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Tamil Nadu HSE First Year Chemistry study materials
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Tamil Nadu HSE First Year Chemistry practice questions

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Question 1Basic Concepts of Chemistry and Chemical Calculations

The relative atomic mass of an element expressed in grams is called its gram atomic mass. What is the gram atomic mass of oxygen, given that its relative atomic mass is 16?

  1. 8 g
  2. 32 g
  3. 16 g Correct answer.
  4. 1.6 g

Explanation

The gram atomic mass is numerically equal to the relative atomic mass but expressed in grams, so for oxygen it is 16 g. This quantity contains one mole of oxygen atoms, that is 6.022 × 10^23 atoms. Multiplying or dividing the value is incorrect because the numerical value must stay the same.

Question 2Basic Concepts of Chemistry and Chemical Calculations

How many significant figures are present in the measured value 0.004560 m?

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

Explanation

Leading zeros that only fix the decimal point are not significant, so the zeros before 4560 are ignored. The digits 4, 5, 6 and the trailing zero after the 6 are significant, giving four significant figures. The trailing zero counts because it follows a non-zero digit in a decimal number.

Question 3Basic Concepts of Chemistry and Chemical Calculations

What is the molar mass of glucose, C6H12O6, using atomic masses C = 12, H = 1 and O = 16?

  1. 200 g mol^-1
  2. 180 g mol^-1 Correct answer.
  3. 150 g mol^-1
  4. 172 g mol^-1

Explanation

Sum the atomic masses: carbon contributes 6 × 12 = 72, hydrogen 12 × 1 = 12 and oxygen 6 × 16 = 96. Adding gives 72 + 12 + 96 = 180 g mol^-1. The molar mass is the mass of one mole of the compound.

Question 4Basic Concepts of Chemistry and Chemical Calculations

Avogadro's law states that equal volumes of all gases under the same conditions of temperature and pressure contain equal numbers of molecules. Which quantity therefore has the same value for 22.4 L of any gas at STP?

  1. Mass of the gas
  2. Density of the gas
  3. Number of moles Correct answer.
  4. Kinetic energy per molecule

Explanation

At STP, 22.4 L of any ideal gas contains one mole, that is 6.022 × 10^23 molecules, regardless of identity. Mass and density depend on the molar mass of the particular gas, so they differ from gas to gas. The number of moles is fixed by volume alone under the stated conditions.

Question 5Basic Concepts of Chemistry and Chemical Calculations

Which of the following is a homogeneous mixture?

  1. Air Correct answer.
  2. Sand and iron filings
  3. A suspension of chalk in water
  4. Oil and water

Explanation

A homogeneous mixture has a uniform composition throughout, and air is a uniform blend of nitrogen, oxygen and other gases. The other examples contain visibly separate phases or settle on standing. Homogeneity is judged by uniformity at the molecular or visible level.

Question 6Basic Concepts of Chemistry and Chemical Calculations

A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula? (C = 12, H = 1, O = 16)

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

Explanation

Divide each percentage by the atomic mass: C gives 40/12 = 3.33, H gives 6.7/1 = 6.7 and O gives 53.3/16 = 3.33. Dividing by the smallest value 3.33 gives the ratio 1 : 2 : 1, so the empirical formula is CH2O. The empirical formula shows the simplest whole-number ratio of atoms.

Question 7Basic Concepts of Chemistry and Chemical Calculations

How many moles of carbon dioxide are present in 11 g of CO2? (C = 12, O = 16)

  1. 1.00 mol
  2. 0.25 mol Correct answer.
  3. 0.50 mol
  4. 2.00 mol

Explanation

The molar mass of CO2 is 12 + 32 = 44 g mol^-1. The number of moles is mass divided by molar mass, so 11/44 = 0.25 mol. This uses the relation n = m/M central to chemical calculations.

Question 8Basic Concepts of Chemistry and Chemical Calculations

What volume of oxygen at STP is required to completely burn 4 g of methane, CH4? (CH4 + 2O2 → CO2 + 2H2O; molar volume at STP = 22.4 L)

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

Explanation

Moles of methane = 4/16 = 0.25 mol. The equation shows that one mole of CH4 needs two moles of O2, so 0.25 mol requires 0.50 mol of oxygen. At STP this occupies 0.50 × 22.4 = 11.2 L.

Question 9Basic Concepts of Chemistry and Chemical Calculations

The molecular mass of a compound is 180 and its empirical formula is CH2O. What is its molecular formula?

