d and f Block Elements

Mixed oxides are those oxides which can be written as a combination of two simpler oxides of the same metal, like $MO \cdot M_2O_3$.

Now, check the given oxides and see which ones can be written in this form:

$ \begin{aligned} & \mathrm{Mn}_3 \mathrm{O}_4=\mathrm{MnO} \cdot \mathrm{Mn}_2 \mathrm{O}_3 \\ & \mathrm{Fe}_3 \mathrm{O}_4=\mathrm{FeO} \cdot \mathrm{Fe}_2 \mathrm{O}_3 \\ & \mathrm{Co}_3 \mathrm{O}_4=\mathrm{CoO} \cdot \mathrm{Co}_2 \mathrm{O}_3\end{aligned} \quad \begin{aligned} & \text { Only three } \\ & \text { mixed oxides }\end{aligned}$

So, the number of mixed oxides in the given list is $3$.

The first reaction is the well-known Chromyl chloride test. When a chloride salt like sodium chloride ($\mathrm{NaCl}$) is heated with potassium dichromate ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$) and concentrated sulfuric acid ($\mathrm{H}_2\mathrm{SO}_4$), deep red vapors of chromyl chloride are evolved.

$ 4\mathrm{NaCl} + \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 6\mathrm{H}_2\mathrm{SO}_4 \rightarrow 2\mathrm{CrO}_2\mathrm{Cl}_2 + 2\mathrm{KHSO}_4 + 4\mathrm{NaHSO}_4 + 3\mathrm{H}_2\mathrm{O} $

Thus, Product A is Chromyl chloride ($\mathrm{CrO}_2\mathrm{Cl}_2$).

When chromyl chloride gas is passed through an aqueous solution of sodium hydroxide ($\mathrm{NaOH}$), a yellow solution of sodium chromate is formed.

$ \mathrm{CrO}_2\mathrm{Cl}_2 + 4\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{CrO}_4 + 2\mathrm{NaCl} + 2\mathrm{H}_2\mathrm{O} $

Thus, Product B is Sodium chromate ($\mathrm{Na}_2\mathrm{CrO}_4$).

When a solution of a chromate acts with hydrogen peroxide ($\mathrm{H}_2\mathrm{O}_2$) in an acidic medium (provided by $\mathrm{H}_2\mathrm{SO}_4$), a deep blue-colored compound known as chromium pentoxide is formed.

$ \mathrm{Na}_2\mathrm{CrO}_4 + \mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{CrO}_5 + \mathrm{Na}_2\mathrm{SO}_4 + 3\mathrm{H}_2\mathrm{O} $

Thus, Product C is Chromium pentoxide ($\mathrm{CrO}_5$).

Chromium pentoxide has a famous "butterfly structure".

• Finding $X$: In its structure, there are two peroxide ($\mathrm{O-O}$) linkages. Each peroxide linkage represents one $\mathrm{O}_2^{2-}$ unit. Therefore, the number of $\mathrm{O}_2^{2-}$ units is $2$.

  So, $X = 2$.

• Finding $Y$: The chemical formula is $\mathrm{CrO}_5$, meaning there are $5$ oxygen atoms in total.

  So, $Y = 5$.

• Finding $Z$: Out of the $5$ oxygen atoms in the butterfly structure, one is bonded by a double bond (normal oxide with an oxidation state of $-2$), and the remaining four form the peroxide linkages (with an oxidation state of $-1$ each).

  Let the oxidation state of Chromium ($\mathrm{Cr}$) be $Z$. We can calculate it as follows:

  $ Z + 1(-2) + 4(-1) = 0 $

  $ Z - 2 - 4 = 0 $

  $ Z = +6 $

  So, $Z = 6$.

