d and f Block Elements

Manganate ion $\rightarrow \mathrm{MnO}_4^{-2}$

Permanganate ion $\rightarrow \mathrm{MnO}_4^{-}$

(A) In both ions, Mn is surrounded by four oxygen atoms. So, the shape of both $\mathrm{MnO}_4^{-2}$ and $\mathrm{MnO}_4^{-}$ is tetrahedral (often written as $\mathrm{d}^3\mathrm{s}$ hybridisation in this context).

(B) For $\mathrm{MnO}_4^{-}$, let oxidation state of Mn be $x$. Then $x+4(-2)=-1 \Rightarrow x=+7$.

For $\mathrm{MnO}_4^{-2}$, $x+4(-2)=-2 \Rightarrow x=+6$.

So, Mn is $+7$ in permanganate and $+6$ in manganate (the statement given is reversed).

(C) When $\mathrm{Mn}^{2+}$ salt is oxidised by peroxodisulphate $\mathrm{S}_2\mathrm{O}_8^{2-}$, the product formed is permanganate ion, not manganate:

$\mathrm{Mn}^{2+}+\mathrm{S}_2 \mathrm{O}_8^{2-} \rightarrow \mathrm{MnO}_4^{-}$ (Permanganate ion)

(D) In $\mathrm{MnO}_4^{-}$, Mn is in $+7$ state, which corresponds to $3d^0$ (no unpaired electrons), so it is diamagnetic.

In $\mathrm{MnO}_4^{-2}$, Mn is in $+6$ state, which corresponds to $3d^1$ (one unpaired electron), so it is paramagnetic.

(E) Acidified permanganate ion acts as a strong oxidising agent, so it oxidises (i.e., “reduces” itself while oxidising) oxalate, nitrite and iodide ions.

Statement I

An ion is coloured in aqueous solution if it has partially filled $d$-orbitals (so that $d\text{-}d$ transitions are possible). Ions with $d^0$ or $d^{10}$ configurations are colourless.

Now check each pair:

1. $\left[\mathrm{Sc}^{3+}, \mathrm{Ti}^{3+}\right]$

• $\mathrm{Sc}^{3+}$ : Sc $(Z=21)$ is $[Ar]\,3d^1 4s^2$; so $\mathrm{Sc}^{3+} = [Ar]\,3d^0$ (colourless)

• $\mathrm{Ti}^{3+}$ : Ti $(Z=22)$ is $[Ar]\,3d^2 4s^2$; so $\mathrm{Ti}^{3+} = [Ar]\,3d^1$ (coloured)

  $\Rightarrow$ Both not coloured

1. $\left[\mathrm{Mn}^{2+}, \mathrm{Cr}^{2+}\right]$

• $\mathrm{Mn}^{2+} = [Ar]\,3d^5$ (coloured)

• $\mathrm{Cr}^{2+} = [Ar]\,3d^4$ (coloured)

  $\Rightarrow$ Both coloured ✅

1. $\left[\mathrm{Cu}^{2+}, \mathrm{Zn}^{2+}\right]$

• $\mathrm{Cu}^{2+} = [Ar]\,3d^9$ (coloured)

• $\mathrm{Zn}^{2+} = [Ar]\,3d^{10}$ (colourless)

  $\Rightarrow$ Both not coloured

1. $\left[\mathrm{Ni}^{2+}, \mathrm{Ti}^{4+}\right]$

• $\mathrm{Ni}^{2+} = [Ar]\,3d^8$ (coloured)

• $\mathrm{Ti}^{4+} = [Ar]\,3d^0$ (colourless)

  $\Rightarrow$ Both not coloured

So, the number of pairs in which both ions are coloured is only $1$, not $3$.

Hence, Statement I is false.

Statement II

A reducing agent is a species that gets oxidised (it donates electrons).

• $\mathrm{Ce}^{4+}$ is a well-known strong oxidising agent (it gets reduced to $\mathrm{Ce}^{3+}$).

• $\mathrm{Th}^{4+}$ is also in a high oxidation state and is not a strong reducing agent.

• $\mathrm{Gd}^{3+}$ is a very stable common oxidation state for Gd; not strongly reducing.

• $\mathrm{Eu}^{2+}$ readily gets oxidised to $\mathrm{Eu}^{3+}$, so it acts as a strong reducing agent among the given species.

Therefore, the strongest reducing agent among $\mathrm{Th}^{4+}, \mathrm{Ce}^{4+}, \mathrm{Gd}^{3+}$ and $\mathrm{Eu}^{2+}$ is $\mathrm{Eu}^{2+}$, not $\mathrm{Th}^{4+}$.

Hence, Statement II is false.

Correct Option

Since both Statement I and Statement II are false, the correct choice is:

Option B: Both Statement I and Statement II are false.

First, write the electronic configurations of Cr and Mn.

$\begin{aligned} & \mathrm{Cr}=(\mathrm{Ar}) 3 \mathrm{~d}^5 4 \mathrm{~s}^1 \\ & \mathrm{Mn}=(\mathrm{Ar}) 3 \mathrm{~d}^5 4 \mathrm{~s}^2 \\ & \mathrm{IE}_1(\mathrm{Cr})<\mathrm{IE}_1(\mathrm{Mn}) \\ & \mathrm{IE}_2(\mathrm{Cr})>\mathrm{IE}_2(\mathrm{Mn}) \\ & \mathrm{IE}_3(\mathrm{Cr})<\mathrm{IE}_3(\mathrm{Mn})\end{aligned}$

For first ionization enthalpy ($\mathrm{IE}_1$): In Cr, the electron is removed from $4s^1$. After losing this one $4s$ electron, Cr becomes $\mathrm{Cr}^+$ with a stable half-filled $3d^5$ configuration. So removing the first electron is comparatively easier, hence $\mathrm{IE}_1(\mathrm{Cr})<\mathrm{IE}_1(\mathrm{Mn})$.

For second ionization enthalpy ($\mathrm{IE}_2$): After losing one electron, $\mathrm{Cr}^+$ already has the stable $3d^5$ configuration. Removing the second electron would disturb this stable half-filled $3d$ arrangement, so it needs more energy. But in Mn, after the first electron loss, removing the next electron can lead to $\mathrm{Mn}^{2+}$ with stable $3d^5$. Therefore, $\mathrm{IE}_2(\mathrm{Cr})>\mathrm{IE}_2(\mathrm{Mn})$.