  1. C2H4O2
  2. C5H10O5
  3. C6H12O6 Correct answer.
  4. C3H6O3

Explanation

The empirical formula mass of CH2O is 12 + 2 + 16 = 30. Dividing the molecular mass by this value gives 180/30 = 6, so every subscript is multiplied by six, giving C6H12O6. The molecular formula is a whole-number multiple of the empirical formula.

Question 10Basic Concepts of Chemistry and Chemical Calculations

A sample of a compound contains 0.2 mol of atoms of element X and 0.3 mol of atoms of element Y. What is the simplest whole-number ratio of X to Y in the compound?

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

Explanation

The mole ratio 0.2 : 0.3 is simplified by dividing both terms by 0.1, giving 2 : 3. Empirical formulas always use the smallest whole-number ratio. Reversing the order would give the ratio of Y to X, not X to Y.

Question 11Basic Concepts of Chemistry and Chemical Calculations

What is the mass percentage of nitrogen in ammonium sulphate, (NH4)2SO4? (N = 14, H = 1, S = 32, O = 16)

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

Explanation

The molar mass of (NH4)2SO4 is 2(14) + 8(1) + 32 + 4(16) = 132 g mol^-1, and it contains two nitrogen atoms, that is 28 g of nitrogen. The mass percentage is (28/132) × 100 ≈ 21.2%. Mass percentage compares the mass of the element in one mole with the total molar mass.

Question 12Basic Concepts of Chemistry and Chemical Calculations

How many molecules are present in 0.5 mol of nitrogen gas? (Avogadro constant = 6.022 × 10^23 mol^-1)

  1. 1.204 × 10^24
  2. 6.022 × 10^23
  3. 3.011 × 10^22
  4. 3.011 × 10^23 Correct answer.

Explanation

One mole contains 6.022 × 10^23 molecules, so 0.5 mol contains half this number, that is 3.011 × 10^23 molecules. The calculation multiplies the number of moles by the Avogadro constant. The result expresses the count of N2 molecules, not atoms.

Question 13Basic Concepts of Chemistry and Chemical Calculations

A gaseous hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass, and its molecular mass is 42. What is its molecular formula? (C = 12, H = 1)

  1. C3H6 Correct answer.
  2. CH2
  3. C2H4
  4. C3H8

Explanation

Dividing by atomic masses gives C: 85.7/12 = 7.14 and H: 14.3/1 = 14.3, a ratio of 1 : 2, so the empirical formula is CH2 with mass 14. Dividing 42 by 14 gives 3, so the molecular formula is C3H6. The molecular formula must also be consistent with the given molecular mass.

Question 14Basic Concepts of Chemistry and Chemical Calculations

In the reaction N2 + 3H2 → 2NH3, 28 g of nitrogen reacts with 6 g of hydrogen. Which reactant is the limiting reagent, and how many moles of ammonia are formed? (N = 14, H = 1)

  1. N2 limits; 4 mol NH3
  2. H2 limits; 2 mol NH3
  3. N2 limits; 2 mol NH3 Correct answer.
  4. H2 limits; 4 mol NH3

Explanation

Moles of N2 = 28/28 = 1 and moles of H2 = 6/2 = 3, exactly the 1 : 3 stoichiometric ratio, so neither is in excess and both are consumed fully. One mole of N2 produces two moles of NH3 according to the equation. Hence 2 mol of ammonia forms.

Question 15Basic Concepts of Chemistry and Chemical Calculations

A 10.0 g sample of an impure carbonate is found to contain 8.0 g of pure carbonate. The sample reacts to release gas, and the percentage purity affects the yield. Which statement correctly describes the relationship between percentage purity and the moles of product obtained?

  1. The number of moles of product depends on the mass of pure substance, so percentage purity scales the expected yield proportionally. Correct answer.
  2. Percentage purity affects only the volume of product gas at STP but not the number of moles.
  3. Percentage purity changes the molar mass of the compound and therefore the moles of product.
  4. The number of moles of product depends only on the total mass of the impure sample, regardless of purity.

Explanation

Percentage purity = (mass of pure substance / total mass) × 100, so here it is 80%. Only the pure carbonate reacts, so the moles of product are calculated from 8.0 g, not 10.0 g, meaning the yield scales with purity. Impurities neither react nor change the molar mass of the pure compound.

Question 16Basic Concepts of Chemistry and Chemical Calculations

Two samples of an ideal gas are held at the same temperature and pressure. Sample A occupies 2.24 L and sample B occupies 4.48 L. If sample B contains 0.20 mol, how many moles does sample A contain?

