JEE d and f-Block Elements Questions

We need to identify three metal chlorides based on their solubility properties.

x: Water soluble at room temperature

y: Sparingly soluble in water at room temperature

z: Soluble in hot water

Option 2: $$CuCl_2, AgCl, PbCl_2$$

- $$CuCl_2$$: Copper(II) chloride is readily soluble in water at room temperature. ✓ (matches x)

- $$AgCl$$: Silver chloride is sparingly soluble (nearly insoluble) in water at room temperature. ✓ (matches y)

- $$PbCl_2$$: Lead(II) chloride is sparingly soluble in cold water but dissolves readily in hot water. ✓ (matches z)

This perfectly matches all three conditions.

Checking other options:

- Option 1: $$AlCl_3$$ is soluble, $$AgCl$$ is sparingly soluble, but $$AlCl_3$$ doesn't specifically need hot water to dissolve.

- Option 3: $$BaCl_2$$ is freely soluble in water at room temperature, not specifically requiring hot water.

- Option 4: The notation $$AlCl$$ is unusual, and $$Hg_2Cl_2$$ is insoluble even in hot water.

Therefore, the correct answer is Option 2: $$CuCl_2, AgCl, PbCl_2$$.

We need to identify the oxoanion X of the lightest element of group 7, with metal in +6 oxidation state.

First, the lightest element of group 7 (Group VIIB / Mn group) is Manganese (Mn).

Next, Mn in +6 oxidation state forms the manganate ion: $$MnO_4^{2-}$$. Check: Mn + 4(-2) = -2, so Mn = +6. ✓

Then, potassium manganate $$K_2MnO_4$$ has a green color (this is different from $$KMnO_4$$ which is purple, and $$K_2Cr_2O_7$$ which is orange).

Therefore, the color is Option 4: green.

For any ion, magnetic behaviour depends on the presence or absence of unpaired electrons.
No unpaired electrons ⇒ diamagnetic, at least one unpaired electron ⇒ paramagnetic.

Lanthanide ground-state configurations are based on the xenon core $$[Xe]$$ followed by $$4f$$, $$5d$$ and $$6s$$ subshells. While writing ionic configurations, electrons are removed first from $$6s$$, then $$5d$$, then $$4f$$.

Case A (La$$^{3+}$$, Ce$$^{4+}$$)

Lanthanum (Z = 57): neutral atom $$[Xe]\,5d^{1}6s^{2}$$.
La$$^{3+}$$ : remove three electrons (two from $$6s$$, one from $$5d$$) ⇒ $$[Xe]\,4f^{0}$$.
All subshells are completely filled; no unpaired electrons ⇒ diamagnetic.

Cerium (Z = 58): neutral atom $$[Xe]\,4f^{1}5d^{1}6s^{2}$$ (the commonly quoted ground state).
Ce$$^{4+}$$ : remove four electrons (two $$6s$$, one $$5d$$, one $$4f$$) ⇒ $$[Xe]\,4f^{0}$$.
Again, no unpaired electrons ⇒ diamagnetic.

Case B (Yb$$^{2+}$$, Lu$$^{3+}$$)

Ytterbium (Z = 70): neutral atom $$[Xe]\,4f^{14}6s^{2}$$ (filled $$4f$$ shell).
Yb$$^{2+}$$ : remove the two $$6s$$ electrons ⇒ $$[Xe]\,4f^{14}$$.
The $$4f$$ subshell is completely filled; no unpaired electrons ⇒ diamagnetic.

Lutetium (Z = 71): neutral atom $$[Xe]\,4f^{14}5d^{1}6s^{2}$$.
Lu$$^{3+}$$ : remove two $$6s$$ and one $$5d$$ electron ⇒ $$[Xe]\,4f^{14}$$.
Filled $$4f^{14}$$ subshell; no unpaired electrons ⇒ diamagnetic.

Case C (La$$^{2+}$$, Ce$$^{3+}$$)

La$$^{2+}$$ : $$[Xe]\,5d^{1}$$ - one electron in $$5d$$ ⇒ unpaired ⇒ paramagnetic.
Ce$$^{3+}$$ : $$[Xe]\,4f^{1}$$ - one electron in $$4f$$ ⇒ unpaired ⇒ paramagnetic.
Hence the pair is not diamagnetic.

Case D (Yb$$^{3+}$$, Lu$$^{2+}$$)

Yb$$^{3+}$$ : $$[Xe]\,4f^{13}$$ - one vacancy (equivalently one unpaired electron) in $$4f$$ ⇒ paramagnetic.
Lu$$^{2+}$$ : $$[Xe]\,4f^{14}5d^{1}$$ - one electron in $$5d$$ ⇒ unpaired ⇒ paramagnetic.
Hence this pair is not diamagnetic.

Therefore, the diamagnetic pairs are:
Option A (La$$^{3+}$$, Ce$$^{4+}$$) and Option B (Yb$$^{2+}$$, Lu$$^{3+}$$).

Option A ( La$$^{3+}$$, Ce$$^{4+}$$ ), Option B ( Yb$$^{2+}$$, Lu$$^{3+}$$ )

List - I contains four familiar salts of the d-block elements and List - II lists the characteristic colours shown by those salts.

Case A (Compound : $$\text{KMnO}_4$$)
The manganate(VII) ion $$\text{MnO}_4^-$$ has an $$n \rightarrow \pi^{*}$$ charge-transfer transition whose absorption lies in the green region of the spectrum; the transmitted (complementary) colour is therefore purple. Hence,
$$\text{KMnO}_4$$  →  Purple colour  (Entry III).

Case B (Compound : $$\text{K}_2\text{Cr}_2\text{O}_7$$)
The dichromate ion $$\text{Cr}_2\text{O}_7^{2-}$$ shows ligand-to-metal charge-transfer bands that make the solution absorb mainly in the violet-blue region; the observed colour is orange. Hence,
$$\text{K}_2\text{Cr}_2\text{O}_7$$  →  Orange colour  (Entry I).

Case C (Compound : $$\text{FeSO}_4\cdot7\text{H}_2\text{O}$$)
In ferrous sulphate the ion present is the high-spin octahedral $$\text{Fe}^{2+}$$ (d6). A weak d-d transition in the red part of the spectrum makes the crystalline salt appear light (pale) green. Hence,
$$\text{FeSO}_4\cdot7\text{H}_2\text{O}$$  →  Light green colour  (Entry IV).

Case D (Compound : $$\text{CuSO}_4\cdot5\text{H}_2\text{O}$$)
Aqueous copper(II) ion, present as $$[\text{Cu(H}_2\text{O})_6]^{2+}$$, undergoes an octahedral d-d transition that absorbs in the orange-red region, imparting an intense blue colour to the salt. Hence,
$$\text{CuSO}_4\cdot5\text{H}_2\text{O}$$  →  Blue colour  (Entry II).

Therefore the required matching is:
(A) − (III),  (B) − (I),  (C) − (IV),  (D) − (II)

Option A which is: (A) − (III), (B) − (I), (C) − (IV), (D) − (II)

To determine which metal ions have a spin-only magnetic moment of $$4.9\text{ B.M.}$$, we use the spin-only magnetic moment formula

$$\mu = \sqrt{n(n+2)}\text{ B.M.}$$

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

For

$$\mu = 4.9\text{ B.M.},$$

the corresponding value of $$n$$ is

$$n = 4.$$

Hence, we need to identify the ions containing 4 unpaired electrons.

The electronic configurations and corresponding numbers of unpaired electrons are summarized below:

For $$Cr^{2+}$$ with configuration $$3d^4$$, there are four unpaired electrons. Therefore,

$$\mu = \sqrt{4(4+2)}=\sqrt{24}\approx 4.9\text{ B.M.}$$

For $$Fe^{2+}$$ with configuration $$3d^6$$, the electrons are distributed as

$$\uparrow\downarrow,\ \uparrow,\ \uparrow,\ \uparrow,\ \uparrow,$$

giving four unpaired electrons. Hence,

$$\mu = \sqrt{24}\approx 4.9\text{ B.M.}$$

For $$Fe^{3+}$$ with configuration $$3d^5$$, there are five unpaired electrons, so

$$\mu = \sqrt{5(5+2)}=\sqrt{35}\approx 5.9\text{ B.M.}$$

For $$Co^{2+}$$ with configuration $$3d^7$$, there are three unpaired electrons, giving

$$\mu = \sqrt{3(3+2)}=\sqrt{15}\approx 3.87\text{ B.M.}$$

For $$Mn^{3+}$$ with configuration $$3d^4$$, there are four unpaired electrons. Therefore,

$$\mu = \sqrt{4(4+2)}=\sqrt{24}\approx 4.9\text{ B.M.}$$

Hence, the ions exhibiting a spin-only magnetic moment of approximately $$4.9\text{ B.M.}$$ are

$$\boxed{Cr^{2+},\ Fe^{2+},\ \text{and}\ Mn^{3+}}.$$

Therefore, the correct answer is A, B and E only.

We need to identify lanthanoid ions with $$4f^7$$ configuration because the lanthanoids fill the 4f orbitals, so we examine which ions have exactly 7 electrons in the 4f subshell.

• (A) Eu²⁺: Eu (Z=63) has configuration [Xe] 4f⁷ 6s². Eu²⁺ loses two 6s electrons: [Xe] 4f⁷. This has 4f⁷ configuration.
• (B) Gd³⁺: Gd (Z=64) has configuration [Xe] 4f⁷ 5d¹ 6s². Gd³⁺ loses 6s² and 5d¹: [Xe] 4f⁷. This has 4f⁷ configuration.
• (C) Eu³⁺: Eu (Z=63) has [Xe] 4f⁷ 6s². Eu³⁺ loses 6s² and one 4f: [Xe] 4f⁶. This does NOT have 4f⁷.
• (D) Tb³⁺: Tb (Z=65) has [Xe] 4f⁹ 6s². Tb³⁺ loses 6s² and one 4f: [Xe] 4f⁸. This does NOT have 4f⁷.
• (E) Sm³⁺: Sm (Z=62) has [Xe] 4f⁶ 6s². Sm³⁺ loses 6s² and one 4f: [Xe] 4f⁵. This does NOT have 4f⁷.

The ions with 4f⁷ configuration are (A) Eu²⁺ and (B) Gd³⁺, so the correct answer is Option 3: (A) and (B) only.

We need to identify the amphoteric oxide between $$V_2O_3$$ and $$V_2O_5$$, and find the oxidation state of V in the oxide anion formed with alkali.

In vanadium oxides, the nature changes with oxidation state: $$V_2O_3$$ (V in +3 state) is a basic oxide because lower oxidation state oxides of transition metals tend to be basic, whereas $$V_2O_5$$ (V in +5 state) is an amphoteric oxide that can react with both acids and bases.

When $$V_2O_5$$ reacts with NaOH (alkali), it forms the metavanadate ion according to $$V_2O_5 + 2NaOH \rightarrow 2NaVO_3 + H_2O$$. The oxide anion formed is $$VO_3^-$$.

To calculate the oxidation state of vanadium in $$VO_3^-$$, let the oxidation state of V be $$x$$; each oxygen contributes $$-2$$ and the overall charge is $$-1$$, so $$x + 3(-2) = -1$$, which simplifies to $$x - 6 = -1$$ and thus $$x = +5$$.

The correct answer is Option D: +5.

We need to identify the strongest oxidising agent among the given lanthanide/actinide ions.

Key Concept: An oxidising agent gains electrons (gets reduced). The stronger the tendency to gain electrons, the stronger the oxidising agent. For lanthanide ions, unusual oxidation states are strongly driven towards the most stable +3 state, and ions that strongly want to reach a stable electronic configuration make the best oxidising agents.

Analysing each ion:

Option 1: $$Eu^{2+}$$ (Z = 63)

$$Eu^{2+}$$: [Xe] 4f$$^7$$. This has a half-filled 4f shell, which is very stable. $$Eu^{2+}$$ is actually relatively stable and is a reducing agent (it wants to lose an electron to form $$Eu^{3+}$$ [Xe]4f$$^6$$, but the half-filled stability resists this). It is not a strong oxidising agent.

Option 2: $$Tb^{4+}$$ (Z = 65)

$$Tb^{4+}$$: [Xe] 4f$$^7$$. Terbium in the +4 state also has a half-filled 4f$$^7$$ configuration. However, $$Tb^{4+}$$ strongly tends to gain an electron to become $$Tb^{3+}$$ [Xe]4f$$^8$$, because the +3 state is the most common and stable oxidation state for lanthanides. The high charge (+4) combined with the strong drive to reach +3 makes $$Tb^{4+}$$ a very strong oxidising agent.

Option 3: $$Lu^{3+}$$ (Z = 71)

$$Lu^{3+}$$: [Xe] 4f$$^{14}$$. This has a completely filled 4f shell and is already in the most stable +3 oxidation state. It has no tendency to gain or lose electrons easily. It is not an oxidising agent.

Option 4: $$Ce^{3+}$$ (Z = 58)

$$Ce^{3+}$$: [Xe] 4f$$^1$$. Cerium in the +3 state has one 4f electron. While $$Ce^{4+}$$ (4f$$^0$$, noble gas configuration) is a good oxidising agent, $$Ce^{3+}$$ itself is stable and not an oxidising agent.

Conclusion: $$Tb^{4+}$$ is the strongest oxidising agent because it has the strongest tendency to gain an electron to achieve the stable +3 state.

The correct answer is Option 2: $$Tb^{4+}$$.

Bronze is an alloy of $$Cu$$ and $$Sn$$, therefore List-I item (A) corresponds to List-II entry (IV).

Brass is an alloy of $$Cu$$ and $$Zn$$, therefore List-I item (B) corresponds to List-II entry (III).

The present-day United Kingdom “silver” coin actually contains no silver; it is a cupronickel alloy of $$Cu$$ and $$Ni$$, so List-I item (C) matches List-II entry (I).

Stainless steel typically contains $$Fe, Cr, Ni$$ and a small amount of $$C$$, hence List-I item (D) matches List-II entry (II).

Thus the correct matching is:
(A) → (IV), (B) → (III), (C) → (I), (D) → (II).

Option B which is: (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

The chromite ore is given as $$FeCr_{2}O_{4}$$. When fused with $$Na_{2}CO_{3}$$ in the presence of oxygen $$(O_{2})$$, the reaction occurs as follows:

The balanced chemical equation is:

$$4FeCr_{2}O_{4} + 8Na_{2}CO_{3} + 7O_{2} \rightarrow 8Na_{2}CrO_{4} + 2Fe_{2}O_{3} + 8CO_{2}$$

The products are sodium chromate $$(Na_{2}CrO_{4})$$, ferric oxide $$(Fe_{2}O_{3})$$, and carbon dioxide $$(CO_{2})$$.

Among these:

• Sodium chromate $$(Na_{2}CrO_{4})$$ is soluble in water.
• Carbon dioxide $$(CO_{2})$$ is a gas and soluble in water to some extent, but it is not a solid insoluble product.
• Ferric oxide $$(Fe_{2}O_{3})$$ is insoluble in water and remains as a solid.

Thus, the water-insoluble product is ferric oxide $$(Fe_{2}O_{3})$$.

To find the molar mass of $$Fe_{2}O_{3}$$:

Atomic mass of iron $$(Fe) = 56$$ g/mol.

Atomic mass of oxygen $$(O) = 16$$ g/mol.

Molar mass of $$Fe_{2}O_{3} = (2 \times 56) + (3 \times 16) = 112 + 48 = 160$$ g/mol.

Therefore, the molar mass of the water-insoluble product is 160 g mol⁻¹.

The enthalpy of atomisation ($$ΔH_\text{atom}$$) of the 3d transition elements first rises from Sc to about Ni and then falls towards Cu and Zn. This behaviour is linked to the strength of metallic bonding, which in turn depends mainly on the number of 3d electrons available for delocalised metallic bonding.

Approximate experimental values (kJ mol−1) illustrate the trend: Sc ≈ 330, Mn ≈ 280, Co ≈ 425, Cu ≈ 335. Hence, Co exhibits the largest enthalpy of atomisation among Sc, Mn, Co and Cu.

For the magnetic moment we consider Co in its +2 oxidation state.

Atomic number of Co = 27, so neutral Co is $$[Ar]\,3d^7\,4s^2$$.

Co2+ is formed by losing the two 4s electrons: $$[Ar]\,3d^7$$.

In an aqueous or other weak-field environment the 3d orbitals remain high-spin. A 3d7 high-spin configuration has the distribution $$t_{2g}^5\,e_g^2$$, giving

Number of unpaired electrons, $$n = 3$$.

The spin-only formula is $$\mu_\text{so} = \sqrt{n(n+2)}\; \text{BM}$$.

Substituting $$n = 3$$:

$$\mu_\text{so} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM} \approx 4 \text{ BM (nearest integer)}.$$

Therefore, the element is Co and its spin-only magnetic moment in the +2 state is 4 BM.

First write the usual laboratory method for making chromyl chloride.
When a dichromate such as $$K_2Cr_2O_7$$ is heated with solid $$NaCl$$ and concentrated $$H_2SO_4$$, deep-red vapours of chromyl chloride are obtained.

Balanced form: $$K_2Cr_2O_7 + 4\,NaCl + 3\,H_2SO_4 \rightarrow 2\,CrO_2Cl_2 + 2\,KHSO_4 + 2\,NaHSO_4 + 3\,H_2O$$

Hence compound $$A$$ is $$K_2Cr_2O_7$$, and the first product of interest is $$CrO_2Cl_2$$ (chromyl chloride).

Next, chromyl chloride undergoes alkaline hydrolysis:
$$CrO_2Cl_2 + 4\,NaOH \rightarrow Na_2CrO_4 + 2\,NaCl + 2\,H_2O$$

Thus compound $$B$$ is $$Na_2CrO_4$$, which contains the chromate ion $$CrO_4^{2-}$$.

On acidifying chromate we get dichromate:
$$2\,CrO_4^{2-} + 2\,H^+ \rightarrow Cr_2O_7^{2-} + H_2O$$

Therefore compound $$C$$ is the dichromate ion $$Cr_2O_7^{2-}$$.

Structure of $$Cr_2O_7^{2-}$$: it consists of two $$CrO_4$$ tetrahedra sharing one oxygen atom (bridging O).
Total O atoms = 7.
Bridging O atoms = 1.
Terminal O atoms = $$7 - 1 = 6$$.

Hence the number of terminal oxygen atoms present in compound $$C$$ is $$6$$.

Electronic Configurations

• Niobium (Nb, $$Z=41$$): Instead of the expected $$[Kr] 4d^3 5s^2$$, the experimental configuration is $$[Kr] 4d^4 5s^1$$
  • Number of electrons in the $$4d$$ orbital ($$x$$) = $$4$$
• Ruthenium (Ru, $$Z=44$$): Instead of the expected $$[Kr] 4d^6 5s^2$$, the experimental configuration is $$[Kr] 4d^7 5s^1$$.
  • Number of electrons in the $$4d$$ orbital ($$y$$) = $$7$$

Calculation

Using the values identified above: $$x + y = 4 + 7 = 11$$

Final Answer: The value of $$x + y$$ is $$11$$.