For third ionization enthalpy ($\mathrm{IE}_3$): $\mathrm{Mn}^{2+}$ is very stable due to $3d^5$. Removing one more electron from this stable state is difficult, so $\mathrm{IE}_3$ for Mn becomes high. In comparison, $\mathrm{Cr}^{2+}$ does not have this extra stability, so removing the third electron is relatively easier. Hence, $\mathrm{IE}_3(\mathrm{Cr})<\mathrm{IE}_3(\mathrm{Mn})$.

Gas evolved is chromyl chloride, formed in the chromyl chloride test.

Step 1: Reaction (NCERT)

On heating $K_2Cr_2O_7$ with a chloride salt (common salt $NaCl$) and conc. $H_2SO_4$, red fumes of chromyl chloride are obtained:

$K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2KHSO_4 + 4NaHSO_4 + 3H_2O$

So, the gas is $CrO_2Cl_2$.

Step 2: Oxidation state of Cr in $CrO_2Cl_2$

Let oxidation state of Cr be $x$.

• Oxygen: $2 \times (-2) = -4$

• Chlorine: $2 \times (-1) = -2$

Sum = $x - 4 - 2 = 0$

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

Correct option

Option C: $CrO_2Cl_2$ and +6

A. Mn forms the oxide $\mathrm{Mn_2O_7}$, in which Mn is in its highest oxidation state.

• To find the oxidation state of Mn in $\mathrm{Mn_2O_7}$, we set the sum of oxidation states to zero. Let Mn's oxidation state be 'x'. Oxygen's oxidation state is typically -2.

• $2(\text{x}) + 7(-2) = 0$

• $2\text{x} - 14 = 0$

• $2\text{x} = 14$

• $\text{x} = +7$

• Manganese is a Group 7 element with an electronic configuration of $[Ar] 3d^5 4s^2$. The highest possible oxidation state for manganese is +7, which involves the loss of all 7 valence electrons.

• Therefore, Mn is indeed in its highest oxidation state (+7) in $\mathrm{Mn_2O_7}$.

• This statement is correct.

B. Oxygen stabilizes the Mn in higher oxidation states by forming multiple bonds with Mn.

• In high oxidation states, transition metals strongly attract electrons, increasing the covalent character of their bonds with oxygen. This is particularly true for elements like Mn in a +7 oxidation state.

• The formation of multiple bonds (e.g., Mn=O) allows for better delocalization of electron density, which helps to stabilize the high positive charge on the central metal atom. This is a common phenomenon for high oxidation state metal oxides (e.g., $\mathrm{CrO_3}$, $\mathrm{V_2O_5}$).

• This statement is correct.

C. $\mathrm{Mn_2O_7}$ is an ionic oxide.

• Ionic oxides are typically formed when the metal has a low oxidation state and a significant electronegativity difference with oxygen leads to electron transfer (e.g., $\mathrm{MnO}$).

• However, as the oxidation state of a metal increases, its ability to polarize electron clouds increases, and the covalent character of its bonds with oxygen becomes prominent.

• $\mathrm{Mn_2O_7}$ is a dark green, viscous liquid with a melting point of -20 °C. These properties are characteristic of a molecular covalent compound, not an ionic solid which would have a high melting point. It is highly covalent due to the high electronegativity of oxygen and the high charge density of $\mathrm{Mn^{7+}}$.

• Therefore, $\mathrm{Mn_2O_7}$ is a covalent oxide, not an ionic oxide.

• This statement is incorrect.

D. The structure of $\mathrm{Mn_2O_7}$ consists of one bridged oxygen.

• The structure of $\mathrm{Mn_2O_7}$ can be represented as $\mathrm{O_3Mn-O-MnO_3}$. It consists of two $\mathrm{MnO_4}$ tetrahedra sharing a common oxygen atom.

• The oxygen atom that links the two manganese atoms is a bridging oxygen. Each Mn atom has three terminal oxygen atoms and one bridging oxygen atom.

• This statement is correct.

Based on the analysis, statements A, B, and D are correct, while statement C is incorrect.

The final answer is $\boxed{\text{B}}$

In $\mathrm{MnO_4^{2-}}$ (manganate ion), oxidation state of Mn is:

$x+4(-2)=-2 \;\Rightarrow\; x-8=-2 \;\Rightarrow\; x=+6$

In acidic medium, it disproportionates, i.e., Mn in $+6$ state is converted into a higher and a lower oxidation state:

• Oxidation: $+6 \to +7$ gives $\mathrm{MnO_4^-}$ (permanganate)

• Reduction: $+6 \to +4$ gives $\mathrm{MnO_2}$

Balanced equation in acidic medium:

$3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O}$

So, $\mathrm{MnO_4^{2-}}$ disproportionates to $\mathrm{MnO_4^-}$ and $\mathrm{MnO_2}$.

Correct option: B

$ \text { Coloured ions have unpaired electron. } $

$ \begin{aligned} &\text { }\\ &\begin{gathered} (i)\,\,\,\,\,\, T_i{ }^{+4}=4 s^{\circ} 3 d^{\circ} \\ n=0 \end{gathered} \end{aligned} $

$ \text { (ii) } V^{+2}=4 s^0 3 d^3 $

$ n=3 $

$ \text { (iii) } \mathrm{Mn}^{+2}=4 s^{\circ} 3 d^5 $

$ n=5 $

$ \text { (iv) } \mathrm{Fe}^{+3}=4 \mathrm{~s}^{\circ} 3 \mathrm{~d}^5 $

$ n=5 $

$ \text { (v) } \mathrm{Cr}^{+2}=4 \mathrm{~s}^{\circ} \mathrm{3d}{ }^4 $

$ n=4 $

The absence of unpaired electron makes the element diamagnetic.

