d and f Block Elements

Let’s determine the color of each aquated ion:

(A) $ \mathrm{Ni^{2+}} $

• In aqueous solution: $[\mathrm{Ni(H_2O)_6}]^{2+}$

• This complex is green in color.

✅ (A) → (V) Green

(B) $ \mathrm{Fe^{3+}} $

• In aqueous solution: $[\mathrm{Fe(H_2O)_6}]^{3+}$

• This complex is yellow to brownish-yellow.

✅ (B) → (III) Yellow

(C) $ \mathrm{Mn^{3+}} $

• In aqueous solution: $[\mathrm{Mn(H_2O)_6}]^{3+}$

• This complex is violet or red-purple.

✅ (C) → (I) Violet

(D) $ \mathrm{V^{4+}} $

• In aqueous solution: $[\mathrm{VO}]^{2+}$ (vanadyl ion)

• This species is blue.

✅ (D) → (II) Blue

✅ Final Matching:

$ A - V, \quad B - III, \quad C - I, \quad D - II $

✅ Correct Option: D

Step 1. Recall that $ 4f^7 $ means half-filled 4f shell (7 electrons).

Step 2. Check each ion

(A) $ \mathrm{Pr^{3+}} $

Praseodymium (Pr) = atomic number 59

Neutral configuration: $[Xe] 4f^3 6s^2$

When it loses 3 electrons → remove $6s^2$ and one $4f$:

$ \mathrm{Pr^{3+}} : [Xe] 4f^2 $

❌ Not $4f^7$.

(B) $ \mathrm{Lu^{3+}} $

Lutetium (Lu) = atomic number 71

Neutral configuration: $[Xe] 4f^{14} 5d^1 6s^2$

Removing three electrons → $6s^2$ and $5d^1$:

$ \mathrm{Lu^{3+}} : [Xe] 4f^{14} $

❌ Not $4f^7$.

(C) $ \mathrm{Eu^{2+}} $

Europium (Eu) = atomic number 63

Neutral configuration: $[Xe] 4f^7 6s^2$

Losing two electrons → remove both $6s$ electrons:

$ \mathrm{Eu^{2+}} : [Xe] 4f^7 $

✅ Matches $4f^7$.

(D) $ \mathrm{Ce^{4+}} $

Cerium (Ce) = atomic number 58

Neutral: $[Xe] 4f^1 5d^1 6s^2$

Losing 4 electrons → $4f^1 5d^1 6s^2$ → remove all →

$ \mathrm{Ce^{4+}} : [Xe] $

❌ Not $4f^7$.

✅ Correct Answer: Option C – $ \mathrm{Eu^{2+}} $

We calculate the number of unpaired electrons in each ion,

(i) $\mathrm{Cr}^{3+} \rightarrow[\mathrm{Ar}] 3 d^3 \rightarrow 3$ unpaired electrons

(ii) $\mathrm{Fe}^{3+} \rightarrow[\mathrm{Ar}] 3 d^5 \rightarrow 5$ unpaired electrons

(iii) $\mathrm{Mn}^{2+} \rightarrow[\mathrm{Ar}] 3 d^5 \rightarrow 5$ unpaired electrons

(iv) $\mathrm{Co}^{2+} \rightarrow[\mathrm{Ar}] 3 d^7 \rightarrow 3$ unpaired electrons

(v) $\mathrm{Ni}^{2+} \rightarrow[\mathrm{Ar}] 3 d^8 \rightarrow 2$ unpaired electrons

Now compare:

• $\mathrm{Cr}^{3+}$ and $\mathrm{Co}^{2+} \rightarrow$ both have 3 unpaired electrons.

• $\mathrm{Fe}^{3+}$ and $\mathrm{Mn}^{2+} \rightarrow$ both have 5 unpaired electrons.

Two distinct pairs have same number of unpaired electrons.

$\mathrm{Fe}^{3+}$ is $3 d^5$, all electrons unpaired (high-spin $d^5$ ): Spin-only moment

$ \begin{aligned} & =\sqrt{[n(n+2)]}=\sqrt{[5 \times 7]} \\ & =\sqrt{35} \approx 5.92 \mathrm{BM} \approx 6 \mathrm{BM} \end{aligned} $

The correct option for the electronic configuration of Einsteinium (atomic number 99) is:

Here's why:

1. Atomic Number and Electrons: Einsteinium has an atomic number of 99, which means it has 99 protons in its nucleus. Since atoms are neutral, it also has 99 electrons surrounding the nucleus.


