Unit 1: Structure, Bonding and Introduction to Organic Chemistry AS

An empirical formula gives the simplest whole-number ratio of atoms, whereas a molecular formula gives the actual number of each type of atom in one molecule. For example, glucose has empirical formula $\ce{CH2O}$ and molecular formula $\ce{C6H12O6}$.

For $0.250\,\mathrm{mol}$ of sodium chloride, the number of formula units is $0.250\times6.02\times10^{23}=1.51\times10^{23}$. It also contains that many $\ce{Na+}$ ions and that many $\ce{Cl-}$ ions, so the total number of ions is twice the number of formula units.

1. Write correct chemical formulae; never change subscripts to balance an equation.
2. Adjust coefficients until each element is conserved.
3. Add $(s)$, $(l)$, $(g)$ and $(aq)$ from the physical states.
4. For an ionic equation, split soluble aqueous ionic substances into ions.
5. Cancel spectator ions that appear unchanged on both sides, then check atoms and charge.

For $\ce{MgCl2}$, the relative formula mass is $24.3+2(35.5)=95.3$, so its molar mass is $95.3\,\mathrm{g\,mol^{-1}}$. Calling this an $M_r$ suggests discrete molecules and is inappropriate for the giant ionic lattice.

Dissolving $5.85\,\mathrm{g}$ of $\ce{NaCl}$ and making the solution up to $500\,\mathrm{cm^3}$ gives $V=0.500\,\mathrm{dm^3}$. With $M(\ce{NaCl})=58.5\,\mathrm{g\,mol^{-1}}$, $n=0.100\,\mathrm{mol}$, so $c=0.200\,\mathrm{mol\,dm^{-3}}$. The mass concentration is $5.85/0.500=11.7\,\mathrm{g\,dm^{-3}}$.

1. Treat percentages as masses in a $100\,\mathrm{g}$ sample, or use the measured masses directly.
2. Divide each mass by the element's $A_r$ to obtain moles.
3. Divide all mole values by the smallest.
4. If necessary, multiply the whole ratio to remove values such as $1.5$ or $1.33$.
5. Write the empirical formula. For a molecular formula, calculate $k=M_r/(\text{empirical-formula mass})$ and multiply every subscript by the integer $k$.

A compound containing $40.0\%$ C, $6.7\%$ H and $53.3\%$ O gives mole values $40.0/12.0$, $6.7/1.0$ and $53.3/16.0$, a ratio close to $1:2:1$. Its empirical formula is $\ce{CH2O}$. If $M_r=180$, then $k=180/30=6$ and the molecular formula is $\ce{C6H12O6}$.

For $4.80\,\mathrm{g}$ Mg, $n(\ce{Mg})=4.80/24.3=0.198\,\mathrm{mol}$. The $1:1$ coefficient ratio gives $0.198\,\mathrm{mol}$ $\ce{MgCl2}$. With $M(\ce{MgCl2})=95.3\,\mathrm{g\,mol^{-1}}$, the theoretical mass is $18.8\,\mathrm{g}$.

At room temperature and pressure, use the stated or accepted molar volume (commonly $24.0\,\mathrm{dm^3\,mol^{-1}}$). For other conditions, use $pV=nRT$: pressure in Pa, volume in $\mathrm{m^3}$, temperature in K and $R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}}$. The same equation can find the molar mass of a volatile liquid from the mass and amount of its vapour.

$2.40\,\mathrm{dm^3}$ of gas at room conditions is $2.40/24.0=0.100\,\mathrm{mol}$. Under specified non-room conditions, $120\,\mathrm{cm^3}=1.20\times10^{-4}\,\mathrm{m^3}$ and $25^\circ\mathrm{C}=298\,\mathrm{K}$ before substitution into $n=pV/(RT)$.

For $\ce{CaCO3 -> CaO + CO2}$, if $\ce{CaO}$ is desired, the atom economy is $56.1/(56.1+44.0)\times100=56.0\%$. If the theoretical yield of $\ce{CaO}$ is $5.00\,\mathrm{g}$ but $4.20\,\mathrm{g}$ is isolated, the percentage yield is $84.0\%$.

