Question

Which of the following options are incorrect?

• A. Decreasing $I.E_2$: F $>$ N $>$ O
• B. Increasing atomic size : Be $<$ B $<$ Li $<$ Na
• C. Metallic bond strength : Na $<$ Zn $<$ Ca
• D. Increasing electron affinity : $O^{2-}<O^{-}<O^{+}<O$

Answer 1

The correct option is A Decreasing $I.E_2$: F $>$ N $>$ O

Second $IE$ of $O$ will be highest because after losing an $e^{-}$ we have the following electronic configuration.

$O^{+}⟶1s^{2}2s^{2}2p^3$

$F^{+}⟶1s^{2}2s^{2}2p^4$

$N^{+}⟶1s^{2}2s^{2}2p^2$

$O^{+}$ has half-filled $p-$ orbital which has extra-stability due to which next removal of $e^{-}$ becomes difficult hence requires high energy.

Then $F^{+}$ will have second highest $IE$ due to its' small size and higher electronegativity than $N^{+}$.