2. Electron Shells: Electrons fill up energy shells around the nucleus. The order of filling follows the Aufbau principle, which generally fills lower energy shells before moving to higher ones.


3. Noble Gas Core Notation: The notation [Rn] represents the filled electron configuration of Radon (atomic number 86). This is a shorthand to avoid writing out the entire configuration of the inner shells, as they are not involved in the chemical behavior of Einsteinium.


4. Filling the f-orbital: Einsteinium is an actinide element, located in the f-block of the periodic table. The 5f subshell fills up after the 6s subshell.


5. Aufbau Principle: According to the Aufbau principle, the 5f subshell can hold up to 14 electrons. In Einsteinium, 11 electrons fill the 5f subshell (5f¹¹).


6. Empty d-orbital: The 6d subshell has a higher energy level than the 5f subshell. Since the 5f subshell isn't completely filled, the 6d subshell remains empty (6d⁰).


7. Outermost Electrons: The remaining two electrons fill the outermost 7s subshell (7s²).

Therefore, the complete electronic configuration of Einsteinium is:

To evaluate the given statements, let's analyze each statement in the context of chemical principles:

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

This statement is true. In transition elements, as we move down the group, the electrons are being added to the n-1 d orbitals, where 'n' represents the principal quantum number. The ability to exhibit higher oxidation states stems from the fact that not only are the s electrons of the outermost shell used in bonding, but the d electrons are also involved. Moreover, down the group, the effective nuclear charge decreases, and the shielding effect increases, so electrons from both s and d orbitals can be easily used for bonding, leading to stable higher oxidation states. This trend is opposite in the case of p-block elements, where the stability of higher oxidation states generally decreases down the group due to the inert pair effect, which makes it hard to remove electrons from the s orbital as we move down the group.

Statement II: Copper can not liberate hydrogen from weak acids.

This statement is also true. Copper (Cu) is less reactive and lies below hydrogen in the electrochemical series, meaning it has a higher reduction potential than hydrogen. For a metal to displace hydrogen from an acid, it must have a lower reduction potential than hydrogen. This is not the case with copper. Therefore, copper does not react with weak acids (like acetic acid) to liberate hydrogen gas. This behavior can be explained through the electrochemical series, where metals placed above hydrogen are capable of displacing hydrogen from acids (exemplified by metals such as zinc or magnesium), whereas copper does not have this ability due to its relative placement in the series.

Based on the analysis, Option C is the correct answer: Both Statement I and Statement II are true.

$\mathrm{Cu}^{2+}$ is more stable.

$\mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$ disproportionation reaction

To determine the correct answer to this question, let's analyze the catalysis mechanism involving Iron (III) in the reaction between iodide ($\mathrm{I^-}$) and persulfate ($\mathrm{S_2O_8^{2-}}$) ions.

Iron (III), or $\mathrm{Fe}^{3+}$, can act as a catalyst by undergoing redox reactions, cycling between $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$. Here’s how it typically works:

1. $\mathrm{Fe}^{3+}$ can oxidize iodide ions to iodine. The reaction will proceed as follows:

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

2. $\mathrm{Fe}^{2+}$, now formed, can be oxidized back to $\mathrm{Fe}^{3+}$ by reducing persulfate ions. The reaction will be:

$\mathrm{2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}}$

Based on this mechanism, $\mathrm{Fe}^{3+}$ is responsible for oxidizing the iodide ions, and $\mathrm{Fe}^{2+}$ is responsible for reducing the persulfate ions. Thus, we can conclude:

A. $\mathrm{Fe}^{3+}$ oxidises the iodide ion: This statement is correct.

B. $\mathrm{Fe}^{3+}$ oxidises the persulphate ion: This statement is incorrect because $\mathrm{Fe}^{3+}$ does not oxidize persulfate ions.

C. $\mathrm{Fe}^{2+}$ reduces the iodide ion: This statement is incorrect because $\mathrm{Fe}^{2+}$ does not reduce iodide ions.

