$$ - This is of the type $$[M(AA)_3]$$, which shows optical isomerism only.

Option 3: $$[Co(NH_3)_3Cl_3]$$ - This is of the type $$[MA_3B_3]$$, which is the classic case for facial-meridional isomerism. The three $$NH_3$$ (and three $$Cl$$) ligands can be arranged in either fac or mer configuration.

Option 4: $$[Co(NH_3)_4Cl_2]^+$$ - This is of the type $$[MA_4B_2]$$, which shows cis-trans isomerism.

The correct answer is Option 3: $$[Co(NH_3)_3Cl_3]$$.

First, write the outer electronic configuration of each metal ion (remember that electrons are removed first from the 4s orbital, then from 3d).
Co (atomic no. 27): [Ar] 3d$$^{7}$$ 4s$$^{2}$$ → Co$$^{3+}$$ : [Ar] 3d$$^{6}$$
Mn (atomic no. 25): [Ar] 3d$$^{5}$$ 4s$$^{2}$$ → Mn$$^{2+}$$ : [Ar] 3d$$^{5}$$
Zn (atomic no. 30): [Ar] 3d$$^{10}$$ 4s$$^{2}$$ → Zn$$^{2+}$$ : [Ar] 3d$$^{10}$$

Next, decide whether the complex is high-spin or low-spin. This depends on the ligand field strength (spectrochemical series).
en (ethylenediamine) is a strong-field ligand, F$$^-$$ and H$$_2$$O are weak-field ligands.

Case 1: $$[Co(en)_3]^{3+}$$
Ligand: en → strong field
Ion: Co$$^{3+}$$ → d$$^{6}$$
Strong field ⇒ low-spin (pairing preferred).
d$$^{6}$$ low-spin octahedral configuration: $$t_{2g}^{6}e_g^{0}$$

Case 2: $$[CoF_6]^{3-}$$
Ligand: F$$^-$$ → weak field
Ion: Co$$^{3+}$$ → d$$^{6}$$
Weak field ⇒ high-spin (no extra pairing).
d$$^{6}$$ high-spin octahedral configuration: $$t_{2g}^{4}e_g^{2}$$

Case 3: $$[Mn(H_2O)_6]^{2+}$$
Ligand: H$$_2$$O → weak field
Ion: Mn$$^{2+}$$ → d$$^{5}$$
Weak field ⇒ high-spin.
d$$^{5}$$ high-spin octahedral configuration: $$t_{2g}^{3}e_g^{2}$$

Case 4: $$[Zn(H_2O)_6]^{2+}$$
Ligand: H$$_2$$O → weak field
Ion: Zn$$^{2+}$$ → d$$^{10}$$
All d-orbitals already filled irrespective of field strength.
Configuration: $$t_{2g}^{6}e_g^{4}$$

Now calculate the crystal field stabilization energy (CFSE) for each octahedral configuration.
Formula: CFSE = number in $$t_{2g}$$ × $$(-0.4\Delta_o)$$ + number in $$e_g$$ × $$(+0.6\Delta_o)$$

$$t_{2g}^{6}e_g^{0}$$ : CFSE = $$6(-0.4\Delta_o)= -2.4\Delta_o$$
$$t_{2g}^{4}e_g^{2}$$ : CFSE = $$4(-0.4\Delta_o)+2(+0.6\Delta_o)= -0.4\Delta_o$$
$$t_{2g}^{3}e_g^{2}$$ : CFSE = $$3(-0.4\Delta_o)+2(+0.6\Delta_o)= 0$$
$$t_{2g}^{6}e_g^{4}$$ : CFSE = $$6(-0.4\Delta_o)+4(+0.6\Delta_o)= 0$$

The most negative value (largest magnitude of stabilisation) is $$-2.4\Delta_o$$ for $$t_{2g}^{6}e_g^{0}$$.

Therefore the complex with the highest CFSE is $$[Co(en)_3]^{3+}$$, and its d-orbital electronic configuration is $$t_{2g}^{6}e_g^{0}$$.

Correct option: Option A

We need to arrange the given cobalt complexes in the increasing order of their Crystal Field Stabilization Energy (CFSE) magnitude by calculating their values in terms of octrahedral splitting ($$\Delta_o$$) and tetrahedral splitting ($$\Delta_t$$).

CFSE Calculations:

• 1. $$[\text{Co}(\text{NH}_3)_4]^{2+}$$ (Tetrahedral, weak-field ligand):
  • $$\text{Co}^{2+} \implies d^7$$ system.
  • Tetrahedral configuration: $$e^4 t_2^3$$
  • $$\text{CFSE} = [-0.6 \times 4 + 0.4 \times 3]\Delta_t = -1.2\Delta_t$$
  • Since $$\Delta_t \approx \frac{4}{9}\Delta_o$$, this corresponds to a very small energy splitting:
    $$\text{CFSE} \approx -1.2 \times \frac{4}{9}\Delta_o \approx -0.53\Delta_o$$
• 2. $$[\text{Co}(\text{NH}_3)_6]^{2+}$$ (Octahedral, weak-to-moderate field ligand):
  • $$\text{Co}^{2+} \implies d^7$$ system (high-spin).
  • Configuration: $$t_{2g}^5 e_g^2$$
  • $$\text{CFSE} = [-0.4 \times 5 + 0.6 \times 2]\Delta_o = -0.8\Delta_o$$
• 3. $$[\text{Co}(\text{NH}_3)_6]^{3+}$$ (Octahedral, strong-field behavior with $$\text{Co}^{3+}$$):
  • $$\text{Co}^{3+} \implies d^6$$ system (low-spin due to high charge on cobalt pairing the electrons).
  • Configuration: $$t_{2g}^6 e_g^0$$
  • $$\text{CFSE} = [-0.4 \times 6]\Delta_o + 2P = -2.4\Delta_o + 2P$$
• 4. $$[\text{Co}(\text{en})_3]^{3+}$$ (Octahedral, strong chelating ligand):
  • $$\text{Co}^{3+} \implies d^6$$ system (low-spin).
  • Configuration: $$t_{2g}^6 e_g^0$$
  • $$\text{CFSE} = [-0.4 \times 6]\Delta_o + 2P = -2.4\Delta_o + 2P$$
  • Because ethylenediamine ($$\text{en}$$) is a stronger chelating ligand than ammonia ($$\text{NH}_3$$), its splitting parameter ($$\Delta_o$$) is significantly larger. Consequently, its absolute CFSE magnitude is the highest.

Conclusion:

Comparing the calculated values and ligand field strengths, the increasing order of CFSE magnitude is:

$$\text{[Co(NH}_3)_4]^{2+} < \text{[Co(NH}_3)_6]^{2+} < \text{[Co(NH}_3)_6]^{3+} < \text{[Co(en)}_3]^{3+}$$

Answer: Option D

A homoleptic complex has only one type of ligand. We need to identify those with odd number of d electrons.

(A) $$[FeO_4]^{2-}$$: Fe is in +6 oxidation state ($$d^2$$). Homoleptic (only O ligands). d-electrons = 2 (even). Not selected.

(B) $$[Fe(CN)_6]^{3-}$$: Fe is in +3 oxidation state ($$d^5$$). Homoleptic (only CN ligands). d-electrons = 5 (odd). ✓

(C) $$[Fe(CN)_5NO]^{2-}$$: Has two types of ligands (CN and NO). Not homoleptic. Not selected.

(D) $$[CoCl_4]^{2-}$$: Co is in +2 oxidation state ($$d^7$$). Homoleptic (only Cl ligands). d-electrons = 7 (odd). ✓

(E) $$[Co(H_2O)_3F_3]$$: Has two types of ligands (H₂O and F). Not homoleptic. Not selected.

The correct answer is Option 3: (B) and (D) only.

