We can deduce that the element is Aluminum because it has an atomic number of 13. The symbol of aluminum is Al.
Actually, we used the number of protons to determine the atomic number because the atomic number of any element is the number of protons in an atom. In any atom, the number of protons is also equal to the number of electrons.
The element is therefore Aluminum. Aluminum is found in Group III of the periodic table and therefore, has 3 electrons in its outermost shell. To gain a stable electronic configuration, every atom needs to have 8 electrons in its last shell. Aluminum has eight electrons in its second last shell. So, it loses the three electrons in its last shell so that its second last electron shell becomes its last shell.
Whenever an atom loses electrons, it gains a positive charge. Since aluminum loses three electrons, the charge on it would be 3+.
When aluminum has lost three electrons, the resulting ion would have only 10 electrons in it (13-3=10). The number of protons would remain 13 because protons are not involved in forming an ion.
Therefore, the symbol of the ion would be:
Al3+
Actually, we used the number of protons to determine the atomic number because the atomic number of any element is the number of protons in an atom. In any atom, the number of protons is also equal to the number of electrons.
The element is therefore Aluminum. Aluminum is found in Group III of the periodic table and therefore, has 3 electrons in its outermost shell. To gain a stable electronic configuration, every atom needs to have 8 electrons in its last shell. Aluminum has eight electrons in its second last shell. So, it loses the three electrons in its last shell so that its second last electron shell becomes its last shell.
Whenever an atom loses electrons, it gains a positive charge. Since aluminum loses three electrons, the charge on it would be 3+.
When aluminum has lost three electrons, the resulting ion would have only 10 electrons in it (13-3=10). The number of protons would remain 13 because protons are not involved in forming an ion.
Therefore, the symbol of the ion would be:
Al3+
