d and f Block Elements

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.

Given the species, we need to determine which have the same magnetic moment by counting their unpaired electrons.

For the species $ \text{Cr}^{2+} $ and $ \text{Mn}^{3+} $, both have the same number of unpaired electrons:

Number of unpaired electrons, $ n = 4 $

The magnetic moment $\mu$ is calculated using the formula:

$ \mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \, \text{BM} $

Thus, $ \text{Cr}^{2+} $ and $ \text{Mn}^{3+} $ have the same magnetic moment.

The spin-only magnetic moment of the $ \mathrm{Fe}^{3+} $ ion is determined by its electronic configuration and the number of unpaired electrons.

Electronic Configuration:

The neutral iron atom ($_{26} \mathrm{Fe}$) has the electronic configuration:

$ [\mathrm{Ar}] 3d^6 4s^2 $

When it forms a $ \mathrm{Fe}^{3+} $ ion, it loses three electrons. The configuration becomes:

$ [\mathrm{Ar}] 3d^5 $

Number of Unpaired Electrons:

In the $ \mathrm{Fe}^{3+} $ ion, there are 5 electrons in the 3d subshell. All these electrons are unpaired, hence:

$ n = 5 $

Calculating the Spin-Only Magnetic Moment:

The formula for the spin-only magnetic moment ($\mu_s$) is:

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

Substituting $ n = 5 $ into the formula:

$ \mu_s = \sqrt{5(5+2)} = \sqrt{35} $

Calculating further gives:

$ \mu_s \approx 5.85 \, \text{BM} $

Therefore, the spin-only magnetic moment of $ \mathrm{Fe}^{3+} $ is approximately 5.85 Bohr Magnetons ($\mathrm{BM}$), which can be rounded to about 6 $\mathrm{BM}$.

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 $\mathrm{Mn_2O_7}$.

Nessler's reagent is $\mathrm{K_2H_g I_4}$

Potassium permanganate ($\mathrm{KMnO}_4$) is a strong oxidizing agent that can oxidize iodide ions ($\mathrm{I}^{-}$) to different products depending on the pH of the solution.

In acidic solutions, $\mathrm{KMnO}_4$ oxidizes $\mathrm{I}^{-}$ to molecular iodine ($\mathrm{I}_2$) according to the following equation:

In neutral or faintly alkaline solutions, the oxidation product is different because the $\mathrm{I}^{-}$ is oxidized to iodate ion ($\mathrm{IO}_3^{-}$). The reaction in a neutral or slightly alkaline solution is as follows:

So according to the reactions above, in acidic solutions, $\mathrm{I}^{-}$ is oxidized to $\mathrm{I}_2$ while in neutral or faintly alkaline solutions, $\mathrm{I}^{-}$ is oxidized to $\mathrm{IO}_3^{-}$. Therefore, the correct answer is:

Option C: $\mathrm{I}_2 ~\&~ \mathrm{IO}_3^{-}$

$\mathrm{S{O_2} + C{r_2}O_7^{2 - }\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} \mathop {C{r^{3 + }}}\limits_{(green)} + SO_4^{2 - }}$

Statement I is correct, Statement II is incorrect as Si forms electron precise hydride.

(i) $\mathrm{Mn_2O_7\to Mn}$ in (+7) oxidation state

(ii) It is also correct

(iii) Sc only shows +3 oxidation state

(iv) $\mathrm{Cr^{+6}}$ is oxidising in nature

The indicator 'X' is starch, and it forms a blue colored complex with iodine (compound 'A'). Therefore, the correct option is:

Option D: Starch and iodine

In this situation, starch is commonly used as an indicator to detect the presence of iodine because starch forms a deep blue complex with iodine. This color change is visible and helps in monitoring the progress of the reaction involving iodide ions and hydrogen peroxide, which ultimately produces iodine. The blue color indicates the presence of iodine in the solution.

The thermal decomposition of ammonium dichromate gives nitrogen gas, chromium (III) oxide and water vapours.

A good reducing agent oxidises itself readily. Eu and Yb act as reducing agent in +2 oxidation state by using two electron and forming $\mathrm{Eu}^{2+}$ and $\mathrm{Yb}^{2+}$, these will get stable half-filled [Xe]4f ${ }^7$ and fully-filled ( $[\mathrm{Xe}], 4 f^{14}$ ) configuration respectively.

$ \begin{aligned} & { }_{25} \mathrm{Mn}=[\mathrm{Ar}] 3 d^5 4 s^2 \\ & \mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 d^5 4 s^0 \end{aligned} $

$ \mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 d^{4+} 4 s^0 $

Hence, $\mathrm{Mn}^{2+}$ due to half-filled orbital is more stable than $\mathrm{Mn}^{3+}$.

Thus, $\mathrm{Zn}^{2+}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}$ and $\mathrm{Cu}^{+}$have zero magnetic moment.

The melting points of given elements are as follows :

$ \begin{array}{ll} \text { Element } & \text { Melting point } \\ \hline \mathrm{W} & 3422^{\circ} \mathrm{C} \\ \hline \mathrm{Cr} & 1907^{\circ} \mathrm{C} \\ \hline \text { Mo } & 2623^{\circ} \mathrm{C} \\ \hline \end{array} $

Hence, correct order of melting point is $\mathrm{W}>\mathrm{Mo}>\mathrm{Cr}$.

To determine the number of $f$-electrons in the +3 oxidation state of gadolinium (Gd) and +2 oxidation state of ytterbium (Yb), we can examine their electronic configurations:

Gadolinium (Gd):

Atomic number: 64

Electronic configuration: $[\mathrm{Xe}] \, 4f^7 \, 5d^1 \, 6s^2$

In the +3 oxidation state ($\text{Gd}^{3+}$), gadolinium loses three electrons. These electrons are typically removed from the outermost orbitals, i.e., $6s$ and $5d$, leaving the $f$-orbital unchanged.

Configuration of $\text{Gd}^{3+}$: $[\mathrm{Xe}] \, 4f^7$

Number of $f$-electrons ($x$): 7

Ytterbium (Yb):

Atomic number: 70

Electronic configuration: $[\mathrm{Xe}] \, 4f^{14} \, 6s^2$

In the +2 oxidation state ($\text{Yb}^{2+}$), ytterbium loses two electrons, typically from the $6s$ orbital, leaving the $f$-orbital full.

Configuration of $\text{Yb}^{2+}$: $[\mathrm{Xe}] \, 4f^{14}$

Number of $f$-electrons ($y$): 14

Finally, the sum of $x$ and $y$ is:

$ x + y = 7 + 14 = 21 $