The number of valence electrons present in the metal among Cr, Co, Fe and Ni which has the lowest enthalpy of…

Q: The number of valence electrons present in the metal among Cr, Co, Fe and Ni which has the lowest enthalpy of atomisation is:

Let's look at the enthalpy of atomisation values for the given 3d transition metals: - **Chromium (Cr)**: 397 kJ mol-1 - **Iron (Fe)**: 416 kJ mol-1 - **Cobalt (Co)**: 425 kJ mol-1 - **Nickel (Ni)**: 430 kJ mol-1 Among the choices, **Chromium (Cr)** has the lowest enthalpy of atomisation (397 kJ mol-1), due to a highly stable half-filled d-subshell configuration which leads to weaker metallic bonding relative to the other metals listed. The valence electronic configuration of Cr is: Total valenc

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Questions	10	17	17	44

Q80 jee_main_2024_29_jan_morning Potassium Permanganate Reactions

In alkaline medium.

MnO₄^-

oxidises

I^-

to

A. $IO₄^-$ $IO_4^-$
B. $IO^-$ $IO^-$
C. $I₂$ $I_2$
D. $IO₃^-$ $IO_3^-$

Solution

Core Logic

The behavior of the permanganate ion ( $MnO₄^-$ $MnO_4^-$) varies with the pH of the medium. In a faintly alkaline or neutral medium, $MnO₄^-$ $MnO_4^-$ oxidizes iodide ( $I^-$ $I^-$) completely to iodate ( $IO₃^-$ $IO_3^-$) while getting reduced to manganese dioxide ( $MnO₂$ $MnO_2$).

The balanced ionic equation is:

2MnO₄^- + H₂O + I^- arrow 2MnO₂ + 2OH^- + IO₃^-

Pattern Recognition

Rule of thumb for $I^-$ $I^-$ oxidation by $KMnO₄$ $KMnO_4$: In acidic medium: $I^- arrow I₂$ $I^- \rightarrow I_2$ In alkaline/neutral medium: $I^- arrow IO₃^-$ $I^- \rightarrow IO_3^-$

Chapter Mix

Class 12 Chemistry: d and f Block Elements

Q jee_main_2024_30_january_evening Compounds of Transition Elements

A and B formed in the following reactions are:

A. $A = Na₂CrO₄, B = CrO₅$
B. $A = Na₂Cr₂O₄, B = CrO₄$
C. $A = Na₂Cr₂O₇, B = CrO₃$
D. $A = Na₂Cr₂O₇, B = CrO₅$

Solution

Core Logic

Step 1: Chromyl chloride ( $CrO₂Cl₂$ $CrO_2Cl_2$) reacts with an alkali like $NaOH$ $NaOH$ to give a yellow solution of sodium chromate ( $Na₂CrO₄$ $Na_2CrO_4$).

CrO₂Cl₂ + 4NaOH arrow Na₂CrO₄ (A) + 2NaCl + 2H₂O

Step 2: Sodium chromate ( $Na₂CrO₄$ $Na_2CrO_4$) reacts with hydrogen peroxide ( $H₂O₂$ $H_2O_2$) in an acidic medium ( $HCl$ $HCl$) to yield the deep blue colored chromium pentoxide ( $CrO₅$ $CrO_5$, also known as chromium(VI) oxide peroxide).

Na₂CrO₄ + 2H₂O₂ + 2HCl arrow CrO₅ (B) + 2NaCl + 3H₂O

Note: $NaCl$ $NaCl$ formation implies the overall balanced reaction uses the acid for neutralization/salt formation.

Pattern Recognition

Chromyl chloride test intermediate: Yellow solution = $Na₂CrO₄$ $Na_2CrO_4$. Reaction of chromate with $H₂O₂$ $H_2O_2$ in acid = Blue peroxide $CrO₅$ $CrO_5$ (butterfly structure).

Chapter Mix

Class 12 Chemistry: The d and f Block Elements Class 11 Chemistry: Redox Reactions

Q70 jee_main_2024_30_january_evening Properties of Transition Metal Compounds

The orange colour of

K₂Cr₂O₇

and purple colour of

KMnO₄

is due to

A. $Charge transfer transition in both.$
B. $d arrow d transition in KMnO₄ and charge transfer transitions in K₂Cr₂O₇$
C. $d arrow d transition in K₂Cr₂O₇ and charge transfer transitions in KMnO₄.$
D. $d arrow d transition in both.$

Solution

Core Logic

In $K₂Cr₂O₇$ $K_2Cr_2O_7$, Chromium is in the $+6$ $+6$ oxidation state, which means its electronic configuration is $d⁰$ $d^0$. Since there are no d-electrons, $d-d$ $d-d$ transitions cannot occur. The orange color is due to ligand-to-metal charge transfer (LMCT) from oxygen to chromium.

Similarly, in $KMnO₄$ $KMnO_4$, Manganese is in the $+7$ $+7$ oxidation state, which also corresponds to a $d⁰$ $d^0$ configuration. Again, no $d-d$ $d-d$ transitions are possible. The intense purple color is due to ligand-to-metal charge transfer (LMCT) from oxygen to manganese.

