d and f Block Elements

(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.

Let's analyze the formation of a simple monoxide (with formula $\text{MO}$) for each metal:

For a metal to form an oxide of the type $\text{MO}$, the metal would need to adopt a +2 oxidation state (since oxygen is $-2$).

Vanadium, chromium, and manganese can adopt a +2 oxidation state:

Vanadium can form $\text{VO}$ (vanadium(II) oxide).

Chromium can form $\text{CrO}$ (chromium(II) oxide), although it is less stable, it is known under certain conditions.

Manganese forms $\text{MnO}$ (manganese(II) oxide) quite readily.

Scandium, on the other hand, almost exclusively forms a +3 oxidation state. Its stable oxide is $\text{Sc}_2\text{O}_3$ (scandium(III) oxide), and there is no stable monoxides like $\text{ScO}$.

Thus, the transition metal that does not form an "MO" type oxide is:

Option D: Scandium.

Assertion A: 5f electrons can participate in bonding to a far greater extent than 4f electrons

This is correct. In actinides (elements in which 5f orbitals are being filled), the 5f electrons can participate in bonding. This contrasts with lanthanides (elements in which 4f orbitals are being filled), where the 4f electrons largely do not participate in bonding.

Reason R: 5f orbitals are not as buried as 4f orbitals

This is also correct. The term "buried" refers to the degree to which an orbital is shielded from the outside environment by other electrons. The 5f orbitals are less shielded, or less "buried," than the 4f orbitals. This means they are more exposed to the outside environment and can more readily interact with other atoms to form bonds. This is the reason why 5f electrons can participate in bonding to a greater extent than 4f electrons.

Therefore, both A and R are true, and R is the correct explanation of A.

Prolonged heating is avoided during the preparation of ferrous ammonium sulfate to prevent oxidation. Ferrous ammonium sulfate is a green crystalline solid that is used as a reducing agent in various chemical reactions. When heated, it can undergo oxidation to form ferric ammonium sulfate, which is a brown crystalline solid.

Here is a chemical equation for the oxidation of ferrous ammonium sulfate:

As you can see, the oxidation of ferrous ammonium sulfate results in the formation of ferric ammonium sulfate, which is a brown crystalline solid. This is why prolonged heating is avoided during the preparation of ferrous ammonium sulfate.

Here are the other options and why they are incorrect:

• Option A: Hydrolysis is the process of a chemical compound reacting with water. Ferrous ammonium sulfate is not hydrolyzed by water, so prolonged heating is not necessary to prevent hydrolysis.


• Option B: Breaking is the process of a chemical compound being physically broken apart. Ferrous ammonium sulfate is not broken apart by heat, so prolonged heating is not necessary to prevent breaking.


• Option D: Reduction is the process of a chemical compound gaining electrons. Ferrous ammonium sulfate is a reducing agent, so it is not reduced by heat.

The given reaction describes the formation of a copper(II) halide and the corresponding diatomic halogen molecule. Not all halogens will cause this reaction to occur.

This reaction is known to occur with iodine (I) because the formation of $\mathrm{I}_{2}$ (a diatomic iodine molecule) is easier due to the relatively low bond energy of the $\mathrm{I}-\mathrm{I}$ bond.

On the other hand, the bond energies of $\mathrm{Cl}-\mathrm{Cl}$ and $\mathrm{Br}-\mathrm{Br}$ are higher, making it more difficult for these diatomic halogen molecules to form. Thus, these halogens would not typically cause this reaction to occur.

Therefore, the correct answer is:

Option B : Only Iodine.

Assertion A is correct. In the complex $\mathrm{Ni(CO)_4}$, if we assume the oxidation state of Ni to be x, we can set up the equation:

$ x + (0 \times 4) = 0 $

From which we find that (x = 0).

Similarly, in $\mathrm{Fe(CO)_5}$, if we let the oxidation state of Fe be y, the equation becomes:

$ y + (0 \times 5) = 0 $

Solving this gives (y = 0).

Therefore, the oxidation states of both Ni and Fe are zero in their respective complexes, verifying that Assertion A is true.

Regarding Reason R, it is incorrect. The low oxidation states in these complexes are stabilized through synergic bonding, which involves the ligand being a $\pi$-acceptor, not a $\pi$-donor. The CO ligand in these complexes accepts electron density from the metal into its antibonding molecular orbitals, characterizing a $\pi$-acid ligand, and not a $\pi$-donor ligand as stated. Consequently, Reason R does not hold true.

In a neutral or weakly alkaline solution, permanganate ions are reduced to MnO₂, with a change in oxidation state from +7 to +4, which is a change of 3 units.

The balanced redox reaction is :

3MnO₄⁻ + 5S₂O₃²⁻ + 2H₂O → 3MnO₂ + 5SO₄²⁻ + 10OH⁻

Strong reducing agents are those that readily donate electrons, and strong oxidizing agents are those that readily accept electrons.

• $\mathrm{Ce}^{4+}$: It can accept an electron to form $\mathrm{Ce}^{3+}$, so it is a strong oxidizing agent.


• $\mathrm{Eu}^{2+}$: It can donate an electron to form $\mathrm{Eu}^{3+}$, so it is a strong reducing agent.

So the correct answer is: $\mathrm{Eu}^{2+}$ and $\mathrm{Ce}^{4+}$.

Nessler’s reagent is K₂[HgI₄]

Therefore, among the options given, oxygen is the only element that is not present in Nessler's reagent.

The correct statements about $\mathrm{Mn_2O_7}$ are:

(A) Mn is tetrahedrally surrounded by oxygen atoms.

(C) Contains Mn-O-Mn bridge.

Manganese (Mn) in $\mathrm{Mn_2O_7}$ exhibits its highest oxidation state of +7. The oxidation state is the charge an atom would have if the electrons in a chemical bond were shared equally.

In $\mathrm{Mn_2O_7}$, the manganese atoms are surrounded by oxygen atoms in a tetrahedral arrangement. This means that each manganese atom is bonded to four oxygen atoms arranged in a pyramid shape. This tetrahedral arrangement of oxygen atoms results in the highest oxidation state for manganese in $\mathrm{Mn_2O_7}$.

$\mathrm{Mn_2O_7}$ also contains Mn-O-Mn bridges, which are bonds between two manganese atoms through a shared oxygen atom. These bonds help to reinforce the structure of the molecule and contribute to its stability.

It is incorrect to state that Mn is octahedrally surrounded by oxygen atoms or that $\mathrm{Mn_2O_7}$ contains a Mn-Mn bond. An octahedral arrangement would consist of six oxygen atoms surrounding a central manganese atom, and a Mn-Mn bond would involve two manganese atoms sharing electrons without an oxygen atom in between. Neither of these are present in $\mathrm{Mn_2O_7}$.