$\therefore $ $ X + Y + Z = 2 + 5 + 6 = 13 $

$\begin{aligned} &\text { Highest Co }\\ &\mathrm{Co}^{+2}=(\mathrm{Ar}) 3 \mathrm{~d}^7 \end{aligned}$

$\begin{aligned} &\begin{aligned} & n=3 \\ & \mu=\sqrt{15}=3.87 \end{aligned}\\ &\text { Nearest integer }=4 \end{aligned}$

$\begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CrO}_2 \mathrm{Cl}_2 \\ & \mathrm{CrO}_2 \mathrm{Cl}_2 \text { (Vapour) }+\mathrm{NaOH} \rightarrow \\ & \mathrm{Na}_2 \mathrm{CrO}_4+\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{Na}_2 \mathrm{CrO}_4+\mathrm{H}^{\oplus} \rightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{O} \\ & (\mathrm{C}) \\ & \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 \rightarrow 2 \mathrm{Na}^{\oplus}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \end{aligned}$

No of terminal "O" = 6

$Fe{C_2}{O_4} + N{a_2}C{O_3}\buildrel {{O_2}} \over \longrightarrow \,?$

The chemical reaction can be written as

The insoluble product is $Fe_2O_3$.

Molar mass of $Fe_2O_3$

Fe (Atomic mass) $\to$ 55.845 amu

Molar mass = 55.845 g mol$^{-1}$

O (Atomic mass) $\to$ 15.999 amu

Molar mass = 15.999 g mol$^{-1}$

Molar mass of $Fe_2O_3$ = 55.845 g mol$^{-1}\times2$ + 15.999 g mol$^{-1}\times3$

$=(111.69+47.997)$ g mol$^{-1}$

$=159.687$ g mol$^{-1}\approx 160$ g mol$^{-1}$

We need to determine the number of electrons each element has in its 4d orbitals and then add them together.

1. Niobium $\mathrm{(Nb, Z = 41)}$

The ground‐state electronic configuration of niobium is:

$ [\mathrm{Kr}]\,4d^4\,5s^1. $

Hence, Niobium has 4 electrons in its 4d orbitals.

2. Ruthenium $\mathrm{(Ru, Z = 44)}$

The ground‐state electronic configuration of ruthenium is:

$ [\mathrm{Kr}]\,4d^7\,5s^1. $

Hence, Ruthenium has 7 electrons in its 4d orbitals.

3. Sum of 4d Electrons

$ x \;=\; 4 \quad (\text{for Nb}), \quad y \;=\; 7 \quad (\text{for Ru}). $

Therefore,

$ x + y \;=\; 4 \;+\; 7 \;=\; 11. $

Answer: 11

Identify which metal (M) has the highest second ionization enthalpy.

We are comparing the elements Sc, Ti, V, Cr, Mn, and Fe in terms of their second ionization enthalpy (IE₂). Recall:

The second ionization enthalpy (IE₂) is the energy required to remove one electron from the singly charged ion $ \mathrm{M}^{+} $ to form $ \mathrm{M}^{2+} $.

For most 3d transition metals, the first electron is removed from the 4s orbital.

Particularly large values of IE₂ often occur if, after removal of the first electron, one is forced to remove an electron from a stable configuration (e.g., half-filled d-orbital).

Let us outline the ground-state (neutral atom) electron configurations and the removal of the first and second electrons:

Sc $\bigl[ \mathrm{Ar} \bigr] 3d^1\,4s^2 $

$\mathrm{Sc} \rightarrow \mathrm{Sc}^{+}$: remove 1 electron from 4s

$\mathrm{Sc}^{+} \rightarrow \mathrm{Sc}^{2+}$: remove the second 4s electron

Final: $\mathrm{Sc}^{2+} = [\mathrm{Ar}]\,3d^1$

Ti $\bigl[ \mathrm{Ar} \bigr] 3d^2\,4s^2 $

1st electron from 4s → $\mathrm{Ti}^{+} = [\mathrm{Ar}]\,3d^2\,4s^1$

2nd electron from 4s → $\mathrm{Ti}^{2+} = [\mathrm{Ar}]\,3d^2$

V $\bigl[ \mathrm{Ar} \bigr] 3d^3\,4s^2 $

1st electron from 4s → $\mathrm{V}^{+} = [\mathrm{Ar}]\,3d^3\,4s^1$

2nd electron from 4s → $\mathrm{V}^{2+} = [\mathrm{Ar}]\,3d^3$

Cr $\bigl[ \mathrm{Ar} \bigr] 3d^5\,4s^1 $

1st electron from 4s → $\mathrm{Cr}^{+} = [\mathrm{Ar}]\,3d^5$ (a stable half-filled d$^{5}$ configuration)

2nd electron now must come from the 3d shell → $\mathrm{Cr}^{2+} = [\mathrm{Ar}]\,3d^4$

Because removing an electron from a half-filled d$^{5}$ orbital costs a lot of energy, $\mathrm{Cr}$ generally has a notably high second IE.