Yield can fall because reaction is incomplete, side reactions occur, or product is lost during separation. Atom economy is fixed by the chosen balanced reaction and desired product; it does not improve merely because technique or catalyst improves yield.

1. Record masses or volumes with suitable precision and identify which measured change belongs to each substance.
2. Repeat heating or reaction to constant mass where appropriate.
3. Convert each quantity to moles using molar mass or gas conditions.
4. Divide by the smallest amount and test whether the ratio is close to small whole numbers.
5. Compare the experimental ratio with the proposed formula or equation, then state whether the data support it.

Evaluate direction as well as size of error. Incomplete reaction can leave too little mass change; loss of solid can make mass loss too large; oxidation by air can add mass; gas leaks or dissolution can reduce the collected volume. Repeats expose random variation, but repeating does not remove a systematic leak or calibration error.

1. Measure the magnesium mass and calculate $n(\ce{Mg})$.
2. Place excess dilute acid in a sealed reaction vessel connected to the gas collector.
3. Add the magnesium, seal immediately and allow the reaction to finish.
4. Record gas volume, room temperature and pressure; use $\ce{Mg + 2H+ -> Mg^{2+} + H2}$ to obtain $n(\ce{H2})$.
5. Convert the volume to $\mathrm{dm^3}$ and calculate $V_m=V/n$ in $\mathrm{dm^3\,mol^{-1}}$. Repeat and compare values.

For displacement, electron transfer changes the metal and ion identities. In the acid reaction, $\ce{H+}$ becomes hydrogen gas. In precipitation, two aqueous ions form an insoluble lattice. Spectator ions remain aqueous and cancel from the ionic equation.

Protons identify the element. Neutrons add nuclear mass without changing the element's identity. Electrons are attracted to the positive nucleus and are arranged in energy levels and sub-shells; their arrangement later explains bonding and chemical behaviour.

A neutral atom has equal numbers of protons and electrons, so its positive and negative charges cancel. Changing the number of electrons makes an ion; changing the number of neutrons makes another isotope of the same element.

An electric field deflects protons and electrons because they are charged, but it does not deflect neutrons. Electron mass is so small compared with nucleon mass that mass number counts protons and neutrons only.

In $^{84}_{36}\ce{Kr}$, $Z=36$, so the element is krypton and its nucleus contains 36 protons. $A=84$ means the nucleus contains 84 nucleons in total. Mass number is always a whole number for one isotope.

Here $q$ is the ion charge in elementary-charge units: $q=+2$ for a $2+$ ion and $q=-1$ for a $1-$ ion. Subtracting a negative charge therefore adds an electron.

$^{35}\ce{Cl}$ and $^{37}\ce{Cl}$ both contain 17 protons. They contain 18 and 20 neutrons respectively. Neutral atoms of both isotopes contain 17 electrons and therefore have the same ground-state electronic configuration.

Because chemical reactions mainly involve electrons, isotopes of an element have very similar chemical properties. Their different masses can produce different physical behaviour and separate peaks in a mass spectrum.

1. Introduce and vaporise the sample where necessary.
2. Ionise particles to form positive ions.
3. Accelerate ions using an electric field.
4. Separate ions because different $m/z$ values respond differently or take different flight times.
5. Detect ions and convert the signal into a spectrum.

Elemental peaks reveal isotope masses and abundances; the weighted mean gives $A_r$, and the equation can be rearranged to find an unknown abundance or mass. For a molecule, the molecular-ion peak gives its relative molecular mass when $z=1$, helping identify the molecule. Fragment peaks represent smaller ions.

For two different isotopes, count both arrangements: $^{35}\ce{Cl}$-$^{37}\ce{Cl}$ and $^{37}\ce{Cl}$-$^{35}\ce{Cl}$. This factor of two is why the mixed-isotope peak is larger than either single arrangement alone.