  1. 0.20 mol
  2. 0.40 mol
  3. 0.10 mol Correct answer.
  4. 0.05 mol

Explanation

By Avogadro's law, volume is directly proportional to the number of moles at constant temperature and pressure. Sample A has half the volume of sample B, so it contains half the moles, that is 0.10 mol. The ratio of volumes equals the ratio of moles.

Question 17Quantum Mechanical Model of Atom

According to the quantum mechanical model of the atom, what does the square of the wave function, |ψ|², represent at a given point in space?

  1. The magnetic moment of the nucleus at that point
  2. The probability density of finding the electron there Correct answer.
  3. The exact speed of the electron at that point
  4. The total energy of the electron in that orbital

Explanation

In the quantum mechanical model the wave function ψ itself has no direct physical meaning, but its square |ψ|² gives the probability density of locating the electron at a particular region around the nucleus. This replaces the fixed Bohr orbit with a region of probable electron presence.

Question 18Quantum Mechanical Model of Atom

Which quantum number determines the principal energy shell in which an electron is found and takes the values 1, 2, 3, ...?

  1. Spin quantum number s
  2. Azimuthal quantum number l
  3. Principal quantum number n Correct answer.
  4. Magnetic quantum number m

Explanation

The principal quantum number n fixes the main shell (K, L, M, ...) and hence the overall size and energy of the orbital. The other quantum numbers describe subshell shape, orientation and spin rather than the shell itself.

Question 19Quantum Mechanical Model of Atom

For an electron in a 3d orbital, what is the value of the azimuthal quantum number l?

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

Explanation

The azimuthal quantum number is fixed by the subshell letter: s corresponds to l = 0, p to l = 1, d to l = 2 and f to l = 3. A 3d electron therefore has l = 2, while n = 3 gives the shell.

Question 20Quantum Mechanical Model of Atom

What is the maximum number of electrons that can be accommodated in the subshell for which l = 2?

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

Explanation

For a given l the number of orbitals is 2l + 1, so l = 2 gives five orbitals, and each orbital holds two electrons by Pauli's exclusion principle. The d subshell therefore holds a maximum of ten electrons.

Question 21Quantum Mechanical Model of Atom

Which statement correctly describes the shape of an s orbital in the quantum mechanical model?

  1. It is double dumbbell-shaped in most cases
  2. It is a flat circular disc in one plane
  3. It is dumbbell-shaped with two lobes
  4. It is spherically symmetrical about the nucleus Correct answer.

Explanation

An s orbital has l = 0, so its probability distribution depends only on distance from the nucleus and is the same in every direction, giving a spherical shape. The dumbbell and double-dumbbell shapes belong to p and d orbitals respectively.

Question 22Quantum Mechanical Model of Atom

The quantum mechanical model differs from the Bohr model chiefly because it describes the electron in terms of which concept?

  1. A definite circular path of fixed radius
  2. A probability distribution obtained from the wave function Correct answer.
  3. A particle moving only in straight lines
  4. A particle at rest at the nucleus

Explanation

The quantum mechanical model treats the electron as a wave described by ψ, and |ψ|² gives the probability of finding it in a region, so no definite path exists. Bohr's model, by contrast, assigned the electron fixed circular orbits of quantised angular momentum.

Question 23Quantum Mechanical Model of Atom

Which set of quantum numbers is not permitted for an electron in an atom?

  1. n = 1, l = 0, m = 0
  2. n = 2, l = 2, m = 0 Correct answer.
  3. n = 3, l = 2, m = 0
  4. n = 2, l = 1, m = -1

Explanation

The azimuthal quantum number can only range from 0 up to n − 1, so for n = 2 the permitted values of l are 0 and 1 only. The set n = 2, l = 2 is therefore forbidden, while the other sets obey the allowed ranges.

Question 24Quantum Mechanical Model of Atom

How many orbitals in total are present in the third shell (n = 3)?

  1. 18
  2. 3
  3. 9 Correct answer.
  4. 5

Explanation

The third shell contains the 3s, 3p and 3d subshells, giving 1 + 3 + 5 = 9 orbitals. Equivalently, the number of orbitals in a shell is n², which for n = 3 gives 9.

Question 25Quantum Mechanical Model of Atom

Which statement about the energies of orbitals in a multi-electron atom is correct?

