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(C) The first ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge.However,there

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$Be, N$

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(C) The first ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge.However,there are exceptions due to stable electronic configurations.In the second period $(Li, Be, B, C, N, O, F, Ne)$:$1$. $Be$ $(1s^2 2s^2)$ has a higher ionization enthalpy than $B$ $(1s^2 2s^2 2p^1)$ because of the fully filled $2s$ orbital.$2$. $N$ $(1s^2 2s^2 2p^3)$ has a higher ionization enthalpy than $O$ $(1s^2 2s^2 2p^4)$ because of the half-filled $2p$ subshell.Thus,$Be$ and $N$ have higher ionization enthalpies than their immediate neighbors.Therefore,the correct option is $C$.

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

In the second period of the modern periodic table,two elements $X$ and $Y$ have higher first ionization enthalpy values than the preceding and succeeding elements. $X$ and $Y$ are respectively

Similar Questions

In which of the following electronic configurations of an element is there a very large difference between the second and third ionization energy?

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

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The atomic numbers of elements $X, Y$,and $Z$ are $19, 37$,and $55$ respectively. Which of the following statements is correct?

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