Solved Examples

Before You Start — Strategy for Problem-Solving

Welcome to the dedicated problem set. This section is a training ground: 30 solved examples covering every concept we've built over the previous 11 sections, arranged roughly in order of increasing difficulty.

How to work through this section

1. Read the question first, then cover the solution. Try to identify the concept being tested before looking at the worked answer.
2. Identify the category: is it a trend problem (rank elements), a numerical problem (compute IE, EN, radii), a formula-prediction problem (oxide/hydride formula), or an anomaly problem (explain Be vs Mg)?
3. Pull the relevant formula from the reference sheet below before plugging in numbers.

What's tested in this section

• Atomic and ionic radii, isoelectronic species (Examples 1-5).
• Ionization enthalpy, trends, anomalies, successive IEs (Examples 6-10).
• Electron gain enthalpy and related numerical (Examples 11-15).
• Electronegativity, bond polarity, % ionic character (Examples 16-20).
• Periodic trends in chemical properties (Examples 21-25).
• Combined / high-order JEE-style problems (Examples 26-30).

If you can solve 25+ of these 30 without peeking, you're fully prepared for any Class-11 Board or NEET/JEE Main question on this chapter.

Solved Examples — A Structured Problem Set

Example 1: Ranking by atomic radius (Period 3)

Arrange Na, Mg, Al, Si, P, S, Cl in order of DECREASING atomic radius.

Solution:

Moving across a period, atomic radius decreases because $Z_{\text{eff}}$ rises while new electrons go into the same shell.

Na (186) > Mg (160) > Al (143) > Si (117) > P (110) > S (104) > Cl (99) — all values in pm.

Answer:

$$Na > Mg > Al > Si > P > S > Cl$$

Takeaway: this given set follows the smooth left-to-right decrease in atomic radius across Period 3.

Example 2: Cation vs anion size

Compare the radii of (i) Na and Na$^{+}$, (ii) Cl and Cl$^{-}$.

Solution:

• Na (186 pm) → Na$^{+}$ (102 pm): loss of the valence $3s$ electron removes the entire outermost shell. The effective nuclear charge on the remaining electrons increases, so the ion shrinks dramatically (nearly half the size).
• Cl (99 pm) → Cl$^{-}$ (181 pm): gain of an electron increases inter-electron repulsion in the $3p$ shell without changing the number of protons. $Z_{\text{eff}}$ per electron drops, so the ion expands.

So cations are always smaller, anions always larger than the parent atom.

Takeaway: the golden rule — cation shrinks, anion expands.

Example 3: Isoelectronic series

Rank the following isoelectronic species by INCREASING radius: N$^{3-}$, O$^{2-}$, F$^{-}$, Na$^{+}$, Mg$^{2+}$, Al$^{3+}$.

Solution:

All have 10 electrons (same as Ne). Radius depends inversely on $Z$:

Species	Z
N$^{3-}$	7
O$^{2-}$	8
F$^{-}$	9
Na$^{+}$	11
Mg$^{2+}$	12
Al$^{3+}$	13

Larger $Z$ → more attractive pull on 10 electrons → smaller ion.

Answer (increasing radius, i.e. decreasing $Z$):

$$Al^{3+} < Mg^{2+} < Na^{+} < F^{-} < O^{2-} < N^{3-}$$

Takeaway: among isoelectronic species, radius decreases as $Z$ increases.

Example 4: Covalent vs van der Waals radius

The covalent radius of Cl is 99 pm and its van der Waals radius is 180 pm. Why are these different?

Solution:

• Covalent radius: half the distance between two bonded atoms of the same element (from the Cl–Cl bond in Cl$_{2}$). The electrons are shared.
• Van der Waals radius: half the distance between two non-bonded atoms of the same element in an adjacent molecule in a solid. The atoms are merely in contact, with no electron sharing.

Covalent bonding shortens the distance (shared electrons pull atoms closer), while vdW contact distance reflects the atom's overall "size" including its outer electron cloud.

