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ChemistryPeriodic Properties
Identify the incorrect statement from the following :
Options
1
The largest and smallest species among Mg, Mg²⁺, Al and Al³⁺ are Al and Mg²⁺ respectively.
2
The IUPAC name of the element with atomic number 107 is Unnilseptium.
3
The similarity in behaviour of Li with Mg is referred to as diagonal relationship.
4
The oxidation state and covalency of Al in [AlCl(H₂O)₅]²⁺ are +3 and 6 respectively.
Correct Answer
Option 1 is INCORRECT
Solution
1

Correct size order among Mg, Mg²⁺, Al, Al³⁺:

Neutral atoms: Mg (Z=12) > Al (Z=13) — Mg is larger (same period, fewer protons)

Ions: Mg²⁺ vs Al³⁺ — both isoelectronic-ish but Mg²⁺ has fewer protons pulling electrons → Mg²⁺ is larger

Overall order: Mg > Al > Mg²⁺ > Al³⁺

2

Option 1 claims: "largest = Al, smallest = Mg²⁺"

Largest is actually Mg (not Al). Al is smaller than Mg.

Smallest IS Al³⁺ (not Mg²⁺). Al³⁺ has same electrons as Mg²⁺ but more protons (13 vs 12) → smaller.

So Option 1 is DOUBLY WRONG. ❌

Correct order: Mg > Al > Mg²⁺ > Al³⁺
Largest = Mg, Smallest = Al³⁺
Option 1 incorrectly says largest=Al, smallest=Mg²⁺
Theory: Periodic Properties
1. Atomic and Ionic Size Trends

Atomic radius decreases left to right across period (increasing nuclear charge). Ionic radius: cations smaller than parent atom (fewer electrons, same/more protons). More positive charge → smaller ion. For isoelectronic species (same electrons): more protons → smaller size. Anions larger than parent atom. Size order: anion > neutral atom > cation.

2. Isoelectronic Series

Species with same number of electrons. Na⁺, Mg²⁺, Al³⁺ are isoelectronic (all have 10 electrons = Ne config). Size: Na⁺(102 pm) > Mg²⁺(72 pm) > Al³⁺(53 pm). More protons → more attraction → smaller. Similarly: N³⁻ > O²⁻ > F⁻ > Ne > Na⁺ > Mg²⁺ > Al³⁺ (all 10 electrons, increasing Z).

3. Diagonal Relationship

Li and Mg show similar properties (diagonal relationship): both form normal oxides (not peroxides), both form nitrides (Li₃N, Mg₃N₂), both decompose carbonates on heating, similar electronegativity. Similarly: Be-Al, B-Si diagonal pairs. Reason: similarity in charge/radius ratio (ionic potential) between diagonally placed elements.

4. Covalency and Coordination Number

Covalency = number of covalent bonds. Coordination number = number of ligands/bonds around central atom. In [AlCl(H₂O)₅]²⁺: Al forms 6 bonds total (1 with Cl + 5 with H₂O). Covalency = 6, oxidation state = +3 (Al is always +3). Option 4 is CORRECT.

5. IUPAC Systematic Names for Heavy Elements

Element 107 = Bohrium (Bh). But the systematic IUPAC name uses Latin/Greek: 1=un, 0=nil, 7=sept → Unnilseptium (Uns). 106=Unnilhexium, 108=Unniloctium. These systematic names were used before official names were given. Option 2 (Unnilseptium for Z=107) is CORRECT.

6. Effective Nuclear Charge (Zeff)

Zeff = Z − σ (where σ = shielding constant). Slater's rules give σ. Higher Zeff → smaller radius, higher ionisation energy, higher electronegativity. Across period: Z increases by 1 each step, shielding increases less → Zeff increases → size decreases. Down group: new electron shells → size increases despite higher Z.

7. Ionisation Energies — Anomalies

IE₁: B < Be (2p electron easier to remove than 2s). O < N (paired 2p electron → extra repulsion → easier to remove). IE₁ trend: LiBOAlS

8. Electronegativity Trends

Increases left to right (across period), decreases top to bottom (down group). Most electronegative: F(3.98). N(3.04) > O(3.44)? No — O(3.44) > N(3.04). Correct order: F > O > N > Cl > Br > C > S > P > H > metals. Diagonal relationship in electronegativity explains Li-Mg, Be-Al similarities.

Frequently Asked Questions
1. What is the correct size order of Mg, Mg²⁺, Al, Al³⁺? ⌄
Mg(160pm) > Al(143pm) > Mg²⁺(72pm) > Al³⁺(53pm). Neutral Mg > Al because same period, fewer protons. Mg²⁺ > Al³⁺ because both have same electrons (10) but Al³⁺ has more protons (13 vs 12) → pulls electrons more strongly → smaller.
2. Why is Al smaller than Mg despite both being in Period 3? ⌄
Both Mg and Al are in Period 3. Al(Z=13) has one more proton than Mg(Z=12). This extra proton increases nuclear attraction on the same number of electron shells → electrons pulled inward → Al is smaller. Atomic radius: Mg=160pm, Al=143pm.
3. Why is Al³⁺ smaller than Mg²⁺? ⌄
Both Al³⁺ and Mg²⁺ have 10 electrons (isoelectronic). Al has 13 protons; Mg has 12 protons. More protons with same electrons → greater effective nuclear charge → electrons pulled inward more → Al³⁺ is smaller. Radii: Mg²⁺=72pm, Al³⁺=53pm.
4. What is the correct answer for Z=107? ⌄
Element 107 = Bohrium (Bh, official name since 1997). Systematic IUPAC name: Unnilseptium (Uns) — un(1)+nil(0)+sept(7)ium. This is CORRECT. So Option 2 is a true statement.
5. What is the diagonal relationship between Li and Mg? ⌄
Li (Period 2, Group 1) and Mg (Period 3, Group 2) are diagonally placed. Similar charge/radius ratio → similar chemical behavior: both form normal oxides (Li₂O, MgO), both form nitrides directly with N₂ (Li₃N, Mg₃N₂), lithium carbonate decomposes like magnesium carbonate on heating, both form organometallic compounds. Option 3 is CORRECT.
6. What is covalency of Al in [AlCl(H₂O)₅]²⁺? ⌄
Al forms coordinate bonds with 1 Cl⁻ and 5 H₂O molecules = 6 coordinate bonds. Covalency (number of bonds from Al) = 6. Oxidation state of Al = +3 (Al always +3 in most compounds). So Option 4 states covalency=6, oxidation state=+3 → CORRECT.
7. What are other examples of diagonal relationships? ⌄
Be(Group 2, Period 2)–Al(Group 13, Period 3): both form amphoteric oxides (BeO, Al₂O₃), both form polymeric hydrides, AlCl₃ behaves like BeCl₂. B(Group 13, Period 2)–Si(Group 14, Period 3): both form acidic oxides (B₂O₃, SiO₂), both form hydrides by reaction with LiH, both form oxoanions (borates, silicates). The pattern: diagonal elements have similar electronegativity and similar ionic potential.
8. Why do paired electrons cause lower IE than expected? ⌄
In N (2p³): all three 2p electrons unpaired — no electron-electron repulsion within same orbital. In O (2p⁴): one pair of electrons in same 2p orbital — repulsion between paired electrons makes them easier to remove. So IE₁(O) < IE₁(N) despite O having higher Z. This is why IE₁ dips at O (and S in Period 3). This is a classic NEET anomaly.
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