The valence electron, which is the electron located in the outermost energy level of an atom, is the most likely to be removed during a chemical reaction. This is because valence electrons are involved in the formation of chemical bonds and determining an atom's reactivity.
Elements in the 1st group have 1 valence electron. So they are likely to donate 1 electron to get more stable. potassium, rubidium and cesium are likely to donate 1 electron.
completely filled valence shells
An atom with only one electron in its valence shell is more likely to lose that electron easily, making it a better conductor. In contrast, an atom with seven electrons in its valence shell is more stable and less likely to lose electrons, thus making it a poor conductor.
Sodium does not have a full outer valence shell. It has one electron in its outermost shell, which makes it very reactive and likely to lose that electron to achieve stability.
Yes, cesium is likely to form an ionic compound with chlorine. Cesium is a metal with 1 valence electron, while chlorine is a nonmetal with 7 valence electrons. They can form an ionic bond by transferring one electron from cesium to chlorine, resulting in the formation of CsCl (Cesium Chloride).
In the formation of an ionic compound, a metal atom is most likely to lose electrons to achieve a stable electron configuration. This results in the formation of positively charged ions known as cations.
The valence electron, which is the electron located in the outermost energy level of an atom, is the most likely to be removed during a chemical reaction. This is because valence electrons are involved in the formation of chemical bonds and determining an atom's reactivity.
This atom lose an electron.
Elements in the 1st group have 1 valence electron. So they are likely to donate 1 electron to get more stable. potassium, rubidium and cesium are likely to donate 1 electron.
The element E is likely barium (Ba), as its electron configuration is Kr 5s2. The fluoride compound of barium would be BaF2, with barium losing its two valence electrons to form a 2+ cation and fluoride gaining one electron to form a 1- anion.
In the formation of an ionic compound, a metal atom is most likely to lose electrons and form a positively charged ion (cation) due to its low electronegativity and tendency to give up electrons to achieve a stable octet electron configuration.
completely filled valence shells
Yes, iodine is likely to form an ionic compound with chlorine due to their large difference in electronegativity. Iodine can easily lose an electron to form a positively charged ion, while chlorine can readily gain an electron to form a negatively charged ion, resulting in the formation of an ionic compound.
An atom with only one electron in its valence shell is more likely to lose that electron easily, making it a better conductor. In contrast, an atom with seven electrons in its valence shell is more stable and less likely to lose electrons, thus making it a poor conductor.
Sodium is a likely candidate to form an ionic compound with fluorine because sodium is a metal with 1 valence electron, while fluorine is a non-metal with 7 valence electrons. When sodium loses its electron and fluorine gains it, an ionic bond is formed between the two elements.
Sodium does not have a full outer valence shell. It has one electron in its outermost shell, which makes it very reactive and likely to lose that electron to achieve stability.