Numerically, vdW radius is much larger than covalent radius. For Cl: 180 pm vs 99 pm.

Takeaway: covalent radius < van der Waals radius for the same atom.

Example 5: Lanthanide contraction

Explain why Hf has almost the same atomic radius as Zr, despite being one period below.

Solution:

Between Zr (Z = 40, period 5) and Hf (Z = 72, period 6) lies the lanthanide series (La to Lu, Z = 57 to 71) in which 14 electrons fill the $4f$ subshell. 4f orbitals are poor at shielding the outer electrons from the nucleus, so effective nuclear charge increases more than usual, contracting atomic radii. This lanthanide contraction cancels the expected radius increase going from period 5 to period 6.

Result: Zr (160 pm) ≈ Hf (159 pm). Similarly Nb ≈ Ta, Mo ≈ W (classic Zr/Hf-type comparison).

Takeaway: poor shielding by $f$-electrons causes the lanthanide contraction, making 5$d$-series elements unusually small.

Example 6: Why Be > B for IE?

IE$_{1}$(Be) = 899 kJ/mol > IE$_{1}$(B) = 801 kJ/mol. Explain.

Solution:

• Be: $[He]\,2s^{2}$. Removing an electron from the fully-filled, stable $2s$ orbital requires extra energy.
• B: $[He]\,2s^{2}\,2p^{1}$. Removing the sole $2p$ electron is easier because (a) $2p$ is higher in energy than $2s$, and (b) taking this electron away leaves the extra-stable $2s^{2}$ configuration.

The expected "across-period" trend (IE increases L→R) is therefore reversed at B.

Takeaway: subshell stability beats the plain $Z$ trend — one of the famous IE anomalies.

Example 7: Successive ionization jump

Given the successive IEs of Na (in kJ/mol): 496, 4562, 6912, 9544, 13352, … Identify where the big jump occurs, and why.

Solution:

The jumps:

• IE$_{1}$ → IE$_{2}$: 496 → 4562 — factor of 9× ← huge jump
• IE$_{2}$ → IE$_{3}$: 4562 → 6912 — factor of 1.5×
• IE$_{3}$ → IE$_{4}$: 6912 → 9544 — factor of 1.4×

The big jump occurs between IE$_{1}$ and IE$_{2}$. Na's first electron is the $3s^{1}$ valence electron (loose). After removing it, Na$^{+}$ has Ne configuration — a stable noble-gas core. Pulling the next electron out of the inner $2p$ shell requires far more energy.

Takeaway: a big jump in successive IEs tells you how many valence electrons an atom has. Na has 1 valence electron; the jump comes at the 2nd ionisation.

Example 8: Noble gas IE comparison

Compare the first IEs of He (2372), Ne (2081), Ar (1521), Kr (1351).

Solution:

Down Group 18, atomic size increases and the outermost electron becomes less tightly bound. IE decreases:

$$\mathrm{IE}_{1}: He > Ne > Ar > Kr$$

Numerical values support this. He's IE is the highest of any element — removing an electron from the tightly-held $1s^{2}$ shell requires enormous energy (2372 kJ/mol).

Takeaway: the helium record. He has the universal maximum first ionization enthalpy.

Example 9: Unit conversion

IE$_{1}$(F) = 17.42 eV/atom. Convert to kJ/mol.

Solution:

$$\mathrm{IE}_{1}(\mathrm{F}) = 17.42 \times 96.485 \approx \mathbf{1681\ kJ/mol}$$

Matches the standard NCERT value.

Takeaway: 1 eV/atom = 96.485 kJ/mol — keep this conversion at your fingertips.

Example 10: Identify from IE data

Element X has IE$_{1}$ = 738 kJ/mol and IE$_{2}$ = 1451 kJ/mol, but IE$_{3}$ = 7732 kJ/mol. Identify X.

Solution:

The small jumps between IE$_{1}$ and IE$_{2}$ but a huge jump from IE$_{2}$ to IE$_{3}$ indicates X has 2 valence electrons (easy to remove) and a noble-gas core after.

