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UEE Chemistry practice questions

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Question 1Atomic Structure and Periodicity

Which quantum number primarily determines the energy of an electron in a one-electron (hydrogen-like) atom?

  1. Spin quantum number m_s
  2. Magnetic quantum number m_l
  3. Azimuthal quantum number l
  4. Principal quantum number n Correct answer.

Explanation

In a hydrogen-like atom the energy depends only on the principal quantum number n, because the single electron experiences an unscreened nuclear charge. The other quantum numbers affect orbital shape, orientation, and spin but not the energy level in this one-electron case.

Question 2Atomic Structure and Periodicity

What is the maximum number of electrons that can occupy the set of orbitals with n = 3 and l = 2?

  1. 6
  2. 14
  3. 10 Correct answer.
  4. 2

Explanation

For l = 2 the subshell is the d subshell, which contains 2l + 1 = 5 orbitals. Each orbital holds two electrons by the Pauli exclusion principle, giving a maximum of 10 electrons.

Question 3Atomic Structure and Periodicity

Which statement correctly describes the trend in first ionisation energy across Period 3 from sodium to argon?

  1. It remains constant because nuclear charge and shielding rise together
  2. It increases overall, with a dip at aluminium and at sulfur Correct answer.
  3. It decreases steadily because atomic radius increases
  4. It increases steadily with no irregularities

Explanation

First ionisation energy rises across a period as effective nuclear charge increases and radius falls. However, aluminium is lower than magnesium because its electron is removed from a 3p orbital that is higher in energy and better shielded, and sulfur is lower than phosphorus because of electron-pair repulsion in a doubly occupied 3p orbital.

Question 4Atomic Structure and Periodicity

Which ground-state electron configuration is correct for a chromium atom (Z = 24)?

  1. [Ar] 4s2 3d5
  2. [Ar] 4s2 3d4
  3. [Ar] 3d6
  4. [Ar] 4s1 3d5 Correct answer.

Explanation

Chromium is an exception to the aufbau order: a half-filled 3d subshell with a singly occupied 4s orbital gives extra exchange stabilisation. The correct configuration is therefore [Ar] 4s1 3d5 rather than the naively predicted 4s2 3d4.

Question 5Atomic Structure and Periodicity

Which singly charged positive ion has the ground-state electron configuration 1s2 2s2 2p6?

  1. Ne+
  2. Na+ Correct answer.
  3. F-
  4. Mg2+

Explanation

Na+ has lost one electron from sodium (Z = 11), leaving 10 electrons arranged as 1s2 2s2 2p6. It is isoelectronic with neon and carries a single positive charge.

Question 6Atomic Structure and Periodicity

Which element has the highest first ionisation energy?

  1. Boron
  2. Lithium
  3. Carbon
  4. Beryllium Correct answer.

Explanation

Beryllium has a filled 2s subshell, and removing an electron from that stable configuration requires more energy than removing the single 2s electron of lithium or the 2p electron of boron. Carbon's higher nuclear charge does not compensate for the easier removal of its 2p electron in this comparison.

Question 7Atomic Structure and Periodicity

Which property increases from left to right across a period of the periodic table?

  1. Metallic character
  2. Number of neutron-rich isotopes
  3. Electronegativity Correct answer.
  4. Atomic radius

Explanation

Electronegativity increases across a period because effective nuclear charge rises and atomic radius falls, so the nucleus attracts bonding electrons more strongly. Atomic radius and metallic character both decrease in the same direction.

Question 8Atomic Structure and Periodicity

Which of these atoms has the largest atomic radius?

  1. Gallium
  2. Potassium Correct answer.
  3. Calcium
  4. Krypton

Explanation

Across Period 4 the atomic radius decreases as effective nuclear charge increases while electrons enter the same shell. Potassium, being furthest left and having the lowest effective nuclear charge in this set, has the largest radius.

Question 9Atomic Structure and Periodicity

Which statement best explains why the first ionisation energy of fluorine is greater than that of oxygen?

  1. Fluorine has a higher effective nuclear charge and a smaller radius Correct answer.
  2. Oxygen has a filled 2p subshell
  3. Oxygen has more neutrons in its nucleus
  4. Fluorine has a larger atomic radius

Explanation

Fluorine has one more proton than oxygen, so its effective nuclear charge is higher and its radius smaller. The outer electron is therefore held more tightly, raising the first ionisation energy despite the pairing in the 2p subshell.

