Which statement is correct -Sn4-is more stable than Sn2-Pb2-is more stable than Pb4-Both -A- and -B-None of these-

The correct option is C Both (A) and (B) 14^th group elements exhibit an oxidation state of +2 amp; +4. Down the group the stability of +4 decreases. Pb^+4 ...

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Standard X

Chemistry

Acidic,Basic Natures of Oxides

Which stateme...

Question

Which statement is correct ?

Sn4+ is more stable than Sn2+

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Pb2+ is more stable than Pb4+

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Both (A) and (B)

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None of these

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Solution

The correct option is C
Both (A) and (B)

$14^{t h}$ group elements exhibit an oxidation state of +2 & +4. Down the group the stability of +4 decreases.

$P b^{+ 4}$ is less stable due to inert pair effect. Tl ( $13^{t h}$ group), Pb ( $14^{t h}$ group), Bi ( $15^{t h}$ group) exhibit inert pair effect. These elements are stable in their lower oxidation states.

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Similar questions

Consider figure and mark the correct option.

(a) Activation energy of forward reaction is $E_{1}$ + $E_{2}$ and product is less stable than reactant

(b) Activation energy of forward reaction is $E_{1}$ + $E_{2}$ and product is more stable than reactant

(c) Activation energy of both forward and backward reaction is $E_{1}$ + $E_{2}$ and reactant is more stable than product

(d) Activation energy of backward reaction is $E_{1}$ and product is more stable than reactant

Q. 8. Why Pb2+ is more stable than Pb4+ ?

Q. The correct code for stability of oxidation states of Sn and Pb is :

[1] P b^{2 +} > P b^{4}, S n^{2 +} > S n^{4 +}

[2] P b^{2 +} < P b^{4}, S n^{2 +} < S n^{4 +}

[3] P b^{2 +} > P b^{4}, S n^{2 +} < S n^{4 +}

[4] P b^{2 +} < P b^{4}, S n^{2 +} > S n^{4 +}

[5] S n^{2 +} < P b^{2 +}, S n^{4 +} > P b^{4 +}

[6] S n^{2 +} < P b^{2 +}, S n^{4 +} < P b^{4 +}

P b^{4 +}

is less stable than

P b^{2 +}

. This is due to:

Q. Why is

P b^{4 +}

less stable than

P b^{2 +}

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