This exception occurs because the energy gap between the $$4d$$ and $$5s$$ orbitals is very small, allowing for electron shifts that minimize inter-electronic repulsions or provide stability through specific exchange energies.

We need to find the spin-only magnetic moment of the compound with the strongest oxidising power among $$Mn_2O_3$$, TiO, and VO.

In $$Mn_2O_3$$, Mn is in the +3 oxidation state as shown by $$2x + 3(-2) = 0 \Rightarrow x = +3$$, while in TiO and VO, Ti and V are both in the +2 oxidation state.

$$Mn^{3+}$$ in $$Mn_2O_3$$ is the strongest oxidising agent because it has the highest reduction potential among the three. It readily accepts an electron to become the stable $$Mn^{2+}$$ (d$$^5$$, half-filled), whereas $$Ti^{2+}$$ and $$V^{2+}$$ act as better reducing agents.

The electronic configuration of Mn is [Ar] 3d$$^5$$ 4s$$^2$$, so for $$Mn^{3+}$$ it becomes [Ar] 3d$$^4$$.

Because the oxide ion ($$O^{2-}$$) is a weak-field ligand, $$Mn^{3+}$$ in $$Mn_2O_3$$ remains high-spin. A high-spin d$$^4$$ configuration has all four electrons unpaired by Hund’s rule, giving $$n = 4$$ unpaired electrons.

The spin-only magnetic moment is given by the formula

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

where $$n$$ is the number of unpaired electrons. Substituting $$n = 4$$ yields

$$ \mu = \sqrt{4(4+2)} = \sqrt{24} = 2\sqrt{6} \approx 4.90 \text{ BM} $$

Nearest integer: 5 BM.

The answer is 5.

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

Counting unpaired electrons for each configuration:

- $$[Ar] \; 3d^7$$: $$\uparrow\downarrow \; \uparrow\downarrow \; \uparrow \; \uparrow \; \uparrow$$ = 3 unpaired

- $$[Ar] \; 3d^8$$: $$\uparrow\downarrow \; \uparrow\downarrow \; \uparrow\downarrow \; \uparrow \; \uparrow$$ = 2 unpaired

- $$[Ar] \; 3d^3$$: $$\uparrow \; \uparrow \; \uparrow \; \_ \; \_$$ = 3 unpaired

- $$[Ar] \; 3d^6$$: $$\uparrow\downarrow \; \uparrow \; \uparrow \; \uparrow \; \uparrow$$ = 4 unpaired

$$[Ar] \; 3d^6$$ has the maximum number of unpaired electrons (4), giving the highest magnetic moment:

$$\mu = \sqrt{4(4+2)} = \sqrt{24} = 2\sqrt{6}$$ BM

The answer corresponds to Option (4).

First note the atomic numbers: $$Sc = 21$$, $$Ti = 22$$, $$V = 23$$, $$Cr = 24$$, $$Mn = 25$$.

Use the Aufbau order $$\text{4s} \rightarrow \text{3d}$$ to write the ground-state electronic configurations:

$$Sc\;(21):\;[Ar]\,3d^{1}\,4s^{2}$$
$$Ti\;(22):\;[Ar]\,3d^{2}\,4s^{2}$$
$$V\;(23):\;[Ar]\,3d^{3}\,4s^{2}$$
$$Cr\;(24):\;[Ar]\,3d^{5}\,4s^{1}$$  (half-filled 3d subshell gives extra stability)
$$Mn\;(25):\;[Ar]\,3d^{5}\,4s^{2}$$

Count the unpaired electrons in each case (Hund’s rule: every degenerate orbital gets one electron before any pairing starts):

Sc : one unpaired electron in $$3d$$ ⇒ $$1$$ unpaired
Ti  : two unpaired electrons in $$3d$$ ⇒ $$2$$ unpaired
V  : three unpaired electrons in $$3d$$ ⇒ $$3$$ unpaired
Cr : five unpaired in $$3d$$ + one in $$4s$$ ⇒ $$6$$ unpaired
Mn : five unpaired in $$3d$$ (the $$4s^{2}$$ pair is paired) ⇒ $$5$$ unpaired

Arrange in increasing order of the number of unpaired electrons:

$$1 \lt 2 \lt 3 \lt 5 \lt 6$$ corresponds to $$Sc \lt Ti \lt V \lt Mn \lt Cr$$.

Using the given symbols: $$(A) \lt (D) \lt (C) \lt (E) \lt (B)$$.

Therefore the correct option is Option C.

We need to identify which species acts as a strong reducing agent among the given lanthanide ions.

Analysing each option:

Lu$$^{3+}$$ (Z = 71): Lu has the configuration $$[Xe]\, 4f^{14}\, 5d^1\, 6s^2$$. Lu$$^{3+}$$ has a completely filled 4f shell ($$4f^{14}$$), which is very stable. It has no tendency to be further oxidized or reduced. It is not a reducing agent.

Gd$$^{3+}$$ (Z = 64): Gd has the configuration $$[Xe]\, 4f^7\, 5d^1\, 6s^2$$. Gd$$^{3+}$$ has a half-filled 4f shell ($$4f^7$$), which is very stable. It has no tendency to lose or gain electrons easily. It is not a strong reducing agent.

Eu$$^{2+}$$ (Z = 63): Eu has the configuration $$[Xe]\, 4f^7\, 6s^2$$. Eu$$^{2+}$$ has the configuration $$[Xe]\, 4f^7$$, which is a half-filled 4f shell. However, the common and stable oxidation state of lanthanides is $$+3$$. Eu$$^{2+}$$ readily loses one more electron to become Eu$$^{3+}$$ ($$4f^6$$), acting as a strong reducing agent.

Ce$$^{4+}$$ (Z = 58): Ce has the configuration $$[Xe]\, 4f^1\, 5d^1\, 6s^2$$. Ce$$^{4+}$$ has the noble gas configuration $$[Xe]$$, which is stable. Ce$$^{4+}$$ tends to gain an electron to become Ce$$^{3+}$$, making it an oxidizing agent, not a reducing agent.

Eu$$^{2+}$$ is a strong reducing agent because it is readily oxidized to the more stable Eu$$^{3+}$$ state.

The correct answer is Option 3: Eu$$^{2+}$$.

We need to find the option where all elements have a $$d^{10}$$ electronic configuration (in their ground state). First, recall that an element exhibits a $$d^{10}$$ configuration when its d-subshell is completely filled, which typically occurs near the end of a transition series because the fully filled d-subshell provides exceptional stability.

To identify such elements, we write the electronic configurations of the listed species and inspect their d-subshell. For $$_{24}Cr$$ the configuration is [Ar] 3d$$^5$$ 4s$$^1$$, which corresponds to $$d^5$$ rather than $$d^{10}$$. Similarly, $$_{26}Fe$$ is [Ar] 3d$$^6$$ 4s$$^2$$ (giving $$d^6$$), $$_{27}Co$$ is [Ar] 3d$$^7$$ 4s$$^2$$ (giving $$d^7$$), and $$_{28}Ni$$ is [Ar] 3d$$^8$$ 4s$$^2$$ (giving $$d^8$$). In contrast, $$_{29}Cu$$ has [Ar] 3d$$^{10}$$ 4s$$^1$$, $$_{30}Zn$$ has [Ar] 3d$$^{10}$$ 4s$$^2$$, $$_{46}Pd$$ has [Kr] 4d$$^{10}$$, $$_{47}Ag$$ has [Kr] 4d$$^{10}$$ 5s$$^1$$, and $$_{48}Cd$$ has [Kr] 4d$$^{10}$$ 5s$$^2$$, each of which possesses the fully filled d-subshell.

Next, we examine the given options to see which one consists exclusively of elements with a $$d^{10}$$ configuration. Option (1) contains Co, Ni, Fe, and Cr, none of which achieve $$d^{10}$$, so it is excluded. Option (2) lists Cu, Zn, Cd, and Ag, all of which have a complete $$d^{10}$$ subshell. Options (3) and (4) include only one such element each. The correct answer is Option (2): $$_{29}Cu$$, $$_{30}Zn$$, $$_{48}Cd$$, $$_{47}Ag$$.

We need to identify which pair of lanthanoid ions are diamagnetic. A species is diamagnetic if it has no unpaired electrons, i.e., all electrons are paired. For lanthanoid ions, this means the 4f subshell must be either completely empty ($$4f^0$$) or completely filled ($$4f^{14}$$).

Lanthanoids have the general configuration $$[Xe] 4f^n$$. When they form $$3+$$ or $$4+$$ ions, electrons are removed first from the 6s and 5d shells, and then from the 4f subshell.

For $$La^{3+}$$, lanthanum (Z = 57) has the configuration $$[Xe] 5d^1 6s^2$$. Removing three electrons yields $$[Xe] = 4f^0$$, which has no unpaired electrons and is therefore diamagnetic.

Cerium (Z = 58) in $$Ce^{4+}$$ has the configuration $$[Xe] 4f^1 5d^1 6s^2$$. Removing four electrons gives $$[Xe] = 4f^0$$, again with no unpaired electrons, so this ion is also diamagnetic.

Neodymium (Z = 60) in $$Nd^{3+}$$ starts from $$[Xe] 4f^4 6s^2$$. After removing three electrons, the configuration becomes $$[Xe] 4f^3$$, which has three unpaired electrons and is paramagnetic. Europium (Z = 63) in $$Eu^{3+}$$ has $$[Xe] 4f^7 6s^2$$ to start, and removing three electrons gives $$[Xe] 4f^6$$. By Hund’s rule, the six electrons occupy six of the seven 4f orbitals singly, resulting in six unpaired electrons and a paramagnetic species.

Lutetium (Z = 71) in $$Lu^{3+}$$ is described by $$[Xe] 4f^{14} 5d^1 6s^2$$ in the neutral atom. Removing three electrons leaves $$[Xe] 4f^{14}$$, which is completely filled and therefore diamagnetic.

Only $$La^{3+}$$, $$Ce^{4+}$$, and $$Lu^{3+}$$ are diamagnetic, but the pair consisting of $$La^{3+}$$ and $$Ce^{4+}$$ (both $$4f^0$$) matches one of the given options.

Answer: Option 2 — $$La^{3+}$$ and $$Ce^{4+}$$

We need to evaluate each statement about Zn, Cd, and Hg (Group 12 elements with completely filled d-orbitals).

Statement A: They exhibit high enthalpy of atomization as the d-subshell is full.

This is incorrect. Zn, Cd, and Hg have low enthalpies of atomization. In these elements, the d-subshell is completely filled ($$d^{10}$$), so the d-electrons do not participate in metallic bonding. The metallic bonding is weaker, resulting in low atomization enthalpies.

Statement B: Zn and Cd do not show variable oxidation state while Hg shows +I and +II.

This is correct. Zn and Cd predominantly show only the +2 oxidation state. Mercury, however, shows both +1 (as $$Hg_2^{2+}$$) and +2 oxidation states.

Statement C: Compounds of Zn, Cd and Hg are paramagnetic in nature.

This is incorrect. Zn$$^{2+}$$, Cd$$^{2+}$$, and Hg$$^{2+}$$ all have $$d^{10}$$ configuration with no unpaired electrons. Their compounds are diamagnetic, not paramagnetic.

Statement D: Zn, Cd and Hg are called soft metals.

This is correct. Due to weak metallic bonding (d-electrons do not participate in bonding), these metals are soft with low melting and boiling points.

The correct statements are B and D.

The correct answer is Option 1: B, D only.

Statement I: Formation of $$Ce^{4+}$$ is favoured by its noble gas configuration. $$Ce$$ has atomic number 58: $$[Xe]4f^15d^16s^2$$. Losing 4 electrons gives $$Ce^{4+}$$: $$[Xe]$$ — noble gas configuration. True.

Statement II: $$Ce^{4+}$$ is a strong oxidant reverting to the common +3 state. True — the +3 state is more stable for lanthanides, so $$Ce^{4+}$$ readily accepts an electron to become $$Ce^{3+}$$.

Both statements are true. The answer corresponds to Option (2).

We need to evaluate the truth of two statements about transition elements.

Statement I: "The higher oxidation states are more stable down the group among transition elements unlike p-block elements."

This statement is true. Here is why:

In transition elements, as we go down a group, the higher oxidation states become increasingly stable. This is because:

- The ionisation enthalpies of 4d and 5d elements are comparable (and sometimes lower than expected) due to poor shielding by intervening electron shells.

- The greater spatial extension of 4d and 5d orbitals compared to 3d orbitals allows better orbital overlap, which stabilises higher oxidation states through stronger bonding.

- For example, in Group 6: Cr commonly shows +3 and +6, but Mo and W preferentially show +6. In Group 8: Fe shows +2 and +3, while Os readily shows +8 (as in $$OsO_4$$).

This is opposite to p-block elements, where the lower oxidation states become more stable down the group due to the inert pair effect.

Statement II: "Copper cannot liberate hydrogen from weak acids."

This statement is true. Here is why:

The ability of a metal to liberate hydrogen from acids depends on its standard electrode potential. A metal can liberate $$H_2$$ only if its standard reduction potential is less than that of the $$H^+/H_2$$ couple ($$E° = 0.00$$ V).

Copper has a positive standard reduction potential: $$E°(Cu^{2+}/Cu) = +0.34$$ V.

Since $$E°_{Cu} > E°_{H^+/H_2}$$, copper is a less reactive (more noble) metal than hydrogen. Therefore, copper cannot reduce $$H^+$$ ions to $$H_2$$ gas, even from dilute strong acids, let alone from weak acids (which have even lower $$H^+$$ concentration). The reaction is thermodynamically unfavourable (positive $$\Delta G$$).

Since both statements are true, the correct answer is Option 1: Both Statement I and Statement II are true.

In $$K_2Cr_2O_7$$: Cr is in +6 oxidation state with electronic configuration $$d^0$$. Since there are no d-electrons, d-d transitions are impossible. The orange color arises from charge transfer transitions (ligand-to-metal charge transfer, LMCT), where electrons from oxygen ligands are transferred to empty d-orbitals of Cr$$^{6+}$$.

In $$KMnO_4$$: Mn is in +7 oxidation state with electronic configuration $$d^0$$. Again, no d-electrons means no d-d transitions. The intense purple color is due to charge transfer transitions (LMCT from oxygen to Mn$$^{7+}$$).

Both colors are due to charge transfer transitions, not d-d transitions.

The correct answer is Option 1: Charge transfer transition in both.

Chromyl chloride $$CrO_2Cl_2$$ contains chromium in the $$+6$$ oxidation state. In alkaline medium chloride ions are replaced by oxide ions, converting the molecule into the chromate ion $$CrO_4^{2-}$$.

Writing the reaction with hydroxide ions first:
$$CrO_2Cl_2 + 4OH^- \rightarrow CrO_4^{2-} + 2Cl^- + 2H_2O$$

On adding sodium ions to both sides, the molecular equation becomes
$$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$$

Thus, $$A = Na_2CrO_4$$ (sodium chromate).

Next, sodium chromate reacts with dilute acid and hydrogen peroxide. Under mildly acidic conditions the chromate ion forms a blue peroxo complex, commonly written as $$CrO_5$$ (chromium(VI) peroxide):

Ionic form of the reaction:
$$CrO_4^{2-} + 2H^+ + 2H_2O_2 \rightarrow CrO_5 + 3H_2O$$

Adding the spectator sodium and chloride ions gives the required molecular equation:
$$Na_2CrO_4 + 2HCl + 2H_2O_2 \rightarrow CrO_5 + 2NaCl + 3H_2O$$

Hence, $$B = CrO_5$$.

Therefore
$$A = Na_2CrO_4,\qquad B = CrO_5$$, which corresponds to Option A.

We need to identify the correct statements about transition metal oxides.

Statement A: $$Mn_2O_7$$ is an oil at room temperature.

$$Mn_2O_7$$ (dimanganese heptoxide) is indeed a dark green oily liquid at room temperature. It is the anhydride of permanganic acid ($$HMnO_4$$). Statement A is CORRECT.

Statement B: $$V_2O_4$$ reacts with acid to give $$VO_2^{2+}$$.

$$V_2O_4$$ is essentially $$VO_2$$ (vanadium(IV) oxide). When $$VO_2$$ reacts with acid, it forms the vanadyl ion $$VO^{2+}$$, not $$VO_2^{2+}$$. The reaction is: $$VO_2 + 2H^+ \rightarrow VO^{2+} + H_2O$$. Statement B is INCORRECT.

Statement C: CrO is a basic oxide.

CrO (chromium(II) oxide) is a basic oxide. In general, lower oxidation state oxides of transition metals are basic. CrO dissolves in acids to form $$Cr^{2+}$$ salts: $$CrO + 2HCl \rightarrow CrCl_2 + H_2O$$. Statement C is CORRECT.

Statement D: $$V_2O_5$$ does not react with acid.

$$V_2O_5$$ (vanadium(V) oxide) is amphoteric. It reacts with both acids and bases. With acid: $$V_2O_5 + 6HCl \rightarrow 2VOCl_3 + 3H_2O$$. Statement D is INCORRECT.

The correct statements are A and C only.

The correct answer is Option 2: A and C only.

Checking each pair:

- Photography - AgBr: Correct. Silver bromide is used in photographic films.

- Polythene preparation - $$TiCl_4$$, $$Al(CH_3)_3$$: Correct. These are Ziegler-Natta catalysts.

- Haber process - Iron: Correct. Iron is the catalyst in the Haber process.

- Wacker process - $$PtCl_2$$: Incorrect. The Wacker process uses $$PdCl_2$$ (palladium chloride), not $$PtCl_2$$.

The incorrect pair corresponds to Option (4).

We need to identify which element shows only one oxidation state other than its elemental form (0).

Cobalt (Co, Z = 27): Shows +2 and +3 oxidation states. Multiple oxidation states.

Titanium (Ti, Z = 22): Shows +2, +3, and +4 oxidation states. Multiple oxidation states.

Nickel (Ni, Z = 28): Shows +2 and +3 oxidation states. Multiple oxidation states.

Scandium (Sc, Z = 21): Electronic configuration is $$[Ar]3d^1 4s^2$$. It has only 3 electrons beyond the noble gas core. When these 3 electrons are removed, it achieves the very stable noble gas configuration of Argon. Therefore, Sc shows only the +3 oxidation state. The +1 and +2 states are not observed because partial removal of electrons does not give any special stability.

The correct answer is Option (4): Scandium.

We need to count how many of the given elements do not belong to the lanthanoid series.

Key Concept: The lanthanoid series consists of 15 elements from Lanthanum (La, Z = 57) to Lutetium (Lu, Z = 71). These elements are characterised by the progressive filling of the 4f orbitals.

