A I ion, specifically iodide (I⁻), has a total of 8 valence electrons. In its neutral state, iodine has 7 valence electrons, but it gains one additional electron when it forms an ion, resulting in 8. Since each pair of valence electrons consists of 2 electrons, there are 4 pairs of valence electrons in an iodide ion.
Every sodium atom has only one valence electron.
Lithium (Li) has 1 valence electron. It is in the same 'family' as Sodium (Na) and Potassium (K)
Sodium has one valence electron in its outermost shell (the third electron shell). To achieve a stable electron configuration, similar to that of the nearest noble gas (neon), sodium needs to lose this single valence electron rather than gain more. Therefore, sodium does not need additional valence electrons; it only needs to lose its one valence electron to achieve stability.
Zinc has two valence electrons.
Sodium has one valence electron.
A I ion, specifically iodide (I⁻), has a total of 8 valence electrons. In its neutral state, iodine has 7 valence electrons, but it gains one additional electron when it forms an ion, resulting in 8. Since each pair of valence electrons consists of 2 electrons, there are 4 pairs of valence electrons in an iodide ion.
Silicon has 4 Calcium has 2 Chlorine has 7 Sodium has 1
It has only 1 valence electron
Every sodium atom has only one valence electron.
2
Sodium's atomic number is 11. To be neutral then, it must have 11 protons and 11 electrons. Since sodium is in group 1, it has 1 valence electron.
electrons in Na2O
Sodium oxide has a total of 8 valence electrons (1 from sodium and 6 from oxygen) because sodium is in group 1 and oxygen is in group 6 of the periodic table.
Sodium has 1 valence electron. It is present in group-1.
Sodium has one valence electron because it is in group 1 of the periodic table.
To make calcium iodide, calcium transfers two valence electrons to iodine. Calcium wants to lose two electrons to achieve a stable octet configuration, while iodine needs two electrons to complete its octet. This transfer results in the formation of CaI2 with a 2:1 ratio of calcium to iodine atoms.