D. $\mathrm{Fe}^{2+}$ reduces the persulphate ion: This statement is correct.

Therefore, the most appropriate answer is:

Option C: A and D only

The number of unpaired electrons in a transition metal is determined by its electronic configuration in the context of its ground state. The elements listed, Scandium (Sc), Chromium (Cr), Vanadium (V), Titanium (Ti), and Manganese (Mn), are transition metals that have different numbers of unpaired electrons. Here is the electronic configuration and the corresponding number of unpaired electrons for each:

Scandium (Sc):

Electronic Configuration: $ [\mathrm{Ar}] \, 3d^1 \, 4s^2 $

Unpaired Electrons: 1

Titanium (Ti):

Electronic Configuration: $ [\mathrm{Ar}] \, 3d^2 \, 4s^2 $

Unpaired Electrons: 2

Vanadium (V):

Electronic Configuration: $ [\mathrm{Ar}] \, 3d^3 \, 4s^2 $

Unpaired Electrons: 3

Chromium (Cr):

Electronic Configuration: $ [\mathrm{Ar}] \, 3d^5 \, 4s^1 $

Unpaired Electrons: 6

Manganese (Mn):

Electronic Configuration: $ [\mathrm{Ar}] \, 3d^5 \, 4s^2 $

Unpaired Electrons: 5

Arranging them in increasing order of number of unpaired electrons:

$ (\mathrm{A})<(\mathrm{D})<(\mathrm{C})<(\mathrm{E})<(\mathrm{B}) $

Therefore, the correct choice is Option B.

To determine the number of elements from the given list that do not belong to the lanthanoids, let's first define what lanthanoids are. The lanthanoids consist of the 15 chemical elements in the periodic table from Lanthanum ($\mathrm{La}$) with atomic number 57 to Lutetium ($\mathrm{Lu}$) with atomic number 71. Thus, lanthanoids include:

• Lanthanum ($\mathrm{La}$)
• Cerium ($\mathrm{Ce}$)
• Praseodymium ($\mathrm{Pr}$)
• Neodymium ($\mathrm{Nd}$)
• Promethium ($\mathrm{Pm}$)
• Samarium ($\mathrm{Sm}$)
• Europium ($\mathrm{Eu}$)
• Gadolinium ($\mathrm{Gd}$)
• Terbium ($\mathrm{Tb}$)
• Dysprosium ($\mathrm{Dy}$)
• Holmium ($\mathrm{Ho}$)
• Erbium ($\mathrm{Er}$)
• Thulium ($\mathrm{Tm}$)
• Ytterbium ($\mathrm{Yb}$)
• Lutetium ($\mathrm{Lu}$)

Given the list of elements: Eu (Europium), Cm (Curium), Er (Erbium), Tb (Terbium), Yb (Ytterbium), and Lu (Lutetium), we can now check which of these are not part of the lanthanoids:

• $\mathrm{Eu}$ (Europium) belongs to the lanthanoids.
• $\mathrm{Cm}$ (Curium) does not belong to the lanthanoids; it is an actinoid.
• $\mathrm{Er}$ (Erbium) belongs to the lanthanoids.
• $\mathrm{Tb}$ (Terbium) belongs to the lanthanoids.
• $\mathrm{Yb}$ (Ytterbium) belongs to the lanthanoids.
• $\mathrm{Lu}$ (Lutetium) belongs to the lanthanoids.

From the given list, only Cm (Curium) does not belong to the lanthanoids, which means the correct answer is:

Option B: 1

While preparing crystal of Mohr's salt, dil. $\mathrm{H}_2 \mathrm{SO}_4$ is added to prevent hydrolysis of ferrous sulphate.

To determine the ability of the ions to liberate hydrogen from a dilute acid, we need to understand the standard electrode potentials of these ions in comparison to hydrogen. The standard electrode potential of hydrogen ($\mathrm{H}^{+}/\mathrm{H}_{2}$) is set at 0 volts. Any metal ion with a negative standard reduction potential compared to hydrogen has the ability to liberate hydrogen gas from dilute acids because they are more likely to donate electrons to the $\mathrm{H}^{+}$ ions found in acids, reducing them to $\mathrm{H}_{2}$ gas.

The standard reduction potentials for the ions mentioned are as follows:

• $\mathrm{Ti}^{2+}/\mathrm{Ti}$ is more negative than that of $\mathrm{H}^{+}/\mathrm{H}_{2}$.
• $\mathrm{Cr}^{2+}/\mathrm{Cr}$ also has a more negative standard reduction potential compared to $\mathrm{H}^{+}/\mathrm{H}_{2}$.
• The standard reduction potential for $\mathrm{V}^{2+}/\mathrm{V}$ is similarly more negative than that of $\mathrm{H}^{+}/\mathrm{H}_{2}$.