(A) $$[\text{CoF}_6]^{3-}$$

• Oxidation State: $$\text{Co}^{3+} \; (3d^6)$$
• Ligand Nature: $$\text{F}^-$$ is a weak field ligand (WFL), so no electron pairing occurs.
• Hybridization: The six $$3d$$ electrons remain unpaired in their original configuration. To accommodate 6 lone pairs from ligands, the outer orbitals (4s, 4p, and 4d) are used, resulting in an outer orbital complex with $$\text{sp}^3\text{d}^2$$ hybridization.
• Match: (A) $$\rightarrow$$ (III)

(B) $$[\text{NiCl}_4]^{2-}$$

• Oxidation State: $$\text{Ni}^{2+} \; (3d^8)$$
• Ligand Nature: $$\text{Cl}^-$$ is a weak field ligand (WFL), so the two unpaired electrons in the 3d subshell do not pair up.
• Hybridization: The inner 3d orbitals are completely full or occupied. The four ligand lone pairs occupy the 4s and 4p orbitals, leading to a tetrahedral geometry with $$\text{sp}^3$$ hybridization.
• Match: (B) $$\rightarrow$$ (II)

(C) $$[\text{Co(NH}_3)_6]^{3+}$$

• Oxidation State: $$\text{Co}^{3+} \; (3d^6)$$
• Ligand Nature: $$\text{NH}_3$$ behaves as a strong field ligand (SFL) with $$\text{Co}^{3+}$$, forcing the six 3d electrons to pair up completely.
• Hybridization: This pairing clears out two vacant inner 3d orbitals. The six coordination slots are filled using these inner orbitals along with 4s and 4p, forming an inner orbital complex with $$\text{d}^2\text{sp}^3$$ hybridization.
• Match: (C) $$\rightarrow$$ (I)

(D) $$[\text{Ni(CN)}_4]^{2-}$$

• Oxidation State: $$\text{Ni}^{2+} \; (3d^8)$$
• Ligand Nature: $$\text{CN}^-$$ is a very strong field ligand (SFL). It forces the two unpaired electrons in the $3d$ orbital to pair up.
• Hybridization: This pairing vacates a single inner 3d orbital. The four ligand pairs fill this 3d orbital, the 4s orbital, and two 4p orbitals, giving a square planar geometry with $$\text{dsp}^2$$ hybridization.
• Match: (D) $$\rightarrow$$ (IV)

In an octahedral complex, six ligand donor atoms occupy the corners of a regular octahedron around the central metal ion.

Case 1: Homoleptic octahedral complex with monodentate ligands
Homoleptic means all the six positions are occupied by identical ligands, say $$L$$, giving the general formula $$[M L_6]$$. Because every ligand and every metal-ligand bond is identical, there is only one possible spatial arrangement. Therefore such a complex cannot give rise to either geometrical (cis / trans) or optical isomerism. Hence Statement I is true.

Case 2: cis-platin and trans-platin
cis-platin and trans-platin have the formula $$[Pt(NH_3)_2Cl_2]$$. They contain two different kinds of monodentate ligands ( $$NH_3$$ and $$Cl^-$$ ), so they are heteroleptic complexes of platinum, not palladium. Since the statement assigns them to Pd, Statement II is false.

Combining the two results:
Statement I is true, Statement II is false ⇒ Option D.

Final answer: Option D.

The crystal-field stabilization energy in an octahedral field is $$CFSE = (-0.4x + 0.6y)\,\Delta_0$$, where $$x$$ electrons occupy the $$t_{2g}$$ level and $$y$$ electrons occupy the $$e_g$$ level.

For a high-spin $$d^5$$ ion the electronic distribution is $$t_{2g}^3e_g^2$$, giving
$$CFSE = [(-0.4)(3) + (0.6)(2)]\Delta_0 = (-1.2 + 1.2)\Delta_0 = 0$$.
Hence a high-spin $$d^5$$ complex has $$\Delta_0 = 0$$.

The spin-only magnetic moment is $$\mu = \sqrt{n(n+2)} \text{ B.M.}$$, where $$n$$ is the number of unpaired electrons.

For high-spin $$d^5$$, $$n = 5$$, so
$$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ B.M.}$$ (≈ 5.96 B.M.).

Therefore we must locate the complex containing a high-spin $$d^5$$ metal ion.

Option A: $$[Fe(CN)_6]^{4-}$$  ⟹  $$Fe^{2+}$$ (since $$x-6=-4$$) is $$d^6$$. CN⁻ is a strong-field ligand (low-spin), not $$d^5$$.

Option B: $$[Co(NH_3)_6]^{3+}$$  ⟹  $$Co^{3+}$$ is $$d^6$$. Not $$d^5$$.

Option C: $$[FeF_6]^{4-}$$  ⟹  $$Fe^{2+}$$, $$d^6$$ (weak-field, high-spin). Still $$d^6$$, not $$d^5$$.

Option D: $$[Mn(SCN)_6]^{4-}$$  ⟹  $$Mn^{2+}$$ (since $$x-6=-4$$) is $$d^5$$. SCN⁻ is a weak/medium-field ligand, so the complex is high-spin $$d^5$$, giving $$\Delta_0 = 0$$ and $$\mu \approx 5.92$$ B.M.

Hence the correct complex is $$[Mn(SCN)_6]^{4-}$$.

Correct answer: Option D.

Step 1 : Find the oxidation state of Mn in $$[MnCl_6]^{3-}$$.
Each chloride ligand is $$Cl^-$$, carrying charge $$-1$$.
Let the oxidation state of Mn be $$x$$.
Therefore $$x + 6(-1) = -3 \;\Rightarrow\; x - 6 = -3 \;\Rightarrow\; x = +3$$.
So the central metal is $$Mn^{3+}$$.

Step 2 : Write the ground-state electronic configuration of $$Mn^{3+}$$.
Atomic number of Mn is 25: $$Mn : [Ar]\,3d^5\,4s^2$$.
For $$Mn^{3+}$$ we remove two $$4s$$ electrons and one $$3d$$ electron: $$Mn^{3+} : [Ar]\,3d^4$$.

Step 3 : Decide whether the complex is inner-orbital or outer-orbital.
Chloride $$(Cl^-)$$ is a weak-field ligand in the spectro-chemical series.
Weak-field ligands do not cause pairing of $$3d$$ electrons.
Hence the complex remains high-spin, and the metal uses the outer (fourth) shell orbitals for hybridisation.

Step 4 : Determine the hybridisation.
An octahedral complex requires six hybrid orbitals.
Because no pairing occurs in $$3d$$, all five $$3d$$ orbitals are occupied by four unpaired electrons and one empty orbital, leaving no two vacant inner $$d$$ orbitals.
Therefore Mn must use $$4s$$, three $$4p$$, and two $$4d$$ orbitals: $$sp^3d^2$$ (outer-orbital octahedral).

Step 5 : Calculate the number of unpaired electrons.
Configuration of $$Mn^{3+}$$ remains $$3d^4$$ with weak-field ligands: $$\uparrow\,\uparrow\,\uparrow\,\uparrow$$ (four unpaired electrons).
Hence the complex is paramagnetic with four unpaired electrons.

Conclusion : The complex $$[MnCl_6]^{3-}$$ exhibits $$sp^3d^2$$ hybridisation and is paramagnetic with four unpaired electrons.

Therefore the correct choice is Option B.

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

We need to identify homoleptic complexes that are low spin from the given list.

(A) $$[Fe(CN)_5NO]^{2-}$$: Contains CN⁻ and NO — heteroleptic. Eliminated.

(B) $$[CoF_6]^{3-}$$: All F⁻ ligands — homoleptic.

(C) $$[Fe(CN)_6]^{4-}$$: All CN⁻ ligands — homoleptic.

(D) $$[Co(NH_3)_6]^{3+}$$: All NH₃ ligands — homoleptic.

(E) $$[Cr(H_2O)_6]^{2+}$$: All H₂O ligands — homoleptic.

(B) $$[CoF_6]^{3-}$$: Co³⁺ is $$d^6$$, F⁻ is a weak field ligand → high spin. Eliminated.

(C) $$[Fe(CN)_6]^{4-}$$: Fe²⁺ is $$d^6$$, CN⁻ is a strong field ligand → low spin. ✓

(D) $$[Co(NH_3)_6]^{3+}$$: Co³⁺ is $$d^6$$, NH₃ is a moderately strong field ligand, and Co³⁺ has high charge density → low spin. ✓

(E) $$[Cr(H_2O)_6]^{2+}$$: Cr²⁺ is $$d^4$$, H₂O is a weak field ligand → high spin. Eliminated.

The homoleptic and low spin complexes are (C) and (D).

The answer is Option B: (C) and (D) only.

The coordination number of chromium in both complexes is $$6$$, so the expected shape is octahedral.

The complex contains three chloride ions and three pyridine molecules: the general type is $$[MA_3B_3]$$ where $$A = Cl^-$$ and $$B = py$$.

For an octahedral $$[MA_3B_3]$$ species two geometrical arrangements are possible:

1. fac-isomer  (all three identical ligands occupy one face of the octahedron).
2. mer-isomer  (the three identical ligands lie in one meridian plane).