$ \begin{aligned} & L a^{+3}=\left[\mathrm{Xe}\right] 5 d^{\circ} 6 s^{\circ} \\ & 4 b^{+2}=\left[\mathrm{Xe}\right] 4 f^{14} 6 s^{\prime 4} \\ & L u^{+3}=\left[\mathrm{X}e\right] 4 f^{14} 5 d^{\circ} 6 s^{\circ} \\ & C e^{+4}=\left[\mathrm{Xe}\right] 4 f^{\circ} 5 d^{\circ} 6 s^{\circ} \\ & A c^{+3}=\left[Rn\right] 6 d^{\circ} 7 s^{\circ} \\ & L_n^{+3}=\left[Rn\right] 5 f^{14} 6 d^{\circ} 7 s^{\circ} \end{aligned} $

Ans. (A)

The first-row transition metals use both their 3d and 4s electrons to form bonds with reactant molecules on the surface of the catalyst.

∴ Statement I is incorrect.

When reactants are adsorbed on the catalytic surface, their concentration on the surface increases, but this adsorption actually weakens the bonds within the reactant molecules. This helps in forming an activated complex and reduces the activation energy of the reaction.

∴ Statement II is incorrect.

• Cr (Z = 24):

  Ground state configuration:

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

  (due to extra stability of half-filled $3d^5$ and $4s^1$ configuration)

• Mn (Z = 25):

  Ground state configuration:

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

First ionisation enthalpy, $\Delta_i H_1$

In the first ionisation process:

• Cr:

  $\mathrm{Cr} \rightarrow \mathrm{Cr}^+ + e^-$

  After removal of one 4s electron:

  $\mathrm{Cr}^+: [\mathrm{Ar}]\,3d^5$ (a very stable half-filled configuration)

• Mn:

  $\mathrm{Mn} \rightarrow \mathrm{Mn}^+ + e^-$

  After removal of one 4s electron:

  $\mathrm{Mn}^+: [\mathrm{Ar}]\,3d^5\,4s^1$ (not as stable as half-filled 3d⁵ only)

Because the resulting $\mathrm{Cr}^+$ ion is particularly stable, less energy is required to form $\mathrm{Cr}^+$, i.e.

$ \Delta_i H_1(\mathrm{Cr}) < \Delta_i H_1(\mathrm{Mn}) $

Thus,

Correct order for first ionisation enthalpy:

$ \Delta_i H_1: \mathrm{Mn} > \mathrm{Cr} $

(Corresponds to option C.)

Second ionisation enthalpy, $\Delta_i H_2$

Now, let us remove the second electron.

• Cr:

  From the already stable $\mathrm{Cr}^+$ ($[Ar]\,3d^5$):

  $\mathrm{Cr}^+ \rightarrow \mathrm{Cr}^{2+} + e^-$

  After this, $\mathrm{Cr}^{2+}$ becomes:

  $[Ar]\,3d^4$ — less stable.

  Therefore, it requires a large amount of energy to remove this second electron.

• Mn:

  From $\mathrm{Mn}^+$ ($[Ar]\,3d^5\,4s^1$):

  $\mathrm{Mn}^+ \rightarrow \mathrm{Mn}^{2+} + e^-$

  After this, $\mathrm{Mn}^{2+}$ becomes:

  $[Ar]\,3d^5$ — a stable half-filled configuration.

  Thus, less energy is needed for the second ionisation.

Hence,

$ \Delta_i H_2(\mathrm{Cr}) > \Delta_i H_2(\mathrm{Mn}) $

which corresponds to option B.

Combine results

• $\Delta_i H_1$: Mn > Cr  → option C

• $\Delta_i H_2$: Cr > Mn  → option B

Therefore, the correct combination is:

$ \boxed{\text{B and C only}} $

Final Answer: Option B

Statement A: Basic vanadium oxide is used in the manufacture of $\mathrm{H}_2 \mathrm{SO}_4$.

1. The manufacture of sulphuric acid ($\mathrm{H}_2 \mathrm{SO}_4$) by the Contact Process is a very important reaction.

2. In this process, the key step is the catalytic oxidation of $\mathrm{SO}_2$ to $\mathrm{SO}_3$:

   $ 2\mathrm{SO}_2(g) + \mathrm{O}_2(g) \xrightarrow{V_2O_5} 2\mathrm{SO}_3(g) $

3. The catalyst used here is Vanadium pentoxide, $V_2O_5$.

4. Now, we need to check the nature of this oxide. Let's find the oxidation state of Vanadium (V) in $V_2O_5$.

   Let the oxidation state of V be $x$.

   $ 2(x) + 5(-2) = 0 $

   $ 2x - 10 = 0 $

   $ x = +5 $

5. As a general rule from the d-block elements chapter, metal oxides in higher oxidation states are acidic or amphoteric in nature. Since Vanadium is in its highest oxidation state (+5), $V_2O_5$ is amphoteric, meaning it reacts with both acids and bases. It is not a basic oxide.

So, statement A is incorrect.

Statement B: The spin-only magnetic moment value of the transition metal halide employed in Ziegler-Natta polymerization is 2.84 BM.

1. Ziegler-Natta catalysts are used for the polymerization of alkenes, like converting ethene to polyethene.

2. A typical Ziegler-Natta catalyst is a mixture of titanium tetrachloride ($TiCl_4$) and triethylaluminium ($Al(C_2H_5)_3$).

3. The transition metal halide here is $TiCl_4$. Let's find the magnetic moment for the metal ion in this compound.

4. In $TiCl_4$, the oxidation state of Titanium (Ti) is +4.

5. The atomic number of Ti is 22. Its electronic configuration is $[Ar] 3d^2 4s^2$.

6. The electronic configuration of the $Ti^{4+}$ ion is $[Ar] 3d^0$.

7. The number of unpaired electrons ($n$) is 0.

8. The spin-only magnetic moment ($\mu$) is calculated using the formula:

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

   For $Ti^{4+}$, since $n=0$,

   $ \mu = \sqrt{0(0+2)} = 0 \text{ B.M.} $

9. The value given in the statement is 2.84 B.M. A value of $\mu \approx 2.84$ B.M. corresponds to 2 unpaired electrons ($n=2$), because $\sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ B.M.

10. Since the magnetic moment of Ti in $TiCl_4$ is 0 B.M., this statement does not hold for the common example of a Ziegler-Natta catalyst.

So, statement B is incorrect.

Statement C: The p-block metal compound employed in Ziegler-Natta polymerization has the metal in +3 oxidation state.