Question 10Atomic Structure and Periodicity

Which orbital designation corresponds to n = 4 and l = 1?

  1. 4s
  2. 4f
  3. 4p Correct answer.
  4. 4d

Explanation

The letter designating a subshell is fixed by the azimuthal quantum number: l = 0 is s, l = 1 is p, l = 2 is d, and l = 3 is f. Therefore n = 4 with l = 1 is the 4p subshell.

Question 11Atomic Structure and Periodicity

How many unpaired electrons are present in the ground state of a manganese atom (Z = 25)?

  1. 3
  2. 4
  3. 7
  4. 5 Correct answer.

Explanation

Manganese has the configuration [Ar] 4s2 3d5. The five 3d electrons occupy the five d orbitals singly by Hund's rule, and the paired 4s electrons contribute none, giving five unpaired electrons.

Question 12Atomic Structure and Periodicity

Which set of quantum numbers is permissible for an electron in an atom?

  1. n = 3, l = 2, m_l = -2, m_s = +1/2 Correct answer.
  2. n = 1, l = 1, m_l = 0, m_s = +1/2
  3. n = 2, l = 2, m_l = 0, m_s = -1/2
  4. n = 4, l = 3, m_l = -4, m_s = +1/2

Explanation

The rules require l to be less than n and m_l to lie between -l and +l. For n = 3, l = 2, m_l = -2 satisfies both conditions, and m_s = +1/2 is allowed.

Question 13Atomic Structure and Periodicity

Which ion has the ground-state configuration [Ar] 3d5?

  1. Fe3+
  2. Cr3+
  3. Fe2+
  4. Mn2+ Correct answer.

Explanation

Mn2+ is formed by removing the two 4s electrons from manganese ([Ar] 4s2 3d5), leaving [Ar] 3d5. This half-filled d subshell is especially stable.

Question 14Atomic Structure and Periodicity

Which statement about shielding and penetration is correct?

  1. An s electron penetrates the nucleus more than a p electron and experiences a higher effective nuclear charge Correct answer.
  2. Shielding increases the effective nuclear charge felt by an outer electron
  3. A p electron penetrates more than an s electron of the same shell
  4. Penetration has no effect on orbital energy in multi-electron atoms

Explanation

Within a shell, s orbitals have greater electron density close to the nucleus than p orbitals, so they penetrate more and are shielded less. The result is that s electrons feel a higher effective nuclear charge and are lower in energy than p electrons of the same shell.

Question 15Atomic Structure and Periodicity

Which element is a transition metal with a partially filled d subshell in its ground-state atom?

  1. Zinc
  2. Copper Correct answer.
  3. Gallium
  4. Calcium

Explanation

Copper has the configuration [Ar] 4s1 3d10 in its ground state, but it is classified as a transition metal because it forms ions with partly filled d subshells, such as Cu2+ with [Ar] 3d9. Zinc and calcium do not form such ions.

Question 16Atomic Structure and Periodicity

Which factor best explains why the second ionisation energy of magnesium is much larger than its first?

  1. The second electron is removed from a more stable, filled 2p subshell
  2. The nucleus loses a proton after the first ionisation
  3. The second electron is removed from a positively charged ion with a smaller radius Correct answer.
  4. The second electron is removed from a higher-energy 3p orbital

Explanation

After the first electron is removed, the remaining Mg+ ion has a higher effective nuclear charge and a smaller radius, so the second electron is held more tightly. Removing an electron from a positively charged species therefore requires substantially more energy.

Question 17Atomic Structure and Periodicity

Which of these elements has the most negative electron affinity (releases the most energy on gaining an electron)?

  1. Fluorine
  2. Bromine
  3. Chlorine Correct answer.
  4. Iodine

Explanation

Electron affinity becomes more exothermic down the halogen group only partly, but chlorine releases more energy than fluorine because fluorine's very small 2p orbitals cause strong electron-electron repulsion when an extra electron is added. Chlorine therefore has the most negative electron affinity of the halogens.

Question 18Atomic Structure and Periodicity

Which group of elements has a general valence electron configuration of ns2 np5?

