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

$\mathrm{Cu}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^{10}$, 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.

			Configuration	No. of unpaired e$^-$
(1)	$\mathrm{V}^{3+}$	$\Rightarrow$	$[\mathrm{Ar}] 3 \mathrm{~d}^3 4 \mathrm{~s}^0$	3
	$\mathrm{Co}^{2+}$	$\Rightarrow$	$[\mathrm{Ar}] 3 \mathrm{~d}^7 4 \mathrm{~s}^0$	3
(2)	$\mathrm{Ti}^{2+}$	$\Rightarrow$	$[\mathrm{Ar}] 3 \mathrm{~d}^2 4 \mathrm{~s}^0$	2
	$\mathrm{Co}^{2+}$	$\Rightarrow$	$[\mathrm{Ar}] 3 \mathrm{~d}^7 4 \mathrm{~s}^0$	3
(3)	$\mathrm{Fe}^{3+}$	$\Rightarrow$	$[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0$	5
	$\mathrm{Cr}^{2+}$	$\Rightarrow$	$[\mathrm{Ar}] 3 \mathrm{~d}^4 4 \mathrm{~s}^0$	4
(4)	$\mathrm{Ti}^{3+}$	$\Rightarrow$	$[\mathrm{Ar}] 3 \mathrm{~d}^1 4 \mathrm{~s}^0$	1
	$\mathrm{Mn}^{2+}$	$\Rightarrow$	$[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0$	5

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.

(6) Diamagnetic - No unpaired electrons. All the electrons are paired. A) $\mathrm{La}^{3+}, \mathrm{Ce}^{4+}$. Element $L a$ election configuration is $[\mathrm{Xe}] 5 \mathrm{~d}^{\prime} 6 \mathrm{~s}^2$ For $\mathrm{La}^{3+}$, three valence electrons are removed So, the electron configuration of $\mathrm{La}^{3+}$ is $\left[\mathrm{Xe}\right]$ The configuration [ $x_e$ ] is the noble gas configuration and all the elections are paired. So, $\mathrm{La}^{3+}$ is diamagnetic. Element $\mathrm{Ce}_{\text { }}$ election configuration is $\left[x_e\right] 4 f^{\prime} 5 d^{\prime} 6 s^2$ For $\mathrm{Ce} ^{4+}$, for valence elections are removed. so, the electron configuration of $\mathrm{Ce}^{4+}$ is $\left[\mathrm{Xe}\right]$. The origination $\left[X_e\right]$ is the noble gas configuration and all the electrons are paired. So, $Ce^{4+}$ is diamagnetic. So, It is a pair of a diamagnetic ions. B) $\mathrm{Yb}^{2+}, \mathrm{Lu}^{3+}$ Element Yb electron configuration is : $\left[x_e\right] 4 f^{14} 6 s^2$ For $y b^{2+}$, two valence electrons are removed from $y b$ - So, the election configuration of ${y} b^{2+}$ is $[x e] \& f^{14}$. The configuration $\left[X_e\right] 4 f^{14}$ has no unpaired electrons. All the electrons are paired and hence $yb^{2+}$ is diamagnetic. Element Lu electron configuration is $\left[x_e\right] 4 f^{\prime 4} 5 d^1 6 s^2$ For $\mathrm{Lu}^{3+}$, three valence electrons we removed from. Lu . So, the electron configuration of $\mathrm{Lu}^{3+}$ is $\left[x_e\right] 4 f^{14}$. The configuration $\left[\mathrm{Xe}\right] 4 f^{14}$ has no unpaired electrons. All the electrons are paired and hence $L u^3 t$ is diamagnetic. So, It is a pair of diamagnetic ions: c) $\mathrm{La}^{2+}, \mathrm{Ce}^{3+}$ Element la electron configuration is [ $x e$ ] $5 d^{\prime} 6 s^2$. For $\mathrm{La}^{2+}$, two valence electrons are removed so, the electron configuration of $\mathrm{La}^{2+}$ is $[\mathrm{Xe}] 5 d^{\prime}$. The configuration $\left[X_e\right] 5 d^{\prime}$ has one unpaired election and hence it is paramagnetic, not diamagnetic. Element $C_e$ electron configuration is $[x e] 4 f^{\prime} 5 d^{\prime} 6 s^2$ For $\mathrm{Ce}^{3+}$, three valence electrons we removed. So, the electron configuration of $\mathrm{Ce}^{3+}$ is $[x e] 4 f^{\prime}$ The configuration $[\mathrm{Xe}]$. $4f^{\prime}$ has one unpaired electron and hence it is paramagnetic, not diamagnetic. So, it is not a pair of diamagnetic ions. D) $\mathrm{Yb}^{3+}, \mathrm{Lu}^{2+}$ yb electron configuration is $[x e] 4 f^{14} 6 s^2$. For $y b^{3+}$, three valence electrons are removed from $y b$. So, the electron configuration of $\mathrm{Yb}^{3+}$ is $\left[x_e\right] 4 f^{13}$ The configuration $\left[x_e\right] 4 f^{13}$ has unpaired election and it is paramagnetic, not diamagnetic. Lu electron configuration is $\left[X_e\right] 4 f^{14} 5 d^1 6 s^2$. For $L u^{2+}$, two valence electrons are removed from LU. So, the electron configuration of $\mathrm{Lu}^{2+}$ is $\left[\mathrm{Xe}_e\right] 4 f^{14} 5 d^1$. The configuration $[x]$. $4 f^{\prime 4} 5 d^{\prime}$ has unpaired electron and it is paramagnetic, not diamagnetic. So, it is not a pair of diamagnetic ions. The pairs of diamagnetic ions are A) $\mathrm{La}^{3+}, \mathrm{Cl}^{4+}$ B) $\mathrm{Yb}^{2+}, \mathrm{Lu}^{3+}$

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+}} $

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

Option C: [Rn] 5f¹¹ 6d⁰ 7s²

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:

[Rn] 5f¹¹ 6d⁰ 7s²

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Therefore, the most appropriate answer is:

Option C: A and D only

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

Scandium (Sc):

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

Unpaired Electrons: 1

Titanium (Ti):

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

Unpaired Electrons: 2

Vanadium (V):

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

Unpaired Electrons: 3

Chromium (Cr):

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

Unpaired Electrons: 6

Manganese (Mn):

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

Unpaired Electrons: 5

Arranging them in increasing order of number of unpaired electrons:

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

Therefore, the correct choice is Option B.

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

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

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

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

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

Option B: 1

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

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

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

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

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

Option C: 3

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

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


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


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


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

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

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

Explanation:

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

Therefore, the correct answer is Option D: Scandium.

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

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

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

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

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