The equations define first, second and third ionisation energies respectively. Values are quoted per mole, usually in $\mathrm{kJ\,mol^{-1}}$, and the starting species must be gaseous in every case.

An empty orbital holds no electrons; a singly occupied orbital holds one; a full orbital holds a pair. When two electrons share an orbital, their spins must be opposite, represented in electron-in-box notation by arrows pointing in opposite directions, $\uparrow\downarrow$.

Orbitals are grouped into sub-shells. An s sub-shell contains one orbital, a p sub-shell three orbitals and a d sub-shell five orbitals. Orbital capacity therefore controls sub-shell capacity.

A 2p electron is higher in energy and more shielded than a 2s electron, explaining why boron's first ionisation energy is lower than beryllium's despite boron having more protons. Within 2p, repulsion in a paired orbital makes an electron easier to remove from oxygen than from nitrogen's singly occupied 2p orbitals.

For one element, successive ionisation energies rise, but a very large jump occurs after all electrons in the outer shell have been removed. The next electron comes from an inner shell, closer to the nucleus and less shielded. For a main-group element, the number of electrons removed before the first large jump identifies its group.

Across successive elements, first ionisation energy generally rises within a period as nuclear charge increases. A large fall after a noble gas shows the start of a new shell. Smaller falls, such as Be to B and Mg to Al, show entry into a higher-energy p sub-shell; the N to O and P to S falls show the effect of pairing within p orbitals.

Each p sub-shell contains three orbitals of the same basic shape and energy, oriented along mutually perpendicular axes. They are labelled $p_x$, $p_y$ and $p_z$. Each orientation is a different orbital and can hold up to two opposite-spin electrons.

The boundary shape represents a region of high probability for locating an electron; it is not a solid surface. The two p lobes belong to one orbital, not two separate orbitals.

Single occupation keeps electrons apart while the orbitals have equal energy. When two electrons share an orbital, opposite spins are required. The order of equivalent boxes does not matter; the occupancy pattern does.

Use superscripts for electron counts: s holds 2, p holds 6 and d holds 10. For example, sulfur is $1s^2 2s^2 2p^6 3s^2 3p^4$; electron-in-box notation shows one paired and two singly occupied 3p orbitals. Noble-gas shorthand may replace completed inner shells.

Chromium and copper use the observed configurations $[\ce{Ar}]3d^5 4s^1$ and $[\ce{Ar}]3d^{10}4s^1$. For ions, add electrons for negative charge and remove them for positive charge from the highest principal shell first: transition-metal ions lose 4s electrons before 3d electrons. Thus $\ce{Fe^{3+}}$ is $[\ce{Ar}]3d^5$ and $\ce{Cl-}$ is $[\ce{Ar}]$.

Elements in the same group have the same pattern of outer-shell electrons. Group 1 atoms have an outer $ns^1$ electron and commonly form $1+$ ions by losing it. Group 17 atoms have outer $ns^2np^5$ configurations and need one more electron for a filled outer shell, so they commonly form $1-$ ions or one covalent bond.

Moving across a period changes the outer configuration one electron at a time, producing a systematic change from metallic to non-metallic behaviour. Moving down a group preserves the valence pattern but adds shells, so chemical behaviour remains related while reactivity can change.

In the first four quantum shells, the relevant sub-shells are $1s$; $2s,2p$; $3s,3p,3d$; and $4s,4p,4d$. Each s, p or d sub-shell always has the same capacity regardless of its shell number.

Use atomic number on the horizontal axis so the elements are in sequence. Label the vertical axis with the property and its unit, choose a scale that uses the plotting area effectively, plot each value accurately, and join points only when the purpose is to display the sequence rather than imply continuous values between elements.

For a logarithmic treatment, calculate $\log_{10}$ of each first-ionisation-energy value and plot that transformed value against atomic number. Label the axis to show both the logarithm and the original energy unit convention. Equal vertical intervals then represent equal ratios in first ionisation energy, not equal raw-energy differences, which can make repeated proportional features easier to compare.