X is in Group 2. IE$_{1}$ = 738 and IE$_{2}$ = 1451 match Magnesium (Mg).

Takeaway: number of easily-ionised electrons = group valence for main-group elements.

Example 11: Sign and order of halogen $\Delta_{eg}H$

Order F, Cl, Br, I by decreasing (more negative first) $\Delta_{eg}H$.

Solution:

Values (kJ/mol): F $-328$, Cl $-349$, Br $-325$, I $-295$.

Ordering:

$$Cl > F > Br > I$$

(where ">" means more negative).

Cl, despite being bigger than F, has the most negative $\Delta_{eg}H$ because F's compact $2p$ shell suffers heavy electron-electron repulsion.

Takeaway: Cl is the champion, not F. This is a top-5 most-asked fact on exams.

Example 12: Numerical — total enthalpy for $O^{2-}$

$\Delta_{eg}H_{1}(O) = -141$ kJ/mol, $\Delta_{eg}H_{2}(O) = +744$ kJ/mol. Calculate the net enthalpy change for $O(g) + 2e^{-} \rightarrow O^{2-}(g)$.

Solution:

$$\Delta H_{\text{net}} = \Delta_{eg}H_{1} + \Delta_{eg}H_{2} = -141 + 744 = +603\ \text{kJ/mol}$$

Strongly endothermic! Forming $O^{2-}$ in the gas phase requires 603 kJ per mole.

$O^{2-}$ is only stable in ionic solids because the lattice energy released upon crystal formation ($\sim -3000$ to $-4000$ kJ/mol) far exceeds this cost.

Takeaway: the second electron-gain enthalpy is always strongly positive. Lattice energy is what makes $O^{2-}$-containing ionic solids possible.

Example 13: Positive $\Delta_{eg}H$ elements

Which period-2 and period-3 elements have POSITIVE $\Delta_{eg}H$?

Solution:

• Period 2: Be ($+66$) and N ($+34$) — positive. Also Ne ($+116$).
• Period 3: Mg ($+67$) — positive. Also Ar ($+96$).

All share full or half-filled subshells:

• Be, Mg: full $s^{2}$ → next electron must go to $p$.
• N: half-filled $2p^{3}$ → extra stability.
• Ne, Ar: full $p^{6}$ → new shell needed.

Takeaway: full/half-filled configurations create "speed bumps" that flip $\Delta_{eg}H$ to positive.

Example 14: Compact-shell repulsion (O vs S)

Why does S have $\Delta_{eg}H = -200$ kJ/mol but O only $-141$ kJ/mol, even though O is higher in the group and should 'want' the extra electron more?

Solution:

Oxygen's $2p$ shell is very compact, so 4 existing $2p$ electrons are packed tightly. When a 5th electron is added (to form $O^{-}$), it suffers strong electron-electron repulsion that partially cancels the nuclear attraction.

Sulfur's $3p$ shell is larger; the existing 4 electrons are more spread out. The 5th electron added to form $S^{-}$ feels less repulsion, so more energy is released.

Same logic: F-Cl reversal.

Takeaway: compact second-period shells suffer anomalous repulsion — a running theme.

Example 15: Percent comparison

By what percentage is $\Delta_{eg}H(Cl) = -349$ kJ/mol more negative than $\Delta_{eg}H(F) = -328$ kJ/mol?

Solution:

Difference: $|{-349 - (-328)}| = 21$ kJ/mol.

Percentage vs F:

$$\frac{21}{328} \times 100 \;=\; 6.4\%$$

Cl is 6.4% more negative than F.

Takeaway: small difference numerically, but conceptually it's the reason Cl sits atop every "most electron-loving" list.

Example 16: Compute % ionic character

For HF, $\Delta\chi = |2.20 - 3.98| = 1.78$. Compute % ionic character using Pauling's relation.

