The incorrect order of 1^{st} IP of the eleme | vedclass

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(A) The $1^{st}$ Ionization Potential $(IP)$ trends are as follows:$1$. For $Al$ $(13)$ and $Ga$ $(31)$,$Ga$ has a higher $IP$ than $Al$ due to the po

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$Al > Ga$

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(A) The $1^{st}$ Ionization Potential $(IP)$ trends are as follows:$1$. For $Al$ $(13)$ and $Ga$ $(31)$,$Ga$ has a higher $IP$ than $Al$ due to the poor shielding effect of $d$-electrons,so $Al $2$. For $P$ $(15)$ and $S$ $(16)$,$P$ has a stable half-filled $p$-orbital configuration $(3s^2 3p^3)$,making its $IP$ higher than $S$ $(3s^2 3p^4)$. Thus,$P > S$ is correct.$3$. For $Cu$ and $Zn$,$Zn$ has a fully filled $d^{10}s^2$ configuration,making its $IP$ higher than $Cu$. Thus,$Cu $4$. For $Zr$ and $Hf$,due to the lanthanoid contraction,the $IP$ of $Hf$ is slightly higher than $Zr$. Thus,$Zr Wait,re-evaluating the options: $Al$ $(577 \ kJ/mol)$ vs $Ga$ $(579 \ kJ/mol)$,so $Al  S$. $Cu$ $(745 \ kJ/mol)$ vs $Zn$ $(906 \ kJ/mol)$,so $Cu Actually,in many textbooks,$Al > Ga$ is considered the incorrect order because $Ga$ is expected to be lower but is actually higher. However,looking at the options provided,$P > S$ is a standard trend. Let's re-check $Al > Ga$. $Al$ is $577$ and $Ga$ is $579$. So $Al  Ga$ is indeed incorrect.

The incorrect order of $1^{st}$ $IP$ of the elements is

Similar Questions

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