Which of the following processes is endothermic? Explain with reason?
A) #\text{S} -> \text{S}^-#
B) #\text{S}^(-) -> \text{S}^(2-)#
C) #\text{Na} -> \text{Na}^-#
D) #\text{P} -> \text{P}^-#
A)
B)
C)
D)
1 Answer
Well, likely
Note that I am using the convention that an atom that becomes more stable due to gaining an electron has a negative electron affinity. My reference page lists these values as the opposite sign, and I am aware of that.
But as physical chemists, we ought to examine the data...
The first electron affinity of sulfur atom is
#bb(-"2.0771043 eV")# , i.e. the atom becomes more stabilized by adding the electron in.Although we are adding electrons into a
#3p^4# valence shell, which adds repulsion, it seems to balance out to be negative due to the increase in atomic radius.
The second electron affinity of sulfur atom is
#bb"4.726 eV"# , i.e. the atom gets more destabilized by adding the electron in.This makes sense, because one would be shoving an electron into an electron-dense region before forming the noble gas configuration.
The first electron affinity of sodium atom is
#bb(-"0.5479263 eV")# .There is some electron repulsion because it would go into the half-filled
#3s# orbital, but apparently it is sufficiently counterbalanced by the increase in atomic radius that results, because it is actually a little negative, not positive value.[This is not obvious. And well, alright, this value is not that small, but it looks small.]
The first electron affinity of phosphorus atom is
#bb(-"0.7466071 eV")# .There is some electron repulsion because it would go into one of the orbitals in the half-filled
#3p# subshell, but apparently it is sufficiently counterbalanced by the increase in atomic radius that results, because it is actually a little negative, not positive value.[This is not obvious. And well, alright, this value is not that small, but it looks small.]
