Explain y sulphur can show valencies of 2, 4, 6 whereas oxygen is restricted to a valency of 2. [O=8, S=16]

Here are the solutions to your questions: b) Explain why Sulphur can show valencies of 2, 4, 6 whereas oxygen is restricted to a valency of 2. [O=8, S=16] Oxygen (atomic number 8) has an electron configuration of 1s^2 2s^2 2p^4. Its valence electrons are in the second shell, which only contains s and p orbitals. It has two unpaired electrons in its 2p orbitals, allowing it to form two bonds. Oxygen does not have available d-orbitals in its valence shell to expand its octet, thus restricting its valency to 2. Sulphur (atomic number 16) has an electron configuration of 1s^2 2s^2 2p^6 3s^2 3p^4. Its valence electrons are in the third shell, which contains s, p, and empty d-orbitals. In its ground state, sulphur has two unpaired electrons in its 3p orbitals, allowing it to form two bonds. However, sulphur can promote electrons from its 3p and 3s orbitals to the empty 3d orbitals. Promoting one electron from 3p to 3d creates four unpaired electrons (valency 4), and promoting one electron from 3s to 3d (in addition to the previous promotion) creates six unpaired electrons (valency 6). This ability to expand its octet using d-orbitals allows sulphur to exhibit valencies of 2, 4, and 6. c) i) Define the term enthalpy of combustion. The enthalpy of combustion is the enthalpy change that occurs when one mole of a substance is completely burned in excess oxygen under standard conditions (usually 298 K and 1 atm). c) ii) Naturally occurring nuclides emit 3 types of radiations. Name each of the radiation. State the nature of each of the radiations. Alpha () radiation: Consists of helium nuclei* (^4_2He^2+), which are positively charged particles. Beta () radiation: Consists of electrons* (^0_-1e), which are negatively charged particles. Gamma () radiation: Consists of high-energy electromagnetic waves*, which have no charge and no mass. d) i) Write down the electronic configuration of Fe +2 and +3 oxidation states of iron. The atomic number of iron (Fe) is 26. Neutral Fe: [Ar] 3d^6 4s^2 Fe²⁺ (loses 2 electrons from 4s): [Ar] 3d^6 Fe³⁺ (loses 2 electrons from 4s and 1 electron from 3d): [Ar] 3d^5 d) ii) State with reason which of the oxidation states would be more stable. The Fe³⁺ oxidation state is more stable. Reason:* The electron configuration of Fe³⁺ is [Ar] 3d^5. A half-filled d-subshell (d^5) is a particularly stable electronic configuration due to its symmetrical distribution of electrons and maximized exchange energy. e) i) The electron configuration of carbon may be written as 1s^2 2s^2 2px^1 2py^1. Explain each of the symbols in 2px^1. 2: This is the principal quantum number* (n), indicating that the electron is in the second electron shell. p: This is the azimuthal or angular momentum quantum number* (l), indicating that the electron is in a p-orbital, which has a dumbbell shape. x: This indicates the orientation* of the p-orbital in space, specifically along the x-axis. 1: This is the number of electrons* occupying that specific 2px orbital. e) ii) Consider the configuration of carbon stated, explain why carbon forms four equivalent bonds. The given ground state electron configuration for carbon (1s^2 2s^2 2px^1 2py^1) shows only two unpaired electrons in the 2p orbitals, which would suggest a valency of 2. However, to form four equivalent bonds, carbon undergoes hybridization. First, one electron from the 2s orbital is promoted to the empty 2pz orbital, resulting in an excited state configuration of 1s^2 2s^1 2px^1 2py^1 2pz^1. Now there are four unpaired electrons. Next, these four atomic orbitals (one 2s and three 2p orbitals) mix and redistribute their energy to form four new, identical hybrid orbitals called sp^3 hybrid orbitals. These sp^3 orbitals are equivalent in energy and shape and are oriented tetrahedrally around the carbon atom. This allows carbon to form four equivalent covalent bonds, as seen in molecules like methane (CH_4). 3 done, 2 left today. You're making progress.