Solution:

$$\%\,\text{ionic} = \left[1 - e^{-(1.78)^{2}/4}\right] \times 100$$ $$= \left[1 - e^{-0.7921}\right] \times 100$$ $$= [1 - 0.4529] \times 100 \approx 54.7\%$$

So HF is ~55% ionic by Pauling's empirical estimate.

Takeaway: a bond with $\Delta\chi$ near 1.7 lies near the ionic/covalent border on Pauling's scale and is strongly polar.

Example 17: Ranking bonds by % ionic character

Rank by INCREASING % ionic character: H-F, H-Cl, Na-Cl, Cs-F.

Solution:

$\Delta\chi$ values: H-F = 1.78, H-Cl = 0.96, Na-Cl = 2.23, Cs-F = 3.19.

Using Pauling's relation:

• H-Cl (0.96): ~21%
• H-F (1.78): ~55%
• Na-Cl (2.23): ~71%
• Cs-F (3.19): ~92%

Answer:

$$H\text{-}Cl < H\text{-}F < Na\text{-}Cl < Cs\text{-}F$$

Takeaway: % ionic character rises quickly with $\Delta\chi$. Cs-F is one of the most ionic everyday compounds.

Example 18: Hybridization and EN

Which hybridization of carbon has the highest electronegativity: $sp$, $sp^{2}$, or $sp^{3}$?

Solution:

More $s$-character → electrons held closer to nucleus → higher EN.

• $sp$-C: 50% $s$
• $sp^{2}$-C: 33% $s$
• $sp^{3}$-C: 25% $s$

Order: $sp > sp^{2} > sp^{3}$.

This explains the acidity of terminal C-H bonds in alkynes: HC≡CH has $sp$-C pulling electrons away from H, weakening the C-H bond and making H$^{+}$ easier to remove.

Takeaway: hybridization affects electronegativity significantly; $sp$-carbon is the most electronegative among the three common hybrid states.

Example 19: Mulliken EN calculation

For an atom with IE = 1600 kJ/mol and EA = 300 kJ/mol, compute $\chi_{M}$.

Solution:

Mulliken formula:

$$\chi_{M} = \dfrac{\mathrm{IE} + \mathrm{EA}}{2} = \dfrac{1600 + 300}{2} = 950\ \text{kJ/mol}$$

To convert to eV:

$$950\ \text{kJ/mol} = 950/96.485 \approx 9.85\ \text{eV}$$

Then an approximate conversion to the Pauling scale gives:

$$\chi_{\text{Pauling}} \approx 9.85/2.8 \approx 3.5$$

This lies close to oxygen's Pauling value (3.44).

Takeaway: use Mulliken when you have IE and EA; it is directly tied to atomic energetics.

Example 20: Bond length via Schomaker-Stevenson

Estimate the C-Cl bond length using $r_{C} = 77$ pm, $r_{Cl} = 99$ pm, $\chi_{C} = 2.55$, $\chi_{Cl} = 3.16$.

Solution:

$$d_{C-Cl} = r_{C} + r_{Cl} - 0.09 \times 100 \times |\chi_{C} - \chi_{Cl}|\ \text{pm}$$

Careful with units — the 0.09 factor is in Å; convert to pm by multiplying by 100.

$$d = 77 + 99 - 9 \times |2.55 - 3.16| = 176 - 9 \times 0.61$$ $$= 176 - 5.5 \approx 170.5\ \text{pm}$$

The estimate is reasonably close to common experimental C-Cl bond lengths.

Takeaway: polar bonds are slightly shorter than the simple sum of covalent radii predicts, because partial ionic character pulls the atoms closer.

Example 21: Oxide formula from group

Write the formulas of the highest oxide, the common hydride, and the common oxoacid of Group-16 element 'X'.

Solution:

Group 16: O-valence = 6, H-valence = 2.

