⚡ Formula Bank

Nitrogen Group (Group 15) Formulas

Hydrides of Group 15: NH₃, N₂H₄, N₃H, PH₃, AsH₃, SbH₃, BiH₃
Oxidation State: Oxidation state of And in NH₃ = -3, in N₂ = 0, in NO = +2
Ammonia Production: Haber process: N₂ + 3H₂ ⇌ 2NH₃
Nitric Acid Production: Ostwald process: 4NH₃ + 5O₂ → 4NO + 6H₂O, 2NO + O₂ → 2NO₂, 3NO₂ + H₂O → 2HNO₃ + NO
Examiner's Trap: Confusing the order of hydrides' basicity.

Phosphorus Group (Group 16) Formulas

Hydrides of Group 16: H₂O, H₂S, H₂Se, H₂Te
Oxidation States: Oxidation state of O in H₂O = -2, in O₂ = 0
Water: Ionization constant, K_w = [H⁺][OH⁻] = 10⁻¹⁴ at 25°C
Sulfuric Acid: H₂SO₄ ⇌ H⁺ + HSO₄⁻, HSO₄⁻ ⇌ H⁺ + SO₄²⁻
Examiner's Trap: Mixing up the properties of H₂O and H₂S.

Chalcogens (Group 17) Formulas

Halogens: F₂, Cl₂, Br₂, I₂, At₂
Hydrohalic Acids: HF, HCl, HBr, HI
Halogen Displacement: F₂ > Cl₂ > Br₂ > I₂ (reactivity)
HF: Weak acid, K_a = 3.5 × 10⁻⁴
Examiner's Trap: Forgetting that HF is a weak acid.

Noble Gases (Group 18) Formulas

Noble Gases: He, Ne, Ar, Kr, Xe, Rn
Xenon Compounds: XeF₂, XeF₄, XeF₆, XeO₃, XeO₄
Xenon Tetrafluoride: Xe + 2F₂ → XeF₄
Examiner's Trap: Overlooking the existence of noble gas compounds.

Which Formula When?

Situation	Relevant Formula
Calculating oxidation state of And in ammonia	-3
Writing Haber process equation	N₂ + 3H₂ ⇌ 2NH₃
Ionization constant of water at 25°C	10⁻¹⁴
Comparing halogen reactivity	F₂ > Cl₂ > Br₂ > I₂
Xenon tetrafluoride synthesis	Xe + 2F₂ → XeF₄

🪤 The 5 Mistakes That Cost Marks

The 5 Mistakes That Cost Marks

Mistake 1 — Incorrect Oxidation States:
🔴 What students write: Oxidation state of nitrogen in NO₂ is +3.
✅ What examiners expect: Oxidation state of nitrogen in NO₂ is +4.
💸 Marks lost: 1 mark
🔧 The fix (30-second trick): Recall that oxygen has an oxidation state of -2, and there are two oxygen atoms, so 2 × (-2) = -4. Since NO₂ is neutral, nitrogen's oxidation state must be +4 to balance.
Mistake 2 — Wrong Formula for Phosphorus Pentoxide:
🔴 What students write: P₄O₆
✅ What examiners expect: P₄O₁₀
💸 Marks lost: 2 marks
🔧 The fix (30-second trick): Remember that phosphorus pentoxide has the prefix "penta" meaning five, and oxide is O²⁻, so it must be P₄O₁₀.
Mistake 3 — Confusing Group 15 Hydrides:
🔴 What students write: NH₃ is less stable than BiH₃.
✅ What examiners expect: NH₃ is more stable than BiH₃ due to increasing down the group, bond strength decreases.
💸 Marks lost: 2 marks
🔧 The fix (30-second trick): Recall that as you go down Group 15, the stability of hydrides decreases due to increasing size and decreasing bond strength.
Mistake 4 — Incorrect Electron Configuration:
🔴 What students write: The electron configuration of Sn (Z=50) is [Kr] 4d¹⁰ 5s² 5p⁶.
✅ What examiners expect: The electron configuration of Sn (Z=50) is [Kr] 4d¹⁰ 5s² 5p².
💸 Marks lost: 3 marks
🔧 The fix (30-second trick): Ensure you write the correct number of electrons in the p-orbitals for the group and period.
Mistake 5 — Forgetting the Correct Order of Acid Strength:
🔴 What students write: H₂O > H₂S > H₂Se > H₂Te (acid strength in group 16 hydrides).
✅ What examiners expect: H₂O < H₂S < H₂Se < H₂Te (acid strength increases down the group).
💸 Marks lost: 2 marks
🔧 The fix (30-second trick): Recall that acid strength increases down a group due to decreasing bond strength between the hydrogen and the chalcogen.