Mn $\bigl[ \mathrm{Ar} \bigr] 3d^5\,4s^2 $

1st electron from 4s → $\mathrm{Mn}^{+} = [\mathrm{Ar}]\,3d^5\,4s^1$

2nd electron from 4s → $\mathrm{Mn}^{2+} = [\mathrm{Ar}]\,3d^5$ (thus still d$^{5}$ in the end)

The second ionization for Mn is large but less dramatic than for Cr because Mn$^{2+}$ ends with a half-filled d$^{5}$. You are not removing from a half-filled d$^{5}$ to get Mn$^{2+}$.

Fe $\bigl[ \mathrm{Ar} \bigr] 3d^6\,4s^2 $

1st electron from 4s → $\mathrm{Fe}^{+} = [\mathrm{Ar}]\,3d^6\,4s^1$

2nd electron from 4s → $\mathrm{Fe}^{2+} = [\mathrm{Ar}]\,3d^6$

From experimental data (and the arguments above), $\mathrm{Cr}$ indeed has the highest second ionization enthalpy among these six metals.

Determine the electronic configuration of $\mathbf{Cr}^{+}$ and its spin-only magnetic moment.

Neutral $\mathrm{Cr}$: $\bigl[ \mathrm{Ar} \bigr]\,3d^5\,4s^1$

$\mathrm{Cr}^{+}$: Removal of the 4s electron ⇒ $\bigl[ \mathrm{Ar} \bigr]\,3d^5$

The $3d^5$ configuration has 5 unpaired electrons.

The spin-only magnetic moment $\mu$ is given by

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

where $n$ = number of unpaired electrons. Here $n = 5$, so

$ \mu = \sqrt{5(5+2)} \;=\; \sqrt{35} \;\approx\; 5.92 \;\text{BM}. $

This is often rounded to about 5.9 or 6.0 BM.

The color of lanthanoid ions in solutions is mainly due to the electronic transitions within the 4f subshell. The lanthanoid ions are more likely to be colorless when they have fully filled (with 14 electrons) or completely empty (with 0 electrons) f orbitals because, in these cases, there are no electrons to undergo f-f transitions, and as a result, no absorption of visible light occurs leading to colorlessness.

Let's consider the electronic configurations of the lanthanoid ions provided:

- $\mathrm{Eu}^{3+}$: Europium (Eu) has an atomic number of 63. Neutral europium ([Xe]4f⁷ 6s^2) loses three electrons to form Eu³⁺, leaving it with an electronic configuration equivalent to [Xe]4f⁶. With 6 electrons in the f orbital, it can undergo f-f transitions, thus it is not colorless.

- $\mathrm{Lu}^{3+}$: Lutetium (Lu) has an atomic number of 71. In its 3+ ionic state, lutetium has lost its 6s and 5d electrons and is left with a completely filled 4f orbital ([Xe] 4f¹⁴). This configuration cannot allow for any f-f transitions, as there are no available energy levels within the f orbital for an electron to jump to, making Lu³⁺ colorless.

- $\mathrm{Nd}^{3+}$: Neodymium (Nd) has an atomic number of 60. In its 3+ state ([Xe] 4f³), it clearly has partially filled f orbitals, which can absorb visible light for f-f transitions, so it is not colorless.

- $\mathrm{La}^{3+}$: Lanthanum (La) has an atomic number of 57. In its 3+ ionic state, it has a configuration of [Xe], meaning that its 4f orbital is completely empty. Since there are no electrons in the f orbital to undergo f-f transitions, La³⁺ is colorless.

- $\mathrm{Sm}^{3+}$: Samarium (Sm) has an atomic number of 62. As a 3+ ion ([Xe] 4f⁵), it too has electrons in the f orbital capable of undergoing f-f transitions, so it is not colorless.