1. As we just discussed, the Ziegler-Natta catalyst contains a compound like triethylaluminium, $Al(C_2H_5)_3$.

2. Aluminium (Al) belongs to Group 13, which is in the p-block of the periodic table.

3. Let's find the oxidation state of Al in $Al(C_2H_5)_3$. The ethyl group ($C_2H_5$) is an alkyl group and carries a charge of -1 when bonded to a metal.

   Let the oxidation state of Al be $x$.

   $ x + 3(-1) = 0 $

   $ x = +3 $

4. So, the p-block metal (Al) is indeed in the +3 oxidation state.

Therefore, statement C is correct.

Statement D: The number of electrons present in the outer most 'd' orbital of metal halide employed in Wacker process is 8.

1. The Wacker process is used for the oxidation of ethene to ethanal (acetaldehyde) using a catalyst.

2. The catalyst used is Palladium(II) chloride, $PdCl_2$, in an aqueous medium.

3. The metal halide here is $PdCl_2$.

4. In $PdCl_2$, the oxidation state of Palladium (Pd) is +2.

5. Palladium (Pd) has atomic number 46. Its electronic configuration is $[Kr] 4d^{10} 5s^0$.

6. To get the configuration of the $Pd^{2+}$ ion, we remove two electrons. The configuration becomes $[Kr] 4d^8$.

7. The outermost 'd' orbital for Palladium is the 4d orbital. The number of electrons in this 4d orbital is 8.

Therefore, statement D is correct.

Conclusion:

We found that statements C and D are correct, while A and B are incorrect.

Looking at the options, we need to choose the one that says "C and D Only".

The correct answer is Option C.

The correct answer is Option D: $ \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 $.

A paper dipped in dilute $ \mathrm{H}_2\mathrm{SO}_4 $ solution of $X$ turns green when treated with $ \mathrm{SO}_2 $ gas.

This means that $X$ must be a substance which changes to green on reduction by $ \mathrm{SO}_2 $.

Now, $ \mathrm{SO}_2 $ is a reducing agent.

Acidified potassium dichromate, that is, $ \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 $ in dilute $ \mathrm{H}_2\mathrm{SO}_4 $, is orange in colour and gets reduced to $ \mathrm{Cr}^{3+} $, which is green.

The reaction is:

$ \mathrm{Cr}_2\mathrm{O}_7^{2-} + 3\mathrm{SO}_2 + 2\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 3\mathrm{SO}_4^{2-} + \mathrm{H}_2\mathrm{O} $

So, the orange dichromate paper turns green.

Now check other options briefly:

• A: $ \mathrm{KI}$-starch

  This is used to detect oxidising agents; it gives blue colour due to iodine-starch complex, not green.

• B: $ \mathrm{KMnO}_4 $

  Acidified $ \mathrm{KMnO}_4 $ gets decolourised by $ \mathrm{SO}_2 $, not green.

• C: $ \mathrm{Pb}(\mathrm{CH}_3\mathrm{COO})_2 $

  Lead acetate paper is used for detecting $ \mathrm{H}_2\mathrm{S} $, giving black colour.

Hence, the compound $X$ is

$ \boxed{\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7} $

The main physical and chemical properties of interstitial compounds are :

(1) They have very high melting points, even higher than the pure metals from which they are formed.

(2) They are extremely hard. For example, some borides can be almost as hard as diamond.

(3) They continue to show metallic conductivity, meaning they can conduct electricity like metals.

(4) They are chemically inactive (inert) and often do not have a fixed composition, which is called being non-stoichiometric.

We need to compare the number of electrons present in the $4f$ orbital for each given element.

For lanthanoids, we write the electronic configurations and count the electrons in $4f$.

Given elements:

• $\text{Eu }(63) = [Xe]\ 4f^7 6s^2$

• $\text{Gd }(64) = [Xe]\ 4f^7 5d^1 6s^2$

• $\text{Dy }(66) = [Xe]\ 4f^{10} 6s^2$

• $\text{Ho }(67) = [Xe]\ 4f^{11} 6s^2$

• $\text{Tm }(69) = [Xe]\ 4f^{13} 6s^2$

• $\text{Yb }(70) = [Xe]\ 4f^{14} 6s^2$

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

• $\text{Hf }(72) = [Xe]\ 4f^{14} 5d^2 6s^2$

Now check each pair:

A. Eu and Gd

• Eu has $4f^7$

• Gd has $4f^7$

So, same number of $4f$ electrons. This is correct.

B. Dy and Ho

• Dy has $4f^{10}$

• Ho has $4f^{11}$

Not same. This is incorrect.

C. Yb and Hf

• Yb has $4f^{14}$

• Hf has $4f^{14}$

Same number of $4f$ electrons. This is correct.

D. Lu and Tm

• Lu has $4f^{14}$

• Tm has $4f^{13}$

Not same. This is incorrect.

So the correct pairs are:

$ A \text{ and } C $

Hence, the correct option is:

$ \boxed{\text{Option D}} $

The highest oxidation state of manganese in its fluoride is in $\mathrm{MnF_4}$, so oxidation state of Mn is $+4$.

The highest oxidation state of manganese in its oxide is in $\mathrm{Mn_2O_7}$, so oxidation state of Mn is $+7$.

Therefore,

$ |x| = |7 - 4| = 3 $

So, we need to find the ions having 3 unpaired electrons.

Now check each ion one by one.

A. $\mathrm{Sc^{3+}}$

Sc has atomic number $21$.

Electronic configuration of Sc:

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

For $\mathrm{Sc^{3+}}$, remove $2$ electrons from $4s$ and $1$ from $3d$:

$ \mathrm{Sc^{3+}} = [Ar]\,3d^0 $

Unpaired electrons $= 0$

So, this is not correct.

B. $\mathrm{Zn^{2+}}$

Zn has atomic number $30$.

$ \mathrm{Zn} = [Ar]\,3d^{10} 4s^2 $

For $\mathrm{Zn^{2+}}$:

$ \mathrm{Zn^{2+}} = [Ar]\,3d^{10} $

Unpaired electrons $= 0$

So, this is not correct.

C. $\mathrm{V^{2+}}$

V has atomic number $23$.

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

For $\mathrm{V^{2+}}$:

$ \mathrm{V^{2+}} = [Ar]\,3d^3 $

In $3d^3$, there are $3$ unpaired electrons.