  1. Alkali metals
  2. Noble gases
  3. Alkaline earth metals
  4. Halogens Correct answer.

Explanation

Halogens such as fluorine and chlorine have seven valence electrons, arranged as ns2 np5. They are one electron short of a filled shell, which explains their strong tendency to form -1 ions.

Question 19Atomic Structure and Periodicity

Which explanation accounts for the very high third ionisation energy of sodium compared with its first and second?

  1. The third electron is removed from a higher-energy 3p orbital
  2. The third electron is removed from a filled 2p subshell close to the nucleus Correct answer.
  3. The third ionisation energy is lower because the ion is already positive
  4. Sodium forms a stable +3 ion with a noble-gas configuration

Explanation

The first two electrons of sodium come from the 3s orbital, but the third must be removed from the filled 2p subshell of the neon-like core. That electron is much closer to the nucleus and better shielded from the outer region, so a large jump in ionisation energy occurs.

Question 20Atomic Structure and Periodicity

Which observation is consistent with the aufbau principle?

  1. Electrons in an atom all occupy the lowest-energy orbital available
  2. The 4s subshell is filled before the 3d subshell in potassium and calcium Correct answer.
  3. The 3d subshell is always filled before the 4s subshell
  4. Electrons occupy degenerate orbitals in pairs before occupying them singly

Explanation

The aufbau principle states that electrons fill the lowest-energy orbitals first. For potassium and calcium, the 4s orbital lies below the 3d orbitals in energy, so it is occupied first.

Question 21Chemical Bonding and Molecular Geometry

Which statement correctly describes the shape of a methane molecule, CH4, according to VSEPR theory?

  1. Square planar, with bond angles of 90°
  2. Trigonal planar, with bond angles of 120°
  3. Tetrahedral, with bond angles of 109.5° Correct answer.
  4. Trigonal pyramidal, with bond angles of 107°

Explanation

Carbon in CH4 has four bonding pairs and no lone pairs, so the four electron domains adopt a tetrahedral arrangement to minimise repulsion. In a regular tetrahedron the H–C–H angle is 109.5°.

Question 22Chemical Bonding and Molecular Geometry

Which molecule contains a polar covalent bond but has zero net dipole moment overall?

  1. NH3
  2. CO2 Correct answer.
  3. HCl
  4. H2O

Explanation

In CO2 each C=O bond is polar, but the molecule is linear so the two bond dipoles point in opposite directions and cancel exactly. The net dipole moment is therefore zero even though individual bonds are polar.

Question 23Chemical Bonding and Molecular Geometry

Which type of attractive force holds together the particles in a crystal of sodium chloride?

  1. Instantaneous dipole–induced dipole dispersion forces
  2. Electrostatic attraction between oppositely charged ions Correct answer.
  3. Delocalised electrons shared between metal cations
  4. Intermolecular hydrogen bonding between molecules

Explanation

Sodium chloride is an ionic lattice built from Na+ and Cl− ions. The lattice is held together by the strong electrostatic attraction between these oppositely charged ions in all directions.

Question 24Chemical Bonding and Molecular Geometry

How many lone pairs of electrons are present on the central atom of a water molecule, H2O?

  1. 1
  2. 0
  3. 3
  4. 2 Correct answer.

Explanation

Oxygen has six valence electrons; two are used in sigma bonds to hydrogen, leaving four electrons arranged as two lone pairs. These two lone pairs compress the H–O–H angle to about 104.5°.

Question 25Chemical Bonding and Molecular Geometry

Which pair of elements would most readily form an ionic compound when reacted together?

  1. Phosphorus and sulfur
  2. Nitrogen and chlorine
  3. Carbon and oxygen
  4. Potassium and fluorine Correct answer.

Explanation

Ionic bonding occurs between elements of very different electronegativity, typically a reactive metal and a reactive non-metal. Potassium is a group 1 metal and fluorine is the most electronegative element, so electron transfer is strongly favoured.

Question 26Chemical Bonding and Molecular Geometry

Which of the following species has a bond order of 1.5?

  1. O2⁺
  2. O2⁻ (the superoxide ion) Correct answer.
  3. O2
  4. N2

Explanation

Superoxide, O2⁻, has 17 valence electrons. Filling the molecular orbitals gives eight bonding and four antibonding electrons, so the bond order is (8 − 4)/2 = 2, but the extra electron occupies a pi antibonding orbital giving an effective order between 1 and 2, conventionally 1.5.

