Further Inorganic Chemistry Assignment Sample

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LO1 Understand the properties of the group 2 elements

Group 2 elements, also known as alkaline earth metals, include beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). These elements share common properties such as being soft, silver-colored metals with relatively low densities and melting points compared to transition metals. As you move down Group 2 in the periodic table, the atomic radius increases due to the addition of electron shells, leading to weaker attraction between the nucleus and outer electrons.

This trend is associated with a decrease in first ionisation energy, making it easier for atoms lower in the group to lose their two outermost electrons and become more reactive. For example, barium is more reactive than magnesium because its larger atomic size and increased electron shielding reduce the energy required to remove these electrons. Group 2 metals conduct electricity well due to the presence of delocalised electrons in their metallic bonds.

Understanding these trends and properties of Group 2 elements is critical for academic success in inorganic chemistry. Students looking for structured and clear explanation can benefit from help with assignment writing to grasp these concepts deeply and present them effectively. Utilizing such assistance can improve comprehension, accuracy, and quality of chemistry assignments related to Group 2 properties.

AC 1.1 Describe and explain the physical properties of Group 2 elements

a)

i) Explanation of metallic bonding

Metallic bonding is the transfer of electrons between 2 or more metal atoms by chemically bonding together. It is one of the 3 types of chemical bonding. However, this type is between metallic atoms- the attraction between positively charged ions and delocalised (negatively charged) electrons.

ii) Explanation of why metals conduct electricity

Metals are good conductors of electricity as they have delocalised electrons holding rows of positive ions in the metallic structure. ‘Delocalised’ or ‘free-flowing’ electrons have a negative charge, therefore conducting electricity.

(Iqra, 2024)

b) Trend in atomic radius down Group 2

Figure 1: Atomic Radii of the Group 2 elements

The atomic radius of Group 2 elements increases progressively down the group. This trend is due to the addition of electron shells with each successive element, causing the outermost electrons to be farther from the nucleus. For example, in the bar chart, beryllium (Be) has the smallest atomic radius, just above 0.1 nm, while barium (Ba) has the largest, slightly above 0.3 nm. This increase in atomic size weakens the electrostatic attraction between the positively charged nucleus and the outermost electrons. As a result, it becomes easier for these atoms to lose their outer electrons during chemical reactions, enhancing their overall reactivity. Therefore, elements like barium are more reactive than magnesium or beryllium because their larger atomic radii and increased electron shielding reduce the energy required to remove outer electrons. This explains the observed increase in reactivity down Group 2.

c)

i) Electron configuration of Mg

Mg: 1s²2s²2p⁶3s²

ii) Electron configuration of Be

Be: 1s²2s²

iii) Electron configuration of Ca

Ca: 1s²2s²2p⁶3s²3p⁶4s²

iv) Electron configuration of Ca²⁺

Ca²⁺: 1s²2s²2p⁶3s²3p⁶

d)
i) Definition of first ionisation energy

First ionisation energy: The first ionisation energy is the energy required to remove one mole of the most loosely held electrons from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+. (Iqra, 2024).

ii) Equations for the first and second ionisation energies of barium

First ionisation energy equation of barium: Ba(g) Ba⁺ (g) + e^-

Second ionisation energy equation of barium: Ba⁺(g) Ba²⁺ (g) + e^-

iii) Description and explanation of the trend in first ionisation energies down Group 2

Figure 2: First ionisation energy of the Group 2 elements

The first ionisation energy of Group 2 elements decreases steadily down the group. This trend is due to the increasing atomic size as additional electron shells are added with each successive element. The outermost electrons are positioned further from the nucleus, resulting in a weaker electrostatic attraction between the nucleus and these electrons. Additionally, increased electron shielding from inner shells reduces the effective nuclear charge experienced by the outer electrons. This makes it easier to remove the outermost electron, requiring less energy. The bar chart illustrates this trend, with beryllium having the highest first ionisation energy and barium the lowest. As the group is descended, the decreasing ionisation energy directly contributes to the increasing reactivity of Group 2 metals, as less energy is needed to lose electrons and form positive ions. This pattern explains why elements like barium are significantly more reactive than lighter elements such as beryllium.

