Respuesta :
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!!!
The octet rule states that the atoms of the elements bond to each other in an attempt to complete their valence shell with eight electrons. In other words, the atoms will tend to give up or share electrons to complete eight electrons in the valence shell through an ionic, covalent or metallic bond.
In other words, the goal is to have the closest noble gas electron configuration, thus having the last complete electron shell and acquiring stability.
So, in this case, to comply with the octet rule:
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 one electron from the 5s subshell.
For an element with its valence electron at 5s¹, this means that there is one valence electron in the outermost shell of this atom and in the outermost suborbitals. To form a monatomic ion, it would be necessary between losing or gaining an electron and that ion is stable. It takes less energy to lose the electron of the suborbital 5s and acquire a charge of of +1, than to acquire an electron, because it forms the 5s² suborbital, which indicates empty 5p orbitals too and is therefore less stable.
Also, in this way, the octet rule is fulfilled.
B. An element with the valence electron configuration 2s²2p⁴ would form a monatomic ion with a charge of -2. In order to form this ion, the element will gain two electron into the 2p subshell(s).
After gain two electron the atom has eight electrons in its valence shell, giving it the same electronic configuration as a noble gas (2s²2p⁶). In this way, the octet rule is fulfilled and the ion is stable.
Learn more:
- https://brainly.com/question/14077115?referrer=searchResults
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