Due to their negative standard reduction potentials relative to hydrogen, all three ions ($\mathrm{Ti}^{2+}, \mathrm{Cr}^{2+}, \mathrm{V}^{2+}$) have the capacity to liberate hydrogen from a dilute acid. Therefore, the correct answer is:

Option C: 3

The metal that shows the highest and maximum number of oxidation states among the options given is manganese (Mn), which is option B. Manganese exhibits a wide range of oxidation states from +2 to +7, which is more diverse than the other metals mentioned. To further detail:

• Co (Cobalt) typically exhibits oxidation states of +2 and +3, with +2 being the most common. Although it can theoretically show +1 and +4 in very specific compounds, these are not as stable or as common as the states displayed by manganese.


• Mn (Manganese) shows a remarkable series of oxidation states from +2 to +7 in its compounds, making it the element with the highest number of stable oxidation states among the options listed. For example, MnO (in the +2 state), Mn₂O₃ (in the +3 state), MnO₂ (in the +4 state), Mn₂O₇ (in the +7 state), and so forth.


• Fe (Iron) commonly exhibits +2 and +3 oxidation states, such as in FeSO₄ (in the +2 state) and Fe₂O₃ (in the +3 state). Though there are compounds with Fe in the +6 oxidation state, like potassium ferrate (K₂FeO₄), these are not as common or stable as the lower oxidation states.


• Ti (Titanium) shows oxidation states of +2, +3, and +4. Among these, the +4 oxidation state (as in TiO₂) is the most stable and common. Though it offers a variety of states, it doesn't reach the variety or the maximum oxidation state shown by manganese.

Therefore, the correct answer is Option B: Mn (Manganese), which can show the highest and maximum number of oxidation states among the options provided.

The element among the given options that shows only one oxidation state other than its elemental form is Scandium.

Explanation:

• Nickel (Ni) can have multiple oxidation states, including +2 and +3, with +2 being the most common.
• Titanium (Ti) also shows various oxidation states, most commonly +4 and +2, due to its ability to lose electrons from both its 3d and 4s subshells.
• Cobalt (Co) commonly exists in the +2 and +3 oxidation states, among others, due to electron transitions in its 3d orbitals.
• Scandium (Sc), however, primarily exhibits a +3 oxidation state in its compounds (excluding its 0 state in the elemental form). This is because scandium has an electronic configuration of [Ar]3d¹4s², and it loses all three outer electrons to achieve a stable configuration, resulting in a +3 oxidation state. It does not commonly exhibit other oxidation states because removing more than three electrons would require removing electrons from the very stable noble gas core configuration ([Ar]), which is energetically unfavorable.

Therefore, the correct answer is Option D: Scandium.

Let us analyze both the Assertion (A) and the Reason (R) to determine the correct answer.

Assertion (A): In aqueous solutions, $\mathrm{Cr}^{2+}$ is reducing while $\mathrm{Mn}^{3+}$ is oxidising in nature. This assertion is true. The $\mathrm{Cr}^{2+}$ ion has a tendency to be oxidized to $\mathrm{Cr}^{3+}$ because $\mathrm{Cr}^{3+}$ has a more stable electronic configuration. Therefore, $\mathrm{Cr}^{2+}$ acts as a reducing agent. On the other hand, $\mathrm{Mn}^{3+}$ tends to stabilize by being reduced to $\mathrm{Mn}^{2+}$, which is a more stable electronic configuration due to having a half-filled d5 subshell ($\mathrm{Mn}^{2+}$: [Ar] 3d⁵). Thus, $\mathrm{Mn}^{3+}$ acts as an oxidizing agent.

Reason (R): Extra stability to half filled electronic configuration is observed than incompletely filled electronic configuration. This reason is also true. Half-filled and fully filled electron configurations, like those found in $\mathrm{Mn}^{2+}$ and $\mathrm{Cr}^{0}$ respectively, are particularly stable due to symmetry and exchange energy considerations. This is a reason why certain ions like $\mathrm{Mn}^{3+}$ want to be reduced to $\mathrm{Mn}^{2+}$, and why $\mathrm{Cr}^{0}$ (with a [Ar] 3d⁵ 4s¹ configuration) is not as common as $\mathrm{Cr}^{3+}$ (which has a [Ar] 3d³ configuration, and while not half-filled, is favored due to crystal field stabilization energy considerations in octahedral complexes).