AC 1.2 Describe and explain the trends in reactivity down Group 

a) Explanation of why Group 2 elements become more reactive down the group

The reactivity of Group 2 elements increases as you move down the group due to changes in ionisation energy. As you go down Group 2, the atomic size increases because additional electron shells are added. This increase in size means that the outermost electrons are further away from the nucleus. As the distance between the outer electrons and the nucleus increases, the attractive force between the positively charged nucleus and the negatively charged electrons decreases. Therefore, it becomes easier to remove these outer electrons, leading to lower ionisation energy. Since the reactivity of group 2 elements is dependent on their ability to lose their outermost electrons to form positive ions, the decrease in ionisation energy down the group results in an increase in reactivity.

b)
i) Word and balanced chemical equation for the reaction of barium with water

Barium _(s) + Water_(l) Barium Hydroxide _(aq) + Hydrogen _(g)

Ba _(s) + 2H₂o _(l)  Ba(OH)_{2 (aq)} + H_{2 (g)}

ii) Observations during the reaction

When barium reacts with water a precipitate/bubbles will form suggesting there is a release of gas, which in this case we know is hydrogen gas. When barium hydroxide is produced, the solution will appear cloudy or milky if there are any undissolved particles. However, when fully dissolved in water, the solution will become clear. 

iii) Explanation of whether the resulting solution is acidic or alkaline

When Group 2 metals react with water, they produce hydroxides and hydrogen gas. In the case of barium, the reaction forms barium hydroxide (Ba(OH)₂), which is a strong base. Barium hydroxide readily dissolves in water, dissociating completely into barium ions (Ba²⁺) and hydroxide ions (OH⁻). The release of hydroxide ions significantly increases the pH of the solution, making it strongly alkaline. This high solubility of barium hydroxide in water ensures a substantial concentration of OH⁻ ions, resulting in a highly basic solution. The alkaline nature of the solution is due to the abundance of hydroxide ions, which neutralize any acidic components and raise the pH well above 7. Compared to other Group 2 hydroxides, barium hydroxide is notably more soluble, contributing to the stronger alkalinity of the solution. The formation of this basic solution is a characteristic property of Group 2 metal hydroxides.

c)
i) Word and balanced chemical equation for the reaction between magnesium and sulfuric acid

Magnesium + Sulphuric Acid à Magnesium Sulphate + Hydrogen

Mg _(s)+ H₂SO_{4 (l)} MgSO_{4 (aq)}+ H_{2 (g)}

ii) Observations during the reaction

When magnesium reacts with sulphuric acid, bubbles will form suggesting hydrogen gas is produced, and fizzing will occur at the surface of the magnesium. As the reaction is exothermic, the solution may feel warmer as the experiment progresses.

Many indigestion medicines contain magnesium carbonate and calcium carbonate. Use balanced chemical equations to show how these carbonates neutralise the hydrochloric acid in your stomach.

Magnesium carbonate (MgCO₃):

MgCO_{3 (s)} + HCl _(l) MgCl_{2 (s)} + 2H₂O_(l)+ CO_2(g)

Calcium carbonate (CaCO₃):

CaCO_{3 (s)} + 2HCl_(l) → CaCl_2(s) + H₂O_(l) + CO_2(g)

LO2 Understand the Properties of Group 7 Elements

AC 2.1 Describe and explain the physical properties of Group 7 elements

“Element	fluorine	chlorine	bromine	iodine
Appearance at room temperature	pale yellow gas	green gas	dark red, volatile liquid	shiny grey/black solid
Melting point (K)	53	172	266	387
Boiling point (K)	85	239	332	457”

The physical properties of halogens show distinct trends as one moves down Group 7 of the periodic table. Fluorine, the lightest halogen, is a pale yellow gas at room temperature with low melting and boiling points. Chlorine, being heavier, is a green gas with higher melting and boiling points. Bromine, a volatile red liquid, has even higher melting and boiling points, while iodine is a shiny grey/black solid with the highest melting and boiling points among the halogens. These trends can be explained by increasing molecular size and intermolecular forces. As atomic size increases, the van der Waals forces (induced dipole-induced dipole interactions) between the molecules become stronger, requiring more energy to overcome these forces. This explains the higher melting and boiling points down the group. Halogens form diatomic, non-polar molecules, with the strength of these intermolecular forces increasing as the size and mass of the halogen atoms increase.

