117 public questions tagged with this topic.
KMnO 4 is used as an oxidant, disinfectant, and in organic synthesis, but not as a reducing agent.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Preparation and Properties of Potassium Dichromate and Permanganate
Interstitial compounds are formed when small atoms (H, C, N) occupy voids in metal lattices (e.g., TiC).
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Preparation and Properties of Potassium Dichromate and Permanganate
KMnO 4 is an oxidizing agent, not used as a reducing agent.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Preparation and Properties of Potassium Dichromate and Permanganate
Zinc (Zn) has a low melting point (420°C) due to completely filled d-orbitals (3d 10 ), reducing metallic bonding strength.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Lanthanoids - Electronic Configuration and Lanthanoid Contraction
Lanthanoid contraction leads to similar atomic sizes of Zr and Hf, but it does not affect ionization energy trends.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Lanthanoids - Electronic Configuration and Lanthanoid Contraction
Lanthanoids primarily exhibit +3 oxidation state. Only a few elements like Eu and Yb show +2 oxidation state. Therefore, the statement that they primarily exhibit +2 oxidation state is incorrect.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Lanthanoids - Electronic Configuration and Lanthanoid Contraction
Tungsten (W) has the highest melting point (~3422°C) among all transition metals.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: General Properties of d-Block - Electronic Configuration and Trends
Mn shows up to +7 oxidation state (e.g., KMnO 4 ), the highest in the 3d series.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: General Properties of d-Block - Electronic Configuration and Trends
Henry’s law: p = KH · x . New p = 200 × 0.03 = 6 bar .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Molality = (moles/mass of solvent in kg) = (0.01/0.5) = 0.02 mol/kg . Cross-check: x = (6/300) = 0.02 , consistent with molality for dilute solutions.
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Moles of methanol = (16/32) = 0.5 mol . Moles of water = (72/18) = 4 mol . Total moles = 0.5 + 4 = 4.5 . Mole fraction = (0.5/4.5) ≈ 0.111 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Initial moles of ethanol = (46/46) = 1 . Moles of water = (54/18) = 3 . Let additional moles of ethanol = x . New mole fraction = (1 + x/1 + x + 3) = 0.4 . 1 + x = 0.4 (4 + x) , 1 + x = 1.6 + 0.4x , 0.6x = 0.6 , x = 1 . Mass added = 1 × 46 = 46 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
