Which of the following involves transfer of five
electrons ?
(a) MnO₄⁻ → Mn²⁺ (b) CrO₄²⁻ → Cr³⁺
(c) MnO₄²⁻ → MnO₂ (d) Cr₂O₇²⁻ → 2Cr³⁺
Answers
Answer:
in option b and d : charge changes from -2 to +3 that can be possible by transfer of 5 electron
The equation (a) MnO₄⁻ → Mn²⁺ involves a transfer of five electrons.
• The transfer of electrons involved a reaction is determined by the change in oxidation state of the metal involved in the reaction.
(a) MnO₄⁻ → Mn²⁺
Let the oxidation state of Mn be denoted as x.
Oxidation state of O is -2.
In MnO₄⁻ ,
x + (-2 × 4) = - 1
=> x + (-8) = - 1
=> x - 8 = - 1
=> x = - 1 + 8
=> x = +7
In Mn²⁺,
x = +2
• Therefore, the change in oxidation state of Mn = +7 - (+2) = 7 - 2 = 5
(b) CrO₄²⁻ → Cr³⁺
Let the oxidation state of Cr be denoted as x.
Oxidation state of O is -2.
In CrO₄²⁻ ,
x + (-2 × 4) = - 2
=> x + (-8) = - 2
=> x - 8 = - 2
=> x = - 2 + 8
=> x = +6
In Cr³⁺,
x = +3
• Therefore, the change in oxidation state of Cr = +6 - (+3) = 6 - 3 = 3
(c) MnO₄²⁻ → MnO₂
Let the oxidation state of Mn be denoted as x.
Oxidation state of O is -2.
In MnO₄²⁻ ,
x + (-2 × 4) = - 2
=> x + (-8) = - 2
=> x - 8 = - 2
=> x = - 2 + 8
=> x = +6
In MnO₂,
x + (-2 × 2) = 0
=> x + (- 4) = 0
=> x - 4 = 0
=> x = +4
• Therefore, the change in oxidation state of Mn = +6 - (+4) = 6 - 4 = 2
(d) Cr₂O₇²⁻ → 2Cr³⁺
Let the oxidation state of Cr be denoted as x.
Oxidation state of O is -2.
In Cr₂O₇²⁻ ,
2x + (-2 × 7) = - 2
=> 2x + (-14) = - 2
=> 2x - 14 = - 2
=> 2x = - 2 + 14
=> 2x = +12
=> x = +12 / 2
=> x = +6
In Cr³⁺,
x = +3
• Therefore, the change in oxidation state of Cr = +6 - (+3) = 6 - 3 = 3
• Therefore, we see that only the reaction MnO₄⁻ → Mn²⁺ involves a change or transfer of five electrons.