Seven valence electrons in iodine.
The same number of valence electrons as xenon.
Iodine atom has 7 valence electrons, whereas Iodide ion has 8 valence electrons.
7 valence electrons in iodine
Iodine has 7 electron-containing orbitals, corresponding to its 7 valence electrons in the p subshell. These orbitals can hold a maximum of 2 electrons each.
Sodium iodide (NaI) consists of sodium (Na) and iodine (I) atoms. Sodium has 1 valence electron, while iodine has 7 valence electrons. Therefore, in total, sodium iodide has 1 + 7 = 8 valence electrons.
Iodine pentafluoride has 7 valence electrons.
The order is: Iodine (7 valence electrons) Carbon (4 valence electrons) Calcium (2 valence electrons) Sodium (1 valence electron)
If an atom of iodine gains one electron, it will have a total of 8 valence electrons, similar to the noble gas xenon. This is because iodine has 7 valence electrons in its neutral state, and gaining one electron allows it to achieve a full outer shell, characteristic of stable noble gases.
Iodine has 7 valence electrons. I- contains 8 valence electron.
In iodine, the valence electrons are found in the outer shell of the atom, specifically in the 5th energy level or the seventh shell (the seventh Group of the periodic table). Iodine is in Group 17, so it has 7 valence electrons.
Iodine gains one electron to achieve a noble gas electron configuration. Its electron configuration is [Kr]5s²4d¹⁰5p⁵, and by gaining one electron, it attains the stable configuration of [Kr]5s²4d¹⁰5p⁶, which is similar to the noble gas xenon.
Iodine typically forms one single bond due to the presence of seven valence electrons in its outer shell. This allows iodine to share one electron with another atom in order to complete its octet and achieve a more stable electron configuration.