If we evaluate these statements together, we see that Assertion (A) is directly related to the electronic configurations and their stability, as asserted by Reason (R). The stability of half-filled and full orbitals contributes to the observed redox behavior of $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ in aqueous solutions.

Therefore, the correct answer is Option B: Both (A) and (R) are true and (R) is the correct explanation of (A).

(A) $\mathrm{Mn}_2 \mathrm{O}_7$ is green oil at room temperature.

(B) $\mathrm{V}_2 \mathrm{O}_4$ dissolve in acids to give $\mathrm{VO}^{2+}$ salts.

(C) $\mathrm{CrO}$ is basic oxide

(D) $\mathrm{V}_2 \mathrm{O}_5$ is amphoteric it reacts with acid as well as base.

A. $\mathrm{CrO}_4{ }^{2-}$ is tetrahedral

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

C. As per NCERT, green manganate is paramagnetic with 1 unpaired electron.

D. Statement is correct

E. Statement is correct

Mn, Ni and Cd metals used in battery industries.

$\left.\begin{array}{l} \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \rightarrow \mathrm{Cr}^{+6} \rightarrow \text { No d - d transition } \\ \mathrm{KMnO}_4 \rightarrow \mathrm{Mn}^{7+} \rightarrow \text { No d - d transition } \end{array}\right\} \text { Charge transfer }$

$\mathrm{Ce}:[\mathrm{Xe}] 4 \mathrm{f}^1 5 \mathrm{~d}^1 6 \mathrm{~s}^2 ; \mathrm{Ce}^{4+}$ diamagnetic

$\mathrm{La}:[\mathrm{Xe}] 4 \mathrm{f}^0 5 \mathrm{~d}^1 6 \mathrm{~s}^2 ; \mathrm{La}^{3+}$ diamagnetic

(A) $\mathrm{Zn}, \mathrm{Cd}, \mathrm{Hg}$ exhibit lowest enthalpy of atomization in respective transition series.

(C) Compounds of $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ are diamagnetic in nature.

$\begin{aligned} & \mathrm{K}_2 \mathrm{MnO}_4 \\ & 2+\mathrm{x}-8=0 \\ & \Rightarrow \mathrm{x}=+6 \end{aligned}$

$\mathrm{O.S.}$ of $\mathrm{Mn}=+6$

IUPAC Name $=$

Potassium tetraoxidomanganate(VI)

$\mathrm{KMnO}_4 \xrightarrow{\Delta} \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2$

Statement (1) is true, $\mathrm{Ce}^{+4}$ has noble gas electronic configuration.

Statement (2) is also true due to high reduction potential for $\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}(+1.74 \mathrm{~V})$, and stability of $\mathrm{Ce}^{3+}, \mathrm{Ce}^{4+}$ acts as strong oxidizing agent.

$\begin{aligned} & {[\mathrm{Cr}]=[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^5} \\ & {[\mathrm{Cd}]=[\mathrm{Kr}] 5 \mathrm{~s}^2 4 \mathrm{~d}^{10}} \\ & {[\mathrm{Cu}]=[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^{10}} \\ & {[\mathrm{Ag}]=[\mathrm{Kr}] 5 \mathrm{~s}^1 4 \mathrm{~d}^{10}} \\ & {[\mathrm{Zn}]=[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^{10}} \end{aligned}$

The magnetic moment of an atom or ion with unpaired electrons is given by the formula $\mu = \sqrt{n(n+2)}$ Bohr magnetons (BM), where $n$ is the number of unpaired electrons. The magnetic moment depends on the number of unpaired electrons: more unpaired electrons lead to a higher magnetic moment.

