It can't sowwy
Carbon is the sixth element with a total of 6 electrons. In writing the electron configuration for carbon the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for C goes in the 2s orbital. The remaining two electrons will go in the 2p orbital. Therefore the C electron configuration will be 1s2 2s2 2p2.
The standard model suggests an electron cofiguration for nickel (element number 28)as :[Ar] 4s2 3d8However, relativistic effects make the more stable configuration [Ar] 4s1 3d91s2,2s2,2p6,3s2,3p6,4s2,3d8
It has two valance electrons. Abbreviated electron configuration. [Ar] 3d^10 4s^2 ( energy wise that would be 4s^2 3d^10 )
Sulfur must lose six electrons to attain noble gas electron configuration (in SO3, H2SO4 etc) but in most of the compounds it will exist as sulphides which is formed when sulphur will gain two electrons.
That is chlorine and it has seven valance electrons.
The electron configuration of 1s22s22p3s1 is not the ground state electron configuration of any element. This configuration contains 8 electrons, which in the ground state would be oxygen. The ground state configuration of oxygen is 1s22s22p4.
Let us assume that we have Sodium (Na), it has the ground state electron configuration of: [Ne]3S1. The ANION is negative, and thereby has more electrons, the Na anion(Na.) would have the following electron configuration: [Ne]3S2. The CATION(which is a positive ion) of Na(Na+) would have [Ne] as it electron configuration(as it loses an electron and becomes "equal" to Neon)
Carbon is the sixth element with a total of 6 electrons. In writing the electron configuration for carbon the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for C goes in the 2s orbital. The remaining two electrons will go in the 2p orbital. Therefore the C electron configuration will be 1s2 2s2 2p2.
1st shell - 2 electrons 2nd shell - 8 electrons 3rd shell - 4 electrons
that is only the electron configuration of nickel, a nickel (II) cation would lose the 2 electrons in the 4s and be 1s2 2s2 2p6 3s2 3p6 3d8
None - it would have to GAIN one.
There are a total of 9 electrons so that would be the element fluorine.
The standard model suggests an electron cofiguration for nickel (element number 28)as :[Ar] 4s2 3d8However, relativistic effects make the more stable configuration [Ar] 4s1 3d91s2,2s2,2p6,3s2,3p6,4s2,3d8
It has two valance electrons. Abbreviated electron configuration. [Ar] 3d^10 4s^2 ( energy wise that would be 4s^2 3d^10 )
When ionised to have no electrons, the usual notation would be 1s0. However, this is almost never needed, as other information and context will usually imply or define that there are no electrons.
Four: All of its valence electrons. If a silicon atom loses four electrons, it has the stable electron configuration of neon, while if the atom gains four electrons it has the stable electron configuration of argon. A silicon atom can also form a stable compound, as contrasted with a stable electron configuration for a single atom, by sharing four electrons with one or more other atoms.
O is atomic number 8, so it has 8 protons and 8 electrons. The electron configuration would be1s2 2s2 2p4. So, it has 2 electrons in the first shell (1s2).