Question 27Chemical Bonding and Molecular Geometry

Which molecule is predicted to have a bent (V-shaped) geometry around its central atom?

  1. BF3
  2. CO2
  3. SO2 Correct answer.
  4. BeCl2

Explanation

Sulfur in SO2 has three electron domains: two bonding regions and one lone pair. The lone pair occupies a position in a trigonal planar arrangement, leaving an angular or bent shape with an O–S–O angle below 120°.

Question 28Chemical Bonding and Molecular Geometry

Which statement best explains why the boiling point of water is much higher than that of hydrogen sulfide, H2S?

  1. Water molecules form hydrogen bonds, whereas H2S molecules only experience weaker dipole and dispersion forces Correct answer.
  2. Water is a linear molecule, which allows tighter packing in the liquid
  3. H2S is held together by ionic bonds that are easier to break than hydrogen bonds
  4. H2S has a larger molar mass, which always lowers the boiling point

Explanation

Oxygen is highly electronegative and small, so O–H bonds allow strong hydrogen bonding between water molecules. Sulfur is larger and less electronegative, so H2S molecules interact mainly through weaker dipole–dipole and dispersion forces, giving a much lower boiling point.

Question 29Chemical Bonding and Molecular Geometry

What is the hybridisation of the carbon atoms in ethene, C2H4?

  1. sp² Correct answer.
  2. sp³
  3. sp³d
  4. sp

Explanation

Each carbon in ethene is bonded to three other atoms (two hydrogens and one carbon) and has no lone pairs, giving three electron domains. Three domains correspond to sp² hybridisation, leaving one unhybridised p orbital for the pi bond.

Question 30Chemical Bonding and Molecular Geometry

Which species is isoelectronic with the neon atom?

  1. Na⁺
  2. Cl⁻
  3. O²⁻
  4. F⁻ Correct answer.

Explanation

Isoelectronic species have the same total number of electrons. Fluorine has 9 electrons; adding one for the negative charge gives 10, exactly matching neon's 10 electrons.

Question 31Chemical Bonding and Molecular Geometry

Which factor best accounts for the very high melting point of magnesium oxide compared with sodium chloride?

  1. MgO contains covalent bonds while NaCl contains ionic bonds
  2. The 2+ and 2− charges on MgO ions produce much stronger electrostatic attraction than the 1+ and 1− charges in NaCl Correct answer.
  3. MgO molecules are held together by hydrogen bonds
  4. MgO has a much larger lattice spacing than NaCl

Explanation

Lattice energy depends strongly on ionic charge: doubling both charges roughly quadruples the electrostatic attraction. MgO has Mg²⁺ and O²⁻ ions, so its lattice is far stronger and melts at a much higher temperature than the Na⁺Cl⁻ lattice.

Question 32Chemical Bonding and Molecular Geometry

What is the shape of the PCl5 molecule?

  1. Square pyramidal
  2. Tetrahedral
  3. Trigonal bipyramidal Correct answer.
  4. Octahedral

Explanation

Phosphorus in PCl5 has five bonding pairs and no lone pairs, giving five electron domains. Five domains adopt a trigonal bipyramidal arrangement with bond angles of 90° and 120°.

Question 33Chemical Bonding and Molecular Geometry

Which molecule is nonpolar despite containing polar bonds?

  1. CCl4 Correct answer.
  2. H2O
  3. NH3
  4. CHCl3

Explanation

In CCl4 the four C–Cl bond dipoles point to the corners of a regular tetrahedron and cancel exactly, so the molecule has no net dipole. Individual bonds are polar, but the overall molecule is nonpolar.

Question 34Chemical Bonding and Molecular Geometry

Which of the following correctly lists the bond angles in order of decreasing size for BF3, CH4 and NH3?

  1. BF3 > NH3 > CH4
  2. BF3 > CH4 > NH3 Correct answer.
  3. NH3 > CH4 > BF3
  4. CH4 > BF3 > NH3

Explanation

BF3 is trigonal planar with 120° angles, CH4 is tetrahedral with 109.5°, and NH3 has a lone pair that compresses its angle to about 107°. So the order is BF3 > CH4 > NH3.

