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UPCAT ChemistryPeriodic Table, Bonding & Chemical LanguageFlash Cards

Flashcards for Periodic Table, Bonding & Chemical Language — the active-recall tool for UPCAT Chemistry aspirants. Each card tests a key concept, formula, or definition from the UPCAT 2026 syllabus. Use them daily in the final month before exam day.

Exam context

For the University of the Philippines College Admission Test, University of the Philippines tests Chemistry under a "Core" label, with Periodic Table, Bonding & Chemical Language in the 3rd slot across 7 chapters. UPCAT candidates must clear the UPG ≤ 2.2 typical cut on the 2026 paper, which draws about 20 Chemistry questions. Date to watch: Mid-2026 (announced by UP Admissions).

Periodic Table, Bonding & Chemical Language - Flashcards

Master the fundamental concepts of chemistry including periodic trends, chemical bonding types, molecular geometry, and chemical nomenclature. These flashcards cover essential topics for UPCAT and other college entrance exams, with balanced coverage of reactions, calculations, and conceptual understanding.

Cards

What is the trend for atomic size as you move across a period from left to right?

Atomic size DECREASES from left to right across a period. This happens because the number of protons increases, creating stronger nuclear attraction that pulls electrons closer to the nucleus. Example: Na > Mg > Al > Si > P > S > Cl

Tags

  • periodic_trends
  • atomic_size
  • conceptual

Topic

Periodic Trends

Card Id

ptbl_001

Difficulty

medium

Image Prompt

Calculate the number of valence electrons in sulfur (S) and predict its ion charge.

Step 1: S is in Group 16 (6A), so it has 6 valence electrons. Step 2: To achieve noble gas configuration, S needs 2 more electrons. Step 3: S gains 2 electrons to form S²⁻ ion. Answer: 6 valence electrons, forms S²⁻ ion

Tags

  • valence_electrons
  • ion_formation
  • numerical

Topic

Ion Formation

Card Id

ptbl_002

Difficulty

easy

Image Prompt

What type of bond forms between Mg and Cl₂? Write the balanced equation.

IONIC BOND forms between metal (Mg) and nonmetal (Cl). Balanced equation: Mg(s) + Cl₂(g) → MgCl₂(s). Mg loses 2 electrons to form Mg²⁺, each Cl gains 1 electron to form Cl⁻. Electrostatic attraction holds ions together.

Tags

  • ionic_bonding
  • reaction_equation
  • reaction_problems

Topic

Chemical Bonding

Card Id

bond_003

Difficulty

medium

Image Prompt

Explain why NH₃ (ammonia) has a bent molecular geometry.

NH₃ has TRIGONAL PYRAMIDAL geometry, not bent. Using VSEPR theory: N has 3 bonding pairs (N-H bonds) and 1 lone pair. The lone pair takes up more space, pushing the H atoms closer together, creating a pyramid shape with bond angle ~107°.

Tags

  • vsepr_theory
  • molecular_geometry
  • conceptual

Topic

Molecular Geometry

Card Id

bond_004

Difficulty

hard

Image Prompt

Name the compound CaCl₂ and write the formula for aluminum oxide.

CaCl₂ = Calcium chloride (Ca²⁺ + 2Cl⁻). For aluminum oxide: Al³⁺ + O²⁻. Using criss-cross method: Al₂O₃. Steps: Al³⁺ needs 3 electrons, O²⁻ provides 2, so need 2 Al and 3 O for balance.

Tags

  • ionic_compounds
  • nomenclature
  • formula_writing

Topic

Chemical Nomenclature

Card Id

name_005

Difficulty

medium

Image Prompt

What type of intermolecular force exists between HF molecules?

HYDROGEN BONDING - the strongest type of dipole-dipole force. HF has highly polar H-F bond due to large electronegativity difference. H (δ+) of one HF molecule attracts F (δ-) of another HF molecule. This explains HF's high boiling point.

