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

NCERT Exemplar · Q16

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 The molecular formula of the compound formed from B and C will be

(i) BC

(ii) B2C

(iii) BC2

(iv) BC3

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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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$B$ is phosphorus (valency 3) and $C$ is chlorine (valency 1), so they combine as $\text{BC}_3$ — option (D).

Identify the elements from their configurations:

B: $1s^2\,2s^2\,2p^6\,3s^2\,3p^3$ has $5$ valence electrons (phosphorus). It needs $3$ more electrons to complete its octet, so its combining valency is $3$ .
C: $1s^2\,2s^2\,2p^6\,3s^2\,3p^5$ has $7$ valence electrons (chlorine). It needs $1$ more electron, so its combining valency is $1$ .

(A, $1s^2\,2s^2\,2p^6$ , is neon — a noble gas — and plays no part.) …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$A_4B_3$
›Reveal solutionSolution
This tests void-counting in close-packed lattices: a ccp arrangement of N atoms has 2N tetrahedral voids, so if all tetrahedral voids are filled by the cation, the formula is $A_2B$ .

Concept and Intuition

In any cubic close packed (ccp/fcc) arrangement of N spheres, the number of octahedral voids equals N, and the number of tetrahedral voids equals 2N (twice the number of close-packed spheres) — this is a standard structural fact used to derive ionic compound stoichiometries from packing descriptions.

Step-by-Step Solution

Let the number of B atoms (forming the ccp lattice) be $N$ .

Number of tetrahedral voids available $= 2N$ .

All tetrahedral voids are occupied by A atoms, so number of A atoms $= 2N$ .

Ratio A : B $= 2N : N = 2 : 1$ . …
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AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$A_3X_2$
›Reveal solutionSolution
Balancing the +2 charge of a group-2 element against the −3 charge of a group-15 element gives the general ionic formula $A_3X_2$ .

Concept and Intuition

Group-2 (alkaline earth) elements typically form +2 cations, while group-15 elements, to attain a stable octet, typically form −3 anions (as in nitrides, phosphides). The formula of the resulting ionic compound is fixed by charge neutrality: the total positive charge must equal the total negative charge.

Step-by-Step Solution

Group-2 element A forms the cation $A^{2+}$ .

Group-15 element X forms the anion $X^{3-}$ .

For charge balance, find the smallest whole numbers $p,q$ such that $2p = 3q$ : the smallest solution is $p=3, q=2$ . …
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AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$Q. The general formula of the compound formed when a metal (M) of group - 1 reacts with non metal$
(x) of group – 16 is (A) $MX_6$ (B) $M_2X_3$ (C) $MX_2$ (D) $M_2X$
›Reveal solutionSolution
Group-1 metals lose 1 electron ( $M^+$ ); group-16 nonmetals gain 2 electrons ( $X^{2-}$ ). Charge balance gives the ionic formula $M_2X$ .

Concept and Intuition

Ionic compound formulas are fixed by charge neutrality: the total positive charge must equal the total negative charge. A group-1 element has one valence electron and forms a $+1$ ion. A group-16 element needs two more electrons to complete its octet and forms a $-2$ ion.

Step-by-Step Solution

Group-1 metal M → $M^{+}$ .

Group-16 nonmetal X → $X^{2-}$ .

To balance one $X^{2-}$ charge, two $M^+$ ions are required: $2(+1) + 1(-2) = 0$ . …
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AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
$AlCl_3$
›Reveal solutionSolution
$AlCl_3$ 's covalent bonding leaves Al electron-deficient, forcing dimerization to $Al_2Cl_6$ ; $AlF_3$ is essentially ionic, so it has no such covalent octet deficiency and doesn't need to dimerize.

Concept and Intuition

Whether $AlX_3$ behaves as "electron deficient" (needing to dimerize to complete Al's octet) versus being a simple ionic lattice depends on the covalent vs. ionic character of the Al–X bond, governed by Fajan's rules. The key contrast is: covalent $AlCl_3$ genuinely leaves aluminum with only 6 electrons around it (electron deficient), so it dimerizes through chlorine-bridging dative bonds; ionic $AlF_3$ doesn't have discrete "molecules" with an incomplete octet at all — it's a 3-D ionic lattice.

Step-by-Step Solution

Compare bond character: $Cl^-$ is larger and less electronegative than $F^-$ , and per Fajan's rules a larger anion favors more covalent character with a given cation — so $Al$ – $Cl$ bonding is significantly covalent, while $Al$ – $F$ bonding (small, highly electronegative $F^-$ ) is predominantly ionic.

In the covalent, monomeric $AlCl_3$ "molecule," Al has only 3 bond pairs (6 electrons) around it — a genuine electron deficiency — which the molecule resolves by dimerizing to $Al_2Cl_6$ , where two chlorine atoms each donate a lone pair to bridge the two Al centers, completing each Al's octet.

$AlF_3$ is an ionic solid ( $Al^{3+}$ and $F^-$ ions in a lattice), not built from discrete electron-deficient molecules, so the "electron deficiency" concept (and the need to dimerize to fix it) simply doesn't apply. …
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AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
$\mathrm{AlCl_3}$
›Reveal solutionSolution
Fajan's rules say covalency is favoured by a small, highly-charged cation paired with a large, polarizable anion — $Al^{3+}$ (highest charge here) with $I^-$ (largest, most polarizable anion) gives the most covalent compound, $AlI_3$ .

Concept and Intuition

An ionic bond gains covalent character when the cation "pulls" electron density from the anion's cloud toward itself (polarizes it). This polarizing ability increases with the cation's charge and decreases with its size; the anion's susceptibility to being polarized increases with its size (more diffuse, loosely held outer electrons).

Step-by-Step Solution

Compare cations: $Al^{3+}$ (charge +3, small) is far more polarizing than $Mg^{2+}$ (+2) or $Na^+$ (+1).

Among the $Al^{3+}$ compounds, compare anions: $I^-$ is much larger and more polarizable than $Cl^-$ . …
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AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$Q. If an element in group 2 formed a compound with an element in group 17 of the periodic table, the compound formed is likely to ______ (A) conduct electricity in the solid state (B) have a low boiling point (C) dissolve in non-polar solvents (D) be a crystalline solid$
›Reveal solutionSolution
Group 2 + Group 17 → an ionic compound, which is a crystalline solid (not a low-boiling, solid-state-conducting, or non-polar-soluble substance). Answer: (D).

Concept and Intuition

Large electronegativity difference between a Group 2 metal (low electronegativity, loses 2 electrons easily) and a Group 17 non-metal (high electronegativity, gains 1 electron readily) favours complete electron transfer — an ionic bond. Ionic solids pack into rigid, ordered crystal lattices.

Step-by-Step Solution

Group 2 element (e.g. $Ca$ ) loses 2 electrons; Group 17 element (e.g. $Cl$ ) gains 1 electron each — compound $CaCl_2$ forms via ionic bonding.

Ionic solids have strong lattice energy → high melting and boiling points (rules out "low boiling point").

Ions are locked in the lattice in the solid state, so no conduction until molten/dissolved (rules out "conduct electricity in solid state"). …
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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 The molecular formula of the compound formed from B and C will be

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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 The molecular formula of the compound formed from B and C will be

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