• Highest oxide: $XO_{3}$ (SO$_{3}$, SeO$_{3}$).
• Common hydride: $H_{2}X$ (H$_{2}$O, H$_{2}$S, H$_{2}$Se, H$_{2}$Te).
• Common oxoacid: $H_{2}XO_{4}$ (H$_{2}$SO$_{4}$, H$_{2}$SeO$_{4}$, H$_{2}$TeO$_{4}$).

Takeaway: group number dictates valence, which dictates common formulas.

Example 22: Why BeCl$_{2}$ is covalent

Using Fajan's rules, explain why BeCl$_{2}$ is largely covalent but MgCl$_{2}$ is ionic.

Solution:

Fajan's rules: covalent character in an "ionic" compound rises when —

• cation is small,
• cation charge is high,
• anion is large and polarizable.

Be$^{2+}$ has a tiny ionic radius and charge +2. Its charge density is extremely high. It polarizes the large Cl$^{-}$ so strongly that the Be-Cl bond becomes predominantly covalent.

Mg$^{2+}$ is significantly larger. Its charge density is lower, so it cannot polarize Cl$^{-}$ enough to produce comparable covalent character. MgCl$_{2}$ remains predominantly ionic.

Takeaway: small, highly-charged cations produce more covalent character (Fajan's rules).

Example 23: Identify an amphoteric oxide

Given oxides CO$_{2}$, Al$_{2}$O$_{3}$, ZnO, Na$_{2}$O, SO$_{3}$, pick the amphoteric one(s).

Solution:

• CO$_{2}$: acidic.
• Al$_{2}$O$_{3}$: amphoteric.
• ZnO: amphoteric.
• Na$_{2}$O: basic.
• SO$_{3}$: acidic.

ZnO reacts: $$ZnO + 2\,HCl \rightarrow ZnCl_{2} + H_{2}O$$ $$ZnO + 2\,NaOH \rightarrow Na_{2}ZnO_{2} + H_{2}O$$

Al$_{2}$O$_{3}$ does the same with HCl and NaOH.

Takeaway: amphoteric oxides sit in the middle EN range. Quick list to memorise: Al$_{2}$O$_{3}$, ZnO, SnO, PbO, BeO.

Example 24: Acidic strength of hydrohalic acids

Explain why HF is a weak acid in water while HCl, HBr, and HI are strong acids.

Solution:

In water, HX dissociates as $HX + H_{2}O \rightleftharpoons H_{3}O^{+} + X^{-}$.

Strength depends mainly on the H-X bond dissociation enthalpy and on solvation effects.

Bond energies (kJ/mol): H-F 569 (very strong), H-Cl 431, H-Br 366, H-I 299.

H-F is so strong that HF dissociates only partially in water.

HI has the weakest bond and gives the strongest acid.

Takeaway: HX acid strength increases as H-X bond energy decreases: HF < HCl < HBr < HI.

Example 25: Diagonal relationship

List three similarities between Be and Al due to the diagonal relationship.

Solution:

Be and Al share these features because Be$^{2+}$ and Al$^{3+}$ have similar charge density and related periodic behavior:

1. Amphoteric oxides: BeO and Al$_{2}$O$_{3}$ both react with acid and base.
2. Covalent chlorides: BeCl$_{2}$ and AlCl$_{3}$ are both covalent.
3. Carbide formation: Be$_{2}$C and Al$_{4}$C$_{3}$ give CH$_{4}$ on hydrolysis.

Takeaway: diagonal pairs Li/Mg, Be/Al, B/Si appear repeatedly in JEE/NEET.

Example 26: Born-Haber-like calculation

For the reaction $Na(g) + Cl(g) \rightarrow NaCl(g)$, use $IE_{1}(Na) = +496$ kJ/mol and $\Delta_{eg}H(Cl) = -349$ kJ/mol to find the total electron-transfer enthalpy. Is this reaction spontaneous from the electron-transfer step alone?

Solution:

Step 1: Na → Na$^{+}$ + e$^{-}$, $\Delta H = +496$ kJ/mol.

Step 2: Cl + e$^{-}$ → Cl$^{-}$, $\Delta H = -349$ kJ/mol.