Neither fac nor mer form is chiral because each possesses a mirror plane (fac) or an inversion centre (mer). Hence there is no optical isomerism.

Therefore the number of stereoisomers for $$[CrCl_3(py)_3]$$ is $$2$$.

The coordination sphere has two monodentate $$Cl^-$$ ligands and two identical bidentate $$ox^{2-}$$ ligands. The general formulation can be written as $$[M(LL)_2X_2]$$.

Step 1 - Geometrical isomerism:
  • cis-isomer - the two $$Cl^-$$ ligands are adjacent (angle $$90^\circ$$).
  • trans-isomer - the two $$Cl^-$$ ligands are opposite (angle $$180^\circ$$).

Step 2 - Optical activity:
  • trans-isomer possesses a centre of symmetry, so it is achiral (only one form).
  • cis-isomer has no symmetry element that relates left and right: the two bidentate chelate rings wind around the metal in a helical manner, giving two non-superimposable mirror images (d- and l-forms).

Hence:

trans → 1 stereoisomer
cis → 2 stereoisomers (enantiomeric pair)

Total stereoisomers for $$[CrCl_2(ox)_2]^{3-}$$  = $$1 + 2 = 3$$.

Therefore, the required numbers of stereoisomers are:

$$[CrCl_3(py)_3]$$ : $$2$$   and   $$[CrCl_2(ox)_2]^{3-}$$ : $$3$$.

The correct option is Option C (2 & 3).

We need to determine the structure of the complex $$Co(NH_3)_5Cl_3$$ using experimental evidence.

The experimental observations are as follows. One mole of the complex gives 3 moles of ions on dissolution in water, and one mole of the complex reacts with excess $$AgNO_3$$ to give 2 moles of $$AgCl$$.

We begin by interpreting the $$AgCl$$ precipitation data. Silver nitrate precipitates only the free chloride ions (those outside the coordination sphere) as $$AgCl$$. Chloride ions directly coordinated to the metal ion inside the complex do not react with $$AgNO_3$$ under normal conditions. Since 2 moles of $$AgCl$$ are formed, there are 2 free $$Cl^-$$ ions outside the coordination sphere.

With a total of 3 chloride ions and 2 free, the number of coordinated chloride ions is $$3 - 2 = 1$$.

Assuming the complex has the formula $$[Co(NH_3)_5Cl]Cl_2$$, it dissociates as:
$$[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^-$$. This gives 3 ions total (1 complex cation + 2 chloride anions), which matches the given data. ✓

In the cation $$[Co(NH_3)_5Cl]^{2+}$$, cobalt is coordinated to 5 $$NH_3$$ molecules and 1 $$Cl^-$$ ion, giving a coordination number of 6, which is consistent with the octahedral geometry stated in the question. ✓

The correct answer is Option 3: $$[Co(NH_3)_5Cl]Cl_2$$.

Mohr’s salt is $$FeSO_4\cdot (NH_4)_2SO_4\cdot 6H_2O$$.

When it dissolves in water, the crystal breaks into ions according to the stoichiometric formula:
$$FeSO_4\cdot (NH_4)_2SO_4 \rightarrow Fe^{2+} + 2\,NH_4^{+} + 2\,SO_4^{2-}$$

Thus only three kinds of ions are present—ferrous ($$Fe^{2+}$$), ammonium ($$NH_4^{+}$$) and sulphate ($$SO_4^{2-}$$).
So, Statement I is correct.

At infinite dilution, Kohlrausch’s Law states that the molar conductance of an electrolyte equals the sum of the ionic conductances, each multiplied by the number of that ion present in one formula unit:
$$\Lambda_m^{\infty}= \sum n_i\,\lambda_i^{\infty}$$

For Mohr’s salt the stoichiometric coefficients are 1 for $$Fe^{2+}$$, 2 for $$NH_4^{+}$$ and 2 for $$SO_4^{2-}$$. If their individual molar ionic conductances are $$x_1$$, $$x_2$$ and $$x_3$$ S cm$$^2$$ mol$$^{-1}$$, then
$$\Lambda_m^{\infty}= 1\,(x_1) + 2\,(x_2) + 2\,(x_3)= x_1 + 2x_2 + 2x_3$$

Statement II proposes $$x_1 + x_2 + 2x_3$$, which neglects the second ammonium ion. Hence Statement II is false.

Therefore, Statement I is true while Statement II is false, which matches Option C.

The complexes are octahedral, so their magnetic behaviour depends on

• oxidation state of the metal → number of $$d$$ electrons
• strength of the ligands (spectro-chemical series)
• whether the complex is high-spin or low-spin → number of unpaired electrons

Case A:

Complex $$[Mn(CN)_6]^{3-}$$

Oxidation state of Mn: $$x-6 = -3 \;\Rightarrow\; x = +3$$, so Mn is $$Mn^{3+}$$.

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

$$CN^-$$ is a strong-field ligand, so the complex is low-spin. For a d4 ion in a low-spin octahedral field we fill $$t_{2g}$$ first: $$t_{2g}^4 e_g^0$$. This arrangement has $$2$$ unpaired electrons, hence the complex is paramagnetic, not diamagnetic.

Case B:

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

Oxidation state of Co: $$x = +3$$ ⇒ $$Co^{3+}$$.

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

$$NH_3$$ is a borderline/medium-field ligand, but with the highly charged $$Co^{3+}$$ centre the complex is low-spin. Thus the six electrons occupy $$t_{2g}$$ only: $$t_{2g}^6 e_g^0$$. All electrons are paired ⇒ diamagnetic.

Case C:

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

Oxidation state of Fe: $$x-6 = -4 \;\Rightarrow\; x = +2$$, so Fe is $$Fe^{2+}$$.

Electronic configuration of $$Fe^{2+}: [Ar]\,3d^6$$.

$$CN^-$$ is a strong-field ligand giving a low-spin complex. For d6: $$t_{2g}^6 e_g^0$$, all paired ⇒ diamagnetic.

Case D:

Complex $$[Co(H_2O)_3F_3]$$ (overall neutral)

Let oxidation state of Co be $$x$$:
$$x + 0\,(3H_2O) - 1\,(3F^-) = 0 \;\Rightarrow\; x = +3$$, so $$Co^{3+}$$ again (d6).

However, $$H_2O$$ and $$F^-$$ are weak-field ligands, so the complex is high-spin. Electron distribution: $$t_{2g}^4 e_g^2$$ with $$4$$ unpaired electrons ⇒ paramagnetic.

Summary:
A ⇒ paramagnetic
B ⇒ diamagnetic
C ⇒ diamagnetic
D ⇒ paramagnetic

Therefore the diamagnetic octahedral complexes are B and C only.

Correct option: Option D (B and C Only).

For acidic‐basic behaviour of transition metal oxides, the general trend is: basic nature at low oxidation state $$\rightarrow$$ amphoteric $$\rightarrow$$ acidic at the highest oxidation state. We first write the oxidation state of the metal in each oxide.

$$VO_2:\; V+2(-2)=0 \Rightarrow V^{+4}$$
$$V_2O_3:\;2V+3(-2)=0 \Rightarrow V^{+3}$$
$$CrO_3:\; Cr+3(-2)=0 \Rightarrow Cr^{+6}$$
$$V_2O_5:\;2V+5(-2)=0 \Rightarrow V^{+5}$$
$$Mn_2O_7:\;2Mn+7(-2)=0 \Rightarrow Mn^{+7}$$

Nature of these oxides:
• $$V^{+3}$$ oxide ($$V_2O_3$$) is basic.
• $$V^{+4}$$ oxide ($$VO_2$$) is amphoteric.
• $$V^{+5}$$ oxide ($$V_2O_5$$) is also amphoteric (it reacts both with acids and with alkalies).
• $$Cr^{+6}$$ oxide ($$CrO_3$$) is the anhydride of $$H_2CrO_4$$ and is strongly acidic.
• $$Mn^{+7}$$ oxide ($$Mn_2O_7$$) is the anhydride of $$HMnO_4$$ and is strongly acidic.

Thus only $$CrO_3$$ and $$Mn_2O_7$$ are counted as acidic. Therefore
$$X = 2$$

Primary valency (Werner) = oxidation state of the metal.

The compound is $$[Co(H_2NCH_2CH_2NH_3)_3]_2(SO_4)_3$$.
Each ligand $$(H_2NCH_2CH_2NH_3)$$ is treated as neutral for the charge calculation (only the unprotonated $$NH_2$$ donates the lone pair; the protonated $$NH_3^+$$ part does not donate and its positive charge is balanced internally by the non-coordinating N).
Let the oxidation state of cobalt be $$y$$.

