Identify the incorrect match regarding Ioniza | vedclass

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(C) Let us analyze each option:$A$. $N (2s^2 2p^3) > O (2s^2 2p^4)$: Nitrogen has a stable half-filled $p$-orbital,making its $I.P.$ higher than Oxyge

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Only $C$

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(C) Let us analyze each option:$A$. $N (2s^2 2p^3) > O (2s^2 2p^4)$: Nitrogen has a stable half-filled $p$-orbital,making its $I.P.$ higher than Oxygen. This is correct.$B$. $Zr $C$. $Na > K$: As we move down a group,the size increases and $I.P.$ decreases. $Na$ is above $K$,so $Na > K$ is correct. However,the reason given is $Z_{eff}$. While $Z_{eff}$ plays a role,the primary reason for the trend in a group is the increase in the number of shells and shielding effect. This is often considered a mismatch in context.$D$. $Al Therefore,the statement $C$ is the incorrect match because the primary reason for $Na > K$ is the increase in atomic size/number of shells,not just $Z_{eff}$.

$I.P.$ \text{ Trend}	Reason
$A$. $N > O$	Half-filled configuration
$B$. $Zr < Hf$	Lanthanide contraction
$C$. $Na > K$	$Z_{eff}$
$D$. $Al < Ga$	Transition contraction

Element	Ionization energy $1^{st}$ $(kJ/mol)$	Ionization energy $2^{nd}$ $(kJ/mol)$
$X$	$495$	$4563$
$Y$	$731$	$1450$

Identify the incorrect match regarding Ionization Potential $(I.P.)$ trends:

$I.P.$ \text{ Trend} Reason

$A$. $N > O$ Half-filled configuration

$B$. $Zr < Hf$ Lanthanide contraction

$C$. $Na > K$ $Z_{eff}$

$D$. $Al < Ga$ Transition contraction

Similar Questions

The correct order of $C$, $N$, $O$ and $F$ in terms of second ionisation potential is

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Element Ionization energy $1^{st}$ $(kJ/mol)$ Ionization energy $2^{nd}$ $(kJ/mol)$

$X$ $495$ $4563$

$Y$ $731$ $1450$

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Identify the incorrect match regarding Ionization Potential $(I.P.)$ trends: $I.P.$ \text{ Trend} Reason $A$. $N > O$ Half-filled configuration $B$. $Zr < Hf$ Lanthanide contraction $C$. $Na > K$ $Z_{eff}$ $D$. $Al < Ga$ Transition contraction