So, this is correct.

D. $\mathrm{Fe^{2+}}$

Fe has atomic number $26$.

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

For $\mathrm{Fe^{2+}}$:

$ \mathrm{Fe^{2+}} = [Ar]\,3d^6 $

In $3d^6$, number of unpaired electrons $= 4$.

So, this is not correct.

E. $\mathrm{Co^{2+}}$

Co has atomic number $27$.

$ \mathrm{Co} = [Ar]\,3d^7 4s^2 $

For $\mathrm{Co^{2+}}$:

$ \mathrm{Co^{2+}} = [Ar]\,3d^7 $

In $3d^7$, number of unpaired electrons $= 3$.

So, this is correct.

Thus, the correct ions are:

$ \mathrm{V^{2+}} \text{ and } \mathrm{Co^{2+}} $

So the correct answer is:

$ \boxed{\text{Option C: C and E Only}} $

Statement A

$\mathrm{Mo(VI)}$ and $\mathrm{W(VI)}$ are less stable than $\mathrm{Cr(VI)}$.

This statement is incorrect.

In group 6 elements:

• $\mathrm{Cr}$, $\mathrm{Mo}$ and $\mathrm{W}$ commonly show the $+6$ oxidation state.

• The higher oxidation states become more stable for heavier members of the group.

So,

• $\mathrm{Cr(VI)}$ is strongly oxidising and less stable comparatively,

• while $\mathrm{Mo(VI)}$ and especially $\mathrm{W(VI)}$ are more stable.

Hence, A is false.

Statement B

$\mathrm{Ce^{4+}}$ and $\mathrm{Tb^{4+}}$ are oxidant while $\mathrm{Eu^{2+}}$ and $\mathrm{Yb^{2+}}$ are reductant.

This statement is correct.

Reason:

• $\mathrm{Ce^{4+}}$ tends to change to $\mathrm{Ce^{3+}}$, so it acts as an oxidising agent.

• $\mathrm{Tb^{4+}}$ also tends to reduce to $\mathrm{Tb^{3+}}$, so it is an oxidant.

• $\mathrm{Eu^{2+}}$ tends to change to $\mathrm{Eu^{3+}}$, so it donates electrons and acts as a reducing agent.

• $\mathrm{Yb^{2+}}$ similarly tends to go to $\mathrm{Yb^{3+}}$, so it is also a reductant.

Hence, B is true.

Statement C

Cm and Am have seven unpaired electrons.

This statement is incorrect.

Let us see the electronic configurations:

• $\mathrm{Am}$ : $[Rn]\,5f^7 7s^2$

• $\mathrm{Cm}$ : $[Rn]\,5f^7 6d^1 7s^2$

Now count unpaired electrons:

• In $\mathrm{Am}$, $5f^7$ gives 7 unpaired electrons

• In $\mathrm{Cm}$, $5f^7$ gives 7 unpaired, and $6d^1$ gives 1 more unpaired

So $\mathrm{Cm}$ has 8 unpaired electrons, not 7.

Hence, C is false.

Statement D

Actinoid contraction is greater from element to element than lanthanoid contraction.

This statement is correct.

Reason:

• In actinoids, the $5f$ electrons shield the nuclear charge even more poorly than the $4f$ electrons do in lanthanoids.

• Therefore, the decrease in atomic and ionic size across actinoids is greater.

Hence, D is true.

Final conclusion

Correct statements are:

$ \boxed{B \text{ and } D \text{ only}} $

So the correct option is:

$ \boxed{\text{Option C}} $

(A) $\mathrm{MnO}_4^{-}+\mathrm{Fe}^{+2}+\mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{+3}+\mathrm{Mn}^{+2}+\mathrm{H}_2 \mathrm{O}$

(B) $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$ generally not used as a primary standard in volumetric analysis because it is deliquescent.

(C) In acidic medium $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ exists and it get converted into $\mathrm{CrO}_4^{2-}$ in basic medium.

$ \begin{aligned} & 2 \mathrm{CrO}_4^{2-}+2 \mathrm{H}^{+} \longrightarrow \mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{Cr}_2 \mathrm{O}_7^{2-}+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{CrO}_4^{2-}+\mathrm{H}_2 \mathrm{O} \end{aligned} $

(D) $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is a good oxidising agent and bond angle of $\mathrm{Cr}-\mathrm{O}-\mathrm{Cr}$ bond in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is $126^{\circ}$

For a positive borax bead test, the ion must generally have unpaired electrons in its $d$-orbitals, because transition metal ions give characteristic bead colours due to electronic transitions.

Let us check each ion:

• $Fe^{2+} : [Ar]\,3d^6$ → has unpaired electrons, so it gives borax bead test.

• $Fe^{3+} : [Ar]\,3d^5$ → has unpaired electrons, so it gives borax bead test.

• $Cr^{2+} : [Ar]\,3d^4$ → has unpaired electrons, so it gives borax bead test.

• $Zn^{2+} : [Ar]\,3d^{10}$ → all electrons are paired, so it does not give a positive borax bead test.

So, among the given ions, we compare $Fe^{2+}, Fe^{3+}, Cr^{2+}$ for highest ionisation enthalpy.

Now, higher positive charge and greater effective nuclear charge generally mean higher ionisation enthalpy. Also, $Fe^{3+}$ has the stable half-filled configuration:

$ Fe^{3+} = [Ar]\,3d^5 $

This makes removal of another electron more difficult.

Therefore, the ion that:

• shows positive borax bead test, and

• has highest ionisation enthalpy

is

$ \boxed{Fe^{3+}} $

So, the correct option is D. $Fe^{3+}$.

In acidic medium, $\mathrm{K_2Cr_2O_7}$ acts as a strong oxidising agent.

Its orange colour is due to $\mathrm{Cr_2O_7^{2-}}$, which gets reduced to green $\mathrm{Cr^{3+}}$:

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

So, the species that will change the orange colour must be reducing agents, because they will reduce dichromate to $\mathrm{Cr^{3+}}$.

Now check each option.