Charge balance for one complex ion:
$$y + 3(0)=\text{charge of }[CoL_3]^{n+}$$

Overall compound is electrically neutral, so

$$2\bigl(\text{charge of }[CoL_3]^{n+}\bigr) + 3(-2)=0 \; -(1)$$

From $$(1)$$: $$2n - 6 = 0 \Rightarrow n = +3$$. Hence $$y = +3$$.

Therefore the primary valency of cobalt is
$$Y = 3$$

Finally, $$X+Y = 2 + 3 = 5$$.

Option A is correct.

We use the colligative property of freezing point depression to determine the van’t Hoff factor $$i$$.

The formula for depression in freezing point is $$\Delta T_f = i \cdot K_f \cdot m$$.

Given: $$\Delta T_f = 0.558°C$$, $$K_f = 1.86$$ K kg mol^{-1}, $$m = 0.1$$ molal.

Substituting these values yields $$0.558 = i \times 1.86 \times 0.1$$ so that $$i = \frac{0.558}{0.186} = 3$$.

A van’t Hoff factor of $$3$$ indicates that the complex dissociates into three ions in solution.

Complex $$[Cr(NH_3)_5Cl]Cl_2$$ dissociates into $$[Cr(NH_3)_5Cl]^{2+}$$ and $$2Cl^-$$, giving three ions (i = 3).

Complex $$[Cr(NH_3)_6Cl]Cl_3$$ would involve a coordination number of seven (which is not possible for Cr since CN = 6).

Complex $$[Cr(NH_3)_3Cl_3]$$ is a neutral species and does not produce free ions (i = 1).

Complex $$[Cr(NH_3)_4Cl_2]Cl$$ dissociates into two ions (i = 2).

Since chromium must have coordination number six and the van’t Hoff factor is three, the only viable structure is $$[Cr(NH_3)_5Cl]Cl_2$$.

The correct answer is Option 1: $$[Cr(NH_3)_5Cl]Cl_2$$.

For $$K_3[Fe(OH)_6]$$: The complex ion is $$[Fe(OH)_6]^{3-}$$. Fe is in +3 oxidation state: $$d^5$$.

$$OH^-$$ is a weak field ligand, so this is high-spin $$d^5$$ with 5 unpaired electrons.

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

For $$K_4[Fe(OH)_6]$$: The complex ion is $$[Fe(OH)_6]^{4-}$$. Fe is in +2 oxidation state: $$d^6$$.

$$OH^-$$ is a weak field ligand, so this is high-spin $$d^6$$ with 4 unpaired electrons.

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

The correct answer is Option 4: 5.92 and 4.90 B.M.

The electronic configuration $$t_{2g}^3 e_g^1$$ corresponds to a $$d^4$$ metal ion.

In a weak field (small $$\Delta_0$$), electrons fill according to Hund's rule, avoiding pairing. So 4 d-electrons go as: $$t_{2g}^3 e_g^1$$ - one electron in each of 3 $$t_{2g}$$ orbitals and 1 in $$e_g$$. This gives maximum spin (high spin complex).

A weak field ligand produces small crystal field splitting, favoring high spin configuration.

The correct answer is Option B: weak field ligand, high spin complex.

First find the oxidation state and the ground-state $$3d$$ electron configuration of the central metal ion in each complex.

$$[FeF_6]^{3-} :$$
Fluoride is $$-1$$, six ligands give $$-6$$.
Let oxidation state of Fe be $$x$$. $$x-6 = -3 \Rightarrow x = +3$$.
Fe atom: $$[Ar]\,3d^6 4s^2$$ ⇒ $$Fe^{3+}: [Ar]\,3d^5$$ (configuration $$d^5$$).

$$[CoF_6]^{3-} :$$
Similarly, $$x-6 = -3 \Rightarrow x = +3$$.
Co atom: $$[Ar]\,3d^7 4s^2$$ ⇒ $$Co^{3+}: [Ar]\,3d^6$$ (configuration $$d^6$$).

$$[Ni(CO)_4] :$$
Carbonyl is neutral. Total charge $$0$$, so Ni is in $$0$$ oxidation state.
Ni atom: $$[Ar]\,3d^8 4s^2$$ ⇒ $$Ni^{0}: [Ar]\,3d^{10} 4s^{0}$$ (configuration $$d^{10}$$).

$$[Ni(CN)_4]^{2-} :$$
Cyanide is $$-1$$, four ligands give $$-4$$.
Let oxidation state of Ni be $$x$$. $$x-4 = -2 \Rightarrow x = +2$$.
Ni atom: $$[Ar]\,3d^8 4s^2$$ ⇒ $$Ni^{2+}: [Ar]\,3d^8$$ (configuration $$d^8$$).

Next decide whether the complex is high-spin or low-spin and its geometry, using ligand field strength:

• $$F^-$$ is a weak-field ligand (spectrochemical series). Octahedral complexes with $$F^-$$ are high-spin.
• $$CO$$ and $$CN^-$$ are strong-field ligands. $$CO$$ gives a tetrahedral, completely filled $$sp^3$$ complex with Ni(0). $$CN^-$$ gives a square-planar, low-spin $$dsp^2$$ complex with Ni(II).

Now place the electrons:

Case 1: $$[FeF_6]^{3-}$$ (octahedral, high-spin, $$d^5$$)

Electrons fill all five $$d$$ orbitals singly ⇒ number of unpaired electrons $$=5$$.

Case 2: $$[CoF_6]^{3-}$$ (octahedral, high-spin, $$d^6$$)

Arrangement: $$t_{2g}^4 e_g^2$$ ⇒ unpaired electrons $$=4$$.

Case 3: $$[Ni(CO)_4]$$ (tetrahedral, $$d^{10}$$, $$sp^3$$)

All ten $$d$$ electrons are paired ⇒ unpaired electrons $$=0$$.

Case 4: $$[Ni(CN)_4]^{2-}$$ (square-planar, low-spin, $$d^8$$, $$dsp^2$$)

Electron distribution: $$d_{x^2-y^2}^0, d_{z^2}^2, d_{xy}^2, d_{xz}^2, d_{yz}^2$$ ⇒ unpaired electrons $$=0$$.

Collecting the results:

$$[FeF_6]^{3-} : 5 \; \gt$$
$$[CoF_6]^{3-} : 4 \; \gt$$
$$[Ni(CN)_4]^{2-} : 0$$ and $$[Ni(CO)_4] : 0$$ (equal).

Therefore the correct descending order of unpaired electrons is
$$[FeF_6]^{3-} \gt [CoF_6]^{3-} \gt [Ni(CN)_4]^{2-} = [Ni(CO)_4]$$.

This corresponds to Option D.

According to Werner’s theory,

• the oxidation state of the central metal atom / ion gives the number of primary valencies (ionisable).
• the total number of donor atoms directly linked with the metal gives the secondary valencies (coordination number).

Complex : $$[Co(en)_2Cl_2]Cl$$

Step-1 Oxidation state of Co (inside the square bracket):
Let it be $$x$$.
$$x + 2(0) + 2(-1) = +1\;$$ (because one $$Cl^-$$ is outside the bracket)
$$x - 2 = +1 \Rightarrow x = +3$$

Primary valencies = $$3$$.

Step-2 Coordination number:
$$en$$ is bidentate → $$2 \; en$$ supplies $$2 \times 2 = 4$$ donor atoms.
$$2 \; Cl^-$$ supplies $$2$$ donor atoms.
Total $$= 4 + 2 = 6$$.

Secondary valencies = $$6$$.

Hence (A) ⟶ (I) (primary 3, secondary 6).

Complex : $$[Pt(NH_3)_2Cl(NO_2)]$$

Step-1 Oxidation state of Pt:
Let it be $$x$$.
$$x + 2(0) + (-1) + (-1) = 0$$
$$x - 2 = 0 \Rightarrow x = +2$$

Primary valencies = $$2$$.

Step-2 Coordination number:
All the four ligands ($$NH_3, NH_3, Cl^- , NO_2^-$$) are monodentate → $$4$$ donor atoms.

Secondary valencies = $$4$$.

Hence (B) ⟶ (IV) (primary 2, secondary 4).

Complex : $$Hg[Co(SCN)_4]$$

Step-1 Find the charge on the complex anion.
Hg is present as $$Hg^+$$ (monovalent mercurous ion).
For electrical neutrality, the anion must carry $$-1$$ charge.