Further Inorganic Chemistry Assignment Sample

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AC 2.2 Describe and explain the trend in reactivity down Group 7

a) Explanation of redox reactions

Redox reactions involve both oxidation and reduction processes, where the oxidation states of the involved elements change. Oxidation is the process in which a substance loses electrons, resulting in an increase in its oxidation state. Conversely, reduction is when a substance gains electrons, causing its oxidation state to decrease. In a redox reaction, one substance is oxidised and another is reduced, as the electrons lost by the oxidised substance are gained by the reduced substance. For example, in the reaction between magnesium and sulphuric acid, magnesium (Mg) is oxidised. It loses two electrons, and its oxidation state increases from 0 to +2, forming Mg²⁺ ions. On the other hand, hydrogen ions (H⁺) from the acid are reduced. They gain electrons, and their oxidation state decreases from +1 to 0, forming hydrogen gas (H₂). Therefore, the oxidation state of magnesium increases, while the oxidation state of hydrogen decreases.

b) Explanation of why halogens are oxidising agents

Halogens behave as oxidising agents due to their electron structure, which makes them highly capable of gaining electrons. Halogens have seven electrons in their outermost shell, making them just one electron short of achieving a stable octet configuration. This electron configuration drives their strong tendency to attract and accept an electron from other elements during reactions. For example, chlorine (Cl), with an electron configuration of [Ne] 3s² 3p⁵, is one electron away from completing its valence shell. When chlorine reacts with sodium, chlorine gains the missing electron, forming chloride ions (Cl⁻), while sodium loses an electron to form sodium ions (Na⁺). The process of chlorine gaining an electron, while sodium loses one, demonstrates the halogen's ability to oxidise another substance by accepting electrons. This electron configuration and the resulting electron affinity make halogens powerful oxidising agents.

c) Explanation of why fluorine is the strongest oxidising agent and the trend in reactivity down Group 7

Fluorine is the strongest oxidising agent in Group 7 due to it having a high electronegativity and a small atomic radius. Being the most electronegative element, fluorine has a very strong attraction for electrons, allowing it to easily gain electrons from other substances during reactions.

While fluorine is the strongest oxidising agent due to its high electronegativity and small size, the reactivity of halogens decreases as you move down the group due to increasing atomic size and shielding effects.

As you go down the group, the atomic size increases due to the addition of electron shells. This increase in size means that the outer electrons are further from the nucleus, resulting in a weaker attraction between the nucleus and electrons. With the addition of more electron shells, there is an increase in electron shielding. This shielding reduces the effective nuclear charge felt by the outer electrons, making it harder for larger halogens like bromine and iodine to attract and gain electrons compared to fluorine.

a) Ionic equation and observations for the reaction between aqueous chlorine and bromide ions

When you add aqueous chlorine to the bromide solution, a colour change would take place. The bromine that forms is reddish-brown. The solution will change colour from clear/colourless to a reddish-brown colour as bromine is produced.

This type of reaction is classified as a redox reaction. Chlorine (Cl₂) acts as an oxidising agent and bromide ions (Br^-) are oxidised to bromine (Br₂).

Cl₂(aq) + 2Br^-(aq) 2Cl^-(aq) + Br₂(aq)

b) Description and ionic equations for testing unknown halide solutions with silver nitrate

To identify an unknown halide solution, you can carry out a test using silver nitrate (AgNO₃). This test is based on the formation of insoluble silver halides when silver nitrate is added to the solution containing halide ions.

