A shortened form of the Periodic Table is shown below. Use it to answer questions (a) and (b)
(b)(i) What type of bond would exist in a compound formed when element D reacts with oxygen?
(ii) Write the formula of the compound formed in (b)(i) above.
Reading the shortened Periodic Table. The columns are the groups, labelled I, II (on the far left), then the wide transition-metal block, then III, IV, V, VI, VII, 0 (on the far right). The rows 1, 2, 3, 4 are the periods. From the diagram the five elements sit as follows:
- A: far top-right, in Group 0 (period 1) - a noble gas.
- B: on the right, in the Group VII region (period 3) - a non-metal.
- C: far left, in Group I (period 3) - an alkali metal.
- D: on the left, in Group II (period 4) - an alkaline earth metal.
- E: in the central transition block (period 4), between Groups II and III.
(a)(i) Transition metal. Transition metals occupy the block wedged between Group II and Group III. The only element sitting in that central block is E.
(a)(ii) Alkaline earth metal. The alkaline earth metals are the Group II elements. The element placed in the second column (Group II) is D.
(a)(iii) Least reactive. The least reactive elements are the noble gases of Group 0, which have stable, filled outer electron shells and rarely form compounds. The Group 0 element is A.
(a)(iv) Most electronegative. Electronegativity (the pull an atom exerts on a shared electron pair) increases across a period towards Group VII and decreases down a group; the halogens of Group VII are the most electronegative reactive elements (noble gases are excluded as they scarcely bond). The Group VII element here is B.
(b)(i) Bond type in D's oxide. Element D is a Group II metal, so it readily loses its two outer electrons to form a
D2+ cation, while oxygen (Group VI) gains two electrons to form the O2- anion. A metal reacting with a non-metal by electron transfer gives an ionic (electrovalent) bond.
(b)(ii) Formula of the compound. Balancing the charges of the ions:
\[ \text{D}^{2+} \;+\; \text{O}^{2-} \;\longrightarrow\; \text{DO} \]
One D2+ exactly balances one O2-, so the formula of the oxide is DO.