Option A: $\mathrm{Fe^{2+},\ Sn^{2+},\ I^-,\ S^{2-}}$

• $\mathrm{Fe^{2+} \rightarrow Fe^{3+}}$ : reducing agent

• $\mathrm{Sn^{2+} \rightarrow Sn^{4+}}$ : reducing agent

• $\mathrm{I^- \rightarrow I_2}$ : reducing agent

• $\mathrm{S^{2-}}$ : strong reducing agent

All these can reduce acidified dichromate, so all will change the orange colour.

Option B: $\mathrm{S^{2-},\ Fe^{3+},\ I^-,\ C_2O_4^{2-}}$

• $\mathrm{S^{2-}}$ : reducing agent

• $\mathrm{Fe^{3+}}$ : not a reducing agent here

• $\mathrm{I^-}$ : reducing agent

• $\mathrm{C_2O_4^{2-}}$ : reducing agent

Since $\mathrm{Fe^{3+}}$ will not reduce dichromate, this option is incorrect.

Option C: $\mathrm{Fe^{2+},\ NO_2^-,\ SO_2,\ Sn^{4+}}$

• $\mathrm{Fe^{2+}}$ : reducing agent

• $\mathrm{NO_2^-}$ : can act as reducing agent

• $\mathrm{SO_2}$ : reducing agent

• $\mathrm{Sn^{4+}}$ : already in higher oxidation state, not a reducing agent

So this option is incorrect.

Option D: $\mathrm{Fe^{3+},\ SO_4^{2-},\ S^{2-},\ Sn^{4+}}$

• $\mathrm{Fe^{3+}}$ : not reducing

• $\mathrm{SO_4^{2-}}$ : not reducing

• $\mathrm{S^{2-}}$ : reducing

• $\mathrm{Sn^{4+}}$ : not reducing

So this is also incorrect.

Therefore, the correct answer is:

$ \boxed{\text{Option A}} $

$\mathrm{Cu}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^{10}$, Spin only magnetic moment $=0$ B.M.

$\mathrm{Cu}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^9$, Spin only magnetic moment $=\sqrt{3}$ B.M.

$\mathrm{Cr}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^4$, Spin only magnetic moment $=\sqrt{24}$ B.M.

$\mathrm{Cr}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^3$, Spin only magnetic moment $=\sqrt{15}$ B.M.

Order of $\mu: \mathrm{Cr}^{+2}>\mathrm{Cr}^{+3}>\mathrm{Cu}^{+2}>\mathrm{Cu}^{+}$

In the 3d transition series, Vanadium (V) has the highest enthalpy of atomization. The ion $\mathrm{V}^{3+}$ is known to exhibit a green color in its aqueous form. The oxide formed by the $\mathrm{V}^{3+}$ ion is $\mathrm{V}_2 \mathrm{O}_3$, which is classified as a basic oxide.

So $\mathrm{V}^{2+} \& \mathrm{Co}^{2+}$ same number of unpaired electron.

Statement I is true while Statement II is false.

Cr(VI) is less stable than Mo(VI). Consequently, CrO₃ more readily reduces to Cr³⁺ compared to MoO₃, exhibiting a stronger oxidizing nature.

To determine which metal ions have a calculated spin-only magnetic moment of 4.9 Bohr Magnetons (B.M.), we can use the formula:

$ \text{Magnetic Moment (M.M)} = \sqrt{n(n+2)} \, \text{B.M.} $

where $ n $ is the number of unpaired electrons.

Given that the magnetic moment is 4.9 B.M., we can set up the equation:

$ 4.9 = \sqrt{n(n+2)} $

Solving this equation, we find that $ n = 4 $.

Let's analyze each metal ion:

(A) ${}_{24} \mathrm{Cr}^{2+}$: Electronic configuration: $[\mathrm{Ar}] 3d^4$. This ion has 4 unpaired electrons.

(B) ${}_{26} \mathrm{Fe}^{2+}$: Electronic configuration: $[\mathrm{Ar}] 3d^6$. This ion also has 4 unpaired electrons.

(C) ${}_{26} \mathrm{Fe}^{3+}$: Electronic configuration: $[\mathrm{Ar}] 3d^5$. This ion has 5 unpaired electrons, which does not match the required $ n $ value of 4.

(D) ${}_{27} \mathrm{Co}^{2+}$: Electronic configuration: $[\mathrm{Ar}] 3d^7$. This ion has 3 unpaired electrons.

(E) ${}_{25} \mathrm{Mn}^{3+}$: Electronic configuration: $[\mathrm{Ar}] 3d^4$. This ion has 4 unpaired electrons.

Based on the number of unpaired electrons, the metal ions with a spin-only magnetic moment of approximately 4.9 B.M. are $\mathrm{Cr}^{2+}$, $\mathrm{Fe}^{2+}$, and $\mathrm{Mn}^{3+}$.

Melting point of the pairs $\left(M_n, F_e\right),\left(T_c, R_u\right)$, and $\left(\mathrm{Re}_{\mathrm{e}}, \mathrm{O}_{\mathrm{s}}\right)$ :

$M n, T_c, \mathrm{Re}$ are same group elements.

Also, $\mathrm{Fe}, \mathrm{Bu}, \mathrm{Os}$, are same group elements.

$\mathrm{Mn}, \mathrm{Fe}$

Fe has a higher melting point than $M_n, M_n< F_e$

$\left[\begin{array}{cc}M_n & F_e \\ T_c & R_u \\ R_e & O_s\end{array}\right]$

"Melting point property is directly proportional to the strength of metallic bond, which proportional to the delocalize electrons. The more electrons are free to move around, the stronger will be the metallic bond."

For Mn and $\mathrm{Fe}, \quad\left(\mathrm{Mn}-d^5 s^2, \mathrm{Fe}-d^6 s^2\right)$.

$M_n$ has a weaker metallic bond due to its electronic configuration (half filled configuration). The half filled configuration (stable) results in less delocalization of electrons and weaker attraction between atoms, requiring less energy to break apart and met. But, Fe has localized electrons and hence stronger metallic bonds, leading to higher melting point.

$T_c, R u$

Tc has lower melting point than $\mathrm{Ru}, \mathrm{Tc}_c<\mathrm{Ru}$ This is because of the electronic configuration of its $d$-orbitals.

$T c-d^5 s^2, R u-d^7 s^1$

Tc has halt filled d-orbitals, which makes its electronic configuration stable. This configuration causes the nucleus to hold the electrons tightly, which weakens the metallic bond and it results in lower melting point.

