Question

Five atoms are labeled A to E.

Atoms	Mass No.	Atomic No.
A	40	20
B	19	9
C	7	3
D	16	8
E	14	7

(a) Which one of these atoms: (i) contains 7 protons (ii) has electronic configuration 2, 7

(b) Write down the formula of the compound formed between C and D.

(c) Predict: (i) metals (ii) non-metals.

Answer 1

(a) From the table given below:

(i) Atom that contains 7 protons is E.

(ii) Atom that has electronic configuration 2, 7 is B.

Atoms	Number of neutrons=Mass No.- Number of protons	Atomic No= Number of electrons= Number of protons	Electronic configuration
A	40- 20= 20	20	2, 8, 8, 2
B	19- 9= 10	9	2, 7
C	7- 3= 4	3	2, 1
D	16- 8= 8	8	2, 6
E	14- 7= 7	7	2, 5

(b) Formula of the compound formed between C and D.

Valency of C= 1

Valency of D= 8- 6= 2

• For elements with valence electrons$\le 4$, valence electrons= valency
• For elements with valence electrons$>4$, valency= 8- valence electrons

Using criss-cross method: $$\begin{gathered} \mathrm{Element}\mathrm{C}\mathrm{D} \\ \searrow \swarrow \\ \mathrm{Valency}12 \end{gathered}$$, the formula comes out to be C₂D.

(c) Prediction of:

(i) metals: A, C

• Metals are electropositive and have valence electrons less than 4.
• From the electronic configuration, A and C have less than 4 electrons and are metals.

(ii) non-metals: B, D, E

• Non-metals are electronegative and have valence electrons more than 4.
• From the electronic configuration, B, D and E have more than 4 electrons and are non-metals.