Checking each element:

Eu (Europium, Z = 63): Falls between La (57) and Lu (71). Configuration: [Xe] 4f$$^7$$ 6s$$^2$$. This is a lanthanoid. ✓

Cm (Curium, Z = 96): Curium has atomic number 96, which falls in the actinoid series (Ac, Z = 89 to Lr, Z = 103). Configuration: [Rn] 5f$$^7$$ 6d$$^1$$ 7s$$^2$$. This is an actinoid, NOT a lanthanoid. ✗

Er (Erbium, Z = 68): Falls between La (57) and Lu (71). Configuration: [Xe] 4f$$^{12}$$ 6s$$^2$$. This is a lanthanoid. ✓

Tb (Terbium, Z = 65): Falls between La (57) and Lu (71). Configuration: [Xe] 4f$$^9$$ 6s$$^2$$. This is a lanthanoid. ✓

Yb (Ytterbium, Z = 70): Falls between La (57) and Lu (71). Configuration: [Xe] 4f$$^{14}$$ 6s$$^2$$. This is a lanthanoid. ✓

Lu (Lutetium, Z = 71): This is the last element of the lanthanoid series. Configuration: [Xe] 4f$$^{14}$$ 5d$$^1$$ 6s$$^2$$. This is a lanthanoid. ✓

Conclusion: Only Curium (Cm) does not belong to the lanthanoids. The number of non-lanthanoid elements is 1.

The correct answer is Option 3: 1.

We need to determine how many of the ions $$Ti^{2+}$$, $$Cr^{2+}$$, and $$V^{2+}$$ can liberate hydrogen from dilute acid.

Key concept: An ion can liberate hydrogen from dilute acid if it can be further oxidized, i.e., if it has a higher oxidation state available and the corresponding reduction potential is sufficiently negative (the ion acts as a reducing agent).

For an ion $$M^{2+}$$ to liberate $$H_2$$ from acid, the $$M^{3+}/M^{2+}$$ couple must have a reduction potential less than 0 V (so $$M^{2+}$$ can reduce $$H^+$$ to $$H_2$$ while being oxidized to $$M^{3+}$$).

$$Ti^{2+}$$: $$E^0(Ti^{3+}/Ti^{2+}) = -0.37$$ V. Since this is negative, $$Ti^{2+}$$ can reduce $$H^+$$ to $$H_2$$. $$Ti^{2+}$$ is a strong reducing agent.

V$$^{2+}$$: $$E^0(V^{3+}/V^{2+}) = -0.26$$ V. Since this is negative, $$V^{2+}$$ can reduce $$H^+$$ to $$H_2$$.

Cr$$^{2+}$$: $$E^0(Cr^{3+}/Cr^{2+}) = -0.41$$ V. Since this is negative, $$Cr^{2+}$$ can reduce $$H^+$$ to $$H_2$$.

All three ions — $$Ti^{2+}$$, $$Cr^{2+}$$, and $$V^{2+}$$ — have the ability to liberate hydrogen from dilute acid because their $$M^{3+}/M^{2+}$$ reduction potentials are all negative.

The correct answer is Option (2): 3.

Alkaline oxidative fusion of MnO$$_2$$:

$$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$$

A = MnO$$_4^{2-}$$ (manganate ion, green color).

Electrolytic oxidation in alkaline solution:

$$MnO_4^{2-} \rightarrow MnO_4^- + e^-$$

B = MnO$$_4^-$$ (permanganate ion, purple color).

The answer is Option (2): $$MnO_4^{2-}$$ and $$MnO_4^-$$.

Cu(II) has configuration $$3d^9$$ and Cu(I) has $$3d^{10}$$.

Although Cu(I) has a fully filled $$3d^{10}$$ configuration (which is stable), Cu(II) is actually more stable in aqueous solution. This is because the higher hydration enthalpy of Cu²⁺ (due to its smaller size and higher charge) more than compensates for the second ionization energy required.

The high hydration energy of Cu²⁺ stabilizes it in aqueous solutions. Cu(I) compounds tend to disproportionate in water: $$2Cu^+ \rightarrow Cu + Cu^{2+}$$.

The correct answer is Option 2: Cu(II) is more stable.

We need to evaluate both statements about manganese compounds.

First, fusion of $$MnO_2$$ with KOH and an oxidising agent gives dark green $$K_2MnO_4$$ by a standard preparation method for potassium manganate.

$$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$$

Here, $$MnO_2$$ (Mn in +4 state) is oxidised to $$K_2MnO_4$$ (Mn in +6 state) and the product is indeed dark green in colour. Therefore, the first statement is true.

Next, the manganate ion $$MnO_4^{2-}$$ can be oxidised to the permanganate ion $$MnO_4^{-}$$ by electrolytic oxidation in an alkaline solution.

$$MnO_4^{2-} \rightarrow MnO_4^{-} + e^{-}$$

This is a well-known method for preparing $$KMnO_4$$ from $$K_2MnO_4$$, so the second statement is also true.

Since both statements are true, the correct answer is Option 4: Both Statement I and Statement II are true.

This is the chromyl chloride test. When NaCl reacts with concentrated $$H_2SO_4$$ and $$K_2Cr_2O_7$$, reddish-brown fumes of chromyl chloride are produced:

$$4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2CrO_2Cl_2\uparrow + 4NaHSO_4 + K_2SO_4 + 3H_2O$$

So compound [B] = $$CrO_2Cl_2$$ (reddish fumes).

When chromyl chloride reacts with NaOH, it produces the yellow solution of sodium chromate:

$$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$$

So compound [C] = $$Na_2CrO_4$$ (yellow solution).

The answer is Option A: $$CrO_2Cl_2, \; Na_2CrO_4$$.

Among the given transition metals, we need to find which shows the highest and maximum number of oxidation states.

Manganese (Mn): Electronic configuration [Ar] $$3d^5 4s^2$$. It shows oxidation states from +2 to +7 (total 6 different oxidation states). The maximum oxidation state is +7 (in $$KMnO_4$$), which equals the total number of 3d + 4s electrons (5 + 2 = 7).

Iron (Fe): [Ar] $$3d^6 4s^2$$. Shows up to +6 (rare). Common: +2, +3.

Cobalt (Co): [Ar] $$3d^7 4s^2$$. Shows up to +4. Common: +2, +3.

Titanium (Ti): [Ar] $$3d^2 4s^2$$. Shows up to +4.

Mn shows the highest maximum oxidation state (+7) and the maximum number of oxidation states among the 3d transition metals.

The correct answer is Option (2): Mn.

We need to identify the transition metal with the highest 3rd ionisation enthalpy among Cr, Mn, V, and Fe.

After removal of two electrons (to form $$M^{2+}$$):

$$V^{2+}$$: $$[Ar]3d^3$$

$$Cr^{2+}$$: $$[Ar]3d^4$$

$$Mn^{2+}$$: $$[Ar]3d^5$$ (half-filled, extra stable)

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

The 3rd IE involves removing an electron from the $$M^{2+}$$ configuration. The $$Mn^{2+}$$ ion has a half-filled $$3d^5$$ configuration, which is exceptionally stable due to maximum exchange energy. Removing an electron from this half-filled configuration requires the most energy.

$$Mn^{2+}$$ ($$3d^5$$) has the highest 3rd IE because disrupting the half-filled d-subshell requires extra energy.

The correct answer is Option B) Mn.

Mohr’s salt is ferrous ammonium sulphate: $$FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$$. When preparing its crystals, a mixture of ferrous sulphate ($$FeSO_4$$) and ammonium sulphate ($$(NH_4)_2SO_4$$) is dissolved in water, and dilute $$H_2SO_4$$ is added.

In aqueous solution, ferrous sulphate undergoes hydrolysis, forming a brown/green basic ferrous sulphate precipitate via the reaction $$ Fe^{2+} + 2H_2O \rightleftharpoons Fe(OH)_2 + 2H^+. $$ Adding dilute $$H_2SO_4$$ increases the $$H^+$$ concentration, shifting the equilibrium to the left (Le Chatelier’s principle) and keeping the ferrous ions in solution.

Furthermore, the acidic medium prevents oxidation of $$Fe^{2+}$$ to $$Fe^{3+}$$. Thus, dilute $$H_2SO_4$$ is added to prevent the hydrolysis of ferrous sulphate, corresponding to Option (1).

A first row transition metal in +2 oxidation state has a spin-only magnetic moment of 3.86 BM. Find the atomic number.

To determine the number of unpaired electrons, we use the spin-only magnetic moment formula: $$\mu = \sqrt{n(n+2)}$$ BM. Substituting $$\mu = 3.86$$ gives $$3.86 = \sqrt{n(n+2)}$$ and thus $$14.9 \approx n(n+2)$$. For $$n = 3$$, $$\sqrt{3 \times 5} = \sqrt{15} \approx 3.87$$ BM, which matches the observed value.

Next, we consider first row transition metals in the +2 oxidation state, which typically lose two 4s electrons, leaving only the 3d electrons. We look for a 3d configuration with three unpaired electrons:

V$$^{2+}$$ (Z=23): [Ar] 3d$$^3$$ - 3 unpaired electrons. Matches.

Cr$$^{2+}$$ (Z=24): [Ar] 3d$$^4$$ - 4 unpaired electrons (high spin). Does not match.

Mn$$^{2+}$$ (Z=25): [Ar] 3d$$^5$$ - 5 unpaired electrons. Does not match.

Fe$$^{2+}$$ (Z=26): [Ar] 3d$$^6$$ - 4 unpaired electrons (high spin). Does not match.

Co$$^{2+}$$ (Z=27): [Ar] 3d$$^7$$ - 3 unpaired electrons, but among the given options only Z=23 is listed.

The correct answer is Option 3: 23 (Vanadium).

We need to identify which statements about chromium and manganese compounds are correct.

Statement A: "The chromate ion is square planar."

In $$CrO_4^{2-}$$, Cr is in the +6 oxidation state with 0 d-electrons. It bonds to 4 oxygen atoms. Using VSEPR theory, with 4 bond pairs and no lone pairs, the geometry is tetrahedral, not square planar. Statement A is INCORRECT.

Statement B: "Dichromates are generally prepared from chromates."

This is correct. Dichromates ($$Cr_2O_7^{2-}$$) are prepared by acidifying chromate ($$CrO_4^{2-}$$) solutions: $$2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O$$. Statement B is CORRECT.

Statement C: "The green manganate ion is diamagnetic."

The manganate ion $$MnO_4^{2-}$$ has Mn in +6 oxidation state: Mn is $$[Ar]3d^1$$ (since Mn normally has 7 electrons beyond Ar, minus 6 gives 1). With 1 unpaired d-electron, it is paramagnetic, not diamagnetic. Statement C is INCORRECT.

Statement D: "Dark green $$K_2MnO_4$$ disproportionates in neutral or acidic medium to give permanganate."

This is correct. The reaction is: $$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$$. Here Mn(+6) disproportionates to Mn(+7) in $$MnO_4^-$$ and Mn(+4) in $$MnO_2$$. Statement D is CORRECT.

Statement E: "With increasing oxidation number of transition metal, ionic character of the oxides decreases."

This is correct. Higher oxidation states lead to greater polarizing power of the cation, increasing covalent character (and decreasing ionic character), consistent with Fajans' rules. For example, $$CrO$$ (Cr²⁺) is more ionic while $$CrO_3$$ (Cr⁶⁺) is covalent. Statement E is CORRECT.

Correct statements: B, D, and E.

The correct answer is Option (4): B, D, E only.

We need to find the oxidation state of chromium in chromium pentoxide ($$CrO_5$$).

Identify the structure of $$CrO_5$$.

Chromium pentoxide ($$CrO_5$$) has the structure with one oxide ion ($$O^{2-}$$) and two peroxide ions ($$O_2^{2-}$$).

$$CrO_5 = CrO(O_2)_2$$

Calculate oxidation state.

Let the oxidation state of Cr be $$x$$.

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

$$x - 2 - 4 = 0$$

$$x = +6$$

Alternatively: One $$O^{2-}$$ contributes $$-2$$, two $$O_2^{2-}$$ contribute $$2 \times (-2) = -4$$. Total oxygen charge = $$-6$$. So $$x = +6$$.

Conclusion.

The oxidation state of chromium in $$CrO_5$$ is $$+6$$, which matches Option A.

Therefore, the answer is Option A.

We need to identify which compound shows colour due to d-d transition among: $$CuSO_4 \cdot 5H_2O$$, $$K_2Cr_2O_7$$, $$K_2CrO_4$$, and $$KMnO_4$$.

d-d transitions involve the excitation of an electron from one d-orbital to another within the same metal ion. These transitions are typically weak (Laporte forbidden) and give rise to pale colours.

Charge transfer (CT) transitions involve transfer of electrons between ligand and metal orbitals. These are much more intense and give rise to deep, vivid colours.

$$CuSO_4 \cdot 5H_2O$$: Contains $$Cu^{2+}$$ with electronic configuration $$[Ar]3d^9$$. The $$Cu^{2+}$$ ion has an incomplete d-shell with one unpaired electron. In the hydrated form, $$[Cu(H_2O)_4]^{2+}$$ (approximately square planar/distorted octahedral), d-d transitions occur, giving the characteristic blue colour. This is a d-d transition.

$$K_2Cr_2O_7$$: Contains $$Cr^{6+}$$ ($$Cr_2O_7^{2-}$$) with configuration $$[Ar]3d^0$$. Since there are no d-electrons, d-d transitions are impossible. The orange colour is due to ligand-to-metal charge transfer (LMCT) from oxygen to chromium.

$$K_2CrO_4$$: Contains $$Cr^{6+}$$ ($$CrO_4^{2-}$$) with $$3d^0$$. Again, no d-electrons means no d-d transitions. The yellow colour is due to LMCT transition.

$$KMnO_4$$: Contains $$Mn^{7+}$$ ($$MnO_4^-$$) with $$3d^0$$. No d-electrons, so no d-d transitions. The intense purple colour is due to LMCT transition from oxygen to manganese.

Only $$CuSO_4 \cdot 5H_2O$$ shows colour due to d-d transition, as $$Cu^{2+}$$ has a $$3d^9$$ configuration allowing d-d electronic transitions.

The correct answer is $$CuSO_4 \cdot 5H_2O$$ (Option A).

We evaluate the Assertion-Reason statement about $$Cr^{2+}$$ and $$Mn^{3+}$$.

Assertion A: In aqueous solutions, $$Cr^{2+}$$ is reducing while $$Mn^{3+}$$ is oxidising.

Analysis of Assertion:

• $$Cr^{2+}$$ has the electronic configuration $$[Ar]3d^4$$. It readily loses an electron to become $$Cr^{3+}$$ ($$[Ar]3d^3$$), which has a half-filled $$t_{2g}$$ sublevel — a very stable configuration. Hence, $$Cr^{2+}$$ acts as a strong reducing agent. $$\checkmark$$
• $$Mn^{3+}$$ has the configuration $$[Ar]3d^4$$. It readily gains an electron to become $$Mn^{2+}$$ ($$[Ar]3d^5$$), which has a half-filled d-shell — an extra stable configuration. Hence, $$Mn^{3+}$$ acts as a strong oxidising agent. $$\checkmark$$

Assertion A is TRUE.

Reason R: Extra stability to half-filled electronic configuration is observed compared to incompletely filled configuration.

Analysis of Reason:

This is a well-known fact. Half-filled configurations ($$d^3$$ in octahedral field and $$d^5$$) provide extra stability due to exchange energy. $$\checkmark$$

Reason R is TRUE.

Is R the correct explanation of A?

Yes. $$Cr^{2+}$$ ($$d^4$$) tends to become $$Cr^{3+}$$ ($$d^3$$, half-filled $$t_{2g}$$), making it reducing. $$Mn^{3+}$$ ($$d^4$$) tends to become $$Mn^{2+}$$ ($$d^5$$, half-filled d-shell), making it oxidising. The driving force in both cases is the extra stability of the half-filled configuration.

The correct answer is Option A: Both A and R are true and R is the correct explanation of A.

List I (Species of the d- and f-block)             List II (Characteristic feature)

A. $$CeO_2$$ (cerium(IV) oxide)             I. Orange-coloured salt

B. $$KMnO_4$$ (permanganate ion)       II. Not classified as a transition element

C. $$K_2Cr_2O_7$$ (dichromate ion)      III. Powerful oxidising agent in acidic medium, reduced to $$Mn^{2+}$$

D. $$Zn$$ (zinc metal)                  IV. Exhibits the stable +4 oxidation state and is used for glass-polishing (variable $$+3/+4$$ states)

We match each species with its correct characteristic one by one.

$$CeO_2$$ belongs to the 4f-block (lanthanoid). Cerium commonly shows $$+3$$ and $$+4$$ oxidation states; the $$+4$$ form $$CeO_2$$ is used industrially for glass-polishing and as an oxidising agent. Hence A → IV.

In $$KMnO_4$$ manganese is in the $$+7$$ oxidation state. In acidic medium it acts as a very strong oxidising agent and gets reduced to $$Mn^{2+}$$. Hence B → III.

$$K_2Cr_2O_7$$ is orange in colour because of a charge-transfer transition in the dichromate ion $$Cr_2O_7^{2-}$$. Hence C → I.

Zinc has completely filled $$3d^{10}$$ subshell in both atomic ($$3d^{10}4s^{2}$$) and ionic ($$Zn^{2+}: 3d^{10}$$) states, so it does not show variable oxidation states and is therefore not considered a transition element. Hence D → II.

The complete matching is

A - IV, B - III, C - I, D - II.

Therefore the correct option is:
Option A which is: A-IV, B-III, C-I, D-II.

We need to find the spin-only magnetic moment of the species with the least oxidising ability among $$VO_2^+$$, $$MnO_4^-$$, and $$Cr_2O_7^{2-}$$.

Oxidising ability is related to the ease with which the species can be reduced. The higher the reduction potential, the stronger the oxidising agent. Among common oxidising agents in acidic medium: $$MnO_4^- > Cr_2O_7^{2-} > VO_2^+$$ in terms of oxidising strength.

The species with the least oxidising ability is $$VO_2^+$$.

Find the oxidation state and d-electron configuration of V in $$VO_2^+$$:

Let the oxidation state of V be $$x$$. Then $$x + 2(-2) = +1$$, so $$x = +5$$.

Vanadium (Z = 23) has the configuration [Ar]3d$$^3$$4s$$^2$$. V$$^{5+}$$ loses all 5 valence electrons, giving [Ar] = 3d$$^0$$.

Calculate the spin-only magnetic moment using the formula:

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

where $$n$$ is the number of unpaired electrons. Since V$$^{5+}$$ has d$$^0$$ configuration, $$n = 0$$:

$$\mu = \sqrt{0(0+2)} = 0 \, \text{BM}$$

The answer is 0.

