Question

Match List I with List II

List I Element	List II Number of Neutron
(A). \(_{12}^{24}Mg\)	(I). \(5\)
(B). \(_{13}^{27}Al\)	(II). \(12\)
(C). \(_4^9Be\)	(III). \(6\)
(D). \(_6^{12}C\)	(IV). \(14\)

Choose the correct answer from the options given below:

• (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
• (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
• (A) - (I), (B) - (II), (C) - (IV), (D) - (III)
• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Hint:

Neutrons = mass number minus atomic number.

Answer 1

The Correct Option is B

Step 1: Count protons first
Mg has $12$ protons, Al $13$, Be $4$, C $6$. Each mass number counts protons plus neutrons.

Step 2: Subtract
Mg: $24\to12$ neutrons. Al: $27\to14$. Be: $9\to5$. C: $12\to6$.
So (A)-(II), (B)-(IV), (C)-(I), (D)-(III).

Step 3: Compare
Only option (B) lists exactly this set.

Final Answer:
Neutron counts $12,14,5,6$ give option (B). \[ \boxed{\text{(B)}} \]