The electronic configurations of three elements, A, B and C are given below. A: 1s² 2s² 2p⁶ B: 1s² 2s² 2p⁶ 3s²

NCERT Exemplar · Q15

Q. The electronic configurations of three elements, A, B and C are given below. 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 C may be represented by the formula :

(i) C

(ii) C2

(iii) C3

(iv) C4

Uttar Pradesh UpmspMCQ· 1mImportance★★★★★est

53% · 59/112 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

Element C has 7 valence electrons (group 17, a halogen) and needs one more electron to complete its octet. Two C atoms share one electron pair to form a diatomic molecule; the stable form is C₂.

The key to this question lies in understanding how many electrons an atom needs to achieve a stable noble-gas configuration, and how atoms satisfy that need through bonding.

Element C has the configuration $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^5$ . Count the electrons: that's 2 + 2 + 6 + 2 + 5 = 17 total, placing C in group 17 of the periodic table. The valence shell (n = 3) holds 7 electrons. To reach the stable octet of the nearest noble gas (argon, with configuration $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6$ ), each C atom needs one more electron.

Atoms in group 17 — the halogens — achieve this stability by sharing electrons. When two C atoms come together, each contributes one unpaired electron to form a shared pair (a single covalent bond). This way, both atoms effectively "see" eight electrons in their valence shell.

\text{C} + \text{C} \longrightarrow \text{C}_2

Let me walk through why the other options don't work:

Option (A): C — A single isolated atom with 7 valence electrons is highly reactive and unstable. It will immediately seek to gain, lose, or share an electron. Halogens do not exist as monoatomic species under normal conditions.
Option (B): C₂ — Two atoms, each with 7 valence electrons, form one single bond (one shared pair). Each atom now has access to 8 electrons: its own 6 unshared electrons plus the 2 in the bond. This is stable and matches the behavior of all halogens (F₂, Cl₂, Br₂, I₂). …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ14 — The electronic configurations of three elements, A, B and C are given …→Next questionQ16 — The electronic configurations of three elements, A, B and C are given …

Q.The electronic configurations of three elements, A, B and C are given below. 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 C may be represented by the formula :

🎓Unlock everything free for 14 days

Q.The electronic configurations of three elements, A, B and C are given below.
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 C may be represented by the formula :

Unlock everything free for 14 days