• Identification of Metal ($$\text{M}$$): Among the 3d transition series, Vanadium ($$\text{V}$$, $$Z=23$$) has the highest enthalpy of atomization.
• Oxides Formed: Vanadium forms $$\text{V}_2\text{O}_3$$, $$\text{V}_2\text{O}_4$$, and $$\text{V}_2\text{O}_5$$.
• Amphoteric Oxide Selection: $$\text{V}_2\text{O}_5$$ is the characteristic amphoteric oxide in this series.
• Magnetic Moment Calculation: In $$\text{V}_2\text{O}_5$$, Vanadium is in its maximum oxidation state of $$+5$$.
  $$\text{V}^{5+} = [\text{Ar}] 3d^0 4s^0$$ (Number of unpaired electrons, $$n = 0$$)
  $$\text{Spin-only magnetic moment } (\mu) = \sqrt{n(n+2)} = 0\text{ BM}$$

CrO: Cr²⁺ (d⁴), basic oxide. μ = √(4×6) = √24 = 4.90 BM.

Cr₂O₃: Cr³⁺ (d³), amphoteric oxide. μ = √(3×5) = √15 = 3.87 BM.

CrO₃: Cr⁶⁺ (d⁰), acidic oxide. μ = 0 BM.

Sum of basic and amphoteric = 4.90 + 3.87 = 8.77 BM = 877 × 10⁻² BM.

The answer is 877.

When chromite ore ($$FeCr_2O_4$$) is fused with $$Na_2CO_3$$ in the presence of air ($$O_2$$), the reaction is:

$$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$$

The products are sodium chromate ($$Na_2CrO_4$$) and ferric oxide ($$Fe_2O_3$$).

In $$Na_2CrO_4$$, chromium is in the +6 oxidation state, so $$Cr^{6+}$$ has the electronic configuration [Ar] $$3d^0$$ with zero unpaired electrons. The spin-only magnetic moment is given by $$\mu = \sqrt{n(n+2)}$$, which here becomes $$\mu = \sqrt{0(0+2)} = 0 \text{ B.M.}$$.

In $$Fe_2O_3$$, iron is in the +3 oxidation state, hence $$Fe^{3+}$$ has [Ar] $$3d^5$$ with five unpaired electrons. Applying the same formula yields $$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ B.M.}$$.

Therefore, the sum of the magnetic moments is $$\mu_A + \mu_B = 0 + 5.92 = 5.92 \approx 6 \text{ B.M.}$$, and the final answer is 6.

We need to find the sum of spin-only magnetic moment values of products B and C.

We identify product A by considering the reaction $$MnO_2 + KOH + O_2 \rightarrow K_2MnO_4 (A) + H_2O$$; the balanced equation is $$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$$. Product A is potassium manganate ($$K_2MnO_4$$), where Mn is in the +6 oxidation state.

In neutral or acidic solution, $$K_2MnO_4$$ (Mn$$^{6+}$$) disproportionates according to $$3K_2MnO_4 + 2H_2O \rightarrow 2KMnO_4 + MnO_2 + 4KOH$$, in which Mn$$^{6+}$$ is oxidised to Mn$$^{7+}$$ in $$KMnO_4$$ and reduced to Mn$$^{4+}$$ in $$MnO_2$$. Thus B = $$KMnO_4$$ and C = $$MnO_2$$.

For $$KMnO_4$$, Mn is in the +7 oxidation state. The Mn atom has the configuration [Ar] 3d$$^5$$ 4s$$^2$$, so Mn$$^{7+}$$ loses all 7 electrons giving [Ar] = 3d$$^0$$, resulting in zero unpaired electrons and a spin-only magnetic moment of $$\mu_B = \sqrt{n(n+2)} = \sqrt{0} = 0 \text{ BM}$$.

In $$MnO_2$$, Mn is in the +4 oxidation state, so Mn$$^{4+}$$ has the configuration [Ar] 3d$$^3$$ (lost 4s$$^2$$ and 1 of 3d$$^5$$, leaving 3 electrons in 3d), resulting in 3 unpaired electrons and a magnetic moment of $$\mu_C = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} = 3.87 \text{ BM}$$.

Adding these values gives $$\mu_B + \mu_C = 0 + 3.87 = 3.87 \approx 4 \text{ BM (nearest integer)}$$.

The answer is 4.

Colourless lanthanoid ions are those with $$4f^0$$ or $$4f^{14}$$ configurations (no unpaired f-electrons to undergo f-f transitions).

$$Eu^{3+}$$: $$[Xe]4f^6$$ — coloured ✗

$$Lu^{3+}$$: $$[Xe]4f^{14}$$ — colourless ✓

$$Nd^{3+}$$: $$[Xe]4f^3$$ — coloured ✗

$$La^{3+}$$: $$[Xe]4f^0$$ — colourless ✓

$$Sm^{3+}$$: $$[Xe]4f^5$$ — coloured ✗

Number of colourless ions = 2 ($$Lu^{3+}$$ and $$La^{3+}$$).

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

KMnO₄: Mn⁷⁺ (d⁰), μ = 0 BM.

In acidic medium titration with oxalic acid: KMnO₄ → Mn²⁺. Mn²⁺ (d⁵), μ = √(5×7) = √35 = 5.92 ≈ 6 BM.

Difference = |0 - 6| = 6 BM.

The answer is 6.

Among Sc, Ti, V, Cr, Mn, Fe, the one with highest second ionization enthalpy is Cr.

Cr: [Ar]3d⁵4s¹. After removing first electron (4s¹): Cr⁺ = [Ar]3d⁵ (half-filled, very stable).

Second IE involves removing from the stable half-filled 3d⁵ configuration, which requires very high energy.

Cr⁺ = [Ar]3d⁵: number of unpaired electrons = 5.

Spin-only magnetic moment = $$\sqrt{n(n+2)} = \sqrt{5 \times 7} = \sqrt{35} = 5.92 \approx 6$$ BM.

The answer is $$\boxed{6}$$ BM.

A strong reducing agent is a species that readily loses electrons (gets oxidized). A strong oxidizing agent readily gains electrons (gets reduced).

Eu$$^{2+}$$ is a strong reducing agent because europium prefers the +3 oxidation state (half-filled 4f⁷ configuration). Eu²⁺ readily loses an electron to become Eu³⁺.

Ce$$^{4+}$$ is a strong oxidizing agent because cerium prefers the +3 oxidation state. Ce⁴⁺ readily gains an electron to become Ce³⁺ (which has a noble gas core [Xe] configuration).

Therefore, the strong reducing agent is Eu$$^{2+}$$ and the strong oxidizing agent is Ce$$^{4+}$$.

The correct answer is Eu$$^{2+}$$ and Ce$$^{4+}$$.

The question deals with the preliminary and refining steps used in winning metals from their ores. We check every statement against the standard principles used in metallurgy.

Case A:

Malachite is $$CuCO_3\cdot Cu(OH)_2$$. On roasting (strong heating in excess air) it undergoes thermal decomposition:
$$CuCO_3\cdot Cu(OH)_2 \;\xrightarrow{\text{roast}}\; 2\,CuO + CO_2 + H_2O.$$
The oxide formed is cupric oxide $$CuO$$, not cuprous oxide $$Cu_2O$$ (cuprite). Hence statement A is incorrect.

Case B:

Calamine is $$ZnCO_3$$. Calcination means heating in the absence or limited supply of air so that volatile components leave:
$$ZnCO_3 \;\xrightarrow{\text{calcine}}\; ZnO + CO_2.$$
The oxide $$ZnO$$ obtained is called zincite. Therefore statement B is correct.

Case C:

Copper pyrites is $$CuFeS_2$$. During smelting, the ore is heated with silica (acidic flux) in a reverberatory furnace:
$$FeO + SiO_2 \;\rightarrow\; FeSiO_3\;(\text{slag}).$$
The iron present is thus removed as fusible ferrous silicate slag. Hence statement C is correct.

Case D:

Impure silver (or gold) can be purified by the cyanide process. The impure metal is treated with aerated aqueous potassium cyanide:
$$4\,Ag + 8\,KCN + O_2 + 2\,H_2O \;\rightarrow\; 4\,K[Ag(CN)_2] + 4\,KOH.$$
The complex $$[Ag(CN)_2]^-$$ is then reduced by adding zinc dust:
$$2\,K[Ag(CN)_2] + Zn \;\rightarrow\; K_2[Zn(CN)_4] + 2\,Ag\downarrow.$$
Thus pure silver is deposited, so statement D is correct.

Therefore the correct statements are:
Option B, Option C, and Option D.

Final Answer: Option A (incorrect), Option B (correct), Option C (correct), Option D (correct).
⟹ Option B, Option C, Option D are the right choices.

The sequence of reactions involved in the scheme (shown in the paper) starts from $$MnO_2$$. In strongly alkaline medium and in the presence of an oxidising agent such as $$O_2$$ (or $$K_2S_2O_8$$), $$MnO_2$$ is first converted to green manganate ion:

$$MnO_2 + 4\,OH^- + O_2 \;\longrightarrow\; MnO_4^{2-} + 2\,H_2O$$

The green manganate ion $$MnO_4^{2-}$$ is unstable in acidic medium. On acidification it undergoes disproportionation (simultaneous oxidation and reduction of the same species) to give purple permanganate ion and $$Mn^{4+}$$ (as $$MnO_2$$):

$$3\,MnO_4^{2-} + 4\,H^+ \;\longrightarrow\; 2\,MnO_4^- + MnO_2 + 2\,H_2O$$

Thus the species produced after acidification, labelled X in the scheme, is the purple permanganate ion $$MnO_4^-$$.

Purple permanganate in acidic medium is a very strong oxidising agent. It can oxidise chloride ions to chlorine gas according to

$$2\,MnO_4^- + 10\,Cl^- + 16\,H^+ \;\longrightarrow\; 2\,Mn^{2+} + 5\,Cl_2 + 8\,H_2O$$

Hence the gaseous product Y obtained in the last step of the scheme is molecular chlorine $$Cl_2$$.

Therefore, X = $$MnO_4^-$$ and Y = $$Cl_2$$, which corresponds to Option C.

Option C which is: $$MnO_4^{-}$$ and $$Cl_2$$

In the extraction of copper from its sulphide ore, the ore is heated in a reverberatory furnace after mixing with silica.

Purpose of adding silica (SiO$$_2$$):

During the smelting process, the ore contains iron sulphide (FeS) as an impurity. The FeS gets oxidized to FeO. Silica acts as a flux and combines with the basic impurity FeO to form iron silicate (slag):

$$\text{FeO} + \text{SiO}_2 \to \text{FeSiO}_3 \text{ (slag)}$$

This slag is immiscible with the matte (Cu$$_2$$S) and floats on top, allowing easy separation. This removes iron impurities from the copper matte.

The correct answer is Option 4: Remove FeO as FeSiO$$_3$$.

We need to identify which ore has a sulphide-based mineral as its major component.

Analyzing each ore:

Calamine: ZnCO$$_3$$ — This is a carbonate ore of zinc.

Siderite: FeCO$$_3$$ — This is a carbonate ore of iron.

Sphalerite: ZnS — This is a sulphide ore of zinc. ✓

Malachite: CuCO$$_3$$ · Cu(OH)$$_2$$ — This is a basic carbonate ore of copper.

Among the given options, only Sphalerite (ZnS) is a sulphide-based mineral.

The correct answer is Option 3: Sphalerite.

Leaching of silver involves oxidation followed by reduction.

During leaching, silver is treated with dilute cyanide solution in the presence of oxygen:

$$4Ag_{(s)} + 8CN^-_{(aq)} + 2H_2O_{(aq)} + O_{2(g)} \rightarrow 4[Ag(CN)_2]^-_{(aq)} + 4OH^-_{(aq)}$$

Silver is oxidized from oxidation state $$0$$ to $$+1$$ and forms the soluble complex $$[Ag(CN)_2]^-$$.

The dissolved silver is then recovered by adding zinc:

$$2[Ag(CN)_2]^-_{(aq)} + Zn_{(s)} \rightarrow [Zn(CN)_4]^{2-}_{(aq)} + 2Ag_{(s)}$$

Silver is reduced from oxidation state $$+1$$ back to $$0$$ and precipitates as metallic silver.

Aluminium is extracted by electrolytic reduction of alumina, copper is mainly extracted through roasting and self-reduction, and iron is extracted by reduction with carbon monoxide in a blast furnace.

Hence, the metal whose extraction involves oxidation followed by reduction is

$$\boxed{\text{Ag (Silver)}}$$

In chromyl chloride (CrO₂Cl₂), we are asked to identify which species has the same number of d-electrons on its central metal.

To determine the oxidation state of chromium in CrO₂Cl₂, let x represent the oxidation state of Cr. Each oxygen atom contributes -2 and each chloride contributes -1, so that $$x + 2(-2) + 2(-1) = 0 \implies x = +6$$.

Next, the electron configuration of Cr⁶⁺ follows from that of neutral Cr: [Ar] 3d⁵ 4s¹. Removing six electrons from the 3d and 4s orbitals gives Cr⁶⁺ = [Ar], which corresponds to 0 d-electrons.

We then compare this with each option. For Option A - V(IV), vanadium has [Ar] 3d³ 4s², and V⁴⁺ becomes [Ar] 3d¹, giving 1 d-electron. For Option B - Mn(VII), manganese has [Ar] 3d⁵ 4s², and Mn⁷⁺ becomes [Ar], giving 0 d-electrons ✓. For Option C - Fe(III), iron is [Ar] 3d⁶ 4s², and Fe³⁺ is [Ar] 3d⁵, giving 5 d-electrons. For Option D - Ti(III), titanium is [Ar] 3d² 4s², and Ti³⁺ is [Ar] 3d¹, giving 1 d-electron.

Since Mn(VII) has 0 d-electrons, the same number as Cr(VI) in chromyl chloride, the correct answer is Option B: Mn(VII).

In the extraction of copper from copper pyrites (CuFeS$$_2$$), the ore first undergoes partial roasting:

$$2\text{Cu}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{Cu}_2\text{O} + 2\text{SO}_2$$

This is followed by self-reduction (Bessemerisation):

$$2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6\text{Cu} + \text{SO}_2$$

In this process, Cu$$_2$$S acts as both a reducing agent and is itself oxidised. The copper obtained contains blisters (bubbles) on the surface due to the evolution of SO$$_2$$ gas during solidification. This product is called blister copper, which is about 98-99% pure copper. It is further refined by electrolysis to obtain 99.9% pure copper.

So, the answer is blister copper.

When $$Cu^{2+}$$ is treated with excess KI:

$$ 2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 \downarrow + I_2 $$

$$Cu^{2+}$$ is reduced to $$Cu^+$$ which forms the white precipitate of cuprous iodide ($$Cu_2I_2$$), and iodide is oxidized to iodine ($$I_2$$).

Therefore, $$X = Cu_2I_2$$ (white precipitate).

The liberated $$I_2$$ is then titrated with sodium thiosulphate:

$$ I_2 + 2Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2NaI $$

The product sodium tetrathionate ($$Na_2S_4O_6$$) is compound Y.

Therefore, $$X = Cu_2I_2$$ and $$Y = Na_2S_4O_6$$.

Good oxidizing agents are species that readily get reduced (gain electrons).

(a) Sm²⁺: Samarium in +2 state can be oxidized to the more stable +3 state. So Sm²⁺ is a reducing agent, not an oxidizing agent.

(b) Ce²⁺: Cerium in +2 state can be oxidized to more stable +3 or +4 state. So Ce²⁺ is a reducing agent.

(c) Ce⁴⁺: Cerium in +4 state tends to get reduced to the stable +3 state. Ce⁴⁺ is a good oxidizing agent.

(d) Tb⁴⁺: Terbium in +4 state tends to get reduced to the more stable +3 state. Tb⁴⁺ is a good oxidizing agent.

The good oxidizing agents are C and D.

The correct answer is Option 4: C and D only.

Statement I: K₂Cr₂O₇ is preferred as a primary standard over Na₂Cr₂O₇.

This is true. K₂Cr₂O₇ is non-hygroscopic and can be obtained in a highly pure state, making it suitable as a primary standard. Na₂Cr₂O₇ is hygroscopic.

Statement II: K₂Cr₂O₇ has higher solubility than Na₂Cr₂O₇.

This is false. Na₂Cr₂O₇ is actually more soluble in water than K₂Cr₂O₇.

The correct answer is Option 3: Statement I is true but Statement II is false.

In $$Mn_2O_7$$, manganese is in its highest oxidation state of +7.

Let us analyze the structure of $$Mn_2O_7$$:

(A) Mn is tetrahedrally surrounded by oxygen atoms - CORRECT. Each Mn atom is bonded to four oxygen atoms in a tetrahedral arrangement (three terminal O atoms and one bridging O atom).

(B) Mn is octahedrally surrounded by oxygen atoms - INCORRECT. The coordination around Mn is tetrahedral, not octahedral.

(C) Contains Mn-O-Mn bridge - CORRECT. The two MnO₄ tetrahedra are connected through a Mn-O-Mn bridge (one shared oxygen atom between the two Mn centers).

(D) Contains Mn-Mn bond - INCORRECT. There is no direct Mn-Mn bond. The two Mn atoms are connected through a bridging oxygen atom.

Therefore, the correct statements are A and C only.

Acidified potassium dichromate is a strong oxidizing agent. The dichromate ion $$Cr_2O_7^{2-}$$ contains chromium in the +6 oxidation state and is orange in colour. When it oxidizes a reducing agent, it is reduced to $$Cr^{3+}$$, which is green in colour.

Therefore, the gas that turns acidified potassium dichromate paper from orange to green must act as a reducing agent.

Evaluating the given gases:

• $$CO_2$$: Carbon is already in its highest oxidation state (+4) and cannot be oxidized further. Hence, it does not reduce dichromate.

• $$SO_3$$: Sulphur is already in its highest oxidation state (+6) and cannot be oxidized further. Hence, it does not reduce dichromate.

• $$H_2S$$: Sulphur is in the -2 oxidation state and readily acts as a reducing agent. It reduces dichromate, but is simultaneously oxidized to elemental sulphur, producing a yellow deposit.

• $$SO_2$$: Sulphur is in the +4 oxidation state and can be oxidized to +6. Therefore, it acts as a reducing agent and reduces dichromate to green $$Cr^{3+}$$ ions. This is the standard qualitative test for sulphur dioxide.

The reaction is:

$$K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$$

The orange dichromate ion is converted into green chromium(III) ions, causing the colour change.

Hence, the gas that turns acidified potassium dichromate paper green is

$$\boxed{SO_2}$$

Therefore, the correct answer is $$\boxed{\text{Sulphur dioxide}}$$.

We need to determine the change in oxidation state of chromium when potassium dichromate acts as a strong oxidizing agent in acidic solution.

First, we find the oxidation state of Cr in $$K_2Cr_2O_7$$. Let the oxidation state be $$x$$. Then the equation $$2(+1) + 2x + 7(-2) = 0$$ leads to $$2 + 2x - 14 = 0$$, so $$2x = 12$$ and $$x = +6$$.