Step-2 Oxidation state of Co:
Let it be $$x$$.
$$x + 4(-1) = -1$$
$$x - 4 = -1 \Rightarrow x = +3$$

Primary valencies = $$3$$.

Step-3 Coordination number:
Each $$SCN^-$$ is monodentate → $$4$$ donor atoms.

Secondary valencies = $$4$$.

Hence (C) ⟶ (II) (primary 3, secondary 4).

Complex : $$[Mg(EDTA)]^{2-}$$

Step-1 Oxidation state of Mg:
Let it be $$x$$.
Charge on $$EDTA$$ ligand $$= -4$$.
Total charge on complex $$= -2$$.
$$x + (-4) = -2 \Rightarrow x = +2$$

Primary valencies = $$2$$.

Step-2 Coordination number:
$$EDTA^{4-}$$ is hexadentate → $$6$$ donor atoms.

Secondary valencies = $$6$$.

Hence (D) ⟶ (III) (primary 2, secondary 6).

Putting the four results together:

$$(A)-(I),\;(B)-(IV),\;(C)-(II),\;(D)-(III)$$

Therefore the correct choice is Option B.

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: $$1,\,5,\,4,\,2$$.

We need to find how many of the given species are involved in $$sp^3d^2$$ hybridization.

The key distinction is between inner orbital complexes ($$d^2sp^3$$, which use inner $$(n-1)d$$ orbitals) and outer orbital complexes ($$sp^3d^2$$, which use outer $$nd$$ orbitals). Strong field ligands cause pairing and form inner orbital complexes, while weak field ligands form outer orbital complexes.

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

$$Co^{3+}$$: electronic configuration is $$[Ar] 3d^6$$.

$$NH_3$$ is a strong field ligand, so all 6 electrons pair up in the three $$3d$$ ($$t_{2g}$$) orbitals, leaving the two $$e_g$$ ($$3d$$) orbitals empty for bonding.

Hybridization: $$d^2sp^3$$ (inner orbital complex). Not $$sp^3d^2$$.

2. $$SF_6$$

$$S$$ has 6 valence electrons and forms 6 bonds with fluorine atoms.

Sulfur uses one $$3s$$, three $$3p$$, and two $$3d$$ orbitals.

Hybridization: $$sp^3d^2$$. Yes.

3. $$[CrF_6]^{3-}$$

$$Cr^{3+}$$: electronic configuration is $$[Ar] 3d^3$$.

The three $$3d$$ electrons occupy the three $$t_{2g}$$ orbitals (one each), leaving the two $$e_g$$ ($$3d$$) orbitals empty.

Even though $$F^-$$ is a weak field ligand, since the $$e_g$$ orbitals are already empty (only 3 d-electrons), $$Cr^{3+}$$ uses these inner $$3d$$ orbitals for bonding.

Hybridization: $$d^2sp^3$$ (inner orbital complex). Not $$sp^3d^2$$.

4. $$[CoF_6]^{3-}$$

$$Co^{3+}$$: electronic configuration is $$[Ar] 3d^6$$.

$$F^-$$ is a weak field ligand, so no pairing occurs. The 6 electrons are distributed as: $$t_{2g}^4 e_g^2$$ (with 4 unpaired electrons).

Since the $$e_g$$ orbitals are occupied, the inner $$3d$$ orbitals cannot be used. The complex uses outer $$4d$$ orbitals.

Hybridization: $$sp^3d^2$$ (outer orbital complex). Yes.

5. $$[Mn(CN)_6]^{3-}$$

$$Mn^{3+}$$: electronic configuration is $$[Ar] 3d^4$$.

$$CN^-$$ is a strong field ligand, so pairing occurs: the 4 electrons pair up into two orbitals in $$t_{2g}$$, leaving the $$e_g$$ orbitals empty.

Hybridization: $$d^2sp^3$$ (inner orbital complex). Not $$sp^3d^2$$.

6. $$[MnCl_6]^{3-}$$

$$Mn^{3+}$$: electronic configuration is $$[Ar] 3d^4$$.

$$Cl^-$$ is a weak field ligand, so no pairing occurs. The 4 electrons are distributed as: $$t_{2g}^3 e_g^1$$.

Since one $$e_g$$ orbital is occupied, the inner $$3d$$ orbitals cannot both be used for bonding. The complex uses outer $$4d$$ orbitals.

Hybridization: $$sp^3d^2$$ (outer orbital complex). Yes.

Summary:

Species with $$sp^3d^2$$ hybridization: $$SF_6$$, $$[CoF_6]^{3-}$$, $$[MnCl_6]^{3-}$$ = 3 species.

Hence, the correct answer is 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 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 $$\mathbf{2}$$.

Availability of the Nitrogen Lone Pair

The basic strength of an amine or a heterocyclic nitrogen compound depends directly on the availability of its lone pair of electrons on the nitrogen atom to accept a proton ($$\text{H}^+$$). If the lone pair is highly localized, the compound is more basic; if it is delocalized due to resonance or held tightly by a more electronegative orbital, the basicity drops significantly.

Detailed Analysis of Each Compound:

• 1. Piperidine

  Piperidine is a cyclic secondary aliphatic amine. The nitrogen atom is completely $$\text{sp}^3$$ hybridized, and its lone pair of electrons is highly localized in an $$\text{sp}^3$$ orbital. Since it does not participate in resonance and benefits from the electron-donating inductive effect ($$+I$$) of the adjacent alkyl ring chain, piperidine is the most basic compound among the three.




• 2. Pyridine

  In pyridine, the nitrogen atom is part of an aromatic ring and is $$\text{sp}^2$$ hybridized. The lone pair of electrons resides in an $$\text{sp}^2$$ hybrid orbital that lies within the plane of the ring, meaning it is perpendicular to the $$\pi$$ system and does not participate in aromaticity. It is available for protonation, but because an $$\text{sp}^2$$ hybridized nitrogen is more electronegative than an $$\text{sp}^3$$ nitrogen, it holds its electrons more tightly. Therefore, pyridine is significantly less basic than piperidine.




• 3. Pyrrole

  In pyrrole, the lone pair on the nitrogen atom is actively involved in the aromatic ring system to satisfy Hückel's rule ($$6\pi$$ electrons). Because these electrons are fully delocalized within the aromatic ring, donating them to a proton would destroy the molecule's aromatic stabilization. This makes pyrrole an exceptionally weak base, placing it as the least basic compound in the group.

Conclusion:

Combining the hybridization and delocalization factors, the correct decreasing order of basic strength is:

$$\mathbf{\text{Piperidine} > \text{Pyridine} > \text{Pyrrole}}$$

Answer: Option D — Piperidine > Pyridine > Pyrrole

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 $$K_2MnO_4$$. To do this, we first determine the oxidation state of manganese by letting it be $$x$$ and solving the equation $$2(+1) + x + 4(-2) = 0$$. Simplifying, we get $$2 + x - 8 = 0$$, so $$x = +6$$, which means manganese is in the +6 oxidation state.

Next, we apply the modern IUPAC 2005 nomenclature rules for oxoanions. The cation is named first as potassium, with no prefix for quantity because its stoichiometry is clear. For the anion $$MnO_4^{2-}$$, the name is formed by stating the number and type of ligands (four oxide ligands become tetraoxido), followed by the metal name with an -ate suffix (manganate), and then the oxidation state in Roman numerals (VI).

Putting these parts together, the IUPAC name of $$K_2MnO_4$$ is Potassium tetraoxidomanganate(VI). Option 1 is incorrect because it uses “tetraoxo” and “permanganate,” Option 3 is wrong in stating the oxidation state as VII, and Option 4 incorrectly uses “manganese” instead of “manganate.” Therefore, the correct answer is Option 2: Potassium tetraoxidomanganate(VI).

We need to identify which statements are correct from (A) through (E).

Statement (A): Ethane-1,2-diamine is a chelating ligand. Ethane-1,2-diamine (ethylenediamine, en) is $$H_2N-CH_2-CH_2-NH_2$$. It has two nitrogen donor atoms that can simultaneously coordinate to the same metal ion, forming a five-membered chelate ring. This is TRUE.

Statement (B): Metallic aluminium is produced by electrolysis of aluminium oxide in the presence of cryolite. In the Hall-Heroult process, $$Al_2O_3$$ is dissolved in molten cryolite ($$Na_3AlF_6$$) to lower the melting point, and then electrolysed to produce metallic aluminium at the cathode. This is TRUE.