Looking at the given options with respect to their electron configurations and calculating their respective magnetic moments based on their unpaired electrons:

• $[\mathrm{Ar}] 3 \mathrm{~d}^7$ has 3 unpaired electrons, yielding a magnetic moment of $\sqrt{15}$ BM.
• $[\mathrm{Ar}] 3 \mathrm{~d}^8$ has 2 unpaired electrons, for a magnetic moment of $\sqrt{8}$ BM.
• $[\mathrm{Ar}] 3 \mathrm{~d}^3$ also has 3 unpaired electrons, similar to $3 \mathrm{~d}^7$, so its magnetic moment is likewise $\sqrt{15}$ BM.
• Finally, $[\mathrm{Ar}] 3 \mathrm{~d}^6$ has 4 unpaired electrons, resulting in the highest magnetic moment among the options, $\sqrt{24}$ BM.

Hence, the electron configuration associated with the highest magnetic moment is $[\mathrm{Ar}] 3 \mathrm{~d}^6$.

The electronic configuration of an element is determined by adding electrons to atomic orbitals following the Aufbau principle, the Pauli exclusion principle, and Hund's rule. For atoms with atomic numbers higher than that of xenon (Xe, atomic number 54), the electrons start filling the outer orbitals following the xenon core. In this case, we're looking at neodymium (Nd), which has an atomic number of 60.

To find the electronic configuration of neodymium, we first note the configuration of xenon, which serves as the core, and then add the additional electrons beyond xenon to the relevant orbitals.

Neodymium has 60 - 54 = 6 electrons more than xenon. These additional electrons will fill the 4f and 6s orbitals. Based on its position in the periodic table, neodymium’s electrons start filling the 4f orbitals before the 5d. Since the 6s orbital is at a lower energy level than the 4f orbital, it gets filled before the 4f. Neodymium will have 2 electrons in the 6s orbital, and the remaining 4 electrons will go into the 4f orbital.

Therefore, the correct electronic configuration for neodymium is:

So, Option A is the correct answer.

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

A. Ostwald's process is an industrial process to make nitric acid. Catalyst used is Pt gauge.

B. Haber's process is industrial process to make ammonia. Iron is used as catalyst in the process.

C. Deacon's process is used to make chlorine. $ \mathrm{CuCl}_{2} $ is used as catalyst.

D. Cracking of hydrocarbon is done in presence of zeolite catalyst.

Total 3 elements, $ \mathrm{Cu}, \mathrm{Gd} $ and Cm have half filled $ f $-orbitals.

The alloy formed by beryllium with copper ($X = \text{Cu}$) is commonly used in the preparation of high-strength springs. This alloy is known as beryllium copper and is prized for its excellent combination of strength, electrical conductivity, and resistance to fatigue and corrosion, making it ideal for various applications including spring manufacturing.

Therefore, the correct option is:

Option C: Cu

In the given options, statement (c) is correct, while the others contain inaccuracies. Here are the correct versions for the incorrect options:

For option (a), the correct order of standard electrode potential for the reaction $ M^{3+}/M^{2+} $ in volts is: $\mathrm{Cr} < \mathrm{Fe} < \mathrm{Mn}\, (-0.41\,V, +0.77\,V, +1.57\,V) $

For option (b), the correct order of magnetic moments is: $\mathrm{Cr}^{2+}(d^4) = \mathrm{Fe}^{2+}(d^6) < \mathrm{Mn}^{2+}(d^5) $

For option (d), the correct order of the first ionization enthalpy in $\mathrm{kJ/mol}$ is: $\mathrm{V} < \mathrm{Cr} < \mathrm{Ti} \, (650 < 653 < 656)$

Hs, Hg, W elements obey $ (n-1) d^{1-10} n s^{2} $ electronic configuration. Hs - $ 108-[\mathrm{Rn}] 5 f^{14} 6 d^{6} 7 s^{2} $

$ \mathrm{Hg}-80-[\mathrm{Xe}] 4 f^{14} 5 d^{10} 6 s^{2} $

$ \mathrm{W}-74-[\mathrm{Xe}] 4 f^{14} 5 d^{4} 6 s^{2} $

The complete reaction is as follows,

$ \begin{gathered} \mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{CO}_2} \underset{\text { (Product } X \text { ) }}{ \mathrm{PH}_3 }\begin{array}{c} \end{array}+3 \mathrm{NaH}_2 \mathrm{PO}_2 \\ 4 \mathrm{PH}_3+6 \mathrm{CuSO}_4 \longrightarrow 6 \mathrm{H}_2 \mathrm{SO}_4+\underset{\text { (Product } Y \text { ) }}{2 \mathrm{Cu}_3 \mathrm{P}_2} \end{gathered} $

Hence, product $Y$ is $\mathrm{Cu}_3 \mathrm{P}_2$.