  1. The 4s orbital always lies above the 3d orbital in energy
  2. All orbitals of a given shell always have identical energy
  3. Orbitals of lower (n + l) value are generally filled first Correct answer.
  4. Orbital energy depends only on the principal quantum number

Explanation

In multi-electron atoms the energy of an orbital depends on both n and l, and the aufbau order follows the (n + l) rule, with the orbital of lower (n + l) value filled first. When two orbitals have equal (n + l), the one with lower n is filled first.

Question 26Quantum Mechanical Model of Atom

What does the Pauli exclusion principle state about the electrons in an atom?

  1. Electrons always occupy the orbital of highest energy first
  2. No two electrons can have the same set of all four quantum numbers Correct answer.
  3. Electrons in degenerate orbitals remain unpaired as far as possible
  4. An orbital can hold a maximum of three electrons

Explanation

The Pauli exclusion principle requires that the four quantum numbers of any two electrons in an atom cannot be identical, which limits each orbital to two electrons of opposite spin. This restriction underlies the electron capacity of 2n² for a shell.

Question 27Quantum Mechanical Model of Atom

Applying Hund's rule, how are the three electrons in a nitrogen atom's 2p subshell arranged?

  1. Two electrons in one 2p orbital with opposite spins and none elsewhere
  2. Two electrons paired in one 2p orbital and one in another
  3. One electron in each of the three 2p orbitals, all with parallel spins Correct answer.
  4. All three electrons paired in a single 2p orbital

Explanation

Hund's rule states that degenerate orbitals are first occupied singly with parallel spins to minimise electron–electron repulsion. Nitrogen's three 2p electrons therefore occupy the three separate p orbitals unpaired.

Question 28Quantum Mechanical Model of Atom

The wave function of an electron has a node. What is true at the position of a node?

  1. The orbital has its maximum electron density
  2. The electron has zero potential energy
  3. The probability density of finding the electron is zero Correct answer.
  4. The electron moves with maximum velocity

Explanation

A node is a point or surface where the wave function ψ passes through zero, so |ψ|² and hence the probability density of finding the electron are also zero there. The number of nodes increases with the energy of the orbital.

Question 29Quantum Mechanical Model of Atom

What is the value of the azimuthal quantum number for the 4f subshell, and how many orbitals does that subshell contain?

  1. l = 3, with 7 orbitals Correct answer.
  2. l = 3, with 14 orbitals
  3. l = 4, with 9 orbitals
  4. l = 2, with 5 orbitals

Explanation

The letter f corresponds to l = 3, and the number of orbitals is 2l + 1 = 7. These seven f orbitals can together accommodate a maximum of fourteen electrons.

Question 30Quantum Mechanical Model of Atom

An electron is described by n = 3 and l = 1. Which of the following correctly identifies the subshell and one permitted value of its magnetic quantum number?

  1. 3s, with m = +1
  2. 3d, with m = +2
  3. 3p, with m = +2
  4. 3p, with m = +1 Correct answer.

Explanation

With n = 3 and l = 1 the subshell is 3p, and the magnetic quantum number may take the values −1, 0 or +1, so m = +1 is permitted. Values of m outside the range −l to +l are not allowed.

Question 31Quantum Mechanical Model of Atom

The ground-state electron configuration of chromium (Z = 24) is best written as which of the following?

  1. [Ar] 3d³ 4s³
  2. [Ar] 3d⁴ 4s²
  3. [Ar] 3d⁵ 4s¹ Correct answer.
  4. [Ar] 3d⁶ 4s⁰

Explanation

A half-filled d subshell with a single 4s electron gives extra stability from exchange energy and a more symmetrical distribution, so chromium adopts [Ar] 3d⁵ 4s¹ rather than the simple aufbau result. This is a recognised exception to the straightforward filling order.

Question 32Quantum Mechanical Model of Atom

Two orbitals have the same value of (n + l). According to the rules for filling orbitals in multi-electron atoms, which orbital receives electrons first?

  1. Both receive electrons simultaneously in equal amounts
  2. The one with the lower value of n Correct answer.
  3. The one with the higher value of n
  4. The one with the greater value of l

Explanation

When two orbitals have equal (n + l) values, the orbital with the smaller principal quantum number n has the lower energy and is filled first. For example, 3d (n + l = 5) is filled after 4s (n + l = 4) but before 4p (n + l = 5) because 3d has the lower n.