Tags

  • hydrogen_bonding
  • intermolecular_forces
  • conceptual

Topic

Intermolecular Forces

Card Id

imf_006

Difficulty

medium

Image Prompt

Draw the Lewis structure for CO₂ and determine if it's polar or nonpolar.

Lewis structure: O=C=O (linear). Steps: C has 4e⁻, each O has 6e⁻ = 16 total. C forms double bonds with each O. Molecular geometry is LINEAR. Although C=O bonds are polar, they cancel out due to symmetry. Result: NONPOLAR molecule.

Tags

  • lewis_structures
  • molecular_polarity
  • structural_problems

Topic

Lewis Structures

Card Id

lewis_007

Difficulty

hard

Image Prompt

Name the acid HClO₃ and write the formula for sulfurous acid.

HClO₃ = Chloric acid (from ClO₃⁻ chlorate ion, -ate becomes -ic). For sulfurous acid: comes from SO₃²⁻ sulfite ion (-ite becomes -ous), so formula is H₂SO₃. Remember: -ate → -ic acid, -ite → -ous acid

Tags

  • acid_nomenclature
  • oxyacids
  • formula_writing

Topic

Acid Nomenclature

Card Id

acid_008

Difficulty

medium

Image Prompt

Why does ionization energy increase across a period?

Ionization energy increases across a period because: 1) Nuclear charge increases (more protons), 2) Atomic size decreases, 3) Electrons are held more tightly, 4) More energy needed to remove an electron. Example: Na < Mg < Al < Si < P < S < Cl

Tags

  • ionization_energy
  • periodic_trends
  • conceptual

Topic

Periodic Trends

Card Id

trend_009

Difficulty

medium

Image Prompt

Name these binary covalent compounds: CO₂, N₂O₄, SF₆

CO₂ = Carbon dioxide. N₂O₄ = Dinitrogen tetroxide. SF₆ = Sulfur hexafluoride. Rules: First element keeps name, second gets -ide ending. Use prefixes: di- (2), tetra- (4), hexa- (6). Drop 'mono-' for first element.

Tags

  • binary_compounds
  • nomenclature
  • molecular_compounds

Topic

Binary Compounds

Card Id

binary_010

Difficulty

easy

Image Prompt

Calculate formal charge of N in NH₃.

Formula: FC = valence e⁻ - nonbonding e⁻ - (bonding e⁻/2). For N in NH₃: Step 1: N has 5 valence e⁻, Step 2: N has 2 nonbonding e⁻ (1 lone pair), Step 3: N has 6 bonding e⁻ (3 bonds). FC = 5 - 2 - (6/2) = 0

Tags

  • formal_charge
  • lewis_structures
  • numerical

Topic

Formal Charge

Card Id

formal_011

Difficulty

hard

Image Prompt

What happens when sodium metal reacts with chlorine gas?

2Na(s) + Cl₂(g) → 2NaCl(s). This is an ionic bond formation reaction. Na loses 1 electron: Na → Na⁺ + e⁻. Each Cl gains 1 electron: Cl + e⁻ → Cl⁻. Results in crystalline salt with high melting point due to strong electrostatic forces.

Tags

  • ionic_bonding
  • reaction_equation
  • reaction_problems

Topic

Ionic Bond Formation

Card Id

metal_012

Difficulty

medium

Image Prompt

Predict the molecular geometry of water (H₂O) using VSEPR theory.

H₂O has BENT molecular geometry. Steps: 1) O has 2 bonding pairs (O-H bonds), 2) O has 2 lone pairs, 3) Total = 4 electron pairs → tetrahedral electronic geometry, 4) 2 lone pairs push bonding pairs closer → bent shape with ~104.5° bond angle.

Tags

  • vsepr_theory
  • molecular_geometry
  • water

Topic

VSEPR Theory

Card Id

vsepr_013

Difficulty

medium

Image Prompt

Write the formula for calcium phosphate using the phosphate ion PO₄³⁻.