Due to inert pair effect, $\mathrm{Tl}(+1)$ is more stable than $\mathrm{Tl}(+3)$. The valence ' $s$ ' electrons acts as inner core electrons due to poor shielding of ' $d$ ' and ' $f$ ' electrons.

This is the reason Tl is more stable in +1 oxidation state.

Let's examine each statement carefully:

(i) "Ti(IV) is more stable than Ti(III) and Ti(II)."

True. Among titanium’s common oxidation states (+2, +3, +4), the +4 state is generally the most stable due to the high lattice or coordination stabilization and the noble-gas-like configuration of Ti^(4+) (3d^0).

(ii) "Among 3d-series elements (Z = 22 to 29), only copper has a positive reduction potential (M^(2+)/M)."

True. Standard reduction potentials for most 3d metals (like Ti, V, Cr, Mn, Fe, Co, Ni, Zn) are negative. Copper is unique with a standard reduction potential of about +0.34 V, making it the only one in that range with a positive reduction potential for the M^(2+)/M couple.

(iii) "Both Sc and Zn exhibit +1 oxidation state."

False.

Sc typically shows +3 oxidation state (it is isoelectronic with Ar when Sc^(3+)). Instances of Sc in +1 or +2 are either extremely rare or not well-established.

Zn commonly shows +2. While there are organometallic complexes where Zn can appear in unusual oxidation states, Zn(+1) is not a stable or typical oxidation state in standard inorganic chemistry contexts.

Conclusion

(i) is correct

(ii) is correct

(iii) is incorrect

Hence, the correct statements are (i) and (ii) only, matching Option A.

The correct order of oxidizing power for the given ions is:

$ \mathrm{VO}_2^{+} < \mathrm{Cr}_2 \mathrm{O}_7^{2-} < \mathrm{MnO}_4^{-} $

Oxidizing power is directly related to the oxidation number of an element; the higher the oxidation number, the greater the oxidizing power.

For $\mathrm{VO}_2^{+}$, if we let the oxidation number of Vanadium be $x$, we can write:

$ x + 2(-2) = 1 \implies x = +5 \tag{i} $

For $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$, the calculation is:

$ 2x + 7(-2) = -2 \implies x = +6 \tag{ii} $

For $\mathrm{MnO}_4^{-}$, the calculation is:

$ x + 4(-2) = -1 \implies x = +7 \tag{iii} $

This shows that $\mathrm{MnO}_4^{-}$ has the highest oxidation number, therefore, it is the strongest oxidizing agent among the given ions.

$ \begin{array}{lcl} \hline \text { Name } & \text { Atomic Number } & \begin{array}{l} \text { Electronic } \\ \text { Configuration } \end{array} \\ \hline \text { Cerium (Ce) } & 58 & {[\mathrm{Xe}] 4 f^1 5 d^1 6 s^2} \\ \hline \text { Gadolinium (Gd) } & 64 & {[\mathrm{Xe}] 4 F^7 5 d^1 6 s^2} \\ \hline \text { Lutetium (Lu) } & 71 & {[\mathrm{Xe}] 4 f^{14} 5 d^1 6 s^2} \\ \hline \end{array} $

Let's match each element from List I with its correct category in List II:

A. Technicium

Technicium (technetium, symbol Tc) is a transition metal.

Match: II. Transition metal

B. Fluorine

Fluorine is a halogen, which is classified as a non-metal.

Match: I. Non-metal

C. Tellurium

Tellurium is known for its properties that are intermediate between metals and non-metals, so it is classified as a metalloid.

Match: IV. Metalloid

D. Dysprosium

Dysprosium is part of the lanthanide series (also known as lanthanoids).

Match: III. Lanthanoid

Thus, the correct matching is:

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

This corresponds to Option C.

The correct answer is Option D: tungsten (W).

Here's a bit of explanation:

Tungsten (W) has a melting point of about $3422^\circ C$, which is the highest among the elements listed.

For comparison:

Rhenium (Re) melts at around $3186^\circ C$.

Molybdenum (Mo) melts at approximately $2623^\circ C$.

Chromium (Cr) melts at roughly $1907^\circ C$.

Thus, tungsten (W) clearly stands out as the transition metal with the highest melting point.