Statement (C): Cyanide ion is used as ligand for leaching of silver. In the MacArthur-Forrest process, silver (and gold) ores are leached using dilute sodium cyanide solution in the presence of air: $$4Ag + 8CN^- + 2H_2O + O_2 \rightarrow 4[Ag(CN)_2]^- + 4OH^-$$ The cyanide ion acts as a ligand forming the soluble dicyanoargentate(I) complex. This is TRUE.

Statement (D): Phosphine acts as a ligand in Wilkinson catalyst. Wilkinson's catalyst is $$[RhCl(PPh_3)_3]$$, where $$PPh_3$$ is triphenylphosphine, not phosphine ($$PH_3$$). Triphenylphosphine and phosphine are different compounds. The ligand is triphenylphosphine, not phosphine itself. This statement is FALSE.

Statement (E): The stability constants of $$Ca^{2+}$$ and $$Mg^{2+}$$ are similar with EDTA complexes. The stability constant (log K) of $$[Ca(EDTA)]^{2-}$$ is approximately 10.7, while that of $$[Mg(EDTA)]^{2-}$$ is approximately 8.7. These values differ significantly (by about 2 orders of magnitude). This is FALSE.

Statements (A), (B), and (C) are correct.

Answer: Option 3 — (A), (B), (C) only

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.

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.

EDTA⁴⁻ is a hexadentate ligand that coordinates through 2 nitrogen atoms and 4 oxygen atoms (total 6 donor atoms).

Ca²⁺ complex with EDTA⁴⁻ has coordination number 6, forming an octahedral geometry.

The correct answer is Option 3: octahedral.

Identify the homoleptic complex from the given options.

Definition: A homoleptic complex is one in which the central metal ion is bonded to only one type of ligand. A heteroleptic complex has two or more different types of ligands.

Analyzing each option:

Option A: $$[Ni(CN)_4]^{2-}$$ - Ni is bonded to 4 $$CN^-$$ ligands only. One type of ligand. This is homoleptic.

Option B: $$[Ni(NH_3)_2Cl_2]$$ - Ni is bonded to $$NH_3$$ and $$Cl^-$$. Two different ligands. Heteroleptic.

Option C: $$[Fe(NH_3)_4Cl_2]^+$$ - Fe is bonded to $$NH_3$$ and $$Cl^-$$. Two different ligands. Heteroleptic.

Option D: $$[Co(NH_3)_4Cl_2]^+$$ - Co is bonded to $$NH_3$$ and $$Cl^-$$. Two different ligands. Heteroleptic.

The correct answer is Option A: $$[Ni(CN)_4]^{2-}$$.

To decide which of the given carbonyl species are isoelectronic with $$Ni(CO)_4$$ we must compare the total number of electrons (all valence and core electrons of the metal, all electrons of the ligands, and any extra electrons arising from charge) present in each species.

1. Compute the total electrons in $$Ni(CO)_4$$.
  • Atomic electrons in Ni : $$28$$ (atomic number).
  • Each CO molecule contains $$14$$ electrons.
  • Four CO ligands contribute $$4 \times 14 = 56$$ electrons.
  • Charge on the complex is zero, so no extra electrons.
Hence, the total electron count is
$$28 + 56 = 84$$ electrons.

2. Evaluate every listed species in the same way.

3. Compare each total with the 84 electrons of $$Ni(CO)_4$$.
Only $$[Cr(CO)_4]^{4-}$$ matches exactly.

Therefore, the number of species that are isoelectronic with $$Ni(CO)_4$$ is 1.

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.

This question asks us to identify the correct order of ligand field strength (spectrochemical series) from the given options.

Key Concept: The Spectrochemical Series

The spectrochemical series ranks ligands by their ability to cause crystal field splitting ($$\Delta$$). A "strong field" ligand causes a large splitting, while a "weak field" ligand causes a small splitting. The well-established order (from strongest to weakest) is:

$$CO > CN^- > NO_2^- > en > NH_3 > NCS^- > H_2O > OH^- > F^- > SCN^- > Cl^- > S^{2-} > I^-$$

Checking each option against this series:

Option 1: $$NCS^- > EDTA^{4-} > CN^- > CO$$

This places $$NCS^-$$ as the strongest ligand, which is incorrect. In the spectrochemical series, $$NCS^-$$ is a weaker field ligand than $$CN^-$$ and $$CO$$. Also, $$CO$$ should be at the top, not the bottom. This order is wrong.

Option 2: $$CO > H_2O > F^- > S^{2-}$$

Let us verify each comparison: $$CO$$ is the strongest field ligand in the series. $$H_2O$$ is a moderate field ligand, weaker than $$CO$$. $$F^-$$ is a weak field ligand, weaker than $$H_2O$$. $$S^{2-}$$ is a very weak field ligand, weaker than $$F^-$$. This exactly follows the spectrochemical series. This order is correct.

Option 3: $$S^{2-} > OH^- > EDTA^{4-} > CO$$

This places $$S^{2-}$$ as the strongest, which is incorrect since $$S^{2-}$$ is actually one of the weakest field ligands. This order is wrong.

Option 4: $$OH^- > F^- > NH_3 > CN^-$$

This places $$OH^-$$ above $$NH_3$$ and $$CN^-$$, which contradicts the series where $$CN^- > NH_3 > OH^-$$. This order is wrong.

The correct answer is Option 2: $$CO > H_2O > F^- > S^{2-}$$.

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.

The given complex is $$[\,Fe(NH_3)_x(CN)_y\,]^{-}$$ and the metal ion is iron(III), that is $$Fe^{3+}$$. For the d-electron count:

$$Fe \;(Z = 26)\; \Rightarrow \; Fe^{3+} : [\,Ar\,]\,3d^{5}$$ Hence the metal centre is a $$d^{5}$$ system.

In an octahedral field the five d-orbitals split into the lower-energy $$t_{2g}$$ set and the higher-energy $$e_{g}$$ set. If a $$d^{5}$$ ion has no electron in its $$e_{g}$$ orbitals, all five electrons must be paired in the lower $$t_{2g}$$ level, giving the configuration $$t_{2g}^{5}\,e_{g}^{0}$$ (low-spin). This is possible only when the ligand field splitting energy $$\Delta_{0}$$ is large enough to overcome the pairing energy, i.e. the complex must contain strong-field ligands.

$$CN^{-}$$ is a strong-field ligand, whereas $$NH_3$$ is of intermediate strength. A sufficiently large number of $$CN^{-}$$ ligands will make the overall splitting strong enough to produce the low-spin configuration.

Step 1 : Determine the number of cyanide ligands from the overall charge. Let the oxidation state of iron be $$+3$$, the charge on each $$CN^{-}$$ be $$-1$$ and on each $$NH_3$$ be 0. For the complex charge $$-1$$ we have $$+3 \;+\; (0)\,x \;+\; (-1)\,y \;=\; -1$$ $$\Rightarrow \; 3 - y = -1$$ $$\Rightarrow \; y = 4$$

Step 2 : Decide the coordination number. Iron(III) with a mix of monodentate ligands normally forms an octahedral complex, so the total number of ligands is 6: $$x + y = 6$$ With $$y = 4$$ already fixed, $$x = 6 - 4 = 2$$

Step 3 : Verify that the ligand set gives a strong field. The presence of four strong-field $$CN^{-}$$ ligands is more than sufficient to generate a large $$\Delta_{0}$$; the two $$NH_3$$ ligands (intermediate field) do not reduce the splitting below the pairing energy. Hence the configuration remains $$t_{2g}^{5}\,e_{g}^{0}$$ as required.

Therefore $$x + y = 2 + 4 = 6$$

Option C is correct.

Decacarbonyldimanganese(0) is Mn$$_2$$(CO)$$_{10}$$.

Each Mn is bonded to 5 CO ligands and one Mn-Mn bond, giving 6 groups around each Mn.

This corresponds to octahedral geometry.

The answer is Option (1): Octahedral.

We need to identify the correct statements about coordination compounds.

Statement A: "The strength of anionic ligands can be explained by crystal field theory."

Crystal Field Theory (CFT) explains the splitting of d-orbitals by ligands based on their electrostatic interactions. However, CFT cannot fully explain why certain anionic ligands (like $$CN^-$$) are strong field while others (like $$Cl^-$$) are weak field. The spectrochemical series ordering of anionic ligands requires Ligand Field Theory (which includes covalent interactions). So Statement A is not correct.

Statement B: "Valence bond theory does not give a quantitative interpretation of kinetic stability of coordination compounds."

