The first ionization enthalpies of C,N,and F | vedclass

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Question

(A) The correct value is $1310 \, kJ \, mol^{-1}$.In the periodic table,the first ionization enthalpy generally increases across a period from left to

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$1310 \, kJ \, mol^{-1}$

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(A) The correct value is $1310 \, kJ \, mol^{-1}$.In the periodic table,the first ionization enthalpy generally increases across a period from left to right.However,due to the stable half-filled $2p^3$ electronic configuration of Nitrogen $(1s^2 2s^2 2p^3)$,it has a higher ionization enthalpy than Oxygen $(1s^2 2s^2 2p^4)$.Therefore,the ionization enthalpy of Oxygen must be lower than that of Nitrogen $(1402 \, kJ \, mol^{-1})$.Thus,$1310 \, kJ \, mol^{-1}$ is the correct value.

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

The first ionization enthalpies of $C$,$N$,and $F$ are $1086$,$1402$,and $1681 \, kJ \, mol^{-1}$ respectively. Which of the following values is the most likely first ionization enthalpy of $O$ ($1310$ or $1510 \, kJ \, mol^{-1}$)?

Similar Questions

Which one of the following statements is incorrect in relation to ionization enthalpy?

The first ionization energy of alkaline earth metals is higher than that of alkali metals. This is due to:

The first ionization potentials in electron volts $(eV)$ of nitrogen and oxygen atoms are respectively given by:

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

Amongst the elements with the following electronic configurations,which one of them may have the highest ionisation energy?

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