$\mathrm{Re}, \mathrm{Os}_{\mathrm{s}}$

$\operatorname{Re}$ has a higher melting point than $\mathrm{Os}_{\mathrm{s}}, \mathrm{Os}<\operatorname{Re}$ $\left(\mathrm{Re}-d^5 s^2, \mathrm{O} s-d s_s^6\right)$

The higher melting point of $R e$ is due to the strength of atomic bonding.

Re trams strong metallic bonds than Us. Due to the langer size of Re , its valence elections are not tightly hooded by the nucleus and hence, the valence elections causes stronger metallic bond. As a result, melting point increases.

Osmium tills weaker metallic bonds than $\operatorname{Re}$ due to its crystal arrangement and electronic Structure.

Comparing Re, and OS,

$R_e$ is slightly smaller in atomic radius which allows atoms to pack more efficiently than is and has stronger bonds and higher meeting point.

So,

$\begin{aligned} & M_n<\mathrm{Fe} \\ & T_c<\mathrm{Ru} \\ & \mathrm{O}_s<\mathrm{Re} \end{aligned}$

Correct answer:

Option 4) $\mathrm{Mn}<\mathrm{Fe}, \mathrm{TC}<\mathrm{Ru}$ and $\mathrm{Os}<\mathrm{Re}$.

$ \begin{aligned} &\mathrm{V}_2 \mathrm{O}_5+\text { alkali } \rightarrow \mathrm{VO}_4^{3-}\\ &\text { In } \mathrm{VO}_4^{3-} \text { ion, vanadium is in }+5 \text { oxidation state } \end{aligned}$

$\mathrm{Ni}^{+2}$ gives violet coloured bead in non-luminous flame under hot conditions. $\mathrm{Ni}^{+2}$ has $\mathrm{d}^8$ configuration which does not depend on nature of ligand present in octahedral complex.

$\mathrm{Ni}^{+2}: \mathrm{t}_{2 \mathrm{~g}}{ }^6 \mathrm{e}_{\mathrm{g}}{ }^2$

$\begin{array}{ll} \mathrm{Sc}^{+3}=3 \mathrm{~d}^{\circ} & \therefore \mu_{\text {spin }}=0 \\ \mathrm{~V}^{+2}=3 \mathrm{~d}^3 & \therefore \mu_{\text {spin }}=3.87 \text { B.M. } \\ \mathrm{Ni}^{+2}=3 \mathrm{~d}^8 & \therefore \mu_{\text {spin }}=2.84 \text { B.M. } \end{array}$

$\mathrm{Ti}^{+3}=3 \mathrm{~d}^1 \quad \therefore \mu_{\mathrm{spin}}=1.73 \mathrm{~B} . \mathrm{M}$

The preparation of potassium permanganate ($\mathrm{KMnO}_4$) from manganese dioxide ($\mathrm{MnO}_2$) involves a two-step process. In the first step, $\mathrm{MnO}_2$ reacts with potassium hydroxide ($\mathrm{KOH}$) and potassium nitrate ($\mathrm{KNO}_3$) under heating. This reaction produces potassium manganate ($\mathrm{K}_2 \mathrm{MnO}_4$) as an intermediate product. The reaction can be depicted as follows:

$ \mathrm{MnO}_2 \xrightarrow[\mathrm{KNO}_3, \mathrm{\Delta}]{\mathrm{KOH}} \mathrm{K}_2 \mathrm{MnO}_4 $

$\mathrm{Tb}^{4+}$ is strongest oxidising agent

To match the items from List I to List II, we need to understand the composition of each material:

Bronze is an alloy composed of copper (Cu) and tin (Sn).

Brass is an alloy made of copper (Cu) and zinc (Zn).

The UK silver coin primarily consists of copper (Cu) and nickel (Ni).

Stainless steel is an alloy that includes iron (Fe), chromium (Cr), nickel (Ni), and carbon (C).

Based on these definitions, the correct matches are:

(A) Bronze corresponds to (IV) Cu, Sn

(B) Brass corresponds to (III) Cu, Zn

(C) UK silver coin corresponds to (I) Cu, Ni

(D) Stainless steel corresponds to (II) Fe, Cr, Ni, C

These pairings illustrate the composition of each material and ensure accurate alignment between List I and List II.

$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{KOH}} \underset{\text { [A] }}{\mathrm{K}_2 \mathrm{CrO}_4} \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{H}_2 \mathrm{~S}_4} \underset{\text { [B] }}{\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}+\mathrm{K}_2 \mathrm{SO}_4$

(1) $\mathrm{V}^{2+}$ (Violet), $\mathrm{Cr}^{3+}$ (Violet), $\mathrm{Mn}^{3+}$ (Violet)

(2) $\mathrm{Zn}^{2+}$ (Colourless), $\mathrm{V}^{3+}$ (Green), $\mathrm{Fe}^{3+}$ (Yellow)

(3) $\mathrm{Ti}^{4+}$ (Colourless), $\mathrm{V}^{4+}$ (Blue), $\mathrm{Mn}^{2+}$ (Pink)

(4) $\mathrm{Sc}^{3+}$ (Colourless), $\mathrm{Ti}^{3+}$ (Purple), $\mathrm{Cr}^{2+}$ (Blue)

Let's analyze the electron configurations of the given lanthanide ions.

Europium ($\mathrm{Eu}$)

The neutral atom of Eu (atomic number 63) typically has the configuration:

$ [\mathrm{Xe}]\,4f^7\,6s^2. $

For $\mathrm{Eu}^{2+}$, two electrons are removed, usually from the $6s$ orbital, resulting in:

$ [\mathrm{Xe}]\,4f^7. $

Thus, $\mathrm{Eu}^{2+}$ has a $4f^7$ configuration.

For $\mathrm{Eu}^{3+}$, three electrons are removed (the two $6s$ electrons and one $4f$ electron), giving:

$ [\mathrm{Xe}]\,4f^6. $

Therefore, $\mathrm{Eu}^{3+}$ does not have a $4f^7$ configuration.