Across each period, first ionisation energy generally increases because proton number rises while added electrons enter the same main shell, so shielding changes little and atomic radius decreases. The Group 13 dip occurs when the electron removed is in a higher-energy p sub-shell; the Group 16 dip occurs because repulsion in a paired p orbital makes removal easier than from the preceding half-filled p sub-shell.

Down a group, first ionisation energy decreases. The outer electron occupies a higher shell, farther from the nucleus, and experiences more inner-shell shielding. These effects outweigh the increased proton number, weakening attraction to the electron removed.

For a coloured solution containing a blue cation and a yellow anion, movement of the two colours towards opposite electrodes is especially direct migration evidence. A complete circuit and an ion-conducting path are needed for the observation.

Balance both the number of atoms and the total charge. The electron is on the product side for electron loss and on the reactant side for electron gain.

First determine the ions and their simplest whole-number ratio. Draw each ion in square brackets, show its outer shell after transfer, and write the charge outside the bracket. Repeat each ion the number of times required by the formula.

In a two-dimensional section, show alternating positive and negative ions in both row and column directions. A three-dimensional lattice continues the alternation through additional layers; the exact coordination depends on the compound.

The chemical formula gives the lowest whole-number ratio of ions needed for electrical neutrality. For example, NaCl means a 1:1 ratio and CaCl$_2$ means a 1:2 ratio across the giant lattice.

Every ion interacts with many surrounding ions. Opposite charges attract and like charges repel, but at the stable lattice spacing the overall arrangement has a strong net attraction and lower energy than widely separated ions.

The attraction acts in all directions through the giant lattice. This helps explain why separating ions requires substantial energy rather than breaking one isolated link.

BaCl$_2$ has stronger ionic attraction than CsCl because Ba$^{2+}$ has a higher charge and is smaller than Cs$^+$. More energy is therefore needed to separate its ions, consistent with its higher melting temperature.

For ions with the same charge down a group, ionic radius increases. Each step adds an occupied electron shell; the greater distance and shielding outweigh the increased nuclear charge.

All ions from N$^{3-}$ to Al$^{3+}$ are isoelectronic: they have ten electrons and the same occupied-shell pattern. As proton number increases, the same electron cloud experiences stronger nuclear attraction, so radius decreases.

The anion's electrons are drawn towards the cation, so the electron distribution is no longer symmetrical. Greater distortion increases electron density between the nuclei and gives the bonding more covalent character.

The cation has polarising power; the anion is polarisable. These terms describe the ability to cause distortion and the ease of being distorted, respectively.

The greatest distortion is expected when a small, highly charged cation is next to a large, highly charged anion. The bonding then shows more covalent character than a simple ionic model predicts.

F$^-$ is difficult to polarise because it is small and carries only a single negative charge. Al$^{3+}$ has much greater polarising power than Na$^+$ because it is smaller and has a higher charge.

Both nuclei attract the negatively charged shared pair. A stable bond length is reached where the attractions lower the energy while nucleus–nucleus and electron–electron repulsions prevent collapse.

Show outer-shell electrons only, using one symbol for electrons from each starting atom. A single, double or triple bond contains one, two or three shared pairs. Add every lone pair needed to complete the outer-shell accounting.

In a dative covalent bond, both electrons in the shared pair originally come from the same donor atom. Once formed, the pair is attracted to both nuclei like any other covalent bond.

Applications follow from structure: continuous strong bonds resist deformation, delocalised electrons carry charge, and weak attractions between graphite layers allow sliding.

If two bonded atoms have different electronegativities, the more electronegative atom attracts the pair more strongly and becomes partially negative, $\delta-$; the other becomes partially positive, $\delta+$.

Electronegativity is used comparatively. The difference between the two bonded atoms, rather than either value by itself, helps predict how unevenly the pair is shared.

Covalent and ionic bonding are limiting models on a continuum. As the electronegativity difference between bonded atoms increases, electron sharing becomes more unequal: bond polarity and ionic character increase.