In acidic solution, the dichromate ion is reduced according to $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$. The oxidation state of chromium in the product $$Cr^{3+}$$ is $$+3$$.

Thus the oxidation state of chromium changes from $$+6$$ to $$+3$$. This matches Option 2, and the answer is $$\boxed{+6 \text{ to } +3}$$.

We need to find the pair of lanthanides where both elements have high third-ionization energy (IE$$_3$$).

The third ionization energy of a lanthanide involves removing the third electron. A high IE$$_3$$ occurs when removing the third electron disrupts an especially stable electron configuration (half-filled or fully-filled 4f subshell).

Europium (Eu): Electronic configuration is [Xe] 4f$$^7$$ 6s$$^2$$. After losing two electrons (from 6s): Eu$$^{2+}$$ = [Xe] 4f$$^7$$. The third ionization requires removing an electron from the half-filled 4f$$^7$$ shell, which is very stable. Hence, Eu has a high IE$$_3$$.

Ytterbium (Yb): Electronic configuration is [Xe] 4f$$^{14}$$ 6s$$^2$$. After losing two electrons (from 6s): Yb$$^{2+}$$ = [Xe] 4f$$^{14}$$. The third ionization requires removing an electron from the completely filled 4f$$^{14}$$ shell, which is very stable. Hence, Yb has a high IE$$_3$$.

Checking the other options:

Option A (Dy, Gd): Gd has configuration [Xe] 4f$$^7$$ 5d$$^1$$ 6s$$^2$$, so Gd$$^{2+}$$ = [Xe] 4f$$^7$$ 5d$$^1$$. The third electron comes from 5d, not disrupting the 4f$$^7$$, so IE$$_3$$ is not exceptionally high. Dy$$^{2+}$$ = [Xe] 4f$$^{10}$$, which has no special stability.

Option B (Lu, Yb): Lu has configuration [Xe] 4f$$^{14}$$ 5d$$^1$$ 6s$$^2$$, so Lu$$^{2+}$$ = [Xe] 4f$$^{14}$$ 5d$$^1$$. The third electron comes from 5d, so IE$$_3$$ is not especially high.

Option D (Eu, Gd): Eu has high IE$$_3$$, but Gd does not (as explained above).

The correct answer is Option C: Eu, Yb.

The Mond process is used for refining nickel. It involves the following steps:

Impure nickel reacts with carbon monoxide at 330-350 K to form volatile nickel tetracarbonyl:

Nickel tetracarbonyl is decomposed at higher temperature (450-470 K) to give pure nickel:

Let us check the other options:

Option B: $$2\text{K[Au(CN)}_2] + \text{Zn} \rightarrow \text{K}_2[\text{Zn(CN)}_4] + 2\text{Au}$$ — This is the MacArthur-Forrest (cyanide) process for gold.

Option C: $$\text{Zr} + 2\text{I}_2 \rightarrow \text{ZrI}_4$$ — This is the Van Arkel process for zirconium/titanium.

Option D: $$\text{ZnO} + \text{C} \rightarrow \text{Zn} + \text{CO}$$ — This is carbon reduction, not a refining process.

The answer is Option A: $$\text{Ni} + 4\text{CO} \xrightarrow{\Delta} \text{Ni(CO)}_4$$.

Half-filled f-orbitals means the 4f subshell contains exactly 7 electrons (f⁷).

Let us check each element's electronic configuration:

A. Sm (Z = 62): [Xe] 4f⁶ 6s² — 6 f-electrons, NOT half-filled

B. Eu (Z = 63): [Xe] 4f⁷ 6s² — 7 f-electrons, half-filled ✔

C. Tb (Z = 65): [Xe] 4f⁹ 6s² — 9 f-electrons, NOT half-filled

D. Gd (Z = 64): [Xe] 4f⁷ 5d¹ 6s² — 7 f-electrons, half-filled ✔

E. Pm (Z = 61): [Xe] 4f⁵ 6s² — 5 f-electrons, NOT half-filled

Elements with half-filled f-orbitals: B (Eu) and D (Gd).

Let us analyze each statement:

(A) M³⁺/M²⁺ reduction potential: Fe³⁺/Fe²⁺ = +0.77 V, Mn³⁺/Mn²⁺ = +1.51 V. Actually Mn³⁺/Mn²⁺ is higher. So this statement is incorrect.

(B) Higher oxidation states of first row d-block elements are stabilized by oxide ion (due to its small size and ability to form multiple bonds). This is correct. Example: CrO₄²⁻, MnO₄⁻.

(C) Cr²⁺ is a strong reducing agent (E° for Cr³⁺/Cr²⁺ = -0.41 V), so it can reduce H⁺ to H₂. Cr²⁺(aq) can liberate hydrogen from dilute acid. This is correct.

(D) V²⁺ has d³ configuration, so 3 unpaired electrons. $$\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87$$ BM. This is NOT between 4.4-5.2 BM. Incorrect.

Correct statements: B and C only.

The correct answer is Option 1: B, C only.

We need to find which element is NOT present in Nessler's reagent.

Key Fact:

Nessler's reagent is K$$_2$$[HgI$$_4$$] — potassium tetraiodomercurate(II).

Elements present: K (potassium), Hg (mercury), I (iodine).

Elements NOT present: N (nitrogen).

The answer is Option 1: N. This matches the stored answer.

Identify which compound's formation is NOT a confirmatory test for Pb$$^{2+}$$ ions.

In qualitative inorganic analysis, the standard confirmatory tests for Pb$$^{2+}$$ involve forming characteristic precipitates:

(i) Lead sulphate (PbSO$$_4$$): White precipitate formed by adding dilute H$$_2$$SO$$_4$$. This is a confirmatory test. ✓

(ii) Lead chromate (PbCrO$$_4$$): Yellow precipitate formed by adding potassium chromate (K$$_2$$CrO$$_4$$). This is a well-known confirmatory test. ✓

(iii) Lead iodide (PbI$$_2$$): Yellow precipitate formed by adding KI. On heating, it dissolves and recrystallises as golden-yellow flakes on cooling ("golden rain" test). This is a confirmatory test. ✓

(iv) Lead nitrate (Pb(NO$$_3$$)$$_2$$): Lead nitrate is a water-soluble salt. It does not form a precipitate and is not produced as a result of any confirmatory reaction. In fact, lead nitrate solution is often the starting material used to test for other ions. Its formation is NOT a confirmatory test for Pb$$^{2+}$$.

Therefore, lead nitrate (Option B) is the compound whose formation is not a confirmatory test for Pb$$^{2+}$$ ions.

Explanation

During the extraction of copper from copper pyrites, the roasted ore is mixed with silica and smelted in a reverberatory furnace.

During smelting, iron oxide reacts with silica to form iron silicate slag.

$$FeO+SiO_2\rightarrow FeSiO_3$$

The slag is removed, while copper is obtained in the form of copper matte.

According to NCERT, copper matte consists of

$$Cu_2S \text{ and } FeS$$

Now, evaluating the given compounds:

$$CuCO_3$$ is a carbonate ore of copper and is not a constituent of copper matte.

Therefore,

$$CuCO_3$$

is not present.

$$Cu_2S$$ is one of the principal constituents of copper matte.

Therefore,

$$Cu_2S$$

is present.

$$Cu_2O$$ is formed later during the conversion process when part of $$Cu_2S$$ is oxidized. It is not a constituent of the initial copper matte.

Therefore,

$$Cu_2O$$

is not present.

$$FeO$$ is produced during roasting but reacts with silica to form iron silicate slag and is removed from the furnace.

$$FeO+SiO_2\rightarrow FeSiO_3$$

Therefore,

$$FeO$$

is not present in copper matte.

Hence, among the given compounds, only

$$Cu_2S$$

is present in copper matte.

Therefore, the total number of compounds present is

$$1$$

Hence, the correct answer is

$$1$$

Chromyl chloride has the formula CrO₂Cl₂.

Let the oxidation state of Cr be $$x$$:

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

$$x - 4 - 2 = 0$$

$$x = +6$$

The oxidation state of chromium is $$\mathbf{+6}$$.

We need to find the oxidation state of Mo in ammonium phosphomolybdate.

The chemical formula of ammonium phosphomolybdate is $$(NH_4)_3PO_4 \cdot 12MoO_3$$, also written as $$(NH_4)_3[PMo_{12}O_{40}]$$.

Let us find the oxidation state of Mo using the second formula $$(NH_4)_3[PMo_{12}O_{40}]$$.

The ammonium ion $$NH_4^+$$ has a charge of $$+1$$, and there are 3 of them, so the total positive charge from cations is $$+3$$.

Therefore, the anion $$[PMo_{12}O_{40}]$$ has an overall charge of $$-3$$.

In the anion $$[PMo_{12}O_{40}]^{3-}$$:

- Phosphorus (P) has an oxidation state of $$+5$$

- Oxygen (O) has an oxidation state of $$-2$$

- Let the oxidation state of Mo be $$x$$

Writing the charge balance equation:

$$(+5) + 12(x) + 40(-2) = -3$$

$$5 + 12x - 80 = -3$$

$$12x - 75 = -3$$

$$12x = 72$$

$$x = +6$$

Therefore, the oxidation state of Mo in ammonium phosphomolybdate is $$+6$$.

The answer is 6.

Isoelectronic species possess the same number of electrons.

Number of electrons is calculated using:

$$\mathrm{Number\ of\ electrons = Atomic\ Number - Charge}$$

For:

$$\mathrm{Ho^{2+}}$$

$$\mathrm{67 - 2 = 65}$$

For:

$$\mathrm{Er^{3+}}$$

$$\mathrm{68 - 3 = 65}$$

Hence, they are isoelectronic.

For:

$$\mathrm{Yb^{2+}}$$

$$\mathrm{70 - 2 = 68}$$

For:

$$\mathrm{Lu^{3+}}$$

$$\mathrm{71 - 3 = 68}$$

Hence, they are isoelectronic.

For:

$$\mathrm{Eu^{2+}}$$

$$\mathrm{63 - 2 = 61}$$

For:

$$\mathrm{Tb^{4+}}$$

$$\mathrm{65 - 4 = 61}$$

Hence, they are isoelectronic.

For:

$$\mathrm{Tb^{2+}}$$

$$\mathrm{65 - 2 = 63}$$

For:

$$\mathrm{Tm^{4+}}$$

$$\mathrm{69 - 4 = 65}$$

Hence, they are not isoelectronic.

Therefore, the pair which is not isoelectronic is:

$$\mathrm{Tb^{2+}\ and\ Tm^{4+}}$$

Correct option:

$$\mathrm{D}$$

Platinum (Pt) has atomic number 78, so its electron configuration can be built from the xenon core (Xe, atomic number 54), leaving $$78 - 54 = 24$$ electrons to be placed beyond Xe.

Those electrons fill the subshells in the order 6s, 4f, 5d, giving an expected configuration:

(total: $$14 + 8 + 2 = 24$$ electrons)

However, platinum exhibits an exception in which one electron shifts from 6s to 5d to achieve greater stability:

(total: $$14 + 9 + 1 = 24$$ electrons)

Because the resulting near-complete d-subshell is more stable, the actual ground-state configuration of platinum is $$[Xe]4f^{14}5d^{9}6s^{1}$$.

Hence, the correct answer is Option D ($$[Xe]4f^{14}5d^{9}6s^{1}$$).

We are given the reaction of zinc with excess aqueous alkali. Zinc is an amphoteric metal, meaning it can react with both acids and bases.

When zinc reacts with excess aqueous NaOH (alkali), the reaction proceeds as:

$$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2 \uparrow$$

Here, zinc dissolves in the excess alkali solution and reduces water to liberate hydrogen gas. The sodium zincate $$Na_2ZnO_2$$ formed dissociates in solution to give the zincate ion $$[ZnO_2]^{2-}$$.

Now, one might wonder why the product is $$[ZnO_2]^{2-}$$ rather than $$Zn(OH)_2$$ or $$[Zn(OH)_4]^{2-}$$. When the alkali is in excess, any $$Zn(OH)_2$$ initially formed further reacts with NaOH:

$$Zn(OH)_2 + 2NaOH \rightarrow Na_2ZnO_2 + 2H_2O$$

This converts the hydroxide completely into the zincate ion $$[ZnO_2]^{2-}$$. The species $$ZnO$$ and $$Zn(OH)_2$$ are only intermediate products and do not persist when alkali is in excess.

Hence, the correct answer is Option D: $$[ZnO_2]^{2-}$$.

We need to evaluate both the Assertion and Reason about the reduction of metal oxides.

Assertion: Magnesium can reduce $$Al_2O_3$$ at a temperature below $$1350°C$$, while above $$1350°C$$ aluminium can reduce MgO.

Analysis of Assertion:

This can be understood from the Ellingham diagram. At temperatures below about $$1350°C$$, the line for MgO formation lies below the $$Al_2O_3$$ line, meaning Mg has a more negative $$\Delta G$$ for oxide formation, so Mg can reduce $$Al_2O_3$$.

Above $$1350°C$$, the $$Al_2O_3$$ line crosses below the MgO line (because Mg boils around $$1090°C$$ and its oxide becomes less stable at higher temperatures). This means Al can now reduce MgO.

The Assertion is correct.

Reason: The melting and boiling points of magnesium are lower than those of aluminium.

Analysis of Reason:

Magnesium: Melting point = $$650°C$$, Boiling point = $$1090°C$$

Aluminium: Melting point = $$660°C$$, Boiling point = $$2519°C$$

Indeed, the melting and boiling points of Mg are lower than those of Al. The Reason is correct.

Is the Reason the correct explanation of the Assertion?

The crossover in the Ellingham diagram at $$1350°C$$ is primarily due to the increase in entropy when Mg boils (at $$1090°C$$), which makes the $$\Delta G$$ for MgO formation less negative. However, the Reason only states that Mg has lower melting and boiling points than Al — while this is related, it does not directly explain the thermodynamic crossover in the Ellingham diagram. The actual explanation involves the change in slope of the $$\Delta G$$ vs $$T$$ line due to the phase transition (boiling) of Mg.

Therefore, both statements are correct, but the Reason is not the correct explanation of the Assertion.

Hence, the correct answer is Option B: Both Assertion and Reason are correct, but Reason is NOT the correct explanation of Assertion.

We need to find the 3d series metal with the highest (most positive) $$M^{2+}/M$$ standard electrode potential. The standard reduction potentials $$E°(M^{2+}/M)$$ for some 3d metals are:

$$Cr: -0.91$$ V
$$Fe: -0.44$$ V
$$Cu: +0.34$$ V
$$Zn: -0.76$$ V

Among these, copper (Cu) has the highest (most positive) standard electrode potential at $$+0.34$$ V. In fact, Cu is the only 3d metal with a positive $$E°(M^{2+}/M)$$ value, which means $$Cu^{2+}$$ ions are relatively easy to reduce to Cu metal.

This is because copper has high atomization enthalpy, high ionization enthalpy, and low hydration enthalpy for $$Cu^{2+}$$, making the overall reduction favorable.

The answer is $$\boxed{\text{Option C: Cu}}$$.

We need to evaluate two statements about the Ellingham diagram.

Evaluate Statement I.

Statement I: According to the Ellingham diagram, any metal oxide with higher $$\Delta G°$$ is more stable than the one with lower $$\Delta G°$$.

This is incorrect. In the Ellingham diagram, $$\Delta G°$$ values are plotted (which are negative for spontaneous oxide formation). A more negative (i.e., lower) $$\Delta G°$$ indicates a more thermodynamically stable oxide. So the oxide with lower $$\Delta G°$$ (more negative) is more stable, not the one with higher $$\Delta G°$$.

Evaluate Statement II.

Statement II: The metal involved in the formation of oxide placed lower in the Ellingham diagram can reduce the oxide of a metal placed higher in the diagram.

This is correct. In the Ellingham diagram, the oxide placed lower has a more negative $$\Delta G°$$, meaning its metal has a greater affinity for oxygen. Therefore, a metal whose oxide line lies lower in the diagram can reduce the oxide of a metal whose line lies higher (less negative $$\Delta G°$$). For example, aluminium (lower in the diagram) can reduce iron oxide (higher in the diagram) — this is the thermite reaction.

Conclusion.

Statement I is incorrect and Statement II is correct.

The correct answer is Option D: Statement I is incorrect but Statement II is correct.

Statement I says: Leaching of gold with cyanide ion in absence of air/$$O_2$$ leads to cyano complex of Au(III). The actual cyanide leaching process (MacArthur-Forrest process) requires oxygen: $$4Au + 8CN^- + 2H_2O + O_2 \rightarrow 4[Au(CN)_2]^- + 4OH^-$$. This produces $$[Au(CN)_2]^-$$, which is a complex of Au(I), not Au(III). Moreover, the reaction requires air/$$O_2$$ as an oxidizing agent. In the absence of air/$$O_2$$, gold cannot be oxidized and leaching does not occur effectively. Therefore, Statement I is incorrect on two counts: it claims Au(III) is formed (actually Au(I)), and it claims the reaction works without air (it requires $$O_2$$).

Statement II says: Zinc is oxidized during the displacement reaction for gold extraction. In the displacement step, zinc dust is added to the gold cyanide solution: $$2[Au(CN)_2]^- + Zn \rightarrow [Zn(CN)_4]^{2-} + 2Au$$. Here, Zn is oxidized from Zn(0) to Zn(II), and Au(I) is reduced to Au(0). Therefore, Statement II is correct.

The answer is $$\boxed{\text{Option D}}$$.

In metallurgy, when an ore is mined from the earth, it is not pure — it comes mixed with various impurities. The term "gangue" (also called matrix) refers specifically to the undesired earthy materials such as sand, clay, mud, limestone, and other rocky substances that are found associated with the ore.

We should distinguish gangue from other types of impurities. Gangue is not about contamination by other metals (that would be a different kind of impurity), nor does it refer to minerals in pure form, nor is it limited to magnetic impurities. Gangue specifically refers to the commercially worthless earthy material surrounding the valuable mineral in an ore deposit.

The process of removing gangue from the ore is called concentration of ore (or ore dressing/beneficiation), and various methods like hydraulic washing, magnetic separation, and froth flotation are used depending on the nature of the gangue and the ore.

Hence, the correct answer is Option A: Contamination of undesired earthy materials.

We need to match the ores in List-I with their chemical formulas in List-II.

(A) Sphalerite: Sphalerite is the chief ore of zinc. Its chemical formula is ZnS, which matches with (IV).

(B) Calamine: Calamine is a carbonate ore of zinc. Its chemical formula is ZnCO$$_3$$, which matches with (III).

(C) Galena: Galena is the principal ore of lead. Its chemical formula is PbS, which matches with (II).

(D) Siderite: Siderite is a carbonate ore of iron. Its chemical formula is FeCO$$_3$$, which matches with (I).

Summary of matching:

(A) - (IV) ZnS

(B) - (III) ZnCO$$_3$$

(C) - (II) PbS

(D) - (I) FeCO$$_3$$

The correct answer is Option B: (A) - (IV), (B) - (III), (C) - (II), (D) - (I).

