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.

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.

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

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).

(A) $\mathrm{Mn}_2 \mathrm{O}_7$ is green oil at room temperature.

(B) $\mathrm{V}_2 \mathrm{O}_4$ dissolve in acids to give $\mathrm{VO}^{2+}$ salts.

(C) $\mathrm{CrO}$ is basic oxide

(D) $\mathrm{V}_2 \mathrm{O}_5$ is amphoteric it reacts with acid as well as base.

A. $\mathrm{CrO}_4{ }^{2-}$ is tetrahedral

B. $2 \mathrm{Na}_2 \mathrm{CrO}_4+2 \mathrm{H}^{+} \rightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{Na}^{+}+\mathrm{H}_2 \mathrm{O}$

C. As per NCERT, green manganate is paramagnetic with 1 unpaired electron.

D. Statement is correct

E. Statement is correct

Mn, Ni and Cd metals used in battery industries.

$\left.\begin{array}{l} \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \rightarrow \mathrm{Cr}^{+6} \rightarrow \text { No d - d transition } \\ \mathrm{KMnO}_4 \rightarrow \mathrm{Mn}^{7+} \rightarrow \text { No d - d transition } \end{array}\right\} \text { Charge transfer }$

$\mathrm{Ce}:[\mathrm{Xe}] 4 \mathrm{f}^1 5 \mathrm{~d}^1 6 \mathrm{~s}^2 ; \mathrm{Ce}^{4+}$ diamagnetic

$\mathrm{La}:[\mathrm{Xe}] 4 \mathrm{f}^0 5 \mathrm{~d}^1 6 \mathrm{~s}^2 ; \mathrm{La}^{3+}$ diamagnetic

(A) $\mathrm{Zn}, \mathrm{Cd}, \mathrm{Hg}$ exhibit lowest enthalpy of atomization in respective transition series.

(C) Compounds of $\mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ are diamagnetic in nature.

$\begin{aligned} & \mathrm{K}_2 \mathrm{MnO}_4 \\ & 2+\mathrm{x}-8=0 \\ & \Rightarrow \mathrm{x}=+6 \end{aligned}$

$\mathrm{O.S.}$ of $\mathrm{Mn}=+6$

IUPAC Name $=$

Potassium tetraoxidomanganate(VI)

$\mathrm{KMnO}_4 \xrightarrow{\Delta} \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2$

Statement (1) is true, $\mathrm{Ce}^{+4}$ has noble gas electronic configuration.

Statement (2) is also true due to high reduction potential for $\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}(+1.74 \mathrm{~V})$, and stability of $\mathrm{Ce}^{3+}, \mathrm{Ce}^{4+}$ acts as strong oxidizing agent.

$\begin{aligned} & {[\mathrm{Cr}]=[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^5} \\ & {[\mathrm{Cd}]=[\mathrm{Kr}] 5 \mathrm{~s}^2 4 \mathrm{~d}^{10}} \\ & {[\mathrm{Cu}]=[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^{10}} \\ & {[\mathrm{Ag}]=[\mathrm{Kr}] 5 \mathrm{~s}^1 4 \mathrm{~d}^{10}} \\ & {[\mathrm{Zn}]=[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^{10}} \end{aligned}$

The magnetic moment of an atom or ion with unpaired electrons is given by the formula $\mu = \sqrt{n(n+2)}$ Bohr magnetons (BM), where $n$ is the number of unpaired electrons. The magnetic moment depends on the number of unpaired electrons: more unpaired electrons lead to a higher magnetic moment.

Looking at the given options with respect to their electron configurations and calculating their respective magnetic moments based on their unpaired electrons:

• $[\mathrm{Ar}] 3 \mathrm{~d}^7$ has 3 unpaired electrons, yielding a magnetic moment of $\sqrt{15}$ BM.
• $[\mathrm{Ar}] 3 \mathrm{~d}^8$ has 2 unpaired electrons, for a magnetic moment of $\sqrt{8}$ BM.
• $[\mathrm{Ar}] 3 \mathrm{~d}^3$ also has 3 unpaired electrons, similar to $3 \mathrm{~d}^7$, so its magnetic moment is likewise $\sqrt{15}$ BM.
• Finally, $[\mathrm{Ar}] 3 \mathrm{~d}^6$ has 4 unpaired electrons, resulting in the highest magnetic moment among the options, $\sqrt{24}$ BM.

Hence, the electron configuration associated with the highest magnetic moment is $[\mathrm{Ar}] 3 \mathrm{~d}^6$.

The electronic configuration of an element is determined by adding electrons to atomic orbitals following the Aufbau principle, the Pauli exclusion principle, and Hund's rule. For atoms with atomic numbers higher than that of xenon (Xe, atomic number 54), the electrons start filling the outer orbitals following the xenon core. In this case, we're looking at neodymium (Nd), which has an atomic number of 60.

To find the electronic configuration of neodymium, we first note the configuration of xenon, which serves as the core, and then add the additional electrons beyond xenon to the relevant orbitals.

Neodymium has 60 - 54 = 6 electrons more than xenon. These additional electrons will fill the 4f and 6s orbitals. Based on its position in the periodic table, neodymium’s electrons start filling the 4f orbitals before the 5d. Since the 6s orbital is at a lower energy level than the 4f orbital, it gets filled before the 4f. Neodymium will have 2 electrons in the 6s orbital, and the remaining 4 electrons will go into the 4f orbital.

Therefore, the correct electronic configuration for neodymium is:

So, Option A is the correct answer.

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

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

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

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

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

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

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

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

Here are the other options and why they are incorrect:

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


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


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