Question

Match the column I with column II and mark the appropriate choice.

	Column I		Column II
(A)	$Fe$ in solid state	(i)	Electrolytic conductor
(B)	$NaCl$ in molten state	(ii)	$p-$type semiconductor
(C)	$C{O}_2$ in solid state	(iii)	Electronic conductor
(D)	$Si$ doped with aluminium	(iv)	Non-polar insulator

• A. (A)$\to$ (iv), (B) $\to$ (ii), (C) $\to$ (i), (D) $\to$ (iii)
• B. (A) $\to$ (ii), (B) $\to$ (iii), (C) $\to$ (i), (D)$\to$ (iv)
• C. (A) $\to$ (iii), (B) $\to$ (i), (C) $\to$ (iv), (D) $\to$ (ii)
• D. (A) $\to$ (i), (B) $\to$ (iv), (C) $\to$ (iii), (D) $\to$ (ii)

Answer 1

The correct option is C (A) $\to$ (iii), (B) $\to$ (i), (C) $\to$ (iv), (D) $\to$ (ii)

$Fe$ is a metallic solid. It has free electrons. So, it conducts electricity is solid state. It is called as electronic conductor.

In molten state, $NaCl$ has free ions i.e. ${Na}^{+}\&{Cl}^{-}$. These ions helps in flow of electric current. Molten $NaCl$ is called as electrolytic conductor.

In ${CO}_2$, $C$ and $O$ are bonded by covalent bonds and there are no free electron in ${CO}_2$. So, it is Non-Polar insulator.

$Si$ is group 14 element having four electrons in valence shell and $Al$ is 13 element having three electrons in valence shell. On doping $Si$ with $Al$, it will form a electron deficit material. It is called as p-type semiconductor.

So, correct answer is option $C$.