We need to evaluate both statements about electrolytic refining of copper.

Evaluating Statement I:

During electrolytic refining, impure blister copper is used as the anode. When the anode dissolves, less reactive precious metals like gold (Au) and silver (Ag) do not dissolve and settle at the bottom as anode mud. Thus, blister copper effectively deposits (releases) precious metals during the refining process.

Statement I is true.

Evaluating Statement II:

In the electrolytic purification of copper, blister copper (impure copper, ~98-99% pure) is indeed used as the anode. A thin strip of pure copper is used as the cathode, and an acidified copper sulphate solution serves as the electrolyte. At the anode, copper dissolves as $$Cu^{2+}$$, and at the cathode, pure copper is deposited.

Statement II is true.

Both Statement I and Statement II are true.

The correct answer is Option A.

We need to find the pair where both transition metal ions are colourless. The colour of a transition metal ion arises from d-d electronic transitions. An ion will be colourless if it has either a $$d^0$$ (empty d-orbitals) or $$d^{10}$$ (completely filled d-orbitals) configuration, because in either case no d-d transition is possible.

Let us examine each option:

Option A: $$Sc^{3+}$$ and $$Zn^{2+}$$

Scandium (Sc) has atomic number 21 with configuration $$[Ar]\,3d^1\,4s^2$$. So $$Sc^{3+}$$ has configuration $$[Ar]$$ — that is $$3d^0$$. With no d-electrons, no d-d transition is possible, so $$Sc^{3+}$$ is colourless.

Zinc (Zn) has atomic number 30 with configuration $$[Ar]\,3d^{10}\,4s^2$$. So $$Zn^{2+}$$ has configuration $$[Ar]\,3d^{10}$$. With completely filled d-orbitals, no d-d transition is possible, so $$Zn^{2+}$$ is colourless.

Both ions are colourless.

Option B: $$Ti^{4+}$$ and $$Cu^{2+}$$

$$Ti^{4+}$$ is $$3d^0$$ (colourless), but $$Cu^{2+}$$ is $$3d^9$$ which allows d-d transitions, making it blue/coloured. Not both colourless.

Option C: $$V^{2+}$$ and $$Ti^{3+}$$

$$V^{2+}$$ is $$3d^3$$ and $$Ti^{3+}$$ is $$3d^1$$. Both have partially filled d-orbitals, so both are coloured.

Option D: $$Zn^{2+}$$ and $$Mn^{2+}$$

$$Zn^{2+}$$ is $$3d^{10}$$ (colourless), but $$Mn^{2+}$$ is $$3d^5$$. Although $$Mn^{2+}$$ is only faintly coloured (all transitions are spin-forbidden), it is still not truly colourless. Moreover, strictly speaking it has partially filled d-orbitals.

Hence, the correct answer is Option A: $$Sc^{3+}, Zn^{2+}$$.

In the metallurgical extraction of copper, the reaction is: $$\text{FeO} + \text{SiO}_2 \rightarrow \text{FeSiO}_3$$.

In metallurgy, gangue refers to the earthy impurities (unwanted material) associated with the ore, flux is a substance added to remove gangue by combining with it, and slag is the fusible product formed when flux reacts with gangue.

In copper extraction, the ore is copper pyrite (CuFeS$$_2$$). During smelting, FeS is converted to FeO, which is the unwanted impurity (gangue) that needs to be removed. SiO$$_2$$ (silica) is added as flux to combine with the gangue (FeO), producing FeSiO$$_3$$ (iron silicate) as the slag, which is easily separable.

Answer: FeO is the gangue and FeSiO$$_3$$ is the slag; hence, the correct answer is Option D.

Let us match each item in List-I with the correct item in List-II.

A. Concentration of gold ore → IV (NaCN)

Gold ore is concentrated by the cyanide leaching process (MacArthur-Forrest process). Gold reacts with dilute NaCN solution in the presence of air to form a soluble gold cyanide complex:

$$4Au + 8NaCN + 2H_2O + O_2 \rightarrow 4Na[Au(CN)_2] + 4NaOH$$

B. Leaching of alumina → II (NaOH)

In Bayer's process, bauxite ore (containing $$Al_2O_3$$) is leached with concentrated NaOH solution. Alumina dissolves as sodium aluminate:

$$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$$

C. Froth stabiliser → I (Aniline)

In the froth floatation process, froth stabilisers like aniline or cresol are added to stabilise the froth so that the ore particles can be collected effectively.

D. Blister copper → III ($$SO_2$$)

Blister copper gets its name from the blisters formed on its surface due to the evolution of $$SO_2$$ gas during the solidification of molten copper. The $$SO_2$$ is produced during the bessemerisation process:

$$Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$$

$$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$$

The correct matching is: A-(IV), B-(II), C-(I), D-(III)

Hence, the correct answer is Option B.

In the Froth Floatation method, depressants are chemicals that are added to selectively prevent certain mineral particles from forming froth, thereby allowing separation of different sulfide ores.

The froth floatation process is used for the concentration of sulfide ores. The ore is mixed with water and a frothing agent (like pine oil). Air is blown through the mixture, and the sulfide ore particles (which are preferentially wetted by oil) rise with the froth, while the gangue (which is preferentially wetted by water) settles down.

When an ore contains two or more sulfide minerals (e.g., ZnS and PbS), depressants are used to selectively prevent one of the sulfide ores from coming to the froth. For example, NaCN is used as a depressant for ZnS when separating it from PbS. NaCN forms a layer of zinc cyanide on ZnS particles, preventing them from coming to the froth while PbS floats.

Option A: Selectively prevent one component of the ore from coming to the froth — this is correct and describes the exact role of depressants.

Option B: Reduce oil consumption — this is not the role of depressants.

Option C: Stabilize the froth — this is the role of froth stabilizers, not depressants.

Option D: Enhance non-wettability — this is the role of collectors, not depressants.

Therefore, the correct answer is Option A.

Cerium(IV), i.e., $$Ce^{4+}$$, has the electronic configuration $$[Xe]$$, which is a noble gas configuration.

Electronic configuration of Cerium: Cerium (Ce) has atomic number 58. Its ground state configuration is $$[Xe] 4f^1 5d^1 6s^2$$.

When Ce loses 4 electrons to form $$Ce^{4+}$$, it achieves the configuration $$[Xe]$$ — a noble gas configuration.

Predicting the chemical behavior of $$Ce^{4+}$$: Although $$Ce^{4+}$$ has achieved a stable noble gas configuration, the +4 oxidation state represents a very high charge density on the cerium ion. This makes $$Ce^{4+}$$ a strong oxidizing agent because:

$$\bullet$$ It has a strong tendency to gain an electron (get reduced) to form $$Ce^{3+}$$.

$$\bullet$$ The standard reduction potential $$E^\circ(Ce^{4+}/Ce^{3+}) = +1.74$$ V is very high, confirming its strong oxidizing power.

$$\bullet$$ In aqueous solution, $$Ce^{4+}$$ readily accepts electrons from reducing agents.

Evaluate the options: Option A: It will prefer to gain electron and act as an oxidizing agent — Correct. $$Ce^{4+}$$ is a well-known strong oxidizing agent in analytical chemistry (ceric ammonium nitrate is widely used as an oxidant).

Option B: It will prefer to give away an electron and behave as reducing agent — Incorrect. $$Ce^{4+}$$ does not lose electrons easily.

Option C: It will not prefer to undergo redox reactions — Incorrect. $$Ce^{4+}$$ readily participates in redox reactions as an oxidant.

Option D: It acts as both, oxidizing and reducing agent — Incorrect. It predominantly acts as an oxidizing agent.

The correct answer is Option A: It will prefer to gain electron and act as an oxidizing agent.

Silver is commonly extracted using the MacArthur-Forrest cyanide process. In this method, metallic silver reacts with sodium cyanide in the presence of water and oxygen to form the soluble complex sodium dicyanoargentate. This reaction can be represented as follows: $$ 4Ag + 8NaCN + 2H_2O + O_2 \rightarrow 4Na[Ag(CN)_2] + 4NaOH $$ Once the silver is in solution, zinc dust is added to precipitate pure silver metal, which demonstrates that silver extraction does involve cyanide.

Similarly, gold extraction employs the cyanide process, in which gold metal reacts with sodium cyanide, water, and oxygen to yield sodium dicyanoaurate as the soluble complex: $$ 4Au + 8NaCN + 2H_2O + O_2 \rightarrow 4Na[Au(CN)_2] + 4NaOH $$ The precipitated gold metal is then recovered by adding zinc dust, confirming that gold extraction also involves cyanide.

Although zinc is typically obtained by roasting zinc sulfide ore (ZnS) to zinc oxide followed by carbon reduction, certain hydrometallurgical methods utilize cyanide. In these processes, zinc forms the soluble complex $$[Zn(CN)_4]^{2-}$$, showing that cyanide can participate in zinc extraction under specific conditions.

In contrast, copper extraction does not rely on cyanide. Copper is obtained from ores such as copper pyrite ($$CuFeS_2$$) through a sequence of roasting, smelting with silica flux in a blast furnace, Bessemerisation to produce blister copper, and electrolytic refining to yield pure copper. Because copper(I) cyanide is insoluble, cyanide leaching is unsuitable for copper.

The correct answer is Option D: copper.

We are asked about the liquation process used for refining tin (Sn).

Liquation is a metallurgical purification method that exploits the difference in melting points between a metal and its impurities. Tin has a relatively low melting point of about 232°C, whereas many of its common impurities have much higher melting points.

In this process, the impure tin ore is placed on the top of a sloping surface (hearth) and heated gently. Since tin melts at a lower temperature than its impurities, the tin melts and flows down the slope under gravity, leaving behind the higher-melting-point impurities on the hearth. The molten tin is collected at the bottom in a pure form.

Now let us check the options. Option A (reacted with acid) describes acid leaching, not liquation. Option B (dissolved in water) is irrelevant since metals like tin do not dissolve in water. Option D (fused with NaOH) describes a fusion process for dissolving amphoteric ores, not liquation. Option C correctly states that the metal is brought to molten form and made to flow on a slope, which is precisely what happens in liquation.

Hence, the correct answer is Option C.

We need to match each ore with its correct composition.

Recall the compositions of each ore. Siderite is an iron ore with the formula FeCO₃, which corresponds to (I). Malachite is a copper ore with the formula CuCO₃·Cu(OH)₂ (basic copper carbonate), matching (II). Sphalerite, also called zinc blende, is a zinc ore with the formula ZnS and corresponds to (III). Calamine is a zinc ore with the formula ZnCO₃, matching (IV).

Therefore, the matching is A-I, B-II, C-III, D-IV. The answer is Option A: A-I, B-II, C-III, D-IV.

We need to identify which compound(s) are removed as slag during copper extraction.

In the extraction of copper from copper pyrites ($$CuFeS_2$$), the ore is roasted and then smelted in a blast furnace or reverberatory furnace.

During smelting, silica ($$SiO_2$$) is added as a flux and the iron impurity in the form of $$FeO$$ reacts with silica via the reaction $$FeO + SiO_2 \rightarrow FeSiO_3 \text{ (ferrous silicate - slag)}$$ to form the slag.

When evaluating the options, CaO is not a typical impurity in copper pyrites extraction; FeO is removed as iron silicate slag ($$FeSiO_3$$); $$Al_2O_3$$ is not typically removed as slag in copper extraction; ZnO is not a primary slag component in copper extraction; and NiO is not typically removed as slag in copper extraction.

The main slag formed during copper extraction is ferrous silicate ($$FeSiO_3$$), formed from FeO.

Hence, the correct answer is Option B.

We need to find the metal ion (in gaseous state) with the lowest spin-only magnetic moment.

$$V^{2+}$$: $$[Ar] 3d^3$$ — 3 unpaired electrons

$$Ni^{2+}$$: $$[Ar] 3d^8$$ — 2 unpaired electrons

$$Cr^{2+}$$: $$[Ar] 3d^4$$ — 4 unpaired electrons

$$Fe^{2+}$$: $$[Ar] 3d^6$$ — 4 unpaired electrons

$$V^{2+}$$: $$\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87$$ B.M.

$$Ni^{2+}$$: $$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83$$ B.M.

$$Cr^{2+}$$: $$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$$ B.M.

$$Fe^{2+}$$: $$\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$$ B.M.

$$Ni^{2+}$$ has the lowest spin-only magnetic moment of $$2.83$$ B.M. with only 2 unpaired electrons.

The correct answer is Option B.

We need to identify which lanthanoid element most easily deviates from the common +3 oxidation state.

The electronic configurations of the lanthanoids are:

Ce (Z = 58): [Xe] 4f$$^1$$ 5d$$^1$$ 6s$$^2$$

La (Z = 57): [Xe] 5d$$^1$$ 6s$$^2$$ (no 4f electrons)

Lu (Z = 71): [Xe] 4f$$^{14}$$ 5d$$^1$$ 6s$$^2$$

Gd (Z = 64): [Xe] 4f$$^7$$ 5d$$^1$$ 6s$$^2$$

Lanthanoids tend to show +3 as the most stable oxidation state. However, deviation occurs when attaining a particularly stable electronic configuration (empty, half-filled, or fully filled 4f orbitals).

Ce$$^{3+}$$: [Xe] 4f$$^1$$ — By losing one more electron, Ce$$^{4+}$$: [Xe] 4f$$^0$$ achieves a noble gas configuration (empty 4f). This makes Ce$$^{4+}$$ quite stable, so Ce easily deviates to +4 oxidation state.

La$$^{3+}$$: [Xe] 4f$$^0$$ — already has empty 4f, so +3 is very stable. No easy deviation.

Lu$$^{3+}$$: [Xe] 4f$$^{14}$$ — fully filled 4f, so +3 is very stable. No easy deviation.

Gd$$^{3+}$$: [Xe] 4f$$^7$$ — half-filled 4f, so +3 is very stable. No easy deviation.

Cerium (Ce) most easily deviates from the +3 oxidation state because Ce$$^{4+}$$ achieves the very stable empty 4f configuration [Xe] 4f$$^0$$.

Therefore, the correct answer is Option A.

We need to identify which of the given methods are not used for refining metals.

We classify each method as follows. (A) Liquation: This is a refining method. It is used to refine metals with low melting points (like tin, lead, bismuth). The impure metal is heated on a sloped surface; the pure metal melts and flows away from the higher melting impurities.

(B) Calcination: This is a concentration/extraction method, NOT a refining method. Calcination involves heating the ore below its melting point in the absence of air to remove moisture, CO$$_2$$, and other volatile impurities from the ore. It is used in the early stages of extraction.

(C) Electrolysis: This is a refining method. Electrolytic refining is used for metals like copper, zinc, nickel, silver, and gold. The impure metal is made the anode and pure metal is deposited at the cathode.

(D) Leaching: This is a concentration method, NOT a refining method. Leaching involves selectively dissolving the ore in a suitable solvent (like NaOH for bauxite in Bayer's process) to separate it from impurities.

(E) Distillation: This is a refining method. Distillation refining is used for metals with low boiling points (like zinc and mercury). The impure metal is evaporated and then condensed to get pure metal.

Among these methods, calcination (B) and leaching (D) are concentration/extraction methods, not refining methods.

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

We need to determine which lanthanoid is most stable in the divalent ($$Ln^{2+}$$) form. Lanthanoids are most stable in the +3 oxidation state. However, some lanthanoids can exhibit the +2 oxidation state, and the stability of the +2 state depends on the electronic configuration of the resulting ion. Ions with empty ($$4f^0$$), half-filled ($$4f^7$$), or completely filled ($$4f^{14}$$) configurations are exceptionally stable.

Option A: Yb (Z = 70) Yb atom: $$[Xe] 4f^{14} 6s^2$$, so $$Yb^{2+}$$ is $$[Xe] 4f^{14}$$ — completely filled 4f shell.

Option B: Sm (Z = 62) Sm atom: $$[Xe] 4f^6 6s^2$$, so $$Sm^{2+}$$ is $$[Xe] 4f^6$$ — neither half-filled nor completely filled.

Option C: Eu (Z = 63) Eu atom: $$[Xe] 4f^7 6s^2$$, thus $$Eu^{2+}$$ has $$[Xe] 4f^7$$ — exactly half-filled 4f shell.

Option D: Ce (Z = 58) Ce atom: $$[Xe] 4f^1 5d^1 6s^2$$, giving $$Ce^{2+}$$ as $$[Xe] 4f^1 5d^1$$ — neither half-filled nor completely filled.

Both $$Eu^{2+}$$ ($$4f^7$$, half-filled) and $$Yb^{2+}$$ ($$4f^{14}$$, completely filled) have stable configurations. However, $$Eu^{2+}$$ is considered the most stable divalent lanthanoid because the half-filled $$4f^7$$ configuration provides exceptional stability due to maximum exchange energy, and europium has the strongest tendency among all lanthanoids to form the +2 state, as evidenced by its highly negative reduction potential for the $$Eu^{3+}/Eu^{2+}$$ couple ($$-0.36$$ V), meaning $$Eu^{2+}$$ strongly resists oxidation to $$Eu^{3+}$$.

The correct answer is Option C: Eu (Atomic Number 63).

We need to identify the strongest oxidizing agent among $$Mn^{3+}$$, $$Ti^{3+}$$, $$Fe^{3+}$$, and $$Cr^{3+}$$.

Understand what makes a strong oxidizing agent: A strong oxidizing agent has a high tendency to gain electrons (get reduced). This corresponds to a high standard reduction potential for the $$M^{3+}/M^{2+}$$ couple.

Analyze the electronic configurations: $$\bullet$$ $$Mn^{3+}$$: $$[Ar] 3d^4$$. On gaining an electron, it becomes $$Mn^{2+}$$: $$[Ar] 3d^5$$ (half-filled, extra stable).

$$\bullet$$ $$Ti^{3+}$$: $$[Ar] 3d^1$$. On gaining an electron, $$Ti^{2+}$$: $$[Ar] 3d^2$$ (no special stability).

$$\bullet$$ $$Fe^{3+}$$: $$[Ar] 3d^5$$ (already half-filled, stable). On gaining an electron, $$Fe^{2+}$$: $$[Ar] 3d^6$$ (no special stability).

$$\bullet$$ $$Cr^{3+}$$: $$[Ar] 3d^3$$ (half-filled $$t_{2g}$$, stable). On gaining an electron, $$Cr^{2+}$$: $$[Ar] 3d^4$$ (no special stability).

Compare reduction potentials: The standard reduction potentials ($$E^\circ$$ for $$M^{3+}/M^{2+}$$) are:

$$\bullet$$ $$Mn^{3+}/Mn^{2+}$$: $$E^\circ = +1.51$$ V (very high, due to extra stability of half-filled $$d^5$$)

$$\bullet$$ $$Fe^{3+}/Fe^{2+}$$: $$E^\circ = +0.77$$ V

$$\bullet$$ $$Cr^{3+}/Cr^{2+}$$: $$E^\circ = -0.41$$ V

$$\bullet$$ $$Ti^{3+}/Ti^{2+}$$: $$E^\circ = -0.37$$ V

$$Mn^{3+}$$ has the highest reduction potential because its reduction product $$Mn^{2+}$$ ($$d^5$$) has the extra stability associated with the half-filled d subshell.