From the analysis, the colorless lanthanoid ions among the ones listed are $\mathrm{Lu}^{3+}$ and $\mathrm{La}^{3+}$.

Therefore, the number of colorless lanthanoid ions among the given options is 2.

In order to determine the spin-only magnetic moment value of the species with the least oxidizing ability, we first need to analyze their oxidation states and electron configurations.

1. $\mathrm{VO}_2^{+}$:

For vanadium in $\mathrm{VO}_2^{+}$, the oxidation state is +5. The electronic configuration of V is $[Ar] \, 3d^3 \, 4s^2$. Thus, in +5 oxidation state, vanadium will have zero d-electrons (since it loses 5 electrons) and, consequently, $\mathrm{VO}_2^{+}$ is diamagnetic (since zero unpaired electrons).

2. $\mathrm{MnO}_4^{-}$:

For manganese in $\mathrm{MnO}_4^{-}$, the oxidation state is +7. The electronic configuration of Mn is $[Ar] \, 3d^5 \, 4s^2$. Thus, in +7 oxidation state, manganese will have zero d-electrons (since it loses 7 electrons) and, consequently, $\mathrm{MnO}_4^{-}$ is also diamagnetic (since zero unpaired electrons).

3. $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$:

For chromium in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$, we have two chromium atoms. Each chromium is in the +6 oxidation state. The electronic configuration of Cr is $[Ar] \, 3d^5 \, 4s^1$. Thus, in +6 oxidation state, each chromium will have zero d-electrons (since it loses 6 electrons) and, consequently, $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is also diamagnetic (since zero unpaired electrons).

From these observations, we note that all the species we analyzed are diamagnetic. However, to identify the species with the least oxidizing ability, we look at their standard reduction potentials (E° values).

Typically, the oxidizing ability increases with increasing positive E° values. Given the nature of these species, we can infer that:

• $\mathrm{MnO}_4^{-}$ is a very strong oxidizing agent (very high E° value).
• $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is also a strong oxidizing agent (high but less than $\mathrm{MnO}_4^{-}$).
• $\mathrm{VO}_2^{+}$ has a moderate oxidizing ability (lesser E° value than both $\mathrm{MnO}_4^{-}$ and $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$).

Therefore, among the given species, $\mathrm{VO}_2^{+}$ has the least oxidizing ability.

As mentioned previously, $\mathrm{VO}_2^{+}$ has zero unpaired electrons, making it diamagnetic. The spin-only magnetic moment is given by the formula:

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

where $n$ is the number of unpaired electrons.

For $\mathrm{VO}_2^{+}$, $n=0$, thus:

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

Given the options, the nearest integer value of the spin-only magnetic moment for the species with the least oxidizing ability (which is $\mathrm{VO}_2^{+}$) is indeed:

0 BM

First, we need to understand the oxidation states of chromium in the given compounds and determine their magnetic moments based on their electronic configurations.

1. $\mathrm{CrO}$: In $\mathrm{CrO}$, the oxidation state of chromium is +2. The electronic configuration of chromium (Cr) is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$. In the +2 oxidation state, two electrons are removed, typically from the 4s and one of the 3d orbitals, leaving us with the configuration $3d^4$.

To find the spin-only magnetic moment, we use the formula:

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

where n is the number of unpaired electrons. For $\mathrm{Cr}^{2+}$, we have 4 unpaired electrons in the 3d orbitals.

Thus,

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

2. $\mathrm{Cr}_2 \mathrm{O}_3$: In $\mathrm{Cr}_2 \mathrm{O}_3$, the oxidation state of chromium is +3. The electronic configuration of $\mathrm{Cr}^{3+}$ is $3d^3$ after losing three electrons.

For $\mathrm{Cr}^{3+}$, there are 3 unpaired electrons.

Thus,

$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \mathrm{BM}$

3. $\mathrm{CrO}_3$: In $\mathrm{CrO}_3$, the oxidation state of chromium is +6. The electronic configuration of $\mathrm{Cr}^{6+}$ is $3d^0$ after losing six electrons. There are no unpaired electrons for $\mathrm{Cr}^{6+}$.

Since $\mathrm{Cr}^{6+}$ has no unpaired electrons, its spin-only magnetic moment is 0 BM.

