In the following table, six elements A,B,C,D,E and F of the modern periodic table with atomic numbers 3 to 18 are given:
i. Which of these is (a) Noble gas, (b) halogen?
ii. Which of these is the most active metal in 3rd period?
iii. Identify the most electronegative element in the third period.
iv. In the compound between B and F, what type of bond will be formed?
v. What would be the nature of oxide formed by C?
i. Noble Gas: G; Halogen: F
Explanation:
• The last group elements (18th) are known as noble gas because their valence electron is filled.
• Hence, G because of electronic configuration 2, 8 is a noble gas.
• The second last group elements (17th) are known as halogens because they need only one electron to completely fill their valence electron shell.
• Hence, F due to electronic configuration 2, 7 is the halogen.
ii. The most active element in the 3rd period is B.
Explanation: The atomic number of B is 11. Hence, the electronic configuration of B is 2, 8, 1. It has only one electron in its valence shell. It loses its one electron very easily and forms bonds with many elements becoming the most reactive element.
iii. The most electronegative element of the 3rd group is F.
Explanation: The electronic configuration of F (17) is 2,8,7. It needs 1 electron to complete its shell. Instead of losing the 7 electrons, it gains 1 electron. Hence, it is the most electronegative element of group 3rd.
***note: electronegativity is defined as the tendency to gain electrons***
iv. In between B and F, the ionic bond will be formed.
Explanation: The E.C. of B (11) is 2,8,1 and E.C. of F (17) is 2,8,7. B will lose its one electron to form B + cation and F will gain this one electron from B to complete its shell and become F- anion. Hence, the bond between B and F will be an ionic bond as ion formation takes place.
v. The nature of oxides form by C (12) will be basic.
Explanation: The EC of C (12) is 2,8,2. It belongs to group 2. Hence, C is a metal and because of a general rule metal oxides are always basic.