No, not always. but sometimes yes.
Take the structure of water (H2O) for instance. Oxygen already has six electrons, which means it has two sets of two electrons, and two sets of one electrons. The two sets of two electrons are non-bonding electrons, which means they cannot bond with anything, the two spots for electrons have already been filled. But since the hydrogens have only one electron, and need another to fill the first shell, and the oxygen molecule needs another two electrons to fill its second shell, the hydrogen shares it's electron with the oxygen. Remember two electrons are needed to form a bond. Something like CH4 (Methane) uses all electrons, with one carbon and four hydrogens.
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All of the valence electrons are used for bonding.
They are shared in-between the atomsTwo atoms can share one or two electrons (covalent bonding), An electron from one atom can be removed and join a separate atom (ionic bonding) or all the outer shell electrons can separate off and form a lattice of positive ins in a sea of negative electrons (metallic bonding)
No, except for hydrogen. Bonding uses outermost electrons (outside full shells).
There are 8 electrons in the n=2 shell of an oxygen atom before bonding. The electronic configuration of oxygen is 1s^2 2s^2 2p^4, with 6 electrons in the 2p subshell.
All chemical bonding and compound formation occurs from the sharing of electrons.
in H-O-H, dihydrogenoxide they are all used for bonding
The nucleus of an atom contains protons and neutrons. Valence electrons are the outermost electrons in an atom that participate in chemical bonding. The electrons in the inner energy levels, excluding the valence electrons, are referred to as core electrons.
The outer energy shell of electrons. The inner energy shells of electrons do NOT take part in chemical bonding. Not all outer energy shell electrons take part in bonding. Those electrons that DO take part in bonding are described as 'oxidation state'. Taking ammonia as an example. Its formula is NH3 Nitrogen's electronic configuration is is 1s2(inner most shell), 2s2(intermediate shell), 2p5(outer most/valence shell). The '5' is the number of electrons in this shell. Nitrogen combines with 3 hydrogens , using up three of these 5 electrons. The other 2 electrons remain as an unused 'lone pair'. Because it has used three electrons in bonding with hydrogen, then its oxidation state can be described as '+3'.
In $\ce{NiCl2}$, the nickel atom typically exhibits a coordination number of 6. This means that there are no lone pairs of electrons on the nickel atom, since all of its electrons are involved in bonding with the chlorine atoms to form the complex.
All the electrons in an atom form the so-called cloud of electrons.
Methane (CH4) does not have any lone pairs of electrons on the central carbon atom. All electrons are involved in bonding with the four hydrogen atoms, resulting in a tetrahedral geometry.
In carbon tetrabromide (CBr₄), the central atom is carbon. Carbon has four valence electrons and forms four single bonds with the four bromine atoms, using all its valence electrons in bonding. Therefore, there are no lone pairs of electrons around the central carbon atom in CBr₄.