Answer:
σ*2pₓ, also called 
Explanation:
I have drawn the MO diagram for fluorine below.
Each F atom contributes seven valence electrons, so we fill the MOs of fluorine with 14 electrons.
We have filled the 
and 
MOs.
They are the highest occupied molecular orbitals (HOMOs).
The next unfilled level (the LUMO) is the σ*2pₓ orbital. If you use the symmetry notation, it is called the orbital.
This is the orbital that fluorine uses when it acts as an electron acceptor.
1. all living things are composed of cells
2. cells are the basic building blocks of life
3. All cells are created by preexisting cells
Answer:
Not balance.
Explanation:
Chemical equation:
K + Cl₂ → KCl
The given equation is not balance because there are one potassium and two chlorine atoms on left side of equation while on right side there are one potassium and one chlorine atom present.
Balance chemical equation:
2K + Cl₂ → 2KCl
the equation is balance now because there are two potassium and two chlorine atoms on left side of equation and two potassium and two chlorine atoms are also present on right side.
Is there suppose to be a picture? Cause I do t see one
1. The reactivity among the alkali metals increases as you go down the group due to the decrease in the effective nuclear charge from the increased shielding by the greater number of electrons. The greater the atomic number, the weaker the hold on the valence electron the nucleus has, and the more easily the element can lose the electron. Conversely, the lower the atomic number, the greater pull the nucleus has on the valence electron, and the less readily would the element be able to lose the electron (relatively speaking). Thus, in the first set comprising group I elements, sodium (Na) would be the least likely to lose its valence electron (and, for that matter, its core electrons).
2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.
3. While the alkali and alkaline earth metals would lose electrons to attain a noble gas configuration, the group VIIA halogens, as we have here, would need to gain a valence electron for an full octet. The trends in the group I and II elements are turned on their head for the halogens: The smaller the atomic number, the less shielding, and so the greater the pull by the nucleus to gain a valence electron. And as the atomic number increases (such as when you go down the group), the more shielding there is, the weaker the effective nuclear charge, and the lesser the tendency to gain a valence electron. Bromine (Br) has the largest atomic number among the halogens in this set, so an electron would feel the smallest pull from a bromine atom; bromine would thus be the least likely here to gain a valence electron.
4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.
