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Identify correct statements from below: A. The chromate ion is square planar. B. Dichromates are generally prepared from chromates. C. The green manganate ion is diamagnetic. D. Dark green coloured K_2MnO_4 disproportionates in a neutral or acidic medium to give permanganate. E. With increasing oxidation number of transition metal, ionic character of the oxides decreases. Choose the correct answer from the options given below:

Solution & Explanation

### Step 1: Statement A Analysis CrO_4^2- (chromate ion) is tetrahedral, not square planar. Statement A is incorrect. ### Step 2: Statement B Analysis 2Na_2CrO_4 + 2H^+ rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O. Dichromates are indeed prepared from chromates. Statement B is correct. ### Step 3: Statement C Analysis The green manganate ion (MnO_4^2-) has manganese in the +6 oxidation state (3d^1). Thus, it contains 1 unpaired electron and is paramagnetic, not diamagnetic. Statement C is incorrect. ### Step 4: Statement D Analysis Dark green coloured K_2MnO_4 undergoes disproportionation in neutral or acidic media to yield permanganate (MnO_4^-) and manganese dioxide (MnO_2). Statement D is correct. ### Step 5: Statement E Analysis Fajans' rule dictates that as the oxidation state increases, polarizing power increases, leading to a decrease in ionic character (increase in covalent character). Statement E is correct. ### Final Conclusion The correct statements are B, D, and E. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

Reference Study Guides

The D And F Block Elements

More The d- and f-Block Elements Previous-Year Questions — Page 6

Q74 jee_main_2024_29_jan_morning Potassium Permanganate
KMnO_4 decomposes on heating at 513mathrmK to form O_2 along with
  • A. textMnO_2text & K_2textO_2
  • B. textK_2textMnO_4text & Mn
  • C. textMn & KO_2
  • D. textK_2textMnO_4text & MnO_2

Solution

### Core Logic Potassium permanganate (KMnO_4) is a strong oxidizing agent. When heated to 513mathrmK, it undergoes thermal decomposition to give potassium manganate (K_2MnO_4), manganese dioxide (MnO_2), and oxygen gas (O_2). The balanced chemical equation is: 2KMnO_4 xrightarrowDelta K_2MnO_4 + MnO_2 + O_2 ### Step 1: Final Identification The products formed along with O_2 are K_2MnO_4 (green) and MnO_2 (black). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements
Q80 jee_main_2024_29_jan_morning Potassium Permanganate Reactions
In alkaline medium. MnO_4^- oxidises I^- to
  • A. IO_4^-
  • B. IO^-
  • C. I_2
  • D. IO_3^-

Solution

### Core Logic The behavior of the permanganate ion (MnO_4^-) varies with the pH of the medium. In a faintly alkaline or neutral medium, MnO_4^- oxidizes iodide (I^-) completely to iodate (IO_3^-) while getting reduced to manganese dioxide (MnO_2). The balanced ionic equation is: 2MnO_4^- + H_2O + I^- rightarrow 2MnO_2 + 2OH^- + IO_3^- ### Pattern Recognition Rule of thumb for I^- oxidation by KMnO_4: In acidic medium: I^- rightarrow I_2 In alkaline/neutral medium: I^- rightarrow IO_3^- ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements
Q70 jee_main_2024_30_january_evening Properties of Transition Metal Compounds
The orange colour of K_2Cr_2O_7 and purple colour of KMnO_4 is due to
  • A. textCharge transfer transition in both.
  • B. textd rightarrow textd transition in KMnO_4 text and charge transfer transitions in K_2textCr_2textO_7
  • C. textd rightarrow textd transition in K_2textCr_2textO_7 text and charge transfer transitions in KMnO_4.
  • D. textd rightarrow textd transition in both.

Solution

### Core Logic In K_2Cr_2O_7, Chromium is in the +6 oxidation state, which means its electronic configuration is d^0. Since there are no d-electrons, d-d transitions cannot occur. The orange color is due to ligand-to-metal charge transfer (LMCT) from oxygen to chromium. Similarly, in KMnO_4, Manganese is in the +7 oxidation state, which also corresponds to a d^0 configuration. Again, no 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 d^0 configurations, like Cr^+6, Mn^+7, V^+5) are deeply colored primarily due to Charge Transfer spectra, NOT d-d transitions. ### Evaluation Rubric / Model Answer null ### 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 MnO_2 gives "A" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are:
  • A. textMn_2textO_7 text and MnO_4^-
  • B. textMnO_4^2- text and MnO_4^-
  • C. textMn_2textO_3 text and MnO_4^2-
  • D. textMnO_4^2- text and Mn_2textO_7

Solution

### Core Logic Step 1: Alkaline oxidative fusion of MnO_2 (pyrolusite ore) with KOH in the presence of O_2 (or an oxidizing agent like KNO_3) yields the green-colored manganate ion (MnO_4^2-). 2mathrmMnO_2 + 4mathrmOH^- + mathrmO_2 rightarrow 2mathrmMnO_4^2- + 2mathrmH_2mathrmO So, A is mathrmMnO_4^2-. Step 2: Electrolytic oxidation of the manganate ion (MnO_4^2-) in an alkaline medium converts it to the purple-colored permanganate ion (MnO_4^-). mathrmMnO_4^2- rightarrow mathrmMnO_4^- + mathrme^- So, B is mathrmMnO_4^-. ### Pattern Recognition Industrial preparation sequence of KMnO_4: MnO_2 xrightarrowtextfusion, KOH, O_2 MnO_4^2- text (green) xrightarrowtextelectrolytic oxidation MnO_4^- text (purple). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements
Q78 jee_main_2024_30_january_evening Compounds of Transition Elements
A and B formed in the following reactions are: mathrmCrO_2mathrmCl_2 + 4mathrmNaOH rightarrow mathrmA + 2mathrmNaCl + 2mathrmH_2mathrmO mathrmA + 2mathrmHCl + 2mathrmH_2mathrmO_2 rightarrow mathrmB + 3mathrmH_2mathrmO
  • A. textA = Na_2textCrO_4text, B = CrO_5
  • B. textA = Na_2textCr_2textO_4text, B = CrO_4
  • C. textA = Na_2textCr_2textO_7text, B = CrO_3
  • D. textA = Na_2textCr_2textO_7text, B = CrO_5

Solution

### Core Logic Step 1: Chromyl chloride (CrO_2Cl_2) reacts with an alkali like NaOH to give a yellow solution of sodium chromate (Na_2CrO_4). CrO_2Cl_2 + 4NaOH rightarrow Na_2CrO_4 (A) + 2NaCl + 2H_2O Step 2: Sodium chromate (Na_2CrO_4) reacts with hydrogen peroxide (H_2O_2) in an acidic medium (HCl) to yield the deep blue colored chromium pentoxide (CrO_5, also known as chromium(VI) oxide peroxide). Na_2CrO_4 + 2H_2O_2 + 2HCl rightarrow CrO_5 (B) + 2NaCl + 3H_2O Note: NaCl formation implies the overall balanced reaction uses the acid for neutralization/salt formation. ### Pattern Recognition Chromyl chloride test intermediate: Yellow solution = Na_2CrO_4. Reaction of chromate with H_2O_2 in acid = Blue peroxide CrO_5 (butterfly structure). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements Class 11 Chemistry: Redox Reactions

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