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Standard XII

Chemistry

Variable Oxidation States

Mn2+ compound...

Question

$M n^{2 +}$ compounds are more stable than $F e^{2 +}$ compounds towards oxidation to their $+ 3$ state, because :

M n^{2 +}

is more stable with high

3^{r d}

Ionisation energy

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M n^{2 +}

is bigger in size

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M n^{2 +}

has completely filled d-orbitals

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M n^{2 +}

does not exist

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Solution

The correct option is A $M n^{2 +}$ is more stable with high $3^{r d}$ Ionisation energy
$M n^{2 +}$ have half-filled orbitals which are more stable than $F e^{2 +}$ with 4 unpaired electrons, hence $M n^{2 +}$ will have higher 3rd ionisation energy.

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Similar questions

Q. Explain why:

M n^{2 +}

is more stable than

F e^{2 +}

towards oxidation to +3 state.
(At. no. of Mn = 25, Fe = 26)

M n^{2 +}

is much more resistant than

F e^{2 +}

towards oxidation. Give reason.

While $F e^{3 +}$ is stable, $M n^{3 +}$ is not stabe in acid solution because:
$(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)$

Q. While

F e^{3 +}

is stable,

M n^{3 +}

is not stabe in acid solution because:

(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)

Q. Explain the following observations giving an appropriate reason for each.
(i) The enthalpies of atomization of transition elements are quite high.
(ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e.

3^{r d}

series).
(iii)

M n^{2 +}

much more resistant than

F e^{2 +}

towards oxidation.

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Mn2+ compounds are more stable than Fe2+ compounds towards oxidation to their +3 state, because :

The correct option is A Mn2+ is more stable with high 3rd Ionisation energyMn2+ have half-filled orbitals which are more stable than Fe2+ with 4 unpaired electrons, hence Mn2+ will have higher 3rd ionisation energy.

$M n^{2 +}$ compounds are more stable than $F e^{2 +}$ compounds towards oxidation to their $+ 3$ state, because :

The correct option is A $M n^{2 +}$ is more stable with high $3^{r d}$ Ionisation energy
$M n^{2 +}$ have half-filled orbitals which are more stable than $F e^{2 +}$ with 4 unpaired electrons, hence $M n^{2 +}$ will have higher 3rd ionisation energy.