The magnetic moment values for each compound are:

• $\mathrm{CrO}$: 4.90 BM (basic oxide)
• $\mathrm{Cr}_2 \mathrm{O}_3$: 3.87 BM (amphoteric oxide)
• $\mathrm{CrO}_3$: 0 BM

The sum of the spin-only magnetic moments of the basic and amphoteric oxides is:

$4.90 + 3.87 = 8.77 \mathrm{BM}$

Expressing in terms of $10^{-2} \mathrm{BM}$,

$8.77 \mathrm{BM} \times 100 = 877 (\times 10^{-2} \mathrm{BM})$

Thus, the sum of spin-only magnetic moment values of basic and amphoteric oxides is approximately 877 $\times 10^{-2} \mathrm{BM}$ (nearest integer).

To determine the spin-only magnetic moments of products A and B formed from the fusion of chromite ore with sodium carbonate in the presence of air, we first need to examine the given reaction and the properties of the products.

The reaction provided is:

$4 \mathrm{FeCr}_2 \mathrm{O}_4 + 8 \mathrm{Na}_2 \mathrm{CO}_3 + 7 \mathrm{O}_2 \rightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4 (\mathrm{A}) + 2 \mathrm{Fe}_2 \mathrm{O}_3 (\mathrm{B}) + 8 \mathrm{CO}_2$

• Product A is sodium chromate, $\mathrm{Na}_2 \mathrm{CrO}_4$.


• Product B is ferric oxide, $\mathrm{Fe}_2 \mathrm{O}_3$.

1. For Product A ($\mathrm{Na}_2 \mathrm{CrO}_4$):

• In $\mathrm{Na}_2 \mathrm{CrO}_4$, the chromium is in the +6 oxidation state.


• The electronic configuration of $\mathrm{Cr}^{6+}$ is $[\mathrm{Ar}] 3d^0$.


• Since there are no unpaired electrons in the $3d$ orbitals, the spin-only magnetic moment is zero.

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

1. For Product B ($\mathrm{Fe}_2 \mathrm{O}_3$):

• In $\mathrm{Fe}_2 \mathrm{O}_3$, the iron is in the +3 oxidation state.


• The electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3d^5$.


• $\mathrm{Fe}^{3+}$ has 5 unpaired electrons in the $3d$ orbitals.


• The spin-only magnetic moment can be calculated using the formula:

$ \mu = \sqrt{n(n+2)} \text{ BM} = \sqrt{5(5+2)} \text{ BM} = \sqrt{35} \text{ BM} \approx 5.9 \text{ BM} $

• The spin-only magnetic moment of Product A is $0 \text{ BM}$.


• The spin-only magnetic moment of Product B is $5.9 \text{ BM}$.

Therefore, the sum of the spin-only magnetic moments of A and B is:

$ 0 + 5.9 \approx 6 \text{ BM}$

The sum of the spin-only magnetic moment values of A and B is approximately 6 B.M. (to the nearest integer).

Green coloured salt bead represents the metal ion taken is $\mathrm{Fe}^{3+}$ so, $\mathrm{Fe}^{3+}:[\mathrm{Ar}] 4 \mathrm{~s}^0 3 d^5$

So, $\mu=\sqrt{5 \times 7}=5.9=6$

Vanadium has highest enthalpy of atomization among first row transition elements.

$\mathrm{V}_2 \mathrm{O}_5$ is amphoteric

In $\mathrm{V}^{5+}$ there are no unpaired electrons.

Thus, $\mu=0$

$\mathrm{A}$ is $\mathrm{K}_2 \mathrm{MnO}_4$

$\mathrm{B}$ and $\mathrm{C}$ are $\mathrm{KMnO}_4$ and $\mathrm{MnO}_2$

$\mathrm{KMnO}_4(\mu=0)$

$\mathrm{MnO}_2(\mathrm{Mn}^{4+})(\mu=3.87)$

Sum $=3.87=4$ (Nearest integer)

To determine which of these ions have a noble gas configuration, we must consider what the phrase "noble gas configuration" means. Atoms or ions with a noble gas configuration have completely filled electron shells, similar to the electron configuration of the noble gases, which are the elements found in Group 18 of the periodic table. Noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).

