The pair of compounds in which both the metals are in the
highest possible oxidation state is
(a) [Fe(CN)₆ ]⁴⁻ ,[Co(CN)₆ ]³⁻
(b) CrO₂Cl₂ , MnO₄⁻
(c) TiO₃, MnO₂
(d) [Co(CN)₆ ]³⁻ , MnO₃
Answers
Answer:
(b) CrO₂Cl₂ , MnO₄⁻
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The pair of compounds in which both the metals are in the highest possible oxidation state is pair (b) CrO₂Cl₂ , MnO₄⁻.
(a) [Fe(CN)₆ ]⁴⁻
Here, the metal is Fe.
Let its oxidation state be x.
CN has an oxidation state of -1.
The overall charge of the complex is -4.
Oxidation state of Fe can be calculated as :
x + { (-1) × 6 } = -4
=> x - 6 = -4
=> x = -4 + 6
=> x = +2
Therefore, the oxidation state of Fe is +2.
[Co(CN)₆ ]³⁻
Here, the metal is Co.
Let its oxidation state be x.
The overall charge of the complex is -3.
Oxidation state of Co can be calculated as :
x + { (-1) × 6 } = -3
=> x - 6 = -3
=> x = -3 + 6
=> x = +3
Therefore, the oxidation state of Co is +3.
(b) CrO₂Cl₂
Here, the metal is Cr.
Let its oxidation state be x.
O has an oxidation state of -2.
Cl has an oxidation state of -1.
The overall charge of the complex is 0.
Oxidation state of Cr can be calculated as :
x + { (-2) × 2 } + { (-1) × 2 } = 0
=> x - 4 - 2 = 0
=> x - 6 = 0
=> x = +6
Therefore, the oxidation state of Cr is +6.
MnO₄⁻
Here, the metal is Mn.
Let its oxidation state be x.
O has an oxidation state of -2.
The overall charge of the complex is -1.
Oxidation state of Mn can be calculated as :
x + { (-2) × 4 } = -1
=> x - 8 = -1
=> x = -1 + 8
=> x = +7
Therefore, the oxidation state of Mn is +7.
(c) TiO₃
Here, the metal is Ti.
Let its oxidation state be x.
O has an oxidation state of -2.
The overall charge of the complex is 0.
Oxidation state of Ti can be calculated as :
x + { (-2) × 3 } = 0
=> x - 6 = 0
=> x = +6
=> x = +6
Therefore, the oxidation state of Ti is +6.
MnO₂
Here, the metal is Mn.
Let the oxidation state of Mn be x.
The overall charge of the complex is 0.
Oxidation state of Mn can be calculated as :
x + { (-2) × 2 } = 0
=> x - 4 = 0
=> x = 4
Therefore, the oxidation state of Mn is +4.
(d) [Co(CN)₆ ]³⁻
Here, the metal is Cr.
Let its oxidation state be x.
CN has an oxidation state of -1.
The overall charge of the complex is -3.
Oxidation state of CN can be calculated as :
x + { (-1) × 6 } = -3
=> x - 6 = -3
=> x = -3 + 6
=> x = +3
Therefore, the oxidation state of Cr is +3.
MnO₃
Here, the metal is Mn.
Let its oxidation state be x.
O has an oxidation state of -2.
The overall charge of the complex is 0.
Oxidation state of Mn can be calculated as :
x + { (-2) × 3 } = 0
=> x - 6 = 0
=> x = +6
Therefore, the oxidation state of Mn is +6.
• In pair (b), Cr is at its highest possible oxidation state of +6, and Mn is at its highest possible oxidation state of +7.