This is correct. VBT deals with bonding and geometry but does not provide quantitative information about the kinetic stability (lability) of coordination compounds.

Statement C: "The hybridization involved in formation of $$[Ni(CN)_4]^{2-}$$ complex is $$dsp^2$$."

$$Ni^{2+}$$ has configuration $$[Ar]3d^8$$. $$CN^-$$ is a strong field ligand, so it forces pairing of the two unpaired electrons in $$3d$$. This frees one $$3d$$ orbital. The hybridization is $$dsp^2$$ (one $$3d$$, one $$4s$$, two $$4p$$ orbitals), giving a square planar geometry. Statement C is correct.

Statement D: "The number of possible isomer(s) of cis-$$[PtCl_2(en)_2]^{2+}$$ is one."

The complex $$[PtCl_2(en)_2]^{2+}$$ is octahedral. It can have cis and trans isomers. The cis isomer itself can exhibit optical isomerism (non-superimposable mirror images). So the cis form has 2 optical isomers (d and l forms). Therefore, the number of possible isomers of the cis form is not one. Statement D is not correct.

The correct statements are B and C.

The correct answer is Option D: B, C only.

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.

For an octahedral complex, whether it is diamagnetic (all electrons paired) or paramagnetic (one or more unpaired electrons) depends on

(i) the metal-ion oxidation state  →  its $$d^{\,n}$$ configuration, and
(ii) the ligand field strength  →  high-spin (weak field) or low-spin (strong field) arrangement.

Weak-field ligands: $$Cl^-,$$ $$F^-$$
Intermediate/strong ligands: $$NH_3,$$ en ($$H_2NCH_2CH_2NH_2$$). For $$Co^{3+},$$ en is strong enough to give a low-spin complex.

Ligands neutral, so oxidation state $$Mn^{3+}$$. Mn (Z = 25) gives $$d^4$$: $$[Ar]\,3d^4$$.
$$NH_3$$ is not strong enough to pair electrons in $$Mn^{3+}$$, hence high-spin $$t_{2g}^{3}e_g^{1}$$  →  4 unpaired electrons. Paramagnetic.

Again $$Mn^{3+} : d^4$$. $$Cl^-$$ is a weak ligand, so high-spin $$t_{2g}^{3}e_g^{1}$$ with 4 unpaired electrons. Paramagnetic.

$$Fe^{3+} : d^5$$. $$F^-$$ is weak, giving high-spin $$t_{2g}^{3}e_g^{2}$$ with 5 unpaired electrons. Paramagnetic.

$$Co^{3+} : d^6$$. $$F^-$$ weak  →  high-spin $$t_{2g}^{4}e_g^{2}$$ with 4 unpaired electrons. Paramagnetic.

$$Fe^{3+} : d^5$$. $$NH_3$$ is not strong enough to force pairing in $$Fe^{3+}$$, so high-spin $$t_{2g}^{3}e_g^{2}$$ with 5 unpaired electrons. Paramagnetic.

$$en$$ is a bidentate strong-field ligand. For $$Co^{3+} : d^6$$ the crystal-field splitting $$\Delta_o$$ is large enough to cause pairing  →  low-spin $$t_{2g}^{6}e_g^{0}$$. All electrons are paired. Diamagnetic.

Counting the diamagnetic species, only Case 6 qualifies.

Total number of diamagnetic complexes = 1.

First recall the spin-only magnetic moment formula for any transition-metal complex:

$$\mu_{\text{spin}} = \sqrt{n(n+2)}\; \text{B.M.}$$
where $$n$$ is the number of unpaired $$d$$-electrons.

We therefore need, for each complex ion,

(i) the oxidation state and hence the $$d$$-electron count of the metal,

(ii) whether the complex is high-spin or low-spin (decided by ligand field strength).

Case 1 : $$P = [\,FeF_6\,]^{3-}$$

Oxidation state of Fe: $$x + 6(-1) = -3 \;\Rightarrow\; x = +3$$ so $$Fe^{3+}$$.

Electronic configuration of $$Fe^{3+}$$: $$[Ar]\;3d^5$$ (five $$d$$-electrons).

Fluoride ($$F^-$$) is a weak-field ligand, so the octahedral complex is high-spin. Hence no pairing occurs and all five $$d$$-electrons remain unpaired.

Therefore $$n_P = 5$$ and $$\mu_P = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\;\text{B.M.}$$

Case 2 : $$Q = [\,V(H_2O)_6\,]^{2+}$$

Oxidation state of V: $$x + 6(0) = +2 \;\Rightarrow\; x = +2$$ so $$V^{2+}$$.

Electronic configuration of $$V^{2+}$$: $$[Ar]\;3d^3$$ (three $$d$$-electrons).

Water ($$H_2O$$) is also a weak-field ligand, giving a high-spin octahedral complex. All three electrons stay unpaired.

Thus $$n_Q = 3$$ and $$\mu_Q = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\;\text{B.M.}$$

Case 3 : $$R = [\,Fe(H_2O)_6\,]^{2+}$$

Oxidation state of Fe: $$x + 6(0) = +2 \;\Rightarrow\; x = +2$$ so $$Fe^{2+}$$.

Electronic configuration of $$Fe^{2+}$$: $$[Ar]\;3d^6$$ (six $$d$$-electrons).

Again, $$H_2O$$ is a weak-field ligand ⇒ high-spin. The high-spin octahedral $$d^6$$ configuration ($$t_{2g}^4e_g^2$$) contains four unpaired electrons.

Hence $$n_R = 4$$ and $$\mu_R = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\;\text{B.M.}$$

Collecting the magnetic moments:

$$\mu_Q \approx 3.87\; \text{B.M.} \lt \mu_R \approx 4.90\; \text{B.M.} \lt \mu_P \approx 5.92\; \text{B.M.}$$

Therefore the required ascending order is

$$Q \lt R \lt P$$

Which corresponds to Option C.

We evaluate two statements about coordination compounds.

Statement I: Dimethyl glyoxime (DMG) forms a six-membered covalent chelate when treated with $$NiCl_2$$ in the presence of $$NH_4OH$$.

Analysis: When DMG reacts with $$Ni^{2+}$$, it forms a bis(dimethylglyoximato)nickel(II) complex. Each DMG molecule coordinates through two nitrogen atoms to the nickel ion, forming a five-membered chelate ring:

$$Ni - N = C - C = N$$ (five atoms in the ring: Ni, N, C, C, N)

The statement says six-membered, which is incorrect. Statement I is FALSE.

Statement II: Prussian blue precipitate contains iron both in +2 and +3 oxidation states.

Analysis: Prussian blue has the formula $$Fe_4[Fe(CN)_6]_3$$ (or $$KFe[Fe(CN)_6]$$). It contains:

• $$Fe^{3+}$$ ions in the lattice positions
• $$Fe^{2+}$$ ions inside the hexacyanoferrate(II) complex $$[Fe(CN)_6]^{4-}$$

So yes, iron is present in both +2 and +3 oxidation states. Statement II is TRUE.

Since Statement I is false and Statement II is true, the correct answer is Option A: Statement I is false but Statement II is true.

Question: In which metal carbonyl does CO form a bridge between metal atoms?

Analyse each metal carbonyl.

$$[Co_2(CO)_8]$$ (Option 1): Dicobalt octacarbonyl exists in two isomeric forms. One isomer has two bridging CO groups connecting the two Co atoms (along with six terminal CO groups), and the other has a direct Co-Co bond with all terminal COs. The bridged form is well-known and commonly depicted in textbooks. This compound does have bridging CO.

$$[Mn_2(CO)_{10}]$$ (Option 2): This has a direct Mn-Mn bond with all ten CO groups in terminal positions. No bridging CO.

$$[Os_3(CO)_{12}]$$ (Option 3): This has a triangular Os₃ framework with all twelve CO groups in terminal positions (four per Os atom). No bridging CO.

$$[Ru_3(CO)_{12}]$$ (Option 4): Similar to the osmium analogue, this has all terminal CO groups. No bridging CO.

The correct answer is Option 1: $$[Co_2(CO)_8]$$.

{image}

1. $$[\text{Co}(\text{NH}_3)_6]^{3+}$$

• Oxidation State: Cobalt is in the +3 oxidation state and has  $$d^6$$ valence electron configuration.
• Ligand Nature: Ammonia acts as a strong field ligand with $$\text{Co}^{3+}$$.
• Spin Behavior: It provides a large crystal field splitting energy ($$\Delta_o$$), forcing all 6 d-electrons to pair up completely in the lower-energy $$t_{2g}$$ orbitals ($$t_{2g}^6 e_g^0$$).
• Classification: Since the electrons are paired up by the ligand field, it is known as a Spin paired complex (or low-spin / inner orbital complex).