The general ionic equation for the reaction between silver nitrate and halide ions (X^-) can be written as follows:

Ag+(aq) + X-(aq) → AgX(s)

Ag⁺ represents the silver ions from silver nitrate, and X^- represents the halide ions. The product, AgX, is a solid precipitate that forms when the halide is present. The colour of the precipitate will help you identify the specific halide present in the solution.

Test Method:

1. Add a few drops of dilute silver nitrate solution to the unknown halide solution.
2. Observe the colour and appearance of any precipitate that forms.

Possible observations (with halide ions):

- Formation of a cream-coloured precipitate, indicating the presence of bromide ions (AgBr).

- Formation of a white coloured precipitate, indicating the presence of chloride ions (AgCl).

- Formation of a yellow precipitate forms, indicating the presence of iodide ions (Agi).

LO3 Understand the Properties of Period 4 Transition Metals

AC 3.1 Define ‘transition metal’, recognising variable oxidation states

a) Explanation of why iron is a transition metal and scandium is not

A transition metal is defined as an element that has an incomplete d-subshell or can form ions with an incomplete d-subshell. Transition metals exhibit characteristics such as the ability to form multiple oxidation states and often form coloured compounds. Iron (Fe) and scandium (Sc) are both located in the d-block of the periodic table, but they are classified differently. Scandium (Sc) has the electron configuration [Ar] 3d¹ 4s². When scandium forms its ion (Sc³⁺), it loses three electrons, resulting in the electron configuration [Ar], with no electrons in the 3d subshell. This lack of d-electrons means scandium does not exhibit the typical properties of transition metals. On the other hand, iron (Fe) has the electron configuration [Ar] 4s² 3d⁶. Even when iron forms its ions (Fe²⁺ and Fe³⁺), it retains d-electrons, allowing it to exhibit the defining properties of transition metals.

b) Electron configurations of vanadium, chromium, nickel, and copper, with explanations for copper and chromium

1. Vanadium (V): [Ar] 3d³ 4s²
2. Chromium (Cr): [Ar] 3d⁵ 4s¹
3. Nickel (Ni): [Ar] 3d⁸ 4s²
4. Copper (Cu): [Ar] 3d¹⁰ 4s¹

Copper and chromium have unexpected electronic configurations due to the stability associated with half-filled and fully filled d subshells.

For chromium, instead of following the expected order of filling (which would be [Ar] 3d⁴ 4s²), it has the configuration [Ar] 3d⁵ 4s¹. This configuration is more stable as having a half-filled d subshell (3d⁵) minimises electron-electron repulsions, therefore providing greater stability.

Similarly, copper is expected to have the configuration [Ar] 3d⁹ 4s². Instead, it has [Ar] 3d¹⁰ 4s¹. The fully filled d subshell (3d¹⁰) provides additional stability, again due to reduced repulsions.

1. c) Electron configurations of Cr³⁺, Mn²⁺, and Cu²⁺ ions
2. i) Cr³⁺: [Ar] 3d³
3. ii) Mn²⁺: [Ar] 3d⁵

iii) Cu²⁺ : [Ar] 3d⁹

AC 3.2 Explain the catalytic activity of transition metals and their compounds

a) Definition of a catalyst

A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative pathway with a lower activation energy without it being used up.

b) Explanation of heterogeneous vs. homogeneous catalysts

A heterogeneous catalyst is a catalyst which is in a different phase than the reactants. E.g. if the reactants are gaseous, the catalyst may be solid.

A homogeneous catalyst is at catalyst which is in the same phase as the reactants. E.g. if the reactants are liquid, the catalyst will also be liquid. The catalyst will react with reactants throughout the solution, allowing for a more consistent reaction.

c) Explanation of why transition metals are effective catalysts

Transition metals are effective catalysts for multiple reasons:

1. Oxidation States: Transition metals can easily change their oxidation states. Allowing them to participate in multiple steps of a reaction facilitating the conversion of reactants to products.
2. Ability to Adsorb Reactants: Transition metals can adsorb reactants onto their surfaces. This increases the concentration of reactants at the surface of the catalyst, increasing the likelihood of successful collisions and reactions.
3. D-Orbitals: The presence of d-orbitals in transition metals allows for a variety of bonding interactions with reactants. This adaptability in bonding can stabilise transition states, further lowering activation energy.

d) Balanced equations and summaries of two industrial processes using transition metal catalysts

Two industrial processes that use transition metal catalysts are the Haber process and the catalytic hydrogenation process.

