Answer:
A) An element with the valence electron configuration 5s¹ would form a monatomic ion with a charge of +1. In order to form this ion, the element will lose electron(s) from/into the 5s subshell(s).
B) An element with the valence electron configuration 2s²2p⁴ would form a monatomic ion with a charge of +1. In order to form this ion, the element will lose electron(s) from/into the 2p (2pₓ specifically) subshell(s).
Explanation:
The secret to this task is to follow those rules for the stability of electronic structures of elements. The rules include
- Electrons are filled firstly into shells or subshells of lower energies first.
- While filling electronic structure or writing electronic structures for elements/ions, electrons are fed singly to the suborbital before pairing occurs, this is because the totally paired up electrons of a suborbital are more stable than the totally unpaired electrons of the same suborbital which is now in turn more stable than the combination of paired and unpaired electrons in the suborbitals.
A) For an element with its valence electron on 5s¹, this means that there is one valence electron on this atom's outermost shell and outermost suborbitals. So, to form a monoatomic ion, it would take between losing and gaining an electron. Gaining an electron leads to a 5s², which indicates empty 5p orbitals too and is therefore less stable than losing an electron which would lead to the loss of the shell 5 and focus on a completely filled 4-shell.
So, losing the electron from the 5s suborbital to become a monotonic ion makes it acquire a charge of +1.
B) Just like the explanation in (A), to form a monoatomic ion would require a loss or gain of an electron. With valence electrons 2s²2p⁴, gaining an electron would have led to a 2s²2p⁵ and a further breakdown as 2s²2pₓ²2pᵧ²2pz¹ which has unpaired and paired electrons in the 2p suborbital. This is evidently less stable than if an electron was lost, the valence electrons are 2s²2p³ and they are positioned in a totally unpaired fashion in the 2p suborbital as 2s²2pₓ¹2pᵧ¹2pz¹.
Hence, the more stable alternative is more likely to occur and the electron is lost from the 2pₓ suborbital to make the monoatomic ion of the element acquire a +1 charge status too because of lost electron too.
Hope this Helps!!!