Step 1: Final Conclusion

Both compounds owe their colors to charge transfer transitions.

Pattern Recognition

Compounds of transition metals in their highest oxidation states (where they have $d⁰$ $d^0$ configurations, like $Cr⁺⁶$ $Cr^{+6}$, $Mn⁺⁷$ $Mn^{+7}$, $V⁺⁵$ $V^{+5}$) are deeply colored primarily due to Charge Transfer spectra, NOT d-d transitions.

Chapter Mix

Class 12 Chemistry: The d and f Block Elements

Q71 jee_main_2024_30_january_evening Preparation and Properties of KMnO4

Alkaline oxidative fusion of

MnO₂

gives "A" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are:

A. $Mn₂O₇ and MnO₄^-$
B. $MnO₄²⁻ and MnO₄^-$ $\text{MnO}_4^{2-} \text{ and MnO}_4^-$
C. $Mn₂O₃ and MnO₄²⁻$
D. $MnO₄²⁻ and Mn₂O₇$

Solution

Core Logic

Step 1: Alkaline oxidative fusion of $MnO₂$ $MnO_2$ (pyrolusite ore) with $KOH$ $KOH$ in the presence of $O₂$ $O_2$ (or an oxidizing agent like $KNO₃$ $KNO_3$) yields the green-colored manganate ion ( $MnO₄²⁻$ $MnO_4^{2-}$).

2MnO₂ + 4OH^- + O₂ arrow 2MnO₄²⁻ + 2H₂O

So, A is $MnO₄²⁻$ $\mathrm{MnO}_4^{2-}$.

Step 2: Electrolytic oxidation of the manganate ion ( $MnO₄²⁻$ $MnO_4^{2-}$) in an alkaline medium converts it to the purple-colored permanganate ion ( $MnO₄^-$ $MnO_4^-$).

MnO₄²⁻ arrow MnO₄^- + e^-

So, B is $MnO₄^-$ $\mathrm{MnO}_4^-$.

Pattern Recognition

Industrial preparation sequence of $KMnO₄$ $KMnO_4$: $MnO₂ fusion, KOH, O₂ MnO₄²⁻ (green) electrolytic oxidation MnO₄^- (purple)$ .

Chapter Mix

Class 12 Chemistry: The d and f Block Elements

Q66 jee_main_2024_30_jan_morning Lanthanoids

Diamagnetic Lanthanoid ions are:

A. $Nd³⁺ and Eu³⁺$ $Nd^{3+}\text{ and }Eu^{3+}$
B. $La³⁺ and Ce⁴⁺$ $La^{3+}\text{ and }Ce^{4+}$
C. $Nd³⁺ and Ce⁴⁺$ $Nd^{3+}\text{ and }Ce^{4+}$
D. $Lu³⁺ and Eu³⁺$ $Lu^{3+}\text{ and }Eu^{3+}$

Solution

Core Logic

An ion is diamagnetic if all its electrons are paired (i.e., zero unpaired electrons). Let's write the electronic configuration for the elements in question.

Step 1: Checking configurations

Cerium ( $Ce$ $Ce$, Z=58): $[Xe] 4f¹ 5d¹ 6s²$ $[Xe] 4f^1 5d^1 6s^2$ $arrow Ce⁴⁺: [Xe] 4f⁰$ $\rightarrow Ce^{4+}: [Xe] 4f^0$ (0 unpaired electrons $arrow$ $\rightarrow$ Diamagnetic)

Lanthanum ( $La$ $La$, Z=57): $[Xe] 4f⁰ 5d¹ 6s²$ $[Xe] 4f^0 5d^1 6s^2$ $arrow La³⁺: [Xe] 4f⁰$ $\rightarrow La^{3+}: [Xe] 4f^0$ (0 unpaired electrons $arrow$ $\rightarrow$ Diamagnetic)

Pattern Recognition

Ions with an empty f-subshell ( $f⁰$ $f^0$, e.g., $La³⁺, Ce⁴⁺$ $La^{3+}, Ce^{4+}$) or a completely filled f-subshell ( $f¹⁴$ $f^{14}$, e.g., $Lu³⁺, Yb²⁺$ $Lu^{3+}, Yb^{2+}$) are invariably diamagnetic.

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements

The number of valence electrons present in the metal among Cr$\mathrm{Cr}$, Co$\mathrm{Co}$, Fe$\mathrm{Fe}$ and Ni$\mathrm{Ni}$ which has the lowest enthalpy of atomisation is:

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More d- and f-Block Elements Previous-Year Questions — Page 8

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Checking configurations

Pattern Recognition

Chapter Mix

More d- and f-Block Elements Questions — jee_main_2025_07_april_morning

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The number of valence electrons present in the metal among $Cr$ $\mathrm{Cr}$, $Co$ $\mathrm{Co}$, $Fe$ $\mathrm{Fe}$ and $Ni$ $\mathrm{Ni}$ which has the lowest enthalpy of atomisation is: