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

NCERT Exemplar · Q17

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 bond between B and C will be

(i) Ionic

(ii) Covalent

(iii) Hydrogen

(iv) Coordinate

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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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The elements B and C are both non-metals, and non-metals typically form covalent bonds by sharing electrons to achieve a stable octet. The bond between B and C will be Covalent.

Chemical bonds form because atoms seek to achieve a more stable electronic configuration, usually by attaining a full outer shell of electrons, similar to that of a noble gas (the octet rule). The type of bond formed depends on the nature of the atoms involved, specifically their tendency to gain, lose, or share electrons.

Ionic bonds typically form between a metal (which tends to lose electrons) and a non-metal (which tends to gain electrons). This involves the complete transfer of electrons, forming ions that are attracted to each other.
Covalent bonds typically form between two non-metals. Since both atoms have a strong tendency to gain electrons, they achieve stability by sharing electrons, allowing both atoms to effectively count the shared electrons towards their octet.

To determine the bond type between B and C, we first need to identify their nature (metal or non-metal) by examining their electronic configurations.

Analyze Element B:

The electronic configuration of B is $1s^2 2s^2 2p^6 3s^2 3p^3$ .
- The outermost shell is the third shell ( $n=3$ ).
- The number of valence electrons (electrons in the outermost shell) is $2 + 3 = 5$ .
- Elements with 5 valence electrons are typically non-metals. To achieve a stable octet (8 valence electrons), element B needs to gain 3 electrons. Gaining 3 electrons is more energetically favorable than losing all 5 valence electrons. Non-metals tend to share electrons when bonding with other non-metals.
Analyze Element C:

The electronic configuration of C is $1s^2 2s^2 2p^6 3s^2 3p^5$ .
- The outermost shell is the third shell ( $n=3$ ). …

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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 bond between 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 bond between B and C will be

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