✏️ 3 Solved PYQs

Q1 (2020 JEE Main): The hybridisation of the central atom in $\ce{PCl5}$ is

🪤 Trap: Students often get confused between $\ce{sp^3}$ and $\ce{dsp^3}$ hybridization.
🧮 Solution (Step-by-step): Step 1: Determine the central atom → Phosphorus (P) in $\ce{PCl5}$ . Step 2: Calculate the total number of valence electrons → P (5) + 5*Cl (7) = 5 + 35 = 40 electrons. Step 3: Draw the Lewis structure → P with 5 single bonds to Cl and no lone pairs on P. Step 4: Determine electron pairs around P → 5 bonding pairs. Step 5: Apply VSEPR theory → $\ce{PCl5}$ has a trigonal bipyramidal geometry. Step 6: Identify hybridization → $\ce{dsp^3}$ or $\ce{sp^3d}$ hybridization. Final Answer: $\ce{sp^3d}$
⚡ Speed trick: Recall that $\ce{PCl5}$ has a trigonal bipyramidal shape which corresponds to $\ce{sp^3d}$ hybridization.

Q2 (2019 NEET): The nitrogen atom in $\ce{NH3}$ has

A. $\ce{sp^2}$ hybrid orbitals and a lone pair
B. $\ce{sp^3}$ hybrid orbitals and a lone pair
C. $\ce{sp}$ hybrid orbitals and two lone pairs
D. $\ce{sp^2}$ hybrid orbitals and two lone pairs
🪤 Trap: Students often confuse between $\ce{sp^2}$ and $\ce{sp^3}$ hybridization.
🧮 Solution (Step-by-step): Step 1: Determine the central atom → Nitrogen (N) in $\ce{NH3}$ . Step 2: Calculate the total number of valence electrons → N (5) + 3*H (1) = 5 + 3 = 8 electrons. Step 3: Draw the Lewis structure → N with 3 single bonds to H and 1 lone pair on N. Step 4: Determine electron pairs around N → 4 (3 bonding + 1 lone). Step 5: Apply VSEPR theory → $\ce{NH3}$ has a tetrahedral electron pair geometry and a trigonal pyramidal shape. Step 6: Identify hybridization → $\ce{sp^3}$ hybridization. Final Answer: B. $\ce{sp^3}$ hybrid orbitals and a lone pair
⚡ Speed trick: Recall that $\ce{NH3}$ has a trigonal pyramidal shape with $\ce{sp^3}$ hybridization.

Q3 (2018 CBSE Boards): The correct order of bond angles in $\ce{H2O}$ , $\ce{H2S}$ , and $\ce{NH3}$ is

A. $\ce{H2O > NH3 > H2S}$
B. $\ce{H2O > H2S > NH3}$
C. $\ce{NH3 > H2O > H2S}$
D. $\ce{H2S > NH3 > H2O}$
🪤 Trap: Students often get confused between the actual bond angles and the expected bond angles based on hybridization.
🧮 Solution (Step-by-step): Step 1: Recall the bond angles → $\ce{H2O}$ (104.5°), $\ce{NH3}$ (107°), $\ce{H2S}$ (92°). Step 2: Compare the bond angles → $\ce{NH3}$ (107°) > $\ce{H2O}$ (104.5°) > $\ce{H2S}$ (92°). Final Answer: C. $\ce{NH3 > H2O > H2S}$
⚡ Speed trick: Recall the trend of bond angles based on electronegativity and lone pair repulsion.

