The incorrect statement about the oxidation s | vedclass

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Question

(C) The group $14$ elements have a valence shell electronic configuration of $ns^2 np^2$. They commonly exhibit $+4$ and $+2$ oxidation states. Due to

Vedclass · Accepted Answer

Lead in $+2$ state acts as a good reducing agent

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(C) The group $14$ elements have a valence shell electronic configuration of $ns^2 np^2$. They commonly exhibit $+4$ and $+2$ oxidation states. Due to the inert pair effect, the stability of the $+2$ oxidation state increases down the group $(C Therefore, $Pb^{2+}$ is the most stable state for lead, and $Pb^{4+}$ is a strong oxidizing agent because it tends to reduce to $Pb^{2+}$. Conversely, $Sn^{2+}$ is a reducing agent as it tends to oxidize to $Sn^{4+}$. Since $Pb^{2+}$ is the most stable state for lead, it does not act as a reducing agent. Thus, the statement that lead in $+2$ state acts as a good reducing agent is incorrect.

The incorrect statement about the oxidation states of group $14$ elements is

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