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Boron must give up 3 electrons in order to achieve a noble-gas electron configuration.
Boron is the only element to have this atomic make up. Boron has a total of 5 electrons making its mapping appear as 1s2 2s2 2p1.
true, just not for Boron witch tries to gain 6 electrons for a stable arrangement
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The atoms of the elements in Group 13 (IIIA), the boron group, have three valence electrons, all of which are unpaired. The atoms of the elements in Group 15 (VA), the nitrogen group, have five valence electrons, three of which are unpaired.
In the element bromine (Br), there is only 1 unpaired electron. It has 7 valence electrons, so 3 pairs, plus an unpaired electron.
1s2 2s2 2p1 is the electron configuration for boron, and it has a total of 5 electron. Just fill the orbital up with the elements total number of electrons until no more are left, then u have your electron configuration
The oxidation state of boron is either three electrons or one electron. Boron has an valence electron configuration of ns2np1.
Boron exists in period 2, group 13 (IIIA) of the periodic table, with valence of 3 electrons in the outer shell. The electron configuration of boron is 1s22s22p1
Boron has 5 electrons per atom. Boron's electron configuration is 1s2 2s2 2p1. Thus, it has 3 electrons in its outer shell.
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Boron must give up 3 electrons in order to achieve a noble-gas electron configuration.
3, the electron configuration of Boron is 1s2 2s2 2p1, so there are 3 in the outer shell.
The electron configuration of boron is: [He]2s2.2p1.
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.
The electron configuration of boron is [He]2s2.2p1.
Boron has an electronic configuration of 1s22s22p1 (it has 5 electrons). In order to reach the stable electron configuration of a noble gas with a completely filled valence shell, boron atom has to lose 3 electrons to obtain a stable duplet structure (i.e. 2 electrons in its first electron shell). After losing 3 electrons, the boron atom forms a B3+ ion, or a so-called tripositive ion.