Question

Explain why :
(i) E° for Mn³⁺/Mn²⁺ couple is more positive than that for Fe³⁺/Fe²⁺. (At. Nos. Mn = 25, Fe = 26).
(ii) Ce³⁺ can be easily oxidized to Ce⁴⁺. (At. No.Ce = 58).
(i) Stable half-filled 3d³ configuration of Mn²⁺ results in high 3rd ionisation enthalpy of Mn. While in case of Fe²⁺, configuration is 3d⁶. Hence, it can easily loose one electron to give stable configuration 3d⁵.
(ii) Ce³⁺, ions having the configuration 4f¹5d⁰6d⁰ can easily lose electron to acquire the configuration 4f⁰5d⁰6s⁰ and form Ce⁴⁺ ion.

Similar Questions From D And F Block Elements:

Explain the following : (i) The enthalpies of atomization of transition metals are quite high. (ii) The transition metals and many of their compounds act as good catalysts.

Explain the following observation : Most of the transition metal ions exhibit characteristic colour in aqueous solution.

When Cu²⁺ ion is treated with KI, a white precipitate is formed. Explain the reaction with the help of chemical equation.

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with (i) an iodide (ii) H₂S.

Metallic radii of some transition elements are given below. Which of these elements will have highest density?

Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number.

Complete and balance the following chemical equations: (a) Fe²⁺ + MnO₄⁻ + H⁺ → (b) MnO₄⁻ + H₂O + I⁻ →

More 2 Marks Questions:

Explain the following observations: (i) Copper atom has completely filled d orbitals (3d¹⁰) in its ground state, yet it is regarded as a transition element. (ii) Cr²⁺ is a stronger reducing agent than Fe²⁺ in aqueous solution.

Calculate pH of following half-cell. Pt, \(H_2\) / \(H_2\)\(SO_4\) , if its electrode potential is 0.03V.

What is meant by ‘disproportionation’ ? Give an example of a disproportionation reaction in aqueous solution.

The conductivity of an aqueous solution of NaCl in a cell is 92 \(Ω^{−1}\) \(cm^{-1}\) the resistance offered by this cell is 247.8 Ω . Calculate the cell constant.

How many electrons flow when a current of 5 amps is passed through a solution for 193 sec ? Given F = 96500 C. \(N_0\) = 6.002 × \(10^{23}\) \(mol^{-1}\).

Define the following terms : (i) Mole fraction (x) (ii) Molality of a solution (m)

Define the following terms: (i) Colligative properties (ii) Molality (m)

Scroll to Top