Let's examine each ion given:

• $\mathrm{Sr}^{2+}(z=38)$: Strontium has atomic number 38 and loses 2 electrons to form a $\mathrm{Sr}^{2+}$ ion, giving it the same electron configuration as krypton (Kr), with atomic number 36. Hence, it has a noble gas configuration.

${\left[\mathrm{Sr}^{2+}\right]=[\mathrm{Kr}]}$


• $\mathrm{Cs}^{+}(z=55)$: Cesium has atomic number 55, and by losing 1 electron to form $\mathrm{Cs}^{+}$, it has the electron configuration of xenon (Xe), with atomic number 54. Thus, it also has a noble gas configuration.

${\left[\mathrm{Cs}^{+}\right]=[\mathrm{Xe}]}$


• $\mathrm{La}^{2+}(z=57)$: Lanthanum has atomic number 57. If it lost 2 electrons to form $\mathrm{La}^{2+}$, it would not have a noble gas configuration because it would have one electron more than xenon (Xe), which means it would not fully match any noble gas electron configuration.

${\left[\mathrm{La}^{2+}\right]=[\mathrm{Xe}] 5 \mathrm{~d}^1}$


• $\mathrm{Pb}^{2+}(z=82)$: Lead has an atomic number of 82, so when it loses 2 electrons to form $\mathrm{Pb}^{2+}$, it has 80 electrons, which is the same electron number as mercury (Hg) and not a noble gas. Therefore, $\mathrm{Pb}^{2+}$ does not have a noble gas configuration.


${\left[\mathrm{~Pb}^{2+}\right]=[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^{10} 6 \mathrm{~s}^2}$


• $\mathrm{Yb}^{2+}(z=70)$: Ytterbium has an atomic number of 70. By losing 2 electrons to form $\mathrm{Yb}^{2+}$, it has 68 electrons, aligning with the electron configuration of Erbium (Er). $\mathrm{Yb}^{2+}$ and not a noble gas.

${\left[\mathrm{Yb}^{2+}\right]=[Xe]4f^{14}}$


• $\mathrm{Fe}^{2+}(z=26)$: Iron has an atomic number of 26. When it becomes $\mathrm{Fe}^{2+}$ by losing 2 electrons, it has 24 electrons. This does not correspond to any noble gas configuration, as the nearest noble gas is argon (Ar) with 18 electrons.

Therefore, the ions $\mathrm{Sr}^{2+}$ and $\mathrm{Cs}^{+}$ have a noble gas configuration. Altogether, there are 2 ions with a noble gas configuration.

Copper matte is a mixture of copper sulfide and iron sulfide formed as a result of smelting of copper ore. It contains mainly copper sulfide (Cu₂S) as the major component and iron sulfide (FeS) as a minor component.

The compounds A. CuCO₃, C. Cu₂O, and D. FeO are not present in copper matte, but can be formed as a result of further processing or refining of the matte.

Therefore, among the given compounds, only "B. Cu₂S" is present in copper matte.

$\mathrm{3{e^ - } + 4{H^ + } + MnO_4^ - \buildrel {} \over \longrightarrow Mn{O_2} + 2{H_2}O}$

Oxidation number of sulphur in SO₄^2- (A) is + 6.

$\mathrm{Xe}+2 \mathrm{O}_{2}{F}_{2} \rightarrow \underset{(\mathrm{P})}{\mathrm{XeF}_{4}}+2 \mathrm{O}_{2}$

$\underset{(\mathrm{P})}{\mathrm{6XeF}_{4}}+12 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{Xe}+2 \mathrm{XeO}_{3}+24 \mathrm{HF}+3 \mathrm{O}_{2}$

So, from the above reaction, it is clear that 6 moles of $\mathrm{XeF}_4$ produces 24 moles of $\mathrm{HF}$.

So, 1 mole of $\mathrm{XeF}_4$ will produce $\frac{24}{6}$ moles of HF, i.e., 4 moles of $\mathrm{HF}$.