Net gas-phase electron-transfer:

$$\Delta H_{\text{e-transfer}} = +496 - 349 = +147\ \text{kJ/mol}$$

Endothermic by 147 kJ/mol — the electron-transfer step alone is not favourable.

But NaCl solid forms spontaneously because the lattice-formation step releases a large negative lattice enthalpy.

Takeaway: always consider the full Born-Haber cycle for ionic-compound formation. Electron transfer alone is not usually exothermic.

Example 27: Formula of a super-heavy oxide

Predict the formula of the highest oxide of element 117 (Ts, tennessine).

Solution:

Element 117 → Group 17 (halogen position, 7th period).

For Group 17, O-valence = 7, so the highest oxide formula is $X_{2}O_{7}$.

Formula: Ts$_{2}$O$_{7}$.

Whether this compound is actually stable is another question, but the group-rule prediction is $Ts_{2}O_{7}$.

Takeaway: group number alone gives you the predicted highest oxide, even for exotic super-heavy elements.

Example 28: Combined identification

An element X has atomic radius ≈ 77 pm, IE$_{1}$ ≈ 1086 kJ/mol, $\Delta_{eg}H$ ≈ $-122$ kJ/mol, and forms an oxide $XO_{2}$ that is strongly acidic. Identify X.

Solution:

• XO$_{2}$ formula → O-valence = 4 → Group 14.
• IE$_{1}$ ≈ 1086 matches C.
• Atomic radius ≈ 77 pm matches C.
• $\Delta_{eg}H$ ≈ -122 kJ/mol matches C.
• CO$_{2}$ is an acidic oxide.

X = Carbon.

Takeaway: for multi-clue problems, use group-valence to narrow down to a column, then match numerical data to pinpoint the element.

Example 29: JEE-style "which statement is wrong"

Which of the following statements is INCORRECT?

(a) Cl has the most negative $\Delta_{eg}H$ of any element.

(b) F has the highest electronegativity of any element.

(c) He has the highest first IE of any element.

(d) Li has the most negative standard electrode potential $E^{\circ}$ among the common metals in aqueous solution.

Solution:

All four statements are CORRECT:

• (a) Cl: $\Delta_{eg}H = -349$ kJ/mol — the most negative of all elements.
• (b) F: Pauling EN = 3.98 — the highest.
• (c) He: IE$_{1}$ = 2372 kJ/mol — the highest.
• (d) Li/Li$^{+}$: $E^{\circ} = -3.04$ V — the most negative among the common metals in aqueous solution.

So none of these is incorrect. If the question presents these four options, the correct response is effectively "none is incorrect".

Watch out for variations where one of the superlatives is altered slightly (for example, "F has the most negative $\Delta_{eg}H$" would be wrong; Cl does).

Takeaway: keep the "four records" memorised: Cl for $\Delta_{eg}H$, F for EN, He for IE, Li for very negative $E^{\circ}$ in aqueous solution.

Example 30: Advanced — Z$_{\text{eff}}$ calculation by Slater's rules

Calculate $Z_{\text{eff}}$ for the $2p$ electron in a nitrogen atom using Slater's rules.

Solution:

Slater's rules for a $2p$ electron in $(1s^{2})(2s^{2}\,2p^{3})$:

• Each other electron in the SAME (2s, 2p) group: $\sigma = 0.35$ per electron. There are (2 + 3) - 1 = 4 other electrons → contribution $4 \times 0.35 = 1.40$.
• Each electron in $n-1$ shell (i.e., 1s): $\sigma = 0.85$ per electron. 2 × 0.85 = 1.70.

Total shielding: $\sigma = 1.40 + 1.70 = 3.10$.

Effective nuclear charge:

$$Z_{\text{eff}} = Z - \sigma = 7 - 3.10 = 3.90$$

This is why the outer electron of N is strongly held.

Takeaway: Slater's rules give an approximate but very useful $Z_{\text{eff}}$.