314 public questions tagged with this topic.
Lanthanoid contraction causes similar atomic sizes of 4d and 5d transition elements.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Lanthanoids - Electronic Configuration and Lanthanoid Contraction
Lanthanoid contraction is the gradual decrease in atomic/ionic radii across the lanthanoid series.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Lanthanoids - Electronic Configuration and Lanthanoid Contraction
Yb 2+ has f 14 configuration which is stable due to completely filled f-orbitals.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Lanthanoids - Electronic Configuration and Lanthanoid Contraction
Let oxidation state of Cr be x. 2(+1) + 2x + 7(-2) = 0 → x = +6.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Lanthanoids - Electronic Configuration and Lanthanoid Contraction
Sc 3+ has no unpaired electrons (3d 0 ), so it does not exhibit d-d transitions.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Lanthanoids - Electronic Configuration and Lanthanoid Contraction
Co 2+ has 3 unpaired electrons (3d 7 configuration), making it paramagnetic.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Lanthanoids - Electronic Configuration and Lanthanoid Contraction
E°cell = (0.059/n) log Kc . 0.46 = (0.059/2) log Kc , log Kc = 15.59 , Kc ≈ 3.9 × 10¹⁵ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
α = (Lambdam/Lambdam°) = (19/380) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.02/1 - 0.05) = (0.0025 × 0.02/0.95) = 5.263 × 10⁻⁵ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
α = (Lambdam/Lambdam°) = (19/380) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.02/1 - 0.05) = (0.0025 × 0.02/0.95) = 5.263 × 10⁻⁵ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
E = E° - (0.059/n) log (1/[H⁺]) , [H⁺] = 10⁻¹⁰ , n = 1 . E = 0 - 0.059 × 10 = -0.59 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
E = E° - (0.059/n) log (1/[H⁺]) , [H⁺] = 10⁻¹⁰ , n = 1 . E = 0 - 0.059 × 10 = -0.59 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Cu: Charge = 0.2 × 9650 = 1930 C , Moles = (0.127/63.5) = 0.002 mol , Charge = 0.002 × 2 × 96500 = 1930 C , matches. Al: Al³⁺ + 3e⁻ → Al , Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C . I = (2895/9650) = 0.3 A .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
