d and f Block Elements

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 order of oxidizing power for the given ions is:

$ \mathrm{VO}_2^{+} < \mathrm{Cr}_2 \mathrm{O}_7^{2-} < \mathrm{MnO}_4^{-} $

Oxidizing power is directly related to the oxidation number of an element; the higher the oxidation number, the greater the oxidizing power.

For $\mathrm{VO}_2^{+}$, if we let the oxidation number of Vanadium be $x$, we can write:

$ x + 2(-2) = 1 \implies x = +5 \tag{i} $

For $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$, the calculation is:

$ 2x + 7(-2) = -2 \implies x = +6 \tag{ii} $

For $\mathrm{MnO}_4^{-}$, the calculation is:

$ x + 4(-2) = -1 \implies x = +7 \tag{iii} $

This shows that $\mathrm{MnO}_4^{-}$ has the highest oxidation number, therefore, it is the strongest oxidizing agent among the given ions.

$ \begin{array}{lcl} \hline \text { Name } & \text { Atomic Number } & \begin{array}{l} \text { Electronic } \\ \text { Configuration } \end{array} \\ \hline \text { Cerium (Ce) } & 58 & {[\mathrm{Xe}] 4 f^1 5 d^1 6 s^2} \\ \hline \text { Gadolinium (Gd) } & 64 & {[\mathrm{Xe}] 4 F^7 5 d^1 6 s^2} \\ \hline \text { Lutetium (Lu) } & 71 & {[\mathrm{Xe}] 4 f^{14} 5 d^1 6 s^2} \\ \hline \end{array} $

Let's match each element from List I with its correct category in List II:

A. Technicium

Technicium (technetium, symbol Tc) is a transition metal.

Match: II. Transition metal

B. Fluorine

Fluorine is a halogen, which is classified as a non-metal.

Match: I. Non-metal

C. Tellurium

Tellurium is known for its properties that are intermediate between metals and non-metals, so it is classified as a metalloid.

Match: IV. Metalloid

D. Dysprosium

Dysprosium is part of the lanthanide series (also known as lanthanoids).

Match: III. Lanthanoid

Thus, the correct matching is:

A-II, B-I, C-IV, D-III

This corresponds to Option C.

The correct answer is Option D: tungsten (W).

Here's a bit of explanation:

Tungsten (W) has a melting point of about $3422^\circ C$, which is the highest among the elements listed.

For comparison:

Rhenium (Re) melts at around $3186^\circ C$.

Molybdenum (Mo) melts at approximately $2623^\circ C$.

Chromium (Cr) melts at roughly $1907^\circ C$.

Thus, tungsten (W) clearly stands out as the transition metal with the highest melting point.

Let's analyze the formation of a simple monoxide (with formula $\text{MO}$) for each metal:

For a metal to form an oxide of the type $\text{MO}$, the metal would need to adopt a +2 oxidation state (since oxygen is $-2$).

Vanadium, chromium, and manganese can adopt a +2 oxidation state:

Vanadium can form $\text{VO}$ (vanadium(II) oxide).

Chromium can form $\text{CrO}$ (chromium(II) oxide), although it is less stable, it is known under certain conditions.

Manganese forms $\text{MnO}$ (manganese(II) oxide) quite readily.

Scandium, on the other hand, almost exclusively forms a +3 oxidation state. Its stable oxide is $\text{Sc}_2\text{O}_3$ (scandium(III) oxide), and there is no stable monoxides like $\text{ScO}$.

Thus, the transition metal that does not form an "MO" type oxide is:

Option D: Scandium.

Isoelectronic species have same number of electrons in them.

(a) $\mathrm{Pr}^{3+}$ number of electron $=59-3=56$

$\mathrm{Nd}^{3+}$ number of electron $=60-3=57$

$\therefore \mathrm{Pr}^{3+}$ and $\mathrm{Nd}^{3+}$ are not isoelectronic.

(b) $\mathrm{Tb}^{3+}$ : Total number of electron $=65-3=62$

$\mathrm{Dy}^{2+}$ : Total number of electron $=66-2=64$

$\therefore \mathrm{Tb}^{3+}$ and $\mathrm{Dy}^{2+}$ are not isoelectronic.

(c) $\mathrm{Eu}^{2+}$ number of electron $=63-2=61$ $\mathrm{Gd}^{3+}$ number of electron $=64-3=61$

$\therefore \mathrm{Eu}^{2+}$ and $\mathrm{Gd}^{3+}$ are isoelectronic.

(d) $\mathrm{Pr}^{3+}$ number of electron $=59-3=56$ $\mathrm{Ce}^{4+}$ number of electron $=58-4=54$

$\therefore \mathrm{Pr}^{3+}$ and $\mathrm{Ce}^{4+}$ are not isoelectronic.

Transition metals and their complexes show catalytic activity because of

I. presence of vacant $d$-orbitals.

II. ability to show variable oxidation valencies.

III. tendency to form complex compounds.

A catalyst is a chemical species which increases the rate of the reaction by lowering the activation energy of the reaction.

Thus, Assertion and Reason are correct but Reason is not correct explanation of Assertion.

