The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M.$4.9\text{ B.M.}$ are: A. Cr²⁺$\mathrm{Cr}^{2+}$ B. Fe²⁺$\mathrm{Fe}^{2+}$ C. Fe³⁺$\mathrm{Fe}^{3+}$ D. Co²⁺$\mathrm{Co}^{2+}$ E. Mn³⁺$\mathrm{Mn}^{3+}$ Choose the correct answer from the options given below:

Core Logic

The reaction sequence for the chromyl chloride test is:

The chromyl chloride gas is then passed through a basic medium (like NaOH$NaOH$) to form a yellow solution of chromate ions:

Acidification of the yellow CrO₄²⁻$CrO_4^{2-}$ solution followed by the addition of H₂O₂$H_2O_2$ and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide (CrO₅$CrO_5$).

CrO₄²⁻ [yellow solution, 1. Acidification CrO₅ (blue compound)$$CrO_4^{2-} \xrightarrow[\text{yellow solution}, {1. \text{Acidification}} CrO_5 \text{ (blue compound)}$$

Step 1: Oxidation State Calculation

The structure of chromium pentoxide (CrO₅$CrO_5$) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen (O²⁻$O^{2-}$) and four peroxide oxygens (O₂²⁻$O_2^{2-}$). Therefore, there are 2 peroxo linkages.

Let the oxidation state of Chromium be x$x$.

x - 2 - 4 = 0$x - 2 - 4 = 0$ x = +6$x = +6$

Thus, the oxidation state of Cr in CrO₅$CrO_5$ is +6$+6$.

Pattern Recognition

A classic oxidation state trap. Calculating simply via formula CrO₅$CrO_5$ yields x - 10 = 0 x = +10$x - 10 = 0 \implies x = +10$, which is impossible for Chromium (max +6). Whenever calculation exceeds the maximum group valency, peroxide bonds are present.

Chapter Mix

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

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:

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

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$Na_2CrO_4$).

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).

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

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

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⁻$MnO_4^{2-}$).

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^-$MnO_4^-$).

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)$MnO_2 \xrightarrow{\text{fusion, } KOH, O_2} MnO_4^{2-} \text{ (green)} \xrightarrow{\text{electrolytic oxidation}} MnO_4^- \text{ (purple)}$.

Chapter Mix

Class 12 Chemistry: The d and f Block Elements