WHAT IS THE ION CONFIGURATION OF
GALLIUM 3+
AND
TIN 4+
AND WHAT ARE THE RULES (WRITTEN CLEARLY)
please help i really dont know
Please do not type in all-caps. That's considered rude online, as if you're SHOUTING.
Here is a mnemonic to help with electron configurations.
https://www.google.com/search?q=mnemonic+for+electron+filling&tbm=isch&imgil=EkNTrJFoFS3EqM%253A%253BEVsKBY8LWm_POM%253Bhttps%25253A%25252F%25252Fwww.slideshare.net%25252FMosheLacson%25252Felectronic-configuration&source=iu&pf=m&fir=EkNTrJFoFS3EqM%253A%252CEVsKBY8LWm_POM%252C_&usg=__qb2BoITRNiDjxX3evylP1ICocOs%3D&biw=1024&bih=612&ved=0ahUKEwjfm7317cXWAhXK31QKHSEgBXEQyjcIPg&ei=itvLWd-uI8q_0wKhwJSIBw#imgrc=kwQxEPVqRAQSeM:
You see from that for 31Ga it will be
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1 for the atom. Take off the last three (the 4s2 and 4p1) to get 31Ga^3+.
I'll let you do the Sn and Sn^4+.
https://upload.wikimedia.org/wikipedia/commons/thumb/1/11/Electron_orbitals.svg/525px-Electron_orbitals.svg.png
see the listing at the bottom by ordered energy level.
The ion configuration of Gallium 3+ (Ga3+) can be determined by understanding the rules for writing electron configurations and considering the number of electrons gained or lost to form the ion.
Rule 1: Electrons occupy the lowest energy orbitals first, following the Aufbau principle.
Rule 2: Each orbital can hold a maximum of two electrons, with opposite spins.
Rule 3: Half-filled and fully-filled orbitals are more stable than partially-filled orbitals.
Now, let's determine the ion configuration of Gallium 3+ (Ga3+):
Gallium (Ga) has an atomic number of 31. The neutral atom of Gallium has the electron configuration of 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^1.
To form a 3+ ion, Gallium loses 3 electrons. Therefore, we remove the three outermost electrons (4s^2 3d^10 4p^1) from the neutral atom configuration:
[Ar] 3d^10
The ion configuration of Gallium 3+ (Ga3+) is [Ar] 3d^10.
Now, let's determine the ion configuration of Tin 4+ (Sn4+):
Tin (Sn) has an atomic number of 50. The neutral atom of Tin has the electron configuration of 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^2.
To form a 4+ ion, Tin loses 4 electrons. Therefore, we remove the four outermost electrons (5s^2 4d^10 5p^2) from the neutral atom configuration:
[Kr] 5s^2 4d^10
The ion configuration of Tin 4+ (Sn4+) is [Kr] 5s^2 4d^10.
To summarize the rules for writing ion configurations:
1. Determine the electron configuration of the neutral atom.
2. Identify the number of electrons gained or lost to form the ion.
3. Remove the outermost electrons based on the gained or lost electrons.
4. Write the resulting electron configuration using the noble gas notation (if applicable) and the remaining orbitals.