$X\buildrel \Delta \over \longrightarrow {O_2} + Y + Z$ and $Z\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{KN{O_3}}^{KOH}} Y$

$\mathop {2KMn{O_4}}\limits_{(X)} \buildrel {} \over \longrightarrow \mathop {{K_2}Mn{O_4}}\limits_{(Y)} + \mathop {Mn{O_2}}\limits_{(Z)} + {O_2}$

$\mathop {Mn{O_2}}\limits_{(Z)} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{KN{O_3}}^{KOH}} \mathop {{K_2}Mn{O_4}}\limits_{(Y)} $

$\therefore$ $X = KMn{O_4},Y = {K_2}Mn{O_4}$ and $Z = Mn{O_2}$

(A) Electronic configuration of $\mathrm{Cu}$ is

$1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1$

$\Delta_i H_1$ is low while $\Delta_i H_2$ is highest because of fully filled $d$-orbital.

(B) Electronic configuration of $\mathrm{Zn}$ is $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2$

First 2 electrons are removed from $s$-orbital while 3rd electron released from fully filled $d$-orbital, which require large amount of energy.

(C) $M$ in $\left[M(\mathrm{CO})_6\right]$ is $\mathrm{Cr}$.

(D) Ni has highest heat of atomisation in the $3 d$-transition series.

(a) Basic character of metal oxides decreases with increase in oxidation number

$\stackrel{+2}{M \mathrm{O}}, \stackrel{+3}{M_2}\mathrm{O}_3, \stackrel{+4}{M} \mathrm{O}_2, \stackrel{+5}{M_2} \mathrm{O}_5$
(Decreasing basics character)

(b) Manganese ($\mathrm{Mn}$) has maximum number of oxidation state.

Sc, V, Cr, Mn (Increasing number of oxidation states)

(c) Magnetic moment $=\sqrt{n(n+1)}$

( $n=$ number of unpaired electron)

Magnetic moment $\propto$ unpaired electron

So, the correct order of magnetic moment is

$d^5>d^4>d^3>d^1$

(d) $\begin{aligned} \mathrm{Mn}^{2+} & \rightarrow 3 d^5 \\ \mathrm{Fe}^{2+} & \rightarrow 3 d^6 \\ \mathrm{Cr}^{2+} & \rightarrow 3 d^4 \\ \mathrm{Co}^{2+} & \rightarrow 3 d^7 \end{aligned}$

$\mathrm{Mn}^{2+}$ is most stable as $d$-subshell is half filled.

The electronic configuration of $\mathrm{Cu}^{2+}$ is

$[\mathrm{Ar}] 3 d^9$

It contains one unpaired electron,

Hence, $\mathrm{CuCl}_2$ is coloured compound. In all other compounds,

$\begin{aligned} & \mathrm{CuCl} \rightarrow\left[\mathrm{Cu}^{+}\right] \Rightarrow 3 d^{10} 4 s^0 \\ & \mathrm{ScCl}_3 \rightarrow\left[\mathrm{Sc}^{3+}\right] \Rightarrow 3 p^6 4 s^2 \\ & \mathrm{TiCl}_4 \rightarrow\left[\mathrm{Ti}^{4+}\right] \Rightarrow 3 p^6 4 s^2 \end{aligned}$

No unpaired electrons are present.

For general electronic configuration of metal,

$\mathrm{Fe}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^6, 4 s^2$

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

$\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6 4 s^0$ [Remove $2 e^{-}$ from $s$-orbital first]

$\Rightarrow \quad \mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 d^6 4 s^0$

$\mathrm{Fe}^{3+}$ is denoted as ferric (III) ion.

Fe(Z = 26) = [Ar]3d$^6$ 4s$^2$

Fe$^{2+}$ = [Ar]3d$^6$ 4s$^0$

Total number of unpaired electrons = 4

i.e., n = 4

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

$=\sqrt{4(4+2)}=\sqrt{4\times6}$

$=\sqrt{24}$

$=4.9$ BM

Hence, magnetic moment of Fe$^{2+}$ is 4.9 BM.

$\mathrm{CrO}_4^{2-}$ is a stronger oxidising agent as compared to $\mathrm{MnO}_4^{2-}$.

$\mathrm{Cr}^{+6} \rightarrow[\mathrm{Ar}] 3 d^0 4 s^0$ which is stable. So, Cr reading loses its 6 electrons to attain stable noble gas and acting as oxidising agent but in $\mathrm{Mn}^{+6}$ [electronic configuration is $[\mathrm{Ar}] 3 d^1 4 s^0$ ] one electron is difficult to remove and acquire stable form. Hence, $\mathrm{MnO}_4^{-}$ is not oxidising agent but $\mathrm{CrO}_4{ }^{2-}$ is good oxidising agent.

The given electronic configuration is [Ar] 3d$^5$ 4s$^2$. To determine the group of the periodic table to which this element belongs, we need to analyze its electron configuration, particularly the electrons in the outermost shells.

In this configuration:

• [Ar] represents the argon core, which is a stable, noble gas with 18 electrons.
• 3d$^5$ indicates there are 5 electrons in the 3d subshell.
• 4s$^2$ indicates there are 2 electrons in the 4s subshell.

Total electrons in the outer shells (valence electrons) = 5 (in 3d subshell) + 2 (in 4s subshell) = 7 valence electrons.

This pattern of electron configuration is characteristic of the elements in Group 7 of the periodic table. The 3d subshell is filling and the outer 4s subshell contains 2 electrons, typical of a transition metal in Group 7. Examples of such elements include manganese (Mn), which has the atomic number 25 and the electron configuration [Ar] 3d$^5$ 4s$^2$.

Therefore, the element with the electronic configuration [Ar] 3d$^5$ 4s$^2$ belongs to:

Option C: Seventh