Two elements A and B have atomic number 16 and 19
(a) which element belongs to s block and which one belongs to p block of Modern periodic table?
(b) What are the common names of the groups to which A & B belongs respectively?
(c) Which element among them is a powerful reducing agent?
(d) Write down the formula of compound formed between A and B.
Answers
Answer:
a) Element A belongs to p block and element B belongs to s block.
Element A = Sulphur S and Element B = Potassium K
b)A belongs to chalcogen family and B belongs to alkali metal.
c)Potassium
d)Potassium sulfide K2S
Answer:
(a) 16 belongs to P block and 19 belongs to s block
(b) for atom B it is group 1 AKA alkali metals and for atom A it is group 16 AKA chalcogens ( Oxygen family).
(c) element B is a more powerful reducing agent than element A.
(d) A i.e. atomic no. 16 = 1s2, 2s2, 2p6, 3s2, 3p4
and for B i.e. atomic no. 19 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1
Explanation:
By considering all the laws and ways to find the group, block and period, we can say that the atom A (16) is Silicon from p block and atom B(19) is Potassium from s block.
For the solution of C)
reducing agent means the element/ atom is being oxidized. Oxidation means the element is showing loss of electrons. And electropositive atoms/ elements are most likely to lose their electrons. As atom B i.e. Potassium is from S block and include only one valence electron, it will show more tendency to lose the electron and become stable i.e. will get oxidized. This tells us that atom B (potassium) > atom A (silicon) w.r.t. their reducing power.
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