All Free Chemistry MCQs with Answers

401. Chromium and copper have anomalous electronic configurations because

• A. their nuclei are unstable
• B. a half filled or completely filled d subshell is more stable
• C. they lose electrons easily
• D. the 4s orbital does not exist for them

Explanation: Promoting one 4s electron gives chromium a 3d5 4s1 arrangement and copper a 3d10 4s1 arrangement, both of which gain stability from the symmetry and reduced repulsion of a half filled or full subshell. These two exceptions are examined far more often than the rule itself. The energy gap between 4s and 3d is small enough for the swap to be worthwhile.

402. The energy required to remove the most loosely bound electron from a gaseous atom is called

• A. electron affinity
• B. ionisation energy
• C. electronegativity
• D. lattice energy

Explanation: Ionisation energy is always endothermic because work must be done against the nuclear attraction, and it rises steeply for each successive electron removed. Electron affinity is the energy change when an electron is added instead, and electronegativity describes the pull on shared electrons within a bond. All three are distinct and are regularly confused.

403. An ionic bond is formed by

• A. the sharing of a pair of electrons between two atoms
• B. the complete transfer of electrons from a metal to a non metal
• C. the overlap of two half filled orbitals
• D. the delocalisation of electrons over a lattice of cations

Explanation: Transfer produces oppositely charged ions held by electrostatic attraction, which is why ionic compounds have high melting points and conduct when molten. Sharing gives a covalent bond and delocalisation gives metallic bonding. Ionic character increases as the electronegativity difference between the two atoms widens.

404. A coordinate or dative covalent bond differs from an ordinary covalent bond in that

• A. both shared electrons come from the same atom
• B. no electrons are shared
• C. it is much weaker than a normal covalent bond
• D. it involves the transfer of electrons

Explanation: One atom supplies the whole electron pair while the other provides an empty orbital, as when ammonia donates its lone pair to a hydrogen ion to form the ammonium ion. Once formed, the bond is indistinguishable from any other covalent bond, which is why all four N-H bonds in ammonium are identical. Complex ion formation depends entirely on this type of bonding.

405. The hybridisation of the carbon atom in carbon dioxide is

• A. sp3
• B. sp2
• C. sp
• D. unhybridised

Explanation: The carbon has only two regions of electron density, since each double bond counts as one region, so two sp hybrid orbitals point in opposite directions and the molecule is linear. The two remaining unhybridised p orbitals form the pi bonds. Counting regions rather than bonds is what makes this quick.

406. In an sp3 hybridised carbon, the angle between the hybrid orbitals is

• A. 90 degrees
• B. 109.5 degrees
• C. 120 degrees
• D. 180 degrees

Explanation: Four equivalent orbitals repel each other to the corners of a tetrahedron, which places them at 109.5 degrees. The sp2 arrangement gives 120 degrees in a plane and sp gives 180 degrees in a line. Each hybridisation is associated with a fixed geometry, which is worth memorising as a set.

407. The number of sigma and pi bonds in a molecule of carbon dioxide is

• A. two sigma and two pi
• B. four sigma and no pi
• C. two sigma and four pi
• D. one sigma and three pi

Explanation: Each of the two carbon to oxygen double bonds contributes one sigma and one pi bond, giving two of each in total. The rule that every double bond is one sigma plus one pi applies universally. A triple bond would contribute one sigma and two pi.

408. A pi bond is weaker than a sigma bond because

• A. it involves fewer electrons
• B. sideways overlap of p orbitals is less effective than head on overlap
• C. it is formed between different elements
• D. it has no electron density

Explanation: Sideways overlap concentrates less electron density between the nuclei than end on overlap does, so the pi bond is easier to break, which is why alkenes are far more reactive than alkanes. Both bond types involve two electrons. This is also why the pi bond of a double bond is the one that opens during addition reactions.

409. According to VSEPR theory, a molecule with four bond pairs and no lone pairs on the central atom is

• A. linear
• B. trigonal planar
• C. tetrahedral
• D. bent

Explanation: Four regions of electron density spread as far apart as possible, which places them at the corners of a tetrahedron with angles of 109.5 degrees, as in methane. Adding a lone pair in place of a bond pair would give a pyramid, and two lone pairs a bent shape. The electron pair geometry and the molecular shape coincide only when there are no lone pairs.

410. The shape of the sulphur hexafluoride molecule, SF6, is

• A. tetrahedral
• B. octahedral
• C. trigonal bipyramidal
• D. square planar

Explanation: Six bond pairs around the central sulphur arrange themselves at 90 degrees to one another at the corners of an octahedron. This requires sulphur to expand its octet using d orbitals, which is possible from period 3 onwards but never for carbon, nitrogen or oxygen. Phosphorus pentachloride, with five pairs, is trigonal bipyramidal by the same reasoning.