Question

Which of the following would require least energy?

• A. $S{i}^{-}\left(g\right)\to Si\left(g\right)+e^{-}$
• B. $P^{-}\left(g\right)\to P\left(g\right)+e^{-}$
• C. $S^{-}\left(g\right)\to S\left(g\right)+e^{-}$
• D. $C{l}^{-}\left(g\right)\to Cl\left(g\right)+e^{-}$
  

Answer 1

The correct option is B $P^{-}\left(g\right)\to P\left(g\right)+e^{-}$

$S{i}^{-}$ has an outer electronic configuration of $3s^{2}3p^3$. Since it is half filled, removing an electron from this species requires more energy.

$P^{-}$ has an outer electronic configuration of $3s^{2}3p^4$. Removing one electron from this species will require very less energy because on removing an electron it attains the stable half filled configuration.

$S^{-}$and $C{l}^{-}$ have an outer electronic configuration of $3s^{2}3p^5$ and $3s^{2}3p^6$ respectively. The energy required to remove one electron from both of these species will be more (especially the chloride ion since it has a noble gas configuration).