A, B, C, D, and E are the elements whose atomic numbers are 4, 9, 11, 15, and 18, respectively.
Which element can be describe as non-reactive?
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Solution
The ability of any atom to lose or gain electrons in order to participate in any chemical reaction is termed as reactivity.
According to the periodic table, Beryllium () has an atomic number of 4. Hence, element A is . Its electronic configuration is 2, 2.
Thus, has two electrons in the valence shell and it can easily lose or donate these two electrons to attain a stable noble gas configuration of Helium (He) and forms a positively charged Beryllium ion ().
According to the periodic table, Fluorine () has an atomic number of 9. Hence, element B is . Its electronic configuration is 2, 7.
Thus, has seven electrons in the valence shell and needs only one electron to attain a stable noble gas configuration of Neon and form a negatively charged Fluoride ion ().
According to the periodic table, Sodium () has an atomic number of 11. Hence, element C is . Its electronic configuration is 2, 8, 1.
Thus, has only one electron in the valence shell and it can easily lose or donate its one electron to attain a stable noble gas configuration of Neon and forms a positively charged Sodium ion ().
According to the periodic table, Phosphorus () has an atomic number of 15. Hence, element D is . Its electronic configuration is 2, 8, 5.
Thus, has five electrons in the valence shell and needs three electrons to attain a stable noble gas configuration of Argon () and forms a negatively charged Phosphide ion ().
Argon () has an atomic number of 18. Hence, element E is . Its electronic configuration is 2,8,8.
Argon has a completely filled outermost or valence shell (complete octet), and therefore, its electronic configuration is highly stable.
Due to the absence of free electrons in its outermost shell, Argon has zero tendency to accept or lose any electron.
Hence, element E (Argon) is chemically non-reactive.