The first ionization enthalpies of $Li$,$Be$,and $C$ are $520$,$899$,and $1086 \ kJ \ mol^{-1}$ respectively. Which of the values $800$ or $900 \ kJ \ mol^{-1}$ is more likely for the first ionization enthalpy of $B$? Why?
- A$800 \ kJ \ mol^{-1}$
- B$900 \ kJ \ mol^{-1}$
- C$850 \ kJ \ mol^{-1}$
- D$950 \ kJ \ mol^{-1}$
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Similar Questions
Which of the following electronic configurations will have the least $I.P.$ value?
Medium
View SolutionWhich of the following pairs has a higher ionization enthalpy?
$(i)$ $Ne$ or $Ar$ $(ii)$ $Cl$ or $F$ $(iii)$ $F$ or $O$ $(iv)$ $N$ or $O$
$(v)$ $Na$ or $K$ $(vi)$ $Cl$ or $S$ $(vii)$ $Kr$ or $Xe$ $(viii)$ $P$ or $S$
$(i)$ $Ne$ or $Ar$ $(ii)$ $Cl$ or $F$ $(iii)$ $F$ or $O$ $(iv)$ $N$ or $O$
$(v)$ $Na$ or $K$ $(vi)$ $Cl$ or $S$ $(vii)$ $Kr$ or $Xe$ $(viii)$ $P$ or $S$
Medium
View SolutionOutermost electronic configurations of four elements $A, B, C, D$ are given below:
$A: 3s^{2}$
$B: 3s^{2} 3p^{1}$
$C: 3s^{2} 3p^{3}$
$D: 3s^{2} 3p^{4}$
The correct order of first ionization enthalpy for them is:
$A: 3s^{2}$
$B: 3s^{2} 3p^{1}$
$C: 3s^{2} 3p^{3}$
$D: 3s^{2} 3p^{4}$
The correct order of first ionization enthalpy for them is:
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Easy
View SolutionThe correct order of second $I.E.$ of $C$,$N$,$O$,and $F$ is:
Difficult
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