Question 33Periodic Classification of Elements

In the modern periodic law as stated by Moseley, the properties of elements are periodic functions of their:

  1. number of neutrons
  2. atomic number Correct answer.
  3. valency
  4. atomic mass

Explanation

Moseley's study of X-ray spectra showed that the fundamental property ordering the elements is the nuclear charge, that is, the atomic number. Arranging elements by atomic number removes the anomalies of Mendeleev's mass-based table, such as the misplaced pairs Ar–K and Te–I.

Question 34Periodic Classification of Elements

The elements of the third period of the periodic table contain electrons in which set of shells?

  1. L and M shells only
  2. K and L shells only
  3. K, L and M shells Correct answer.
  4. K, L, M and N shells

Explanation

The period number equals the number of shells occupied in the ground-state atom. A third-period element such as sodium or chlorine fills the K and L shells and places its differentiating electron in the M shell, so three shells are in use.

Question 35Periodic Classification of Elements

Which of the following is the correct general electronic configuration of the halogens?

  1. ns² np⁴
  2. ns² np⁶
  3. ns² np⁵ Correct answer.
  4. ns² np³

Explanation

Halogens belong to group 17 and have seven electrons in their outermost shell, one short of the stable octet. Their general outer configuration is therefore ns² np⁵, as in fluorine (2s² 2p⁵) and chlorine (3s² 3p⁵).

Question 36Periodic Classification of Elements

Across a period from left to right, the atomic radius of the elements generally decreases. The best explanation is that:

  1. the mass of the nucleus decreases steadily
  2. the nuclear charge increases while electrons enter the same shell Correct answer.
  3. the number of shells increases by one at each step
  4. the shielding by inner electrons increases sharply

Explanation

Moving across a period, each element adds one proton and one electron, but the extra electron enters the same valence shell. The growing nuclear charge pulls this shell inward while shielding stays nearly constant, so the radius contracts.

Question 37Periodic Classification of Elements

The IUPAC-recommended long form of the periodic table places the lanthanoids and actinoids:

  1. in the main body between groups 2 and 13
  2. inside group 3 as the only members
  3. at the bottom of group 18
  4. in a separate block below the main table Correct answer.

Explanation

In the long form, the f-block elements are shown separately at the foot of the table so that the main body stays compact. They are understood to belong between groups 2 and 3, filling the 4f and 5f subshells, but they are not printed inside the main grid.

Question 38Periodic Classification of Elements

Which of the following orders of first ionisation enthalpy is correct?

  1. Na > Al > Mg
  2. Al > Mg > Na
  3. Na > Mg > Al
  4. Mg > Al > Na Correct answer.

Explanation

Ionisation enthalpy generally rises across a period, so sodium is lowest and magnesium higher. Aluminium falls slightly below magnesium because removing its 3p electron leaves the stable 3s² pair, so the order is Mg > Al > Na.

Question 39Periodic Classification of Elements

The electron gain enthalpy of fluorine is less negative than that of chlorine mainly because:

  1. fluorine has a larger atomic radius than chlorine
  2. the small size of the fluorine atom causes strong electron–electron repulsion in the compact 2p subshell Correct answer.
  3. chlorine has a higher nuclear charge than fluorine
  4. fluorine has a stable half-filled p subshell

Explanation

In the very small fluorine atom the incoming electron must enter a compact 2p orbital already crowded with electrons, so repulsion offsets much of the energy released. Chlorine's larger 3p orbital accommodates the extra electron with less repulsion, making its electron gain enthalpy more negative.

Question 40Periodic Classification of Elements

Which property increases steadily on moving down group 1 from lithium to caesium?

  1. density
  2. first ionisation enthalpy
  3. metallic character Correct answer.
  4. electronegativity

Explanation

Down group 1 the valence electron lies farther from the nucleus and is more shielded, so it is lost more readily and the elements become more metallic. Caesium, at the foot of the group, is the most reactive and most metallic alkali metal.

Question 41Periodic Classification of Elements

An element has the ground-state electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵. To which group and period does it belong?

  1. group 7, period 4
  2. group 18, period 4
  3. group 17, period 4 Correct answer.
  4. group 15, period 4

Explanation

Four occupied shells place the element in period 4, and the seven valence electrons (4s² 4p⁵) identify a halogen in group 17. The element is bromine, which lies below chlorine in the halogen family.

Question 42Periodic Classification of Elements

The correct order of increasing electronegativity among the elements O, F, N and C is:

  1. O < F < C < N
  2. N < C < O < F
  3. F < O < N < C
  4. C < N < O < F Correct answer.

Explanation

Electronegativity increases across a period as effective nuclear charge grows and the radius shrinks. These four elements lie in period 2 in the order C, N, O, F, so the trend rises in exactly that sequence.

