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allochka39001 [22]
3 years ago
7

According to the balanced equation shown below, 1.00 mole of oxalic acid, h2c2o4, reacts with _________ moles of permanganate io

n, mno4-
Chemistry
1 answer:
zheka24 [161]3 years ago
6 0
The balanced equation that shows the reaction between oxalic acid and permanganate ion in an acidic medium is: 2MnO4- + 5H2C2O4 + 6H+ -> 2Mn(2+) + 10CO2 + 8H2O. Thus, 1 mole of oxalic acid reacts with 0.4 mole of permanganate ion. This was obtained using stoichiometry:

1 mol H2C2O4 x (2 mol MnO4-/ 5 mol H2C2O4) = 0.4 mol MnO4- 

In this redox reaction, the permanganate is reduced to manganese(II) ion.
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Rank these transition metal ions in order of decreasing number of unpaired electrons.
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Answer: The given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

Explanation:

In atomic orbitals, the distribution of electrons of an atom is called electronic configuration.

The electronic configuration in terms of noble gases for the given elements are as follows.

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Fe^{3+} - [Ar] 3d^{5}

So, there is only 1 unpaired electron present in Fe^{3+}.

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Mn^{4+} - [Ar]3d^{3}

So, there are only 3 unpaired electrons present in Mn^{4+}.

  • Atomic number of V is 23.

V^{3+} - [Ar] 3d^{2}

So, there are only 2 unpaired electrons present in V^{3+}.

  • Atomic number of Ni is 28.

Ni^{2+} - [Ar] 3d^{8}

So, there will be 2 unpaired electrons present in Ni^{2+}.

  • Atomic number of Cu is 29.

Cu^{+} - [Ar] 3d^{10}

So, there is no unpaired electron present in Cu^{+}.

Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

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