Explanation​:
All statements are correct, but only D is a reason for solubility in water

Explanation:​
Hybridization is a system for explaining the geometry of molecules.
Hybrid orbitals are therefore a conceptual tool for understanding the formation of molecular orbitals, and do not truly exist within atoms or molecules. They must therefore be equal in number to the original atomic orbitals. Paired electrons in s orbitals may occupy separate sp^3 orbitals, as in the hybrid orbitals of the C atom in methane.
Hybrid orbitals overlap end-on to produce σ bonds; π bonds are formed by the sideways overlap of p orbitals.
This question is categorized as hard as the student must understand several key features of hybrid orbitals and correctly distinguish between σ and π bonding.

Explanation​:
The student must recall the appearance of a π bond and the way in which it is formed. This information is stated clearly in the syllabus, so the question is categorized as easy.
Option A describes the formation of a $\sigma$ bond.

Explanation:​
To answer this question, the student must use the elements’ positions on the periodic table and the number of valence electrons to conclude on the melting point and hardness.
Metals with high melting points are those with stronger metallic bonds. The strongest metallic bonds are those that involve the largest number of delocalized electrons.
Calcium has only 2 delocalized electrons compared to nickel’s 10 valence electrons, which are capable of delocalization. As a result, nickel has a stronger metallic bond than calcium, resulting in a higher melting point.
Hardness is also related to metallic bond strength, where stronger bonds equal a harder material.

This question is categorized as medium difficulty because there are many factors to consider in arriving at the final answer. Importantly, the student must recall and apply many definitions such as metallic bonding, melting point, conductivity, and delocalized electrons.

Explanation​:
NH3​ has all of the bonds/forces named above, but its high boiling point is due to the presence of hydrogen bonds which are the strongest intermolecular forces. Hydrogen bonding is possible due to the presence of N−H bonds.

Explanation​:
Non-metal elements bond to each other covalently. Ca3N2 is made up of a metal and a nonmetal (with a large difference in electronegativity) and has ionic bonding.

Explanation:​

Lithium is in group 1 and so forms Li^+ ions. The formula of the nitride ion is N^3−as nitrogen is in group 15. For the positive and negative charges to cancel out, three Li^+ ions are needed for each N^3− ion.

Explanation:​
Graphite is made up of individual graphene layers. Both have carbon atoms with three electron domains and trigonal planar shapes. Each carbon atom in diamond has four electron domains resulting in tetrahedral arrangements.

Explanation:
Volatility refers to the ease at which a substance vaporizes. Ionic bonds are very strong and, therefore, a lot of energy is required to turn ionic compounds into gases.

Polar solvents are able to form electrostatic forces of attraction with ions.

Explanation​:
Oppositely charged ions attract, it is an ionic bonding;

Choice A describes metallic bonding;

Choice B describes a simple molecular lattice;

Choice D describes a covalent bond.

Explanation:​
Electrostatic forces of attraction between the positive ions and delocalized electrons (metallic bonds) must be broken for metal to melt. The stronger the metallic bond, the higher the melting point.

Explanation:​
Atom 1 has four electron domains and so a tetrahedral shape (109.5$°$).
Atom 2 has three electron domains and so a trigonal planar shape (120$°$).
Atom 3 has four electron domains of which one is a lone pair giving it a trigonal pyramidal shape (107$°$).

Explanation:
The charges of the positive and negative ions must cancel out to give an overall neutral compound. Group 2 metals form ions with a 2+ charge. The formulae of the counterions above are: $.$

Explanation:​
Each carbon atom in ethene forms three σ bonds and one π bond. The s electrons are originally paired but are conceptually unpaired during the formation of the three sp^2 orbitals, allowing the σ bonds to form. The p electrons are unpaired in the original atomic orbitals and remain unpaired in the hybrid orbitals. One p orbital in each carbon atom is used to produce the π bond.
This question is moderately difficult as the student must recall the concept of electron unpairing/promotion in hybridization and differentiate between σ and π orbitals without the aid of a diagram.

Explanation:​
In an sp^2 carbon atom, there are three sp^2 orbitals of equal energy and one p orbital of higher energy.
The formation of three σ bonds and one π bond can be understood as being derived from three sp^2 orbitals and one p orbital, respectively, each containing an unpaired electron.
This question is moderately difficult: although the orbital energies of an sp^2 carbon atom is covered explicitly in the syllabus, the student must recognize this distribution without the aid of orbital labels or additional diagrams.

Explanation:​
Alloys are homogeneous mixtures of metals with other metals or nonmetals.

Explanation:
There is no polyatomic ion with the formula of CO4^2−. The correct formula for the carbonate ion is CO3^2−. The formulas and charges of the other ions are correct.

Explanation:​
To successfully answer this question, the student must recall that electrical conductivity requires freely moving electrons.
Metals are able to conduct electricity because the valence electrons are able to move freely through the lattice in the sea of delocalized electrons.
The size and mass of the atom do not factor into the ability of the material to conduct electricity. While transition metal salts can conduct electricity, the ability to form ionic bonds does not explain conductivity.

This question is categorized as easy because it only involves the recall of basic information about metals and conductivity.

Explanation​:
Fluorine is the most electronegative element on the periodic table. Therefore, the greatest difference in electronegativity between $\mathrm{C}$ and $\mathrm{F}$ results in the most polar bond.

Explanation:​
Across the same period, metals have lower ionization energies than non-metals due to lower nuclear charges with the same number of shells (shielding effect).

Explanation:​
C3H6​ belongs to the alkene class with the general formula CnH2n​, and has a three carbon chain.

Explanation:
The higher the charge of the ions, the stronger the electrostatic forces of attraction and, therefore, the greater the lattice enthalpy.
The smaller the ion, the greater the charge density and the stronger the electrostatic attraction between oppositely charged ions.
Statement I is incorrect since the smaller the lattice enthalpy, the weaker the electrostatic attraction and the lower the melting point is.

Explanation​:
N2 has a triple bond, O2​ has a double bond, and H2 only has a single bond.
The greater the number of shared pairs of electrons, the stronger the bond.

Explanation​:
There is a non-bonding (lone) pair of electrons on the nitrogen atom and three on each chlorine. There are three N−Cl bonding pairs of electrons. Only the outermost electrons are shown for each element

Explanation:
Strontium is in group 2 and so forms Sr^2+ ions. The formula of the sulfide ion is S^2− as sulfur is in group 16. For the positive and negative charges to cancel out, one of each ion is required.