The electronic configurations of three elements, A, B and C are given below. Answer questions 14 to 17 on the

NCERT Exemplar · Q14

Q. The electronic configurations of three elements, A, B and C are given below. Answer questions 14 to 17 on the basis of these configurations. A: 1s^2 2s^2 2p^6 B: 1s^2 2s^2 2p^6 3s^2 3p^3 C: 1s^2 2s^2 2p^6 3s^2 3p^5 Stable form of A may be represented by the formula :

(i) A

(ii) A2

(iii) A3

(iv) A4

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Concept understanding — Ionic Compound Prediction

Ionic Compound Prediction: From Intuition to Rule

Imagine you have a bag of positively charged magnets (cations) and negatively charged magnets (anions). If you just dump them together, they'll snap into a neutral clump — but only if the total positive charge exactly cancels the total negative charge. That's the core idea behind ionic compound formation: the compound must be electrically neutral overall.

The Intuition

Sodium ( $\text{Na}$ ) wants to lose one electron to become $\text{Na}^+$ . Chlorine ( $\text{Cl}$ ) wants to gain one electron to become $\text{Cl}^-$ .

If you put one $\text{Na}^+$ and one $\text{Cl}^-$ together, the charges cancel: $+1 + (-1) = 0$ . That's why sodium chloride is $\text{NaCl}$ — one sodium ion for every chloride ion.

But what about magnesium ( $\text{Mg}$ ) and chlorine? Magnesium loses two electrons to become $\text{Mg}^{2+}$ . One $\text{Mg}^{2+}$ needs two $\text{Cl}^-$ ions to balance: $+2 + 2(-1) = 0$ . So the formula is $\text{MgCl}_2$ .

The rule is simple: the total positive charge must equal the total negative charge. You're just finding the smallest whole-number ratio of ions that makes this happen.

The Precise Statement

X^{m+}

In plain language: the subscript on the cation ( $a$ ) times its charge ( $m$ ) must equal the subscript on the anion ( $b$ ) times its charge ( $n$ ).

How to Apply It — Step by Step

Write the ions with their charges.

Example: calcium ( $\text{Ca}^{2+}$ ) and phosphate ( $\text{PO}_4^{3-}$ ).
Find the smallest numbers that balance the charges.

The charges are $+2$ and $-3$ . The least common multiple of 2 and 3 is 6.
- To get $+6$ from $\text{Ca}^{2+}$ , you need 3 calcium ions: $3 \times (+2) = +6$ .
- To get $-6$ from $\text{PO}_4^{3-}$ , you need 2 phosphate ions: $2 \times (-3) = -6$ .
Write the formula with those numbers as subscripts.

$\text{Ca}_3(\text{PO}_4)_2$ — the parentheses around phosphate show it's a polyatomic ion taken as a unit.

Tip

A quick shortcut: swap the charges (without the signs) and use them as subscripts.

For $\text{Ca}^{2+}$ and $\text{PO}_4^{3-}$ , swap 2 and 3 → $\text{Ca}_3(\text{PO}_4)_2$ .

For $\text{Al}^{3+}$ and $\text{O}^{2-}$ , swap 3 and 2 → $\text{Al}_2\text{O}_3$ .

This always works because $a \times m = b \times n$ is exactly the cross-multiplication of the charges.

Common Pitfalls

Watch out

Never change the charge on an ion. The charge is fixed — $\text{Na}$ is always $+1$ , $\text{O}$ is always $-2$ . You only change how many of each ion you use.

Another trap: forgetting to reduce the ratio. If you get $\text{Ca}_2\text{O}_2$ , that's wrong — it should be $\text{CaO}$ (the smallest whole numbers are 1 and 1). Always simplify.

Why This Works …

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Element A has a complete octet in its outermost shell, making it a noble gas. Noble gases are stable as individual atoms, so the stable form of A is $\boxed{A}$ .

The stability of an element is fundamentally linked to its electronic configuration, specifically its tendency to achieve a noble gas configuration. This means having a completely filled outermost electron shell, which typically corresponds to 8 valence electrons (an octet) for most elements, or 2 valence electrons (a duplet) for elements like hydrogen and helium. Elements achieve this stability by gaining, losing, or sharing electrons to form chemical bonds. However, if an element already possesses a stable noble gas configuration, it will exist as a monatomic species, meaning individual atoms rather than molecules composed of multiple atoms.

Let's apply this understanding to element A.

Identify the electronic configuration of A:

The given electronic configuration for element A is $1s^2 2s^2 2p^6$ .
Determine the valence shell and valence electrons:

The valence shell is the outermost electron shell. For element A, the highest principal quantum number is $n=2$ . The electrons in this shell are $2s^2 2p^6$ .

The total number of valence electrons is $2 + 6 = 8$ .
Evaluate stability based on the octet rule:

Element A has 8 electrons in its outermost shell. This is a complete octet. An atom with a complete octet is exceptionally stable and corresponds to the electronic configuration of a noble gas. …

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Q.The electronic configurations of three elements, A, B and C are given below. Answer questions 14 to 17 on the basis of these configurations. A: 1s^2 2s^2 2p^6 B: 1s^2 2s^2 2p^6 3s^2 3p^3 C: 1s^2 2s^2 2p^6 3s^2 3p^5 Stable form of A may be represented by the formula :

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Q.The electronic configurations of three elements, A, B and C are given below. Answer questions 14 to 17 on the basis of these configurations.
A: 1s^2 2s^2 2p^6
B: 1s^2 2s^2 2p^6 3s^2 3p^3
C: 1s^2 2s^2 2p^6 3s^2 3p^5 Stable form of A may be represented by the formula :

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