Question

The electronic configurations of three elements A, B and C are as follows:

A	2, 8 1
B	2, 8, 7
C	2, 4

1. Which of these elements is a metal?
2. Which of these elements are non-metals?
3. Which two elements will combine to form an ionic bond?
4. Which two elements will combine to form a covalent bond?
5. Which element will form an anion of valency 1?

Answer 1

1. Element A (2, 8, 1) is sodium (Na). It behaves as a metal because it has a low ionisation energy and readily loses its single valence electron to achieve stability.
2. Element B (2, 8, 7) is Chlorine (Cl), and element C (2, 4) is Carbon (C). Your point about their flexibility to complete octets via gaining or sharing is completely correct.
3. An electropositive metal (A) reacting with an electronegative non-metal (B) creates a perfect electron-transfer match, yielding the ionic compound AB sodium chloride (NaCl).
4. Non-metals B and C share electrons. Since Carbon needs 4 electrons and chlorine needs 1, one atom of C shares its 4 valence electrons with 4 individual atoms of B to form CB₄ carbon tetrachloride, CCl₄).
5. Element B needs exactly 1 electron to jump from 7 to a stable octet of 8. Gaining that single negative charge creates a stable halide anion (B⁻) with a valency of 1.