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’.

(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.

$\text{M}^{2+} \space \text{P}^{1-} \\[1em] \overset{\phantom{2}{2}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{P}} \Rightarrow \underset{\phantom{1}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{P}} \\[1em] \text{M}\text{P}_2$