Question 43Periodic Classification of Elements

Which pair of elements shows an anomaly in Mendeleev's mass-based arrangement because the element of higher atomic mass was placed first?

  1. Carbon and silicon
  2. Fluorine and chlorine
  3. Lithium and sodium
  4. Argon and potassium Correct answer.

Explanation

Argon has a higher atomic mass than potassium, yet argon is a noble gas and must precede the alkali metal potassium. Mendeleev placed them by chemical similarity rather than mass, an anomaly later explained when the table was based on atomic number.

Question 44Periodic Classification of Elements

The ionic radius of Na⁺ is smaller than that of the Na atom, while the ionic radius of Cl⁻ is larger than that of the Cl atom. This behaviour is best explained by the fact that:

  1. cations lose a shell and anions gain electron–electron repulsion with unchanged nuclear charge Correct answer.
  2. the atomic radius is unaffected by gain or loss of electrons
  3. both ions lose one shell during ionisation
  4. anions have a higher nuclear charge after gaining an electron

Explanation

A sodium atom loses its single 3s electron, leaving the neon-like Na⁺ with one shell fewer, so the ion contracts. A chlorine atom gains an electron into its existing valence shell, increasing electron–electron repulsion while the nuclear charge stays the same, so the cloud expands.

Question 45Periodic Classification of Elements

Which of the following oxides is amphoteric in nature?

  1. Al₂O₃ Correct answer.
  2. P₄O₁₀
  3. Na₂O
  4. SO₃

Explanation

Aluminium oxide reacts with acids to give aluminium salts and with alkalis to give aluminates, so it shows both acidic and basic behaviour. This amphoteric character is typical of oxides near the metal–non-metal diagonal in the p-block.

Question 46Periodic Classification of Elements

In period 3, the oxides change from basic to acidic across the series. Which sequence correctly reflects this change?

  1. SO₃, Na₂O, Al₂O₃
  2. Al₂O₃, Na₂O, SO₃
  3. SO₃, Al₂O₃, Na₂O
  4. Na₂O, Al₂O₃, SO₃ Correct answer.

Explanation

Sodium on the left is a metal and gives a basic oxide, aluminium in the middle gives an amphoteric oxide, and sulphur on the right is a non-metal giving an acidic oxide. The sequence Na₂O, Al₂O₃, SO₃ therefore follows the basic to acidic progression.

Question 47Periodic Classification of Elements

Mendeleev left gaps in his periodic table for elements not yet discovered. The element he named eka-silicon was later identified as:

  1. germanium Correct answer.
  2. scandium
  3. aluminium
  4. gallium

Explanation

Eka-silicon was the predicted element below silicon in group 14, and germanium was discovered in 1886 with properties close to Mendeleev's forecast. The close match between predicted and observed properties was strong evidence for the periodic law.

Question 48Periodic Classification of Elements

The second ionisation enthalpy of an element is always greater than its first ionisation enthalpy. The most appropriate reason is that the second electron is removed from:

  1. an atom with a larger radius than the neutral atom
  2. a shell closer to the nucleus that has more shielding electrons
  3. a species that already carries a positive charge, so more energy is needed to overcome the extra nuclear attraction Correct answer.
  4. a subshell of higher energy than the one involved in the first ionisation

Explanation

After the first electron is lost, the remaining species is a cation with the same nuclear charge but one fewer electron, so each remaining electron is held more firmly. Removing a second electron therefore requires more energy than removing the first from the neutral atom.

Question 49Chemical Bonding and Molecular Structure

Which of the following molecules has a linear shape according to VSEPR theory?

  1. NH3
  2. BeCl2 Correct answer.
  3. SO2
  4. H2O

Explanation

Beryllium in BeCl2 is bonded to only two chlorine atoms and carries no lone pair on the central atom, so the two electron domains arrange themselves at 180 degrees to minimise repulsion, giving a linear molecule.

Question 50Chemical Bonding and Molecular Structure

The number of sigma and pi bonds present in a molecule of ethyne (C2H2) is

  1. 3 sigma and 2 pi Correct answer.
  2. 2 sigma and 3 pi
  3. 4 sigma and 1 pi
  4. 5 sigma and 0 pi

Explanation

Each carbon in ethyne forms one sigma bond to hydrogen and one sigma bond to the other carbon, giving three sigma bonds in total, while the two carbons share two additional electron pairs as two pi bonds.