Gadolinium ($\mathrm{Gd}$)

The neutral atom of Gd (atomic number 64) is usually represented as:

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

For $\mathrm{Gd}^{3+}$, three electrons are removed (typically the $5d^1$ and the two $6s$ electrons), resulting in:

$ [\mathrm{Xe}]\,4f^7. $

So, $\mathrm{Gd}^{3+}$ has a $4f^7$ configuration.

Terbium ($\mathrm{Tb}$)

The neutral atom of Tb (atomic number 65) typically has the configuration:

$ [\mathrm{Xe}]\,4f^9\,6s^2. $

For $\mathrm{Tb}^{3+}$, removal of three electrons gives:

$ [\mathrm{Xe}]\,4f^8. $

Hence, $\mathrm{Tb}^{3+}$ does not have a $4f^7$ configuration.

Samarium ($\mathrm{Sm}$)

The neutral atom of Sm (atomic number 62) generally has:

$ [\mathrm{Xe}]\,4f^6\,6s^2. $

For $\mathrm{Sm}^{2+}$, after removal of two electrons (likely from the $6s$ orbital), the configuration is:

$ [\mathrm{Xe}]\,4f^6. $

Therefore, $\mathrm{Sm}^{2+}$ does not have a $4f^7$ configuration.

Thus, the ions with a $4f^7$ configuration are:

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

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

The correct answer is:

Option D – (A) and (B) only.

Among given, $\mathrm{Eu}^{2+}, \mathrm{Gd}^{3+}$ has same number of electron and are paramagnetic in nature.

$\mathrm{Eu}(\mathrm{Z}=63) \rightarrow \mathrm{Eu}^{2+}=63-2=61$ electron.

$\mathrm{Gd}(\mathrm{Z}=64) \rightarrow \mathrm{Gd}^{3+}=64-3=61$

electrons.

Both are paramagnetic (unpaired $f$-electrons)

Among the given ions, $\mathrm{Fe}^{2+}, \mathrm{Cu}^{2+}$ and $\mathrm{V}^{3+}$ are correctly matched with their colour.

The colour of $\mathrm{Fe}^{3+}$ (aquated) is yellow.

Atomic radius generally increases down a group. However, due to the lanthanoid contraction, elements of the $5 d$ series have nearly the same atomic radii as their counterparts in the $4 d$ series. Lanthanoid contraction is the poor shielding effect of the $4 f$ electrons, which increases the effective nuclear charge and pulls the valence electrons closer to the nucleus.

• $\operatorname{Zr}(4 d)$ and Hf ( $5 d$ ) are in the same group, and size of Hf is reduced by the lanthanoid contraction, making their radii nearly identical.

• Mo (4d) and W (5d) are also a pair in the same group where the size of $W$ is reduced by the lanthanoid contraction, resulting in almost identical radii.

• The other pairs, Y-La and Cr-Mo, do not exhibit this effect to the same extent, and their atomic radii differ significantly.

Therefore, the pairs with nearly the same atomic radius are $\mathrm{Zr}-\mathrm{Hf}$ and $\mathrm{Mo}-\mathrm{W}$.

$E_{M^{3+} / M^{2+}}^{\circ}$ is highest for Mn i.e., $=+1.57 \mathrm{~V}$

It is due to stability of $\mathrm{Mn}^{2+}$ ion which has a half filled $d$-orbitals.

Statements I and IV are correct, while II and III are incorrect. Their correct form is,

II $\mathrm{Eu}^{2+}$ and $\mathrm{Yb}^{2+}$ acts as reducing agent.

IV Misch metal is primarily composed of lanthanoids metal ( $\sim 95 \%$ ) and iron ( $\sim 5 \%$ ) and traces of S, C, Ca and Al.

Among the given option $\mathrm{Cr}^{3+}$ ion's magnetic moment does not match.

$ \begin{aligned} \mathrm{Cr}[24] & =[\mathrm{Ar}] 3 d^5 4 s^1 \\ \mathrm{Cr}^{3+} & =[\mathrm{Ar}] 3 d^3 \end{aligned} $

Unpaired electron = 3

Magnetic moment

$ \begin{aligned} & =\mu=\sqrt{n(n+2)} \\ & =\sqrt{3(3+2)} \\ \mu & =\sqrt{15} \text { or } 3.87 \text { B.M. } \end{aligned} $

Option A

CrO is basic but Cr₂O₃ is amphoteric.

• Cr oxidation states:

• In CrO, Cr is in +2 oxidation state → basic oxide.

• In Cr₂O₃, Cr is in +3 oxidation state → amphoteric oxide.

✅ This statement is correct.

Option B

Nitrite is oxidised to nitrate in acidic medium by KMnO₄.

• In acidic medium, $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$ (oxidizing agent).

• Nitrite ($\text{NO}_2^-$, N in +3) can be oxidized to nitrate ($\text{NO}_3^-$, N in +5)).

✅ This is correct.

Option C

PdCl₂ is the catalyst in Wacker process.

• The Wacker process (oxidation of ethylene to acetaldehyde) uses PdCl₂–CuCl₂ as a catalyst system.

✅ Correct.

Option D

The reactivity of the earlier members of lanthanide series is similar to that of aluminium.

• Lanthanides are strongly electropositive and more reactive than aluminum.

• They readily react with water and acids even at moderate temperatures, unlike Al (which is passivated by oxide film).

❌ This statement is not correct.

✅ Final Answer: Option D is not correct.

Amphoteric oxide of vanadium

$ =\mathrm{V}_2 \mathrm{O}_5 $

$\mathrm{V}_2 \mathrm{O}_5$ with alkali

$ \mathrm{V}_2 \mathrm{O}_5+6 \mathrm{NaOH} \longrightarrow 2 \mathrm{Na}_3 \mathrm{VO}_4+3 \mathrm{H}_2 \mathrm{O} $

oxidation state $=+5$ in $\mathrm{Na}_3 \mathrm{VO}_4$

$\mathrm{V}_2 \mathrm{O}_5$ with acid

$ \mathrm{V}_2 \mathrm{O}_5+6 \mathrm{HCl} \longrightarrow 2 \mathrm{VO}_2 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 $

Oxidation state $=+5$ in $\mathrm{VO}_2 \mathrm{Cl}$

Among given lanthanides, total five elements gives +4 oxidation states, they are

Ce, Pr, Tb, Dy, Nd

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