The correct answer is Option A: $$Mn^{3+}$$.

We need to evaluate two statements about pig iron and cast iron.

Assessing the claim that "Pig iron is obtained by heating cast iron with scrap iron" reveals that it is incorrect. The correct facts are as follows:

- Pig iron is obtained directly from the blast furnace by smelting iron ore. It contains about 4% carbon.

- Wrought iron (not pig iron) is obtained by heating cast iron/pig iron with scrap iron and haematite in a reverberatory furnace (pudding process).

- Cast iron is obtained by melting pig iron with scrap iron and coke in a cupola furnace.

Turning to the second statement—that "Pig iron has a relatively lower carbon content than that of cast iron"—we again find it to be incorrect. Pig iron has about 4% carbon, while cast iron has about 3%, indicating that pig iron actually has a higher carbon content.

Thus, the correct choice is Option B.

We need to identify which property makes a metal most suitable for refining by the liquation method.

Liquation is a metallurgical method used to refine metals. In this method, the impure metal is placed on the sloping hearth of a reverberatory furnace and heated gently. The metal with a low melting point melts and flows down the slope, leaving behind the higher-melting impurities on the hearth.

The essential condition for liquation to work is that the metal must have a low melting point compared to its impurities. This way, the metal melts at a relatively low temperature while the impurities remain solid.

Liquation is commonly used for refining metals like Tin (Sn) with melting point $$232°C$$, Bismuth (Bi) with melting point $$271°C$$, and Lead (Pb) with melting point $$327°C$$. All these metals have relatively low melting points.

Option A: Low melting point — Correct. This is the key requirement for liquation.

Option B: High boiling point — Not specifically relevant to liquation.

Option C: High electrical conductivity — This is relevant to electrolytic refining, not liquation.

Option D: Less tendency to be soluble in melts than impurities — This is not the correct description of liquation.

Hence, the correct answer is Option A: Low melting point.

We need to find lanthanide ions where the f orbitals are half-filled and completely filled, respectively.

The 4f subshell can hold a maximum of 14 electrons. It is half-filled at 7 electrons and completely filled at 14 electrons.

Determine the electron configuration of each ion in Option C.

Tb (Z = 65): Ground state configuration is $$[Xe] 4f^9 6s^2$$.

For $$Tb^{4+}$$, we remove 4 electrons (2 from 6s, 1 from 5d if present, and remaining from 4f):

$$Tb^{4+}: [Xe] 4f^7$$

This gives 7 electrons in 4f, which is a half-filled configuration.

Yb (Z = 70): Ground state configuration is $$[Xe] 4f^{14} 6s^2$$.

For $$Yb^{2+}$$, we remove 2 electrons (both from 6s):

$$Yb^{2+}: [Xe] 4f^{14}$$

This gives 14 electrons in 4f, which is a completely filled configuration.

Verify other options are incorrect.

Option A: $$Eu^{2+}$$: Eu (Z = 63) has $$[Xe] 4f^7 6s^2$$. $$Eu^{2+}: [Xe] 4f^7$$ (half-filled). $$Tm^{3+}$$: Tm (Z = 69) has $$[Xe] 4f^{13} 6s^2$$. $$Tm^{3+}: [Xe] 4f^{12}$$ (neither half nor completely filled). So this pair does not satisfy the condition.

Option B: $$Sm^{2+}$$: Sm (Z = 62) has $$[Xe] 4f^6 6s^2$$. $$Sm^{2+}: [Xe] 4f^6$$ (not half-filled). This fails.

Option D: $$Dy^{3+}$$: Dy (Z = 66) has $$[Xe] 4f^{10} 6s^2$$. $$Dy^{3+}: [Xe] 4f^9$$ (not half-filled). This fails.

Therefore, Option C is correct: $$Tb^{4+}$$ has half-filled 4f orbitals ($$4f^7$$) and $$Yb^{2+}$$ has completely filled 4f orbitals ($$4f^{14}$$).

We need to identify which metal is not extracted from its sulphide ore.

Let us examine each option:

Option A: Aluminium

Aluminium is primarily extracted from bauxite ($$Al_2O_3 \cdot 2H_2O$$), which is an oxide ore, not a sulphide ore. The extraction is done by the Bayer's process followed by the Hall-Heroult electrolytic process.

Option B: Iron

While iron is commonly extracted from oxide ores like haematite ($$Fe_2O_3$$), iron pyrite ($$FeS_2$$) is a sulphide ore of iron. However, iron pyrite is mainly used for making sulphuric acid rather than extracting iron. Nevertheless, iron can be associated with sulphide ores.

Option C: Lead

Lead is extracted from its sulphide ore galena ($$PbS$$). The ore is roasted to convert it to $$PbO$$ and then reduced with carbon.

Option D: Zinc

Zinc is extracted from its sulphide ore zinc blende ($$ZnS$$). The ore is roasted to form $$ZnO$$ and then reduced with carbon.

Aluminium does not have any commercially important sulphide ore. It is always extracted from its oxide ore (bauxite).

Therefore, the correct answer is Option A: Aluminium.

The Hall-Heroult process is the industrial method used for the extraction of aluminium from its ore (alumina, $$Al_2O_3$$).

Let us examine each option:

Option A: $$Cr_2O_3 + 2Al \to Al_2O_3 + 2Cr$$ - This is the thermite reaction (aluminothermic process) used for the reduction of chromium oxide.

Option B: $$2Al_2O_3 + 3C \to 4Al + 3CO_2$$ - This represents the Hall-Heroult process. In this process, purified alumina ($$Al_2O_3$$) is dissolved in molten cryolite ($$Na_3AlF_6$$) and electrolyzed using carbon electrodes. At the cathode, aluminium is deposited, and at the carbon anode, oxygen is released which reacts with the carbon anode to form $$CO_2$$. The overall reaction is as shown.

Option C: $$FeO + CO \to Fe + CO_2$$ - This is the reduction of iron oxide in the blast furnace during iron extraction.

Option D: $$2[Au(CN)_2]^-(aq) + Zn(s) \to 2Au(s) + [Zn(CN)_4]^{2-}$$ - This is the MacArthur-Forrest process (cyanide process) used for gold extraction.

The correct answer is Option B: $$2Al_2O_3 + 3C \to 4Al + 3CO_2$$.

We need to evaluate both statements about d-block element chemistry.

The first statement asserts that "Iron (III) catalyst, acidified $$K_2Cr_2O_7$$ and neutral $$KMnO_4$$ have the ability to oxidise $$I^-$$ to $$I_2$$ independently." In this context, $$Fe^{3+}$$ can oxidise $$I^-$$ to $$I_2$$, as shown by the reaction:

$$ 2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2 $$

This is correct since $$E^\circ(Fe^{3+}/Fe^{2+}) = +0.77$$ V and $$E^\circ(I_2/I^-) = +0.54$$ V. Similarly, dichromate in acidic medium is a strong oxidising agent and can oxidise $$I^-$$ to $$I_2$$:

$$ Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O $$

This reaction is also correct. In a neutral medium, $$KMnO_4$$ acts as a moderate oxidiser ($$MnO_4^- \rightarrow MnO_2$$) and oxidises $$I^-$$ to $$I_2$$:

$$ 2MnO_4^- + 6I^- + 4H_2O \rightarrow 2MnO_2 + 3I_2 + 8OH^- $$

Thus each oxidising agent can convert iodide to iodine. However, the term "Iron (III) catalyst" is misleading: $$Fe^{3+}$$ functions as an oxidising agent rather than a catalyst because it is reduced to $$Fe^{2+}$$ in the process. Therefore, the first statement is false.

The second statement claims that the manganate ion is paramagnetic and involves $$p\pi - p\pi$$ bonding. The manganate ion, $$MnO_4^{2-}$$, has manganese in the +6 oxidation state with the electronic configuration $$[Ar] 3d^1$$, which indeed gives one unpaired electron and makes it paramagnetic. However, the bonding in $$MnO_4^{2-}$$ involves $$d\pi - p\pi$$ overlap between Mn d-orbitals and O p-orbitals, not $$p\pi - p\pi$$ bonding. Since this description is incorrect, the second statement is also false.

Consequently, both statements are false, so the correct answer is Option B: Both Statement I and Statement II are false.

We need to find the total number of $$Mn=O$$ bonds in $$Mn_2O_7$$ (dimanganese heptoxide). To determine the structure of $$Mn_2O_7$$, note that $$Mn_2O_7$$ is the anhydride of permanganic acid ($$HMnO_4$$). In its structure:

- Two $$MnO_3$$ units are linked by a bridging oxygen atom (Mn-O-Mn).

- Each Mn atom is in the +7 oxidation state.

- Each Mn is tetrahedrally coordinated with 4 oxygen atoms.

Identifying the types of Mn-O bonds, for each Mn atom:

- One oxygen is the bridging oxygen (Mn-O-Mn) — a single bond (Mn-O) shared between both Mn atoms.

- Three oxygen atoms are terminal — these form $$Mn=O$$ double bonds.

Counting the total $$Mn=O$$ bonds, each Mn has three terminal $$Mn=O$$ bonds. With two Mn atoms, the total $$Mn=O$$ bonds = $$3 \times 2 = 6$$.

The structure can be represented as:

$$ O=Mn(=O)(=O)-O-Mn(=O)(=O)=O $$

The correct answer is Option C: 6.

In a blast furnace the charge contains iron ore ($$Fe_2O_3$$ / $$Fe_3O_4$$), coke ($$C$$) and limestone ($$CaCO_3$$). The principal reactions that occur between $$900 - 1500 \text{ K}$$ are summarised below, followed by the assessment of each option.

Reaction of limestone and removal of silica
$$CaCO_3 \;\xrightarrow{\,T\,}\; CaO + CO_2$$
$$CaO + SiO_2 \;\rightarrow\; CaSiO_3$$ (fusible slag)
Thus limestone supplies $$CaO$$ which combines with the acidic impurity $$SiO_2$$ to give calcium silicate slag.

Reactions involving coke
(i) $$C + O_2 \;\rightarrow\; CO_2 + \text{heat}$$
(ii) $$C + CO_2 \;\rightarrow\; 2\,CO$$ (Boudouard reaction)
The carbon monoxide produced in (ii) rises up and reduces the iron oxides.

Composition of molten iron (pig iron)
The iron tapped at the bottom (pig iron) dissolves a considerable amount of carbon along with traces of $$Si, P, S$$ etc. The carbon content is roughly $$3 - 5\%$$, the textbook average being $$\approx 4\%$$.

Nature of exhaust (top) gas
The gases leaving the throat mainly contain $$CO$$ (≈ 30-35 %), $$N_2$$ (from blast air, ≈ 60-65 %) and a small amount of $$CO_2$$. Oxides of nitrogen such as $$NO_2$$ are not produced in any significant amount under the strongly reducing conditions of the furnace.

Option-wise verification

Option A: Limestone is used to remove silicate impurity.
Correct, because $$CaO$$ formed from limestone reacts with $$SiO_2$$ to give slag $$CaSiO_3$$.

Option B: Pig iron obtained from blast furnace contains about 4 % carbon.
Correct; the molten iron indeed contains roughly $$3 - 5\%$$ carbon.

Option C: Coke ($$C$$) converts $$CO_2$$ to $$CO$$.
Correct, via the Boudouard reaction $$C + CO_2 \rightarrow 2\,CO$$ which predominates between $$900 - 1500 \text{ K}$$.

Option D: Exhaust gases consist of $$NO_2$$ and $$CO$$.
Incorrect; the main constituents are $$CO$$ and $$N_2$$, not $$NO_2$$.

Hence the correct choices are
Option A (Limestone removes silicate impurity),
Option B (Pig iron contains ≈ 4 % C) and
Option C (Coke converts $$CO_2$$ to $$CO$$).

Option A, Option B and Option C are correct.

We need to find the spin-only magnetic moment of the compound with the strongest oxidizing ability among $$MnF_4$$, $$MnF_3$$, and $$MnF_2$$.

Since fluorine is always -1, in $$MnF_2$$ manganese is in the +2 oxidation state ($$3d^5$$ configuration), in $$MnF_3$$ it is in the +3 oxidation state ($$3d^4$$ configuration), and in $$MnF_4$$ it is in the +4 oxidation state ($$3d^3$$ configuration).

Among these, $$MnF_3$$ is the strongest oxidizing agent because $$Mn^{3+}$$ ($$3d^4$$) has a very strong tendency to gain one electron and form $$Mn^{2+}$$ ($$3d^5$$), which has the exceptionally stable half‐filled $$d$$‐subshell configuration. The large $$Mn^{3+}/Mn^{2+}$$ reduction potential ($$E^\circ = +1.51$$ V) confirms this.

Manganese (Z = 25) in its ground state is $$[Ar]\,3d^5\,4s^2$$.
Removing two electrons from 4s and one from 3d gives $$Mn^{3+} = [Ar]\,3d^4$$.

Because fluoride ($$F^-$$) is a weak field ligand, it does not cause pairing of electrons, so the $$3d^4$$ configuration remains high spin:
$$\uparrow\;|\;\uparrow\;|\;\uparrow\;|\;\uparrow\;|\;\_$$
Thus the number of unpaired electrons is $$n = 4$$.

The spin‐only magnetic moment is given by $$\mu_s = \sqrt{n(n+2)}\text{ B.M.}$$
Substituting $$n=4$$ gives $$\mu_s = \sqrt{4(4+2)} = \sqrt{24} = 2\sqrt{6} = 4.899\text{ B.M.}$$
Rounding to the nearest integer yields $$\mu_s \approx 5\text{ B.M.}$$

Therefore, the spin‐only magnetic moment of the strongest oxidizing compound is $$\boxed{5}$$ B.M.

We need to identify which of the given minerals are sulphide-based.

(a) Baryte: The chemical formula of baryte is $$BaSO_4$$ (barium sulphate). This is a sulphate mineral, not a sulphide mineral.

(b) Galena: The chemical formula of galena is $$PbS$$ (lead sulphide). This is a sulphide mineral.

(c) Zinc blende: The chemical formula of zinc blende is $$ZnS$$ (zinc sulphide). This is a sulphide mineral.

(d) Copper pyrites: The chemical formula of copper pyrites is $$CuFeS_2$$ (copper iron sulphide). This is a sulphide mineral.

Therefore, among the four minerals, 3 are sulphide-based (galena, zinc blende, and copper pyrites).

We need to identify which ion among $$ Co^{3+} $$, $$ Ti^{2+} $$, $$ V^{2+} $$, and $$ Cr^{2+} $$ cannot liberate $$ H_2 $$ from dilute mineral acid, and then find its spin-only magnetic moment in the gaseous state.

To liberate $$ H_2 $$ from acid, the ion must be a reducing agent (must be oxidized). The standard reduction potentials are:

$$ Ti^{2+} $$, $$ V^{2+} $$, and $$ Cr^{2+} $$ are all good reducing agents that can reduce $$ H^+ $$ to $$ H_2 $$.

However, $$ Co^{3+} $$ is a strong oxidizing agent ($$ E^\circ_{Co^{3+}/Co^{2+}} = +1.82 \text{ V} $$). It will oxidize water rather than liberate $$ H_2 $$. Therefore, $$ Co^{3+} $$ cannot be used to liberate $$ H_2 $$ from dilute acid.

Cobalt (Co) has atomic number 27. Its electronic configuration is $$ [Ar] 3d^7 4s^2 $$.

For $$ Co^{3+} $$: remove 2 electrons from 4s and 1 from 3d, giving $$ [Ar] 3d^6 $$.

In the gaseous state (no ligand field), the $$ 3d^6 $$ configuration follows Hund's rule:

$$\uparrow\downarrow \quad \uparrow \quad \uparrow \quad \uparrow \quad \uparrow$$

This gives 4 unpaired electrons.

$$\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} = 2\sqrt{6} \approx 4.9 \text{ B.M.}$$

Rounding to the nearest integer:

$$\mu \approx 5 \text{ B.M.}$$

The spin-only magnetic moment is 5 B.M.

We need to identify the principal ores of iron from the given list of ores.

Let us analyze each ore:

1. Bauxite ($$Al_2O_3 \cdot 2H_2O$$): Principal ore of Aluminium, not iron.

2. Siderite ($$FeCO_3$$): This is an ore of iron. $$\checkmark$$

3. Cuprite ($$Cu_2O$$): Principal ore of Copper, not iron.

4. Calamine ($$ZnCO_3$$): Principal ore of Zinc, not iron.

5. Haematite ($$Fe_2O_3$$): This is a principal ore of iron. $$\checkmark$$

6. Kaolinite ($$Al_2Si_2O_5(OH)_4$$): This is an ore of Aluminium (a clay mineral), not iron.

7. Malachite ($$Cu(OH)_2 \cdot CuCO_3$$): Principal ore of Copper, not iron.

8. Magnetite ($$Fe_3O_4$$): This is a principal ore of iron. $$\checkmark$$

9. Sphalerite ($$ZnS$$): Principal ore of Zinc, not iron.

10. Limonite ($$2Fe_2O_3 \cdot 3H_2O$$): This is an ore of iron. $$\checkmark$$

11. Cryolite ($$Na_3AlF_6$$): Used in extraction of Aluminium, not iron.

Iron ores identified: Siderite, Haematite, Magnetite, Limonite = 4

Hence, the answer is 4.

Step 1: Identify product A from the first reaction.

The chromite ore reaction with sodium carbonate and oxygen (fusion):

$$4\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2$$

So product A is sodium chromate ($$\text{Na}_2\text{CrO}_4$$).

Step 2: Identify product B from the second reaction.

When sodium chromate reacts with acid:

$$2\text{Na}_2\text{CrO}_4 + 2\text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{O} + 2\text{Na}^+$$

So product B is sodium dichromate ($$\text{Na}_2\text{Cr}_2\text{O}_7$$).

Step 3: Count terminal oxygen atoms in $$\text{Cr}_2\text{O}_7^{2-}$$.

In the dichromate ion, there are 7 oxygen atoms total. One oxygen atom bridges the two chromium atoms, while the remaining 6 oxygen atoms are terminal (3 on each Cr).

Therefore, the number of terminal oxygen atoms in product B is 6.

We need to find the spin-only magnetic moment of the most basic oxide of vanadium.

Identify the most basic oxide.

The oxides given are $$V_2O_3$$, $$V_2O_4$$ (i.e., $$VO_2$$), and $$V_2O_5$$.

The oxidation states of vanadium are +3, +4, and +5 respectively.

Lower oxidation state oxides are more basic. Therefore, $$V_2O_3$$ (with V in +3 state) is the most basic oxide.