Ca²⁺ + PO₄³⁻ → Ca₃(PO₄)₂. Steps: 1) Ca has +2 charge, PO₄ has -3 charge, 2) Need 3 Ca²⁺ and 2 PO₄³⁻ for charge balance, 3) 3(+2) + 2(-3) = 0. Parentheses needed around polyatomic ion with subscript >1.

Tags

  • polyatomic_ions
  • ionic_compounds
  • formula_writing

Topic

Polyatomic Ions

Card Id

polyatomic_014

Difficulty

hard

Image Prompt

Why do larger molecules have stronger London dispersion forces?

Larger molecules have MORE ELECTRONS that can shift to create temporary dipoles. More electrons = stronger temporary dipoles = stronger induced dipoles in neighboring molecules = stronger attractions. This explains why boiling points increase: CH₄ < C₂H₆ < C₃H₈ < C₄H₁₀.

Tags

  • london_dispersion
  • molecular_size
  • intermolecular_forces

Topic

London Dispersion Forces

Card Id

dispersion_015

Difficulty

medium

Image Prompt

Calculate the number of bonding and nonbonding electrons in NH₃.

Step 1: Total valence electrons = N(5) + 3H(1) = 8 electrons. Step 2: NH₃ has 3 N-H bonds = 6 bonding electrons. Step 3: Nonbonding electrons = 8 - 6 = 2 electrons (1 lone pair on N). Answer: 6 bonding, 2 nonbonding electrons.

Tags

  • electron_counting
  • lewis_structures
  • numerical

Topic

Electron Counting

Card Id

electron_016

Difficulty

medium

Image Prompt

Predict the bond type between H and F based on electronegativity difference.

H (2.1) and F (4.0): Δχ = 4.0 - 2.1 = 1.9. Since Δχ > 1.7, this forms an IONIC BOND (in compounds) or highly polar covalent bond (in HF molecule). F is much more electronegative, so electrons spend more time near F atom.

Tags

  • electronegativity
  • bond_polarity
  • numerical

Topic

Bond Polarity

Card Id

electronegativity_017

Difficulty

medium

Image Prompt

A compound has molecular formula C₆H₁₂O₆. What is its empirical formula?

Step 1: Find the ratio C₆:H₁₂:O₆ = 6:12:6. Step 2: Divide by GCD (6): 6÷6 : 12÷6 : 6÷6 = 1:2:1. Step 3: Empirical formula = CH₂O. This represents the simplest whole-number ratio of atoms.

Tags

  • empirical_formula
  • molecular_formula
  • numerical

Topic

Chemical Formulas

Card Id

empirical_018

Difficulty

medium

Image Prompt

Name the compounds FeO and Fe₂O₃ using proper nomenclature.

FeO = Iron(II) oxide. Fe₂O₃ = Iron(III) oxide. Since Fe is a transition metal with variable charges, use Roman numerals. In FeO: Fe has +2 charge. In Fe₂O₃: Fe has +3 charge. Always balance with O²⁻ to find metal charge.

Tags

  • transition_metals
  • roman_numerals
  • nomenclature

Topic

Transition Metal Compounds

Card Id

transition_019

Difficulty

hard

Image Prompt

How does temperature affect the rate of evaporation according to kinetic molecular theory?

Higher temperature increases evaporation rate because: 1) Molecules have higher kinetic energy, 2) More molecules have enough energy to overcome intermolecular forces, 3) More molecules escape from liquid surface to gas phase. KE ∝ Temperature (KMT Postulate 4).

Tags

  • kinetic_theory
  • evaporation
  • temperature
  • conceptual

Topic

Kinetic Molecular Theory

Card Id

kinetic_020

Difficulty

medium

Image Prompt

Tag Distribution

Numerical

6

Conceptual

7

Nomenclature

6

Bonding Theory

5

Formula Writing

4

Periodic Trends

3

Reaction Problems

4

Molecular Geometry

3

Topic Distribution

Kinetic Theory

1

Periodic Trends

3

Chemical Bonding

4

Chemical Formulas

2

Molecular Geometry

3

Chemical Nomenclature

4

Intermolecular Forces

3

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