Question 35Chemical Bonding and Molecular Geometry

Which of the following species has a bond angle closest to 180°?

  1. CO2 Correct answer.
  2. SO2
  3. H2O
  4. NH3

Explanation

CO2 has two bonding domains and no lone pairs on carbon, so it is linear with an O–C–O angle of 180°. The other species have lone pairs that reduce their bond angles well below 180°.

Question 36Chemical Bonding and Molecular Geometry

Which statement about the bonding in a metal such as copper is correct?

  1. Positive metal ions are held in a lattice by attraction to delocalised electrons Correct answer.
  2. Metal atoms transfer electrons to form alternating cations and anions
  3. Metal ions are held together by hydrogen bonds between nuclei
  4. Each metal atom forms a single covalent bond to one neighbour

Explanation

In the metallic model, valence electrons are delocalised over the whole structure and are attracted to the fixed positive ions. This sea of electrons accounts for electrical conductivity and malleability.

Question 37Chemical Bonding and Molecular Geometry

Which of the following correctly orders the species O2, O2⁺ and O2⁻ by bond length, from shortest to longest?

  1. O2 < O2⁺ < O2⁻
  2. O2 < O2⁻ < O2⁺
  3. O2⁺ < O2 < O2⁻ Correct answer.
  4. O2⁻ < O2 < O2⁺

Explanation

Bond length decreases as bond order increases. Removing an antibonding electron in O2⁺ raises the bond order above 2, while adding an electron in O2⁻ lowers it below 2, so O2⁺ has the shortest bond and O2⁻ the longest.

Question 38Chemical Bonding and Molecular Geometry

A molecule XY4 has a square planar shape rather than tetrahedral. Which explanation is consistent with VSEPR theory?

  1. The central atom has four bonding pairs and no lone pairs
  2. The central atom has four bonding pairs and two lone pairs, giving six electron domains Correct answer.
  3. The central atom has four bonding pairs and three lone pairs
  4. The central atom has four bonding pairs and one lone pair

Explanation

Square planar geometry arises when a central atom has six electron domains arranged octahedrally, with two lone pairs occupying opposite (axial) positions. The four remaining bonds then lie in a plane at 90° to each other.

Question 39Chemical Bonding and Molecular Geometry

Which statement best explains why the first ionisation energy of nitrogen is greater than that of oxygen?

  1. Oxygen has a larger nuclear charge, which always lowers ionisation energy
  2. Oxygen has more shielding electrons than nitrogen
  3. Nitrogen has a half-filled 2p subshell, which is more stable than the partially filled 2p subshell of oxygen Correct answer.
  4. Nitrogen has a smaller atomic radius than oxygen

Explanation

Nitrogen's 2p subshell is exactly half-filled (one electron in each of three p orbitals), a configuration of extra stability. Removing an electron from oxygen is easier because oxygen has a paired electron in one p orbital, so the general increase across the period is reversed at this point.

Question 40Chemical Bonding and Molecular Geometry

Which of the following species is expected to be paramagnetic?

  1. NO Correct answer.
  2. CO
  3. H2O
  4. N2

Explanation

Nitric oxide, NO, has an odd number of valence electrons (11), so one electron remains unpaired in a pi antibonding orbital. A species with an unpaired electron is attracted into a magnetic field and is paramagnetic.

Question 41Stoichiometry and Chemical Calculations

How many moles of oxygen atoms are present in 0.50 mol of Al2(SO4)3?

  1. 12 mol
  2. 6.0 mol Correct answer.
  3. 0.50 mol
  4. 1.5 mol

Explanation

Each formula unit of Al2(SO4)3 contains three sulfate groups, and each sulfate group holds four oxygen atoms, giving 12 oxygen atoms per formula unit. Multiplying 0.50 mol by 12 mol O per mol of compound yields 6.0 mol of oxygen atoms.

Question 42Stoichiometry and Chemical Calculations

What is the molar mass of calcium nitrate, Ca(NO3)2, in g/mol? (Ar: Ca = 40.1, N = 14.0, O = 16.0)

  1. 164.1 Correct answer.
  2. 132.1
  3. 102.1
  4. 150.1

Explanation

Calcium nitrate contains one calcium, two nitrogen and six oxygen atoms per formula unit. Summing 40.1 + (2 x 14.0) + (6 x 16.0) gives 40.1 + 28.0 + 96.0 = 164.1 g/mol.