Find the electronic configuration of $$V^{3+}$$.

V: [Ar] 3d$$^3$$ 4s$$^2$$

$$V^{3+}$$: [Ar] 3d$$^2$$ (2 unpaired electrons)

Calculate the spin-only magnetic moment.

$$\mu = \sqrt{n(n+2)} \text{ B.M.}$$

where $$n$$ = number of unpaired electrons = 2

$$\mu = \sqrt{2(2+2)} = \sqrt{8} = 2\sqrt{2} \approx 2.83 \text{ B.M.}$$

Rounding to the nearest integer: $$\mu \approx \textbf{3}$$ B.M.

Step 1: Write the disproportionation reaction of manganate ion in acidic medium.

The manganate ion ($$\text{MnO}_4^{2-}$$) has Mn in the +6 oxidation state. In acidic solution, it disproportionates:

$$3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^{-} + \text{MnO}_2 + 2\text{H}_2\text{O}$$

The two products are:

- Permanganate ion ($$\text{MnO}_4^{-}$$): Mn is in +7 oxidation state (higher)

- Manganese dioxide ($$\text{MnO}_2$$): Mn is in +4 oxidation state (lower)

Step 2: Find the spin-only magnetic moment of the product with higher oxidation state.

In $$\text{MnO}_4^{-}$$, Mn is in the +7 state with electronic configuration [Ar] 3d$$^0$$.

Since there are no unpaired electrons ($$n = 0$$):

$$\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \text{ B.M.}$$

The spin-only magnetic moment is 0 B.M.

In the manganate ion:

$$\mathrm{MnO_4^{2-}}$$

let the oxidation state of manganese be:

$$\mathrm{x}$$

Then:

$$\mathrm{x + 4(-2) = -2}$$

$$\mathrm{x - 8 = -2}$$

$$\mathrm{x = +6}$$

Thus, manganese in manganate ion has oxidation state:

$$\mathrm{+6}$$

In acidic medium, manganate ion undergoes disproportionation to form:

$$\mathrm{MnO_4^-}$$

and

$$\mathrm{MnO_2}$$

For permanganate ion:

$$\mathrm{MnO_4^-}$$

let oxidation state of manganese be:

$$\mathrm{y}$$

$$\mathrm{y + 4(-2) = -1}$$

$$\mathrm{y = +7}$$

For manganese dioxide:

$$\mathrm{MnO_2}$$

let oxidation state of manganese be:

$$\mathrm{z}$$

$$\mathrm{z + 2(-2) = 0}$$

$$\mathrm{z = +4}$$

Since oxidation state in compound $$\mathrm{B}$$ is smaller:

$$\mathrm{B = MnO_2}$$

In $$\mathrm{MnO_2}$$, manganese exists as:

$$\mathrm{Mn^{4+}}$$

Electronic configuration of neutral manganese:

$$\mathrm{[Ar]\ 3d^5 4s^2}$$

Electronic configuration of:

$$\mathrm{Mn^{4+}}$$

is:

$$\mathrm{3d^3}$$

Hence, number of unpaired electrons:

$$\mathrm{n = 3}$$

Spin-only magnetic moment:

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

$$\mathrm{\mu = \sqrt{3(3+2)}}$$

$$\mathrm{\mu = \sqrt{15}}$$

$$\mathrm{\mu \approx 3.87\ BM}$$

Rounding to nearest integer:

$$\mathrm{4}$$

Let us analyze each statement about Copper:

Statement (A): Cu(II) complexes are always paramagnetic

Cu(II) has the electronic configuration [Ar] 3d$$^9$$. With 9 electrons in the d-orbitals, there is always one unpaired electron regardless of the geometry or ligand field strength (whether the complex is octahedral, tetrahedral, or square planar).

Therefore, Cu(II) complexes are always paramagnetic. This statement is correct.

Statement (B): Cu(I) complexes are generally colourless

Cu(I) has the electronic configuration [Ar] 3d$$^{10}$$. Since all d-orbitals are completely filled, there are no d-d transitions possible. The colour of transition metal complexes arises from d-d electronic transitions, and since these cannot occur in a completely filled d-subshell, Cu(I) complexes are generally colourless.

This statement is correct.

Statement (C): Cu(I) is easily oxidized

Cu(I) with d$$^{10}$$ configuration is not very stable in aqueous solution and readily undergoes disproportionation or oxidation to Cu(II). The standard reduction potential $$E^\circ(Cu^{2+}/Cu^{+}) = +0.15$$ V indicates Cu(I) can be easily oxidized to Cu(II).

This statement is correct.

Statement (D): In Fehling solution, the active reagent has Cu(I)

Fehling's solution contains Cu(II) ions complexed with sodium potassium tartrate (Rochelle salt). The deep blue colour of Fehling's solution is due to the Cu(II)-tartrate complex. During the reaction with reducing sugars, Cu(II) gets reduced to Cu(I) as Cu$$_2$$O (red precipitate), but the active reagent itself contains Cu(II), not Cu(I).

This statement is incorrect.

The correct statements are (A), (B), and (C).

Therefore, the number of correct statements is 3.

We need to find the spin-only magnetic moment of the $$M^{3+}$$ ion (from the given pairs) that has a negative standard electrode potential for the $$M^{3+}/M^{2+}$$ couple.

Identify the standard electrode potentials of the given pairs

The standard electrode potentials $$E°(M^{3+}/M^{2+})$$ are:

- $$Cr^{3+}/Cr^{2+}$$: $$E° = -0.41 \text{ V}$$ (negative)

- $$Mn^{3+}/Mn^{2+}$$: $$E° = +1.51 \text{ V}$$ (positive)

- $$Fe^{3+}/Fe^{2+}$$: $$E° = +0.77 \text{ V}$$ (positive)

- $$Co^{3+}/Co^{2+}$$: $$E° = +1.81 \text{ V}$$ (positive)

Identify the ion with negative $$E°$$

Only $$Cr^{3+}/Cr^{2+}$$ has a negative standard electrode potential. So we need the magnetic moment of $$Cr^{3+}$$.

Find the electronic configuration of $$Cr^{3+}$$

$$Cr$$: $$[Ar] 3d^5 4s^1$$ (anomalous configuration)

$$Cr^{3+}$$: $$[Ar] 3d^3$$ (3 unpaired electrons)

Calculate the spin-only magnetic moment

$$\mu = \sqrt{n(n+2)} \text{ B.M.}$$

where $$n$$ = number of unpaired electrons = 3

$$\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87 \text{ B.M.}$$

Round to the nearest integer

$$\mu \approx 4 \text{ B.M.}$$

The answer is 4 B.M.

When acidified potassium permanganate ($$KMnO_4$$) oxidises oxalic acid ($$H_2C_2O_4$$), the balanced reaction is:

$$2KMnO_4 + 3H_2SO_4 + 5H_2C_2O_4 \to 2MnSO_4 + K_2SO_4 + 10CO_2 + 8H_2O$$

In this reaction, manganese goes from +7 oxidation state (in $$KMnO_4$$) to +2 oxidation state (in $$MnSO_4$$).

The manganese product is $$Mn^{2+}$$.

Electronic configuration of $$Mn^{2+}$$:

Mn has atomic number 25. Its electronic configuration is [Ar] 3d$$^5$$ 4s$$^2$$.

$$Mn^{2+}$$ loses 2 electrons from 4s: [Ar] 3d$$^5$$

With 5 electrons in the 3d subshell, each orbital gets one electron (by Hund's rule):

$$d_{xy}^1 \; d_{xz}^1 \; d_{yz}^1 \; d_{x^2-y^2}^1 \; d_{z^2}^1$$

Number of unpaired electrons ($$n$$) = 5

The spin-only magnetic moment formula is:

$$\mu = \sqrt{n(n+2)} \text{ B.M.}$$

Substituting $$n = 5$$:

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

$$= \sqrt{5 \times 7}$$

$$= \sqrt{35}$$

$$= 5.916 \text{ B.M.}$$

Rounding to the nearest integer:

$$\mu \approx$$ 6 B.M.

For transition-metal ions the spin-only magnetic moment $$\mu_{\text{spin}}$$ (in Bohr Magnetons, BM) is obtained from the well-known formula

$$\mu_{\text{spin}}=\sqrt{n\,(n+2)}\;\text{BM},$$

where $$n$$ is the number of unpaired electrons present in the ion. We must therefore write the electronic configuration of each ion, count the unpaired electrons, and then substitute the value of $$n$$ in the above expression.

Ion 1 : Ti$$^{3+}$$

Neutral titanium (Ti, $$Z=22$$) has the configuration

$$\text{Ti} : [\text{Ar}]\,3d^{2}\,4s^{2}.$$

For the $$+3$$ oxidation state we remove two electrons first from the $$4s$$ orbital and one electron from the $$3d$$ orbital:

$$\text{Ti}^{3+} : [\text{Ar}]\,3d^{1}.$$

This single $$3d$$ electron is unpaired, so $$n=1$$. Substituting in the formula,

$$\mu_{\text{spin}}=\sqrt{1\,(1+2)}=\sqrt{3}=1.732\;\text{BM}\;(\text{≈}1.73\;\text{BM}).$$

Ion 2 : V$$^{2+}$$

Neutral vanadium (V, $$Z=23$$) has the configuration

$$\text{V} : [\text{Ar}]\,3d^{3}\,4s^{2}.$$

In the $$+2$$ state we remove the two $$4s$$ electrons:

$$\text{V}^{2+} : [\text{Ar}]\,3d^{3}.$$

The three $$3d$$ electrons occupy different orbitals with parallel spins (Hund’s rule), so all three are unpaired. Thus $$n=3$$. Putting this into the formula,

$$\mu_{\text{spin}}=\sqrt{3\,(3+2)}=\sqrt{15}=3.873\;\text{BM}\;(\text{≈}3.87\;\text{BM}).$$

Ion 3 : Sc$$^{3+}$$

Neutral scandium (Sc, $$Z=21$$) has the configuration

$$\text{Sc} : [\text{Ar}]\,3d^{1}\,4s^{2}.$$

Removing three electrons (two from $$4s$$ and one from $$3d$$) for the $$+3$$ state gives

$$\text{Sc}^{3+} : [\text{Ar}],$$

which is a noble-gas core with no electrons in the valence shell. Therefore $$n=0$$ and

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

Collecting the results in the same order as asked (Ti$$^{3+}$$, V$$^{2+}$$, Sc$$^{3+}$$):

$$1.73\;\text{BM},\;3.87\;\text{BM},\;0\;\text{BM}.$$

These values match exactly with the entries of Option B.

Hence, the correct answer is Option B.

When $$Cu^{2+}$$ salt reacts with potassium iodide (KI), a redox reaction takes place. $$Cu^{2+}$$ is a sufficiently strong oxidising agent to oxidise $$I^-$$ to $$I_2$$, while being itself reduced to $$Cu^+$$.

The reaction is:

$$2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2$$

The $$Cu^+$$ ions combine with $$I^-$$ to form cuprous iodide ($$Cu_2I_2$$), which is a white precipitate. This can also be written as $$2CuI$$, but the formula of cuprous iodide in its crystal form is $$Cu_2I_2$$.

Therefore, the product is $$Cu_2I_2$$.

We begin by recalling that transition metals, especially those belonging to the earlier part of the d-block, can absorb hydrogen atoms into the voids (interstices) of their metallic lattices, thereby giving what are called interstitial hydrides. In such hydrides, the small $$H$$ atoms occupy the tetrahedral or octahedral holes of the metal lattice without markedly disturbing the close-packed metal framework.

The general representation of an interstitial hydride is written as

$$M + \dfrac{x}{2}\,H_2 \;\longrightarrow\; MH_x,$$

where $$M$$ stands for the metal and $$x$$ is a non-stoichiometric, often fractional, value because the exact number of hydrogen atoms accommodated depends on temperature, pressure and the radius of the interstices.

For the formation of such hydrides two main factors are crucial:

1. The metal lattice must possess sufficiently large holes so that the radius ratio $$\dfrac{r_H}{r_M}$$ is small enough (about 0.4 or less) to let the hydrogen atoms slip in.
2. The metal should have a relatively low electronegativity so that the bonding is predominantly metallic with a partial $$M \, \delta^{+}$$ and $$H \, \delta^{-}$$ character.

Metals situated toward the left of the first transition series (Sc, Ti, V, Cr, etc.) meet these requirements better than the later members (Fe, Co, Ni, Cu, Zn) because their atomic radii are larger and their lattices provide more spacious interstitial sites.

Among the four given choices—$$\text{Cr}$$, $$\text{Fe}$$, $$\text{Mn}$$ and $$\text{Co}$$—chromium is the earliest element in the series, lying in Group 6 with a comparatively large atomic radius (128 pm) and a bcc lattice that can readily take up hydrogen. Text-book data show that chromium forms non-stoichiometric hydrides approximately of composition $$\text{CrH}_{0.8}$$ under moderate conditions, whereas manganese, iron and cobalt require much more severe pressures and still absorb less hydrogen.

So, when ease of hydride formation is compared, chromium stands out:

$$\text{Cr} + x\,H_2 \;\longrightarrow\; \text{CrH}_x \quad\text{(occurs most readily)}.$$

Hence, the correct answer is Option A.

We begin by recalling the standard chemical formulas of the common ores mentioned.

First, calamine is a zinc ore. The most familiar zinc carbonate ore carries the formula $$\text{ZnCO}_3$$. Therefore,

$$\text{Calamine} \;=\; \text{ZnCO}_3$$

Now, malachite is the well-known basic carbonate of copper. In books it is always written as a double formula combining one molecule of copper(II) carbonate with one molecule of copper(II) hydroxide:

$$\text{Malachite} \;=\; \text{CuCO}_3\cdot\text{Cu(OH)}_2$$

Next, siderite is an iron ore. The very name comes from the Greek word for iron. Siderite is nothing but iron(II) carbonate, so its formula is

$$\text{Siderite} \;=\; \text{FeCO}_3$$

Finally, sphalerite is another important zinc ore, but unlike calamine it is the sulphide, not the carbonate. The sulphide of zinc is written symbolically as:

$$\text{Sphalerite} \;=\; \text{ZnS}$$

We now match each mineral with the correct formula from List-II:

$$(a)\;\text{Calamine}\;\longrightarrow\;(iii)\;\text{ZnCO}_3 (b)\;\text{Malachite}\;\longrightarrow\;(iv)\;\text{CuCO}_3\cdot\text{Cu(OH)}_2 (c)\;\text{Siderite}\;\longrightarrow\;(ii)\;\text{FeCO}_3 (d)\;\text{Sphalerite}\;\longrightarrow\;(i)\;\text{ZnS$$

Scanning the four answer choices, the one that lists the very same sequence—(a) with (iii), (b) with (iv), (c) with (ii) and (d) with (i)—is Option D.

Hence, the correct answer is Option D.

First, let us recall the basic fact that different naturally occurring minerals serve as the chief sources (ores) of various metals. An ore is a mineral from which the metal can be extracted economically.

We have the mineral Calamine. Chemically, calamine is represented as $$\text{ZnCO}_3$$, that is, zinc carbonate. Because the formula contains the metal zinc (Zn), it is classified as the principal ore of zinc.

Now we consider Malachite. The chemical formula of malachite is $$\text{CuCO}_3\cdot\text{Cu(OH)}_2$$, often written more compactly as $$\text{Cu}_2(\text{OH})_2\text{CO}_3$$. In both notations, the symbol Cu stands for copper. Thus malachite is a carbonate-hydroxide ore of copper.

So, summarising the identifications:

$$\text{Calamine} \; ( \text{ZnCO}_3 ) \longrightarrow \text{Ore of Zinc}$$

$$\text{Malachite} \; ( \text{Cu}_2(\text{OH})_2\text{CO}_3 ) \longrightarrow \text{Ore of Copper}$$

Looking at the options, the pair “Zinc and Copper” appears in Option B, which exactly matches the order of the ores given in the question (“Calamine and Malachite, respectively”).

Hence, the correct answer is Option B.

Statement I says the $$E°$$ value of $$Ce^{4+}/Ce^{3+}$$ is $$+1.74$$ V. This is a well-known standard reduction potential value and is correct. The high positive value indicates that $$Ce^{4+}$$ is a strong oxidising agent, meaning it readily gets reduced to $$Ce^{3+}$$.

Statement II says Ce is more stable in the $$Ce^{4+}$$ state than in the $$Ce^{3+}$$ state. This is incorrect. The high positive $$E°$$ value of $$+1.74$$ V for the $$Ce^{4+}/Ce^{3+}$$ couple means that $$Ce^{4+}$$ has a strong tendency to get reduced to $$Ce^{3+}$$. This indicates that $$Ce^{3+}$$ is actually the more stable oxidation state. Although $$Ce^{4+}$$ achieves a noble gas core configuration ($$[Xe]$$), in aqueous solution $$Ce^{3+}$$ is thermodynamically more stable, and $$Ce^{4+}$$ acts as a powerful oxidant precisely because it readily converts to $$Ce^{3+}$$.

Therefore, Statement I is correct but Statement II is incorrect. The correct answer is Option (4).

We need to match each ore with the metal it contains.

Siderite is $$FeCO_3$$, which is an ore of iron (Fe). So (a) matches with (iii).

Calamine is $$ZnCO_3$$, which is an ore of zinc (Zn). So (b) matches with (v).

Malachite is $$Cu_2(OH)_2CO_3$$ (basic copper carbonate), which is an ore of copper (Cu). So (c) matches with (i).

Cryolite is $$Na_3AlF_6$$, which is an ore of aluminium (Al). So (d) matches with (iv).

Therefore the correct matching is: (a)$$\rightarrow$$(iii), (b)$$\rightarrow$$(v), (c)$$\rightarrow$$(i), (d)$$\rightarrow$$(iv).

We have to analyse what happens when solid potassium permanganate, whose formula is $$KMnO_4$$, is heated to about $$513\;{\rm K}$$. Potassium permanganate is well known to undergo thermal decomposition. First, let us write the balanced thermal decomposition reaction that is found experimentally:

$$2\,KMnO_4 \;\xrightarrow{513\;{\rm K}}\; K_2MnO_4 \;+\; MnO_2 \;+\; O_2$$

Now we look at the individual products one by one. The first product on the right‐hand side is $$K_2MnO_4$$, which is called potassium manganate. The second is $$MnO_2$$ (manganese dioxide) and the third is molecular oxygen $$O_2$$.

The question asks about “the product”, and in the context of colour and magnetism the visually dominant solid that forms is $$K_2MnO_4$$, because $$MnO_2$$ is a dark brown or black solid that often masks nothing, while $$K_2MnO_4$$ imparts a distinct green colour in the melt or in aqueous solution. So we examine the properties of $$K_2MnO_4$$ in detail.

First, let us determine the oxidation state of manganese in $$K_2MnO_4$$. We know the rule:

Sum of oxidation states in a neutral compound $$= 0.$$

In $$K_2MnO_4$$: