Chemistry
You are provided with the table given below. Answer the questions that follow :
(a) Which element has the largest size ?
(b) Name the most non metallic element.
(c) Name the element with the lowest ionization potential.
(d) Write the formula of the sulphate of the element with atomic no. 13.
(e) What is the electronic configuration of the element in the fourth period that forms a dipositive ion ?
(f) Write the formula of the phosphate of the element in fourth period, second group.
Periodic Table
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Answer
(a) Potassium (K).
(b) Fluorine (F).
(c) Potassium (K).
(d) The formula is Al2(SO4)3.
(e) The electronic configuration is 2, 8, 8, 2.
(f) The formula is Ca3(PO4)2.
Reason
(a) Atomic size increases down a group (due to more shells) and decreases left to right across a period (due to higher nuclear charge). Potassium, being at the bottom-left, is the largest.
(b) Non-metallic character increases toward the top-right of the periodic table (excluding noble gases). Fluorine is the most electronegative and reactive non-metal.
(c) Ionization potential decreases down a group and increases across a period. Potassium, being at the bottom-left, has the lowest ionization potential.
(d) Atomic number 13 is Aluminium (Al), which has a valency of +3. The sulphate radical (SO4) has a valency of -2. Using the criss-cross method, we get Al2(SO4)3.
(e) The element in the fourth period that forms a dipositive ion M2+ must have 2 valence electrons. This element is Calcium (Ca), located in Period 4, Group 2.
(f) The element is Calcium (Ca) with a valency of +2. The phosphate radical (PO4) has a valency of -3. Criss-crossing the valencies gives us Ca3(PO4)2.
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