Study the table given below. Use only the letters given in the table to answer the questions. Do not identify the elements.
| IA | IIA | IIIA | IVA | VA | VIA | VIIA | 0 |
|---|---|---|---|---|---|---|---|
| E | J | Q | |||||
| L | G | ||||||
| M | D | P | |||||
| N |
(a) State the valency of element ‘G’.
(b) Which element can exhibit catenation?
(c) Write the formula of the compound formed between ‘M’ and ‘P’.
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(a) Element G
Reason — Element G belongs to group VA (group 15). Group 15 elements have 5 valence electrons in their outermost shell they can gain 3 electrons to complete the octet hence, the valency will be 3.
(b) Element E
Reason — Catenation (ability to form long chains of its own atoms) is strongest in the group IVA (14) element shown, i.e., E.
(c) Element m
Reason — M is in group IIA they have 2 valence electrons and forms M2+ and P is in group VIIA have 7 valence electrons and forms P-, so one molecule of M combines with two molecules of P. Hence, the neutral formula is MP2.
$$\begin{aligned}\text{M}^{2+} \space \text{P}^{1-} \\[1em] \overset{\phantom{2}{2}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{P}} \Rightarrow \underset{\phantom{1}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{P}} \\[1em] \text{M}\text{P}_2\end{aligned}$$