Question

An unknown element 'E' forms two compounds: \(EO_{2}\) and \(EX_{4}\). To which group does element 'E' belong?

• A. 13th group (B, Al)
• B. 14th group (C, Si)
• C. 15th group (N, P)
• D. 16th group (O, S)

Hint:

Remember these common compounds: \[ \boxed{ \begin{aligned} CO_2,\;CCl_4\\ SiO_2,\;SiCl_4 \end{aligned} } \] Both carbon and silicon belong to \[ \boxed{\text{Group 14}.} \] If an element forms compounds of the type \(EO_2\) and \(EX_4\), it generally belongs to Group 14.

Answer 1

The Correct Option is B

Concept: The valency and oxidation state of an element can often be determined from the formulas of its compounds. The oxidation state of oxygen is generally \[ \boxed{-2.} \] Similarly, halogens such as fluorine or chlorine generally have an oxidation state of \[ \boxed{-1.} \] The compounds formed by an element help identify its group in the periodic table.

Step 1: Determine the oxidation state of \(E\) in \(EO_{2}\).
Let the oxidation state of \(E\) be \(x\). Since oxygen has oxidation state \(-2\), \[ x+2(-2)=0. \] Therefore, \[ x-4=0, \] \[ x=+4. \] Thus, \[ \boxed{E\text{ exhibits oxidation state }+4.} \]

Step 2: Determine the oxidation state of \(E\) in \(EX_{4}\).
Assuming \(X\) is a halogen, \[ X=-1. \] Hence, \[ x+4(-1)=0. \] Therefore, \[ x=+4. \] Again, \[ \boxed{E\text{ has oxidation state }+4.} \]

Step 3: Identify the group of the element.
Elements of Group 14 commonly exhibit a valency of four and form compounds such as \[ CO_2,\;SiO_2,\;CCl_4,\;SiCl_4. \] Thus, the unknown element belongs to \[ \boxed{\text{Group 14}.} \] Hence, \[ \boxed{\textbf{Option (B)}} \] is the correct answer.