🧠 The One Thing Most Students Get Wrong

The One Thing Most Students Get Wrong

The misconception (what 85% believe): Most students think that the stability of the hydrides of Group 15 elements (NH₃, PH₃, AsH₃, SbH₃, BiH₃) increases down the group due to the decrease in bond dissociation energy.
The reality (what 99% know): The stability of the hydrides actually decreases down the group due to the increase in bond length and decrease in bond strength. This is because as we move down the group, the size of the central atom increases, leading to a longer and weaker bond with hydrogen.

Diagnostic Question

What is the correct order of stability of the hydrides of Group 15 elements?
A) NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃
B) BiH₃ > SbH₃ > AsH₃ > PH₃ > NH₃
C) PH₃ > NH₃ > AsH₃ > SbH₃ > BiH₃
D) NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃
If you answered B, C, or D: you have the misconception → fix: Remember that NH₃ has the strongest bond due to the smallest size of N, resulting in the highest stability.
If you answered A: you are in the top 5% → now extend this: The correct order of stability (NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃) is due to the fact that as we move down the group, the bond length increases and the bond angle decreases, making the hydrides less stable.

How to Never Forget This

Use the mnemonic "No Physical Action Should Be Ignored" to remember the decreasing stability order of the hydrides: NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃. Visualize a person (N) performing physical actions (P) on a surface (A) with stones (S) and bricks (B) in a decreasing order of stability.

Additional Insights

The basicity of the hydrides decreases down the group due to the decrease in the electronegativity of the central atom and the increase in the size of the central atom.
The boiling points of the hydrides increase from PH₃ to NH₃ due to the increase in hydrogen bonding, with NH₃ having the highest boiling point.
The toxicity of the hydrides increases down the group, with BiH₃ being the most toxic.

Key Points to Remember

The stability of the hydrides decreases down the group.
The bond length increases and bond strength decreases down the group.
NH₃ has the strongest bond and highest stability due to the smallest size of N.

👁️ Ayush's Note

🔮 The Hidden Pattern

The P-Block Elements (Group 15-18) have a non-obvious connection with Coordination Compounds, which appears in 30%+ of papers.
Specifically, questions often combine concepts of p-block elements with coordination chemistry, such as the formation of complexes with elements like nitrogen (N₂), oxygen (O₂), and halogens.

🎯 The "Always Check" Rule

Always verify the oxidation state of the p-block element in a given compound, as it's a common edge case tested by examiners.
For example, in Group 15, nitrogen can exhibit oxidation states from -3 (in NH₃) to +5 (in NF₅), and similar variations are seen in other groups.

📊 PYQ Frequency Intel

Nitrogen group (Group 15):
Properties of nitrogen (2019, 2023)
Oxides of nitrogen and their environmental impact (2021)
Chalcogens (Group 16):
Allotropes of sulfur (2019, 2021)
Acidic nature of oxides (2023)
Halogens (Group 17):
Reactivity trends and halogen displacement reactions (2019, 2021, 2023)
Noble gases (Group 18):
Uses and properties of noble gases (2021)

⚡ The 30-Second Shortcut

For questions involving the downward trend in a property (e.g.
bond energy, electronegativity) in a p-block group:
Recall that as you move down a group, atomic size increases, and bond energy generally decreases due to increased distance between nuclei and shared electrons.
Apply this trend to quickly eliminate incorrect options and arrive at the answer within 30 seconds.

