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

List-$I$ (Electronic configuration of neutral atom where $n=2$)	List-$II$ ($1^{st}$ Ionization Energy in $\text{kJ mol}^{-1}$)
$A. ns^2$	$I. 2080$
$B. ns^2np^1$	$II. 899$
$C. ns^2np^3$	$III. 800$
$D. ns^2np^6$	$IV. 1402$

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 ionisation energy of $C$,$N$,$O$,and $F$ is

What is the decreasing order of the ionization energy for the following elements?

Match List-$I$ with List-$II$:
List-$I$ (Electronic configuration of neutral atom where $n=2$) List-$II$ ($1^{st}$ Ionization Energy in $\text{kJ mol}^{-1}$)
$A. ns^2$ $I. 2080$
$B. ns^2np^1$ $II. 899$
$C. ns^2np^3$ $III. 800$
$D. ns^2np^6$ $IV. 1402$

Ionization potential is lowest for

The atomic number of the element from the following with the lowest $1^{st}$ ionisation enthalpy is:

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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