Question

The second ionization enthalpy $IE_{2}$ of the elements of the second period are given below: Element: Li Be B C N O F Ne $IE_{2}$(kJ/mol): 7294 1756 2430 2354 2856 3396 3377 3966 Explain: (a) Why is $IE_{2}$ of Li so much higher than for all other elements of this period? (b) What is the general trend from Be to Ne of increasing $IE_{2}$? (c) Why $IE_{2}$ of F is just about the same or very slightly less than that of oxygen?

Answer 1

"(a) For lithium, the outer electron is present in 2s orbital but for ${ Li }^{ + }$ the outer electron present in 1s orbital.

1s orbital is “closer” to the “nucleus” than the “2s orbital”.

It is very difficult to remove the electron from ${ Li }^{ + }$ ion because $i{ Li }^{ + }$ has the noble gas configuration.

Hence, $I{ E }_{ 2 }$for Li is so much higher than for all other elements of their period.

b) The effective nuclear charge is increased by the removal of an electron from an outer shell.

Therefore, the size of ${ M }^{ + } ion$ is lesser than the M. The electrons are tightly packed with the nucleus.

Hence,${ IE }_{ 2 }$ increases from Be to Ne.

c) The electronic configuration of ${ O }^{ + } is { 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 3 }$ and the electronic configuration of ${ F }^{ + } is { 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 4 }$.

The removal of an electron from partially filled subshell.

Therefore, more energy required for the removal of an electron from ${ O }^{ + } than { F }^{ + }$"

Answer 2

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$\huge{\bold{\black{O^+}}}$