Question

The general formula of the compound formed when a metal (M) of group - 1 reacts with non metal (x) of group - 16 is

• A. MX$_6$
• B. M$_2$X$_3$
• C. MX$_2$
• D. M$_2$X

Hint:

Group 1 metals (alkali metals) form M$^+$ ions (charge +1).
Group 16 non-metals (chalcogens) typically form X$^{2-}$ ions (charge -2) when reacting with metals.
To form a neutral ionic compound, balance the charges. If M has charge $+c_1$ and X has charge $-c_2$, the formula is often M$_{c_2}$X$_{c_1}$ (after simplifying subscripts to smallest integers).
Here, M$^{+1}$ and X$^{-2}$, so M$_2$X$_1$ or M$_2$X.

Answer 1

The Correct Option is D

Metal (M) is from Group 1 of the periodic table (alkali metals). Alkali metals have one valence electron ($ns^1$). They tend to lose this electron to form a cation with a +1 charge (M$^+$). Example: Na $\rightarrow$ Na$^+$ + e$^-$ Non-metal (X) is from Group 16 of the periodic table (chalcogens). Chalcogens have six valence electrons ($ns^2 np^4$). They need two more electrons to complete their octet ($ns^2 np^6$). They tend to gain two electrons to form an anion with a -2 charge (X$^{2-}$). Example: O + 2e$^-$ $\rightarrow$ O$^{2-}$ When M and X react to form an ionic compound, the total positive charge must balance the total negative charge to achieve electrical neutrality. Let the formula of the compound be M$_a$X$_b$. The total positive charge is $a \times (+1) = +a$. The total negative charge is $b \times (-2) = -2b$. For neutrality, $+a = |-2b|$, so $a = 2b$. The simplest whole number ratio for $a$ and $b$ that satisfies this is $a=2$ and $b=1$. So, the formula of the compound is M$_2$X$_1$, which is written as M$_2$X. Example: Sodium (Na, Group 1) reacts with Oxygen (O, Group 16) to form Sodium Oxide, Na$_2$O. $2 \text{Na}^+ + \text{O}^{2-} \rightarrow \text{Na}_2\text{O}$. Another example: Potassium (K, Group 1) reacts with Sulfur (S, Group 16) to form Potassium Sulfide, K$_2$S. \[ \boxed{\text{M}_2\text{X}} \]