Haber Process: This process synthesises ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases, using an iron catalyst. The reaction is as follows:

The iron catalyst lowers the activation energy, enabling the reaction to take place efficiently at lower temperatures and pressures, which is essential for large-scale ammonia production.

Catalytic Hydrogenation: In this process, unsaturated hydrocarbons (such as alkenes) react with hydrogen (H₂) in the presence of a transition metal catalyst, typically palladium (Pd), platinum (Pt), or nickel (Ni). The reaction can be represented as:

This process converts alkenes to alkanes by adding hydrogen across double bonds, which is important in producing margarine and other hydrogenated oils. The catalyst provides an alternative pathway with a lower activation energy.

AC 3.3 Explain the formation of complex ions

A complex ion is a metal ion bonded to one or more molecules or ions called ligands. Ligands are species that have lone pairs of electrons which they can donate to form coordinate covalent bonds with the metal ion. The coordination number refers to the number of bonds or ligands directly attached to the central metal ion in a complex.

Transition metal ions are able to form complexes because they have vacant orbitals in their d-subshells that can accept electron pairs from ligands. These metal ions also have relatively small sizes and high charges, which makes them capable of forming strong bonds with ligands.

Examples of transition metal complexes include:

1. [Cu(H₂O)₆]²⁺: This is a copper(II) complex with six water molecules as ligands. The shape is octahedral, with a coordination number of 6. It appears blue.
2. [Fe(CN)₆]⁴⁻: This is a ferrocyanide complex, where cyanide ions are the ligands. The shape is octahedral, with a coordination number of 6. It appears yellow.
3. [NiCl₄]²⁻: This is a nickel(II) complex with four chloride ions as ligands. The shape is tetrahedral, with a coordination number of 4. It appears green.

These examples highlight how transition metals can form various shapes and colours due to the coordination of different ligands.

AC 3.4 Explain the origins of colour in transition metal complexes

a) Electronic structure of a titanium (Ti) atom

The electronic structure of a neutral titanium (Ti) atom, which has an atomic number of 22, is:

This means titanium has 2 electrons in the 3d subshell and 2 in the 4s subshell in its ground state.

b) Electronic structure of Ti³⁺ ion

The Ti³⁺ ion forms when titanium loses 3 electrons. The electronic configuration for the Ti³⁺ ion, with 19 electrons, is:

The loss of the 2 electrons from the 4s subshell and 1 electron from the 3d subshell results in the Ti³⁺ ion having one electron in the 3d subshell.

c) Explanation of energy changes in 3d orbitals when forming the [Ti(H₂O)₆]³⁺ complex ion

When water molecules approach Ti³ to form the [Ti(H₂O)₆]³⁺ complex ion, the Ti³⁺ ion experiences a ligand field. The water molecules, acting as ligands, create an electric field that causes the d-orbitals of the titanium ion to split into two energy levels. The crystal field splitting results in a higher energy for some d-orbitals and a lower energy for others, creating a split in the d-orbital energies. This splitting depends on the geometry and nature of the ligands involved.

d) Explanation of the violet colour of the [Ti(H₂O)₆]³⁺ complex ion

The violet colour of the [Ti(H₂O)₆]³⁺ complex ion arises from the absorption of specific wavelengths of light. The splitting of the 3d orbitals in the presence of the water ligands leads to the absorption of energy corresponding to visible light, which excites electrons from the lower-energy d-orbitals to the higher-energy d-orbitals. The wavelength of light absorbed corresponds to the energy gap between the split d-orbitals. The remaining transmitted light, which is the complementary colour, is observed as violet. The absorption of light and the specific energy gap between the d-orbitals is what gives the complex its characteristic colour.

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