Question 43Stoichiometry and Chemical Calculations

In the balanced equation 2C2H6 + 7O2 -> 4CO2 + 6H2O, what is the mole ratio of O2 consumed to CO2 produced?

  1. 7:4 Correct answer.
  2. 4:7
  3. 2:7
  4. 7:6

Explanation

The coefficients in a balanced equation give the molar ratio directly. Oxygen has a coefficient of 7 and carbon dioxide has a coefficient of 4, so the ratio of O2 to CO2 is 7:4.

Question 44Stoichiometry and Chemical Calculations

What volume does 0.25 mol of an ideal gas occupy at standard temperature and pressure (273 K, 1 atm), taking the molar volume as 22.4 L/mol?

  1. 11.2 L
  2. 44.8 L
  3. 22.4 L
  4. 5.6 L Correct answer.

Explanation

At STP one mole of an ideal gas occupies about 22.4 L, so volume equals moles multiplied by molar volume. Multiplying 0.25 mol by 22.4 L/mol gives 5.6 L.

Question 45Stoichiometry and Chemical Calculations

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula? (Ar: C = 12, H = 1, O = 16)

  1. C2H2O
  2. CH2O Correct answer.
  3. C2H4O
  4. CHO

Explanation

Dividing each percentage by its atomic mass gives 40.0/12 = 3.33 mol C, 6.7/1 = 6.7 mol H and 53.3/16 = 3.33 mol O. Dividing by the smallest value 3.33 gives the ratio 1:2:1, so the empirical formula is CH2O.

Question 46Stoichiometry and Chemical Calculations

How many grams of NaCl are needed to prepare 250 mL of a 0.20 mol/L solution? (Mr of NaCl = 58.5)

  1. 1.46 g
  2. 11.7 g
  3. 5.85 g
  4. 2.93 g Correct answer.

Explanation

Moles equal concentration multiplied by volume in litres: 0.20 mol/L x 0.250 L = 0.050 mol. Multiplying by the molar mass 58.5 g/mol gives 2.93 g.

Question 47Stoichiometry and Chemical Calculations

In the reaction N2 + 3H2 -> 2NH3, 28 g of N2 reacts with excess H2. What mass of NH3 is produced? (Mr: N2 = 28, NH3 = 17)

  1. 17 g
  2. 34 g Correct answer.
  3. 51 g
  4. 28 g

Explanation

28 g of N2 is 1.0 mol, and the equation shows 1 mol N2 yields 2 mol NH3. Two moles of ammonia have a mass of 2 x 17 = 34 g.

Question 48Stoichiometry and Chemical Calculations

A 1.00 g sample of an impure carbonate is found to contain 0.80 g of pure CaCO3. What is the percentage purity of the sample?

  1. 80% Correct answer.
  2. 0.80%
  3. 125%
  4. 20%

Explanation

Percentage purity is the mass of pure substance divided by the total sample mass, multiplied by 100. Here 0.80 g / 1.00 g x 100 = 80%.

Question 49Stoichiometry and Chemical Calculations

What is the percentage by mass of nitrogen in ammonium sulfate, (NH4)2SO4? (Mr = 132; Ar: N = 14)

  1. 10.6%
  2. 28.0%
  3. 14.0%
  4. 21.2% Correct answer.

Explanation

Ammonium sulfate contains two nitrogen atoms, giving 28 g of nitrogen per 132 g of compound. Dividing 28 by 132 and multiplying by 100 gives approximately 21.2%.

Question 50Stoichiometry and Chemical Calculations

25.0 mL of 0.100 mol/L NaOH exactly neutralises 20.0 mL of HCl. What is the concentration of the HCl?

  1. 0.080 mol/L
  2. 0.125 mol/L Correct answer.
  3. 0.200 mol/L
  4. 0.100 mol/L

Explanation

NaOH and HCl react in a 1:1 ratio, so moles of acid equal moles of base. Moles of NaOH = 0.100 x 0.0250 = 0.00250 mol, and dividing by 0.0200 L gives 0.125 mol/L.