$ 2 \mathrm{AgNO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{NO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) $

Both the $\mathrm{NO}_2$ and $\mathrm{O}_2$ gases are paramagnetic. $\mathrm{NO}_2(\mathrm{~g})$ has 1 unpaired electron and $\mathrm{O}_2(\mathrm{~g})$ has 2 unpaired electrons.

According to MOT,

Among the given species only K₂CrO₄ is diamagnetic as central metal atom Cr in it has [Ar]3d⁰ electronic configuration i.e., all paired electrons. The structure and oxidation state of central metal atom of this compound are as follows

Structure

Oxidation state Cr⁶⁺

Rest all the compounds are paramagnetic. Reasons for their paramagnetism are given below.

(i) H-atom have 1s¹ electronic configuration i.e., 1 unpaired electron.

(ii) NO₂ i.e.

in itself is an odd electron species.

(iii) O$_2^ - $ (Superoxide) has one unpaired electron in $\pi$* molecular orbital.

(iv) S₂ in vapour phase has O₂ like electronic configuration i.e., have 2 unpaired electrons in $\pi$* molecular orbitals.

(v) Mn₃O₄ has following structure

Thus, Mn is showing +2 and +4 oxidation states. The outermost electronic configuration of elemental Mn is 3d⁵4s². Hence, in both the above oxidation states it has unpaired electron as

(vii) (NH₄)₂NiCl₄ has Ni as central metal atom with +2 oxidation state. The electronic configuration of Ni²⁺ in the complex is

(viii) In K₂MnO₄ central metal atom Mn has +6 oxidation state with following structure

Electronic configuration of Mn⁶⁺ is

(i) Lead in +2 oxidation state as $\mathrm{Pb}^{2+}$ reacts with hydrogen sulphide gas $\left(\mathrm{H}_2 \mathrm{~S}\right)$ to form black coloured precipitate of lead sulphide $(\mathrm{PbS})$.

$\begin{aligned} & \mathrm{Pb}^{2+}(a q)+\mathrm{H}_2 \mathrm{~S}(g) \rightarrow \mathrm{PbS}(s) \downarrow\text { (Black } \text { ppt) }+2 \mathrm{H}^{+}(a q) \\\\ & \mathrm{PbS} \text { is black coloured precipitate. } \\ & \end{aligned}$

(ii) Silver in +1 oxidation state as $\mathrm{Ag}^{+}$reacts with hydrogen sulphide in neutral or acidic medium to form black coloured precipitate of silver sulphide $\left(\mathrm{Ag}_2 \mathrm{~S}\right)$.

$ 2 \mathrm{Ag}^{+}(a q)+\mathrm{H}_2 \mathrm{~S}(g) \xrightarrow[\text { or acidic medium }]{\text { Neutral }} \mathrm{Ag}_2 \mathrm{~S}(s) $(Black ppt)

$\mathrm{Ag}_2 \mathrm{~S}$ is black coloured precipitate.

(iii) Mercury in +2 oxidation state as $\mathrm{Hg}^{2+}$ reacts with hydrogen sulphide in dilute hydrochloric acid to form black coloured precipitate of mercury sulphide (HgS).

$ \mathrm{Hg}^{2+}(a q)+\underset{\begin{array}{c} \text { (saturated aq. } \\ \text { solution or gas) } \end{array}}{\mathrm{H}_2 \mathrm{~S}} \xrightarrow{\text { dil. } \mathrm{HCl}} \underset{\begin{array}{c} \text { Black } \\ \text { ppt } \end{array}}{\mathrm{HgS}(s)} $

HgS is black coloured precipitate.

(iv) Copper in +2 oxidation state as $\mathrm{Cu}^{2+}$ reacts with dilute hydrochloric acid to form black coloured precipitate of copper sulphide (CuS).

$ \begin{aligned} & \mathrm{Cu}^{2+}(a q)+\mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \text { (saturated aq. solution) } \xrightarrow[\text { HCl }]{\text { dil }} \mathrm{CuS}(\mathrm{s}) \text { (black ppt) } \\ & \end{aligned} $

CuS is black coloured precipitate.

(v) Cobalt in +2 oxidation state as $\mathrm{Co}^{2+}$ reacts with hydrogen sulphide in neutral or alkaline solution to form black coloured precipitate of cobalt sulphide.