🔁 Last 5 Minutes Box

⚡ Core Formulas

Molar Volume of an Ideal Gas: PV = nRT, where P is pressure, V is volume, n is number of moles, R is gas constant, and T is temperature in Kelvin.
Van der Waals Equation: (P + n²a/V²)(V
nb) = nRT, where a and b are Van der Waals constants.
Ostwald's Dilution Law: α = √(Kₐ/C), where α is degree of dissociation, Kₐ is acid dissociation constant, and C is concentration.
Henderson-Hasselbalch Equation: pH = pKₐ + log([A⁻]/[HA]), where [A⁻] is concentration of conjugate base and [HA] is concentration of acid.
Nernst Equation: E = E⁰
(RT/nF) * ln(Q), where E is cell potential, E⁰ is standard cell potential, R is gas constant, T is temperature, n is number of electrons, F is Faraday's constant, and Q is reaction quotient.

🧠 Must-Know Facts

The p-block elements are characterized by the filling of electrons in the p-orbitals.
Group 15 elements are also known as the nitrogen family, with nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi).
The oxidizing power of the elements decreases down the group in p-block elements.

🚫 Never Forget

❌ Assuming all p-block elements form stable hydrides → ✅ Not all p-block elements form stable hydrides; stability decreases down the group.
❌ Thinking that all oxides of p-block elements are acidic → ✅ Some oxides are basic, some are amphoteric, and some are acidic.

🎯 If you can only remember ONE thing

The p-block elements exhibit a wide range of oxidation states, and their chemistry is largely influenced by the ability of these elements to form stable ions and molecules with varying oxidation states.

📝 Practice MCQs

1. The bond angle in the molecule of NF₃ is A) 102° B) 107° C) 120° D) 109°

Answer: A) The bond angle in NF₃ is 102° due to the presence of a lone pair on the nitrogen atom, which exerts greater repulsion on the bonding pairs, reducing the bond angle from the ideal tetrahedral angle of 109°. Options B, C, and D are incorrect because they do not accurately reflect the effect of the lone pair on the bond angle.

2. The oxidation state of Bi in NaBiO₃ is A) +3 B) +5 C) +1 D) +2

Answer: B) In NaBiO₃, the oxidation state of Bi is +5 because sodium (Na) has an oxidation state of +1 and oxygen (O) has an oxidation state of -2. The sum of the oxidation states must equal zero: +1 (Na) + x (Bi) + 3(-2) (O) = 0, so x = +5. Options A, C, and D are incorrect because they do not accurately reflect the oxidation state of Bi in this compound.

3. The number of electrons in the outermost shell of the element with atomic number 85 (At) is A) 5 B) 7 C) 3 D) 2

Answer: B) The element with atomic number 85 is astatine (At), which is a halogen. Halogens have 7 electrons in their outermost shell. Therefore, the correct answer is 7. Options A, C, and D are incorrect because they do not accurately reflect the number of electrons in the outermost shell of At.

4. The reaction of XeF₄ with H₂O gives A) Xe + O₂ + HF B) XeF₂ + XeO₃ C) XeO₃ + Xe + HF D) XeF₂ + O₂ + H₂F₂

Answer: A) XeF₄ reacts with H₂O to produce Xe, O₂, and HF: 6XeF₄ + 8H₂O → 2Xe + 4XeO₃ + 12HF, but a more simplified and correct representation for a major product is Xe + O₂ + HF. Options B, C, and D are incorrect because they do not accurately reflect the products of this reaction.

5. If 1 mole of Pb(NO₃)₂ is heated, what is the mole ratio of PbO to O₂? A) 1 B) 1 C) 2 D) 1

Answer: A) When Pb(NO₃)₂ is heated, it decomposes according to the reaction: 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂. From the equation, 2 moles of PbO are produced for every 1 mole of O₂. Therefore, the mole ratio of PbO to O₂ is 2:1. Options B, C, and D are incorrect because they do not accurately reflect this ratio.

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This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.