2. $$[\text{CoF}_6]^{3-}$$

• Oxidation State: Cobalt is again in the +3 oxidation state ($$\text{Co}^{3+}$$, $$d^6$$).
• Ligand Nature: Fluoride ($$\text{F}^-$$) is a classic weak field ligand.
• Spin Behavior: The crystal field splitting energy ($$\Delta_o$$) is small, meaning it is energetically cheaper for electrons to remain unpaired and occupy higher energy $$e_g$$ orbitals according to Hund's rule ($$t_{2g}^4 e_g^2$$).
• Classification: Because the electrons remain free from forced pairing, it is known as a Spin free complex (or high-spin / outer orbital complex).

We need to identify the reagent that gives a brilliant red precipitate with nickel ions ($$Ni^{2+}$$) in basic medium.

Dimethylglyoxime (DMG) is a classic analytical reagent used for the qualitative and quantitative detection of nickel ions.

When dimethylglyoxime reacts with $$Ni^{2+}$$ ions in a basic (ammoniacal) medium, it forms a brilliant red (scarlet) precipitate known as nickel dimethylglyoximate:

$$Ni^{2+} + 2 \text{DMG} \xrightarrow{\text{NH}_4\text{OH}} \underset{\text{(brilliant red ppt)}}{\text{Ni(DMG)}_2}$$

The complex has a square planar geometry, and the red colour arises from the chelation of nickel with the two dimethylglyoxime molecules through the nitrogen atoms, with intramolecular hydrogen bonding between the hydroxyl groups.

Let us check the other options:

• Sodium nitroprusside is used to test for sulfide ions ($$S^{2-}$$), giving a violet colour.
• Neutral FeCl$$_3$$ is used to test for phenols and enols.
• Meta-dinitrobenzene is not a standard reagent for nickel detection.

Therefore, the correct answer is Option 4: Dimethyl glyoxime.

• (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.

The reaction mentioned in A) is reaction of Aniline with Nitrous acid at low temperature first forming Diazonium salt which then gets substituted by OH- ion from water. Resulting in Phenol-(II)

The Reaction mentioned in B) is a standard oxidation reaction specifically mentioned in NCERT. Phenol on reaction with sodium di-chromate in presence of sulphuric acid produces Benzoquinone-(IV)

The Reaction mentioned in C) is a named reaction i.e. Reimer-Tiemann reaction. The product is Salicylaldehyde-(I)

The Reaction mentioned in D) is a named reaction i.e. Kolbe's reaction. The product formed is Salicylic acid-(III)

We need to evaluate both statements about the Hinsberg test (reaction with p-toluenesulfonyl chloride, TsCl).

Statement (I): All the following compounds react with TsCl: $$C_6H_5NH_2$$ (aniline, primary amine), $$(C_6H_5)_2NH$$ (diphenylamine, secondary amine), $$(C_6H_5)_3N$$ (triphenylamine, tertiary amine).

In the Hinsberg test:

- Primary amines react with TsCl to form sulfonamides (N-monosubstituted).

- Secondary amines react with TsCl to form sulfonamides (N,N-disubstituted).

- Tertiary amines do not react with TsCl because they have no N-H bond to undergo substitution.

Since $$(C_6H_5)_3N$$ (a tertiary amine) does not react with TsCl, Statement I is false.

Statement (II): Their products in the above reaction are soluble in aqueous NaOH.

For primary amines, the product (N-monosubstituted sulfonamide) has an acidic N-H and is soluble in NaOH. For secondary amines, the product (N,N-disubstituted sulfonamide) has no acidic N-H and is insoluble in NaOH. Since not all products are soluble in NaOH, Statement II is also false.

The correct answer is Option 4: Both Statement I and Statement II are false.

Aniline reacts with bromine water to give 2,4,6-tribromoaniline (white precipitate).

$$ C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr $$

Moles of aniline.

Molar mass of aniline = 93 g/mol. Moles = $$\frac{9.3}{93} = 0.1$$ mol.

Theoretical yield of 2,4,6-tribromoaniline.

Molar mass of $$C_6H_2Br_3NH_2$$ = 72 + 4 + 14 + 240 = 330 g/mol.

Theoretical yield = $$0.1 \times 330 = 33$$ g.

Percentage yield.

$$ \text{\% yield} = \frac{26.4}{33} \times 100 = 80\% $$

The answer is 80.

We need to find the molar mass of amine X, prepared by ammonolysis of benzyl chloride, that gives a clear solution with the Hinsberg reagent (p-toluenesulphonyl chloride).

The Hinsberg test distinguishes between primary, secondary, and tertiary amines: (i) Primary amines react with p-toluenesulphonyl chloride to form a sulphonamide that is soluble in NaOH (acidic NH, dissolves in alkali). (ii) Secondary amines form a sulphonamide that is insoluble in NaOH (precipitate). (iii) Tertiary amines do not react — the solution remains clear.

Since the solution remains clear with p-toluenesulphonyl chloride, the amine X is a tertiary amine.

In ammonolysis of benzyl chloride (C$$_6$$H$$_5$$CH$$_2$$Cl) with excess benzyl chloride, the following sequence occurs: $$C_6H_5CH_2Cl + NH_3 \rightarrow C_6H_5CH_2NH_2$$ (primary), then $$C_6H_5CH_2Cl + C_6H_5CH_2NH_2 \rightarrow (C_6H_5CH_2)_2NH$$ (secondary), and finally $$C_6H_5CH_2Cl + (C_6H_5CH_2)_2NH \rightarrow (C_6H_5CH_2)_3N$$ (tertiary).

Therefore, the tertiary amine is tribenzylamine: $$(C_6H_5CH_2)_3N$$.

The molecular formula of tribenzylamine is $$C_{21}H_{21}N$$, and substituting into the molar mass calculation gives $$M = 21(12) + 21(1) + 14 = 252 + 21 + 14 = 287 \, \text{g/mol}$$.

Option A: 287 g mol$$^{-1}$$.

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 $$\boxed{28}$$.

Find the mass of aniline yellow formed from 279 g of aniline with one equivalent of benzenediazonium chloride.

Since aniline reacts with benzenediazonium chloride to give p-aminoazobenzene (aniline yellow) and HCl, the reaction can be written as $$C_6H_5NH_2 + C_6H_5N_2^+Cl^- \rightarrow C_6H_5N=NC_6H_4NH_2 + HCl.$$

The molar mass of aniline ($$C_6H_5NH_2$$) is 93 g/mol, so the number of moles in 279 g is $$\mathrm{Moles} = \frac{279}{93} = 3\ \text{mol}.$$

One equivalent of benzenediazonium chloride reacts with each mole of aniline in a 1:1 ratio, so 3 mol of aniline yields 3 mol of the azo product.

The molar mass of aniline yellow ($$C_{12}H_{11}N_3$$) is $$12\times12 + 11\times1 + 3\times14 = 144 + 11 + 42 = 197\ \text{g/mol},$$ and hence the mass of product is $$3 \times 197 = 591\ \text{g}.$$

The correct answer is 591.

We have 9.3 g of pure aniline which undergoes diazotisation followed by coupling with phenol to give an orange dye.

In the diazotisation step aniline reacts with nitrous acid (generated in situ from $$NaNO_2 + HCl$$) at 0-5 °C to form benzenediazonium chloride:

$$C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$$

In the coupling step the diazonium salt couples with phenol in a weakly basic medium to produce p-hydroxyazobenzene (an orange azo dye):

$$C_6H_5N_2^+Cl^- + C_6H_5OH \rightarrow C_6H_5-N=N-C_6H_4-OH + HCl$$

To calculate the moles of aniline, we use its molar mass: $$6(12) + 5(1) + 14 + 2(1) = 93\text{ g/mol}$$, which gives $$\text{Moles of aniline} = \frac{9.3}{93} = 0.1\text{ mol}$$.

Since the stoichiometry is 1:1, one mole of aniline yields one mole of dye. The molar mass of p-hydroxyazobenzene ($$C_{12}H_{10}N_2O$$) is $$12(12) + 10(1) + 2(14) + 16 = 198\text{ g/mol}$$. Thus, the mass of dye produced is $$0.1 \times 198 = 19.8 \approx 20\text{ g}$$. Therefore, 20 g of the orange dye is obtained (to the nearest integer).

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.