$\mathrm{Co}^{2+}(a q) \text { (ammonical hydrogen sulphide) }+\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{CoS}(\mathrm{s})\text { (Black ppt.) }$

$(\mathrm{CoS})$ is black coloured precipitate.

(vi) Nickel in +2 oxidation state as $\mathrm{Ni}^{2+}$ reacts with hydrogen sulphide in neutral or slightly alkaline solution to form, black coloured precipitate of nickel sulphide.

$ \mathrm{Ni}^{2+}(a q)+\mathrm{H}_2 \mathrm{~S} \longrightarrow \underset{\substack{\text { Black } \\ \text { ppt. }}}{\mathrm{NiS}(\mathrm{s})} $

NiS is black coloured precipitate.

(vii) Bismuth in +3 oxidation state as $\mathrm{Bi}^{3+}$ reacts with hydrogen sulphide in cold dilute hydrochloric acid to form crystalline dark brown coloured precipitate, but appears black coloured solid of $\mathrm{Bi}_2 \mathrm{~S}_3$.

$\mathrm{Bi}^{3+}(a q) \text { (Saturated solution) }+\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{Bi}_2 \mathrm{~S}_3(\mathrm{~s})\text { (Dark brown or Black precipitate brownish) }$

$\mathrm{Bi}_2 \mathrm{~S}_3$ may appear black in colours.

Either compounds of sulphur are black in colour.

(viii) In mildly acidic, medium tin in +2 state, i.e., $\mathrm{Sn}^{2+}$ reacts with Hydrogen sulphide $\left(\mathrm{H}_2 \mathrm{~S}\right)$ to form brown coloured tin (II) sulphide which further reacts with excess of hydrogen sulphide to form light yellow coloured tin (IV) sulphide $\left(\mathrm{SnS}_2\right)$.

$\begin{aligned} \mathrm{Sn}^{2+}(a q)+\mathrm{H}_2 \mathrm{~S} \rightleftharpoons \underset{\text { Brown }}{\mathrm{SnS}(s)} \\ \mathrm{SnS}(\mathrm{s})+\mathrm{H}_2 \mathrm{~S} \rightleftharpoons \underset{\text { Yellow }}{\mathrm{SnS}_2(\mathrm{~s})}+2 \mathrm{H}^{+}\end{aligned}$

Hence, tin (IV) sulphide or $\mathrm{SnS}_2$ is yellow in colour.

(ix) MnS is known to be dirty pink coloured.

The reaction for alkaline oxidative fusion is

2MnO$_2$ + 4KOH + O$_2$ $\to$ 2K$_2$MnO$_4$ + 2H$_2$O

In potassium manganite formed as product :

K$_2$MnO$_4$

2(+1) + $x$ + 4($-$2) = 0

$2+x-8=0$

$x-6=0\Rightarrow x=+6$

In the development of photographic film, the black and white photographic film is developed as follows:

The silver bromide reacts with hydroquinone producing black coloured silver particles along with HBr and quinone. During this process, silver bromide gets reduced to silver and hydroquinone is oxidised to quinone.

The hydroquinone acts as a developer during this process.

The unreacted silver bromide then reacts with sodium thiosulphate to produce soluble complex and sodium bromide.

$\mathrm{AgBr + \mathop {2N{a_2}{S_2}{O_3}}\limits_{hypo\,solution} \to \mathop {N{a_3}[Ag{{({S_2}{O_3})}_2}]}\limits_{so{\mathop{\rm lub}} le} + NaBr}$

The reaction occurs in acidic medium. In acidic medium, sulphur from sodium thiosulphate will precipitate out. The colloidal sulphur is obtained which gives milky white turbidity.

$\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+2 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{H}_2 \mathrm{SO}_3+\underset{\substack{\text { colloidal } \\ \text { sulphur }}}{\mathrm{S} \downarrow}$

Final Answer :

$\mathrm{AgBr + \mathop {2N{a_2}{S_2}{O_3}}\limits_{hypo\,solution} \to \mathop {N{a_3}[Ag{{({S_2}{O_3})}_2}]}\limits_{so{\mathop{\rm lub}} le} + NaBr}$