Using Ba 0.9, Be 1.5 and Cl 3.0, $\Delta EN$ is 2.1 for Ba–Cl but 1.5 for Be–Cl. BaCl$_2$ therefore has more ionic character, while BeCl$_2$ has more covalent character.

CO$_2$ has two polar C=O bonds but is linear, so equal opposing dipoles cancel and the molecule is non-polar. H$_2$O is bent, so its O–H dipoles do not cancel and the molecule is polar. Symmetrical BF$_3$ and CCl$_4$ are also non-polar despite polar bonds.

Repulsion strength follows lone pair–lone pair > lone pair–bond pair > bond pair–bond pair because a lone pair is held by only one nucleus and occupies more space near the central atom.

Bond angle is the angle between two bonds that meet at the same central atom. The three atoms defining it must be identified; for H–O–H, oxygen is the vertex.

A bond length is an equilibrium value because attraction and repulsion balance at the minimum-energy separation. Bond angles describe the spatial arrangement produced by electron-region repulsions.

For two to six regions with no lone pairs, maximum separation gives linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral arrangements. Lone pairs compress bond angles: CH$_4$ 109.5° becomes NH$_3$ 107° and H$_2$O 104.5° as lone pairs are added.

Include the ion's charge when counting electrons. After arranging all electron regions, ignore lone-pair positions when naming the molecular shape.

Outer electrons are no longer attached to one particular atom or one ion–ion pair. They move throughout the lattice, while the total negative charge of the electrons balances the total positive charge of the ions.

In a particle diagram, show closely packed positive ions in repeating layers and many electrons distributed between them. The diagram represents a continuous giant structure, not separate molecules.

Each ion is attracted to the shared electron sea around many ions. The attraction is non-directional and extends throughout the giant lattice.

More delocalised electrons per ion and a higher ion charge can strengthen the attraction; smaller metal ions also bring charge centres closer, when other structural factors are comparable.

Magnesium usually has stronger metallic bonding than sodium because each Mg atom contributes two delocalised electrons and forms Mg$^{2+}$, whereas sodium contributes one and forms Na$^+$; Mg$^{2+}$ is also smaller. The attraction is therefore stronger and more energy is required to overcome it.

Metals conduct as solids and when molten because their charge carriers are electrons, which remain mobile in both states. This contrasts with an ionic solid, whose ions are fixed until it melts or dissolves.

Concentrated hydrochloric acid is corrosive: that is its hazard. Using a large open beaker near eye level creates a higher risk than handling a few drops in a tray while wearing eye protection, because exposure is more likely or consequential.

Risk assessment therefore asks: what harm can occur, who or what could be exposed, how likely and severe is it, and which controls reduce that risk to an acceptable level?

A risk assessment identifies the substances, quantities, concentrations, temperatures, apparatus, exposure routes and people involved. It then selects controls, records emergency action and considers disposal before work begins.

Organic compounds are often volatile and flammable, so vapour can spread beyond the vessel and meet an ignition source. The same chemical can present very different risks at microscale and at bulk scale.

Prefer controls that remove or contain the hazard before relying only on personal protective equipment. Check ventilation, heating method, secure apparatus, spill response and waste route for the actual procedure.

For carbon monoxide, improve containment and ventilation, use a fume cupboard or suitable extraction, monitor where necessary, and limit exposure time. Gloves alone do not control an inhalation hazard.

A functional group is the atom or group of atoms responsible for the characteristic reactions of an organic compound. Examples include C=C in alkenes, –OH in alcohols and –COOH in carboxylic acids.

Choose the longest parent chain containing the principal functional group and any required multiple bond. Number from the end giving the lowest relevant locants. Name and alphabetise substituents, use di-, tri- or tetra- for repeats, then add the suffix and locant for the functional group.

CH$_3$C(CH$_3$)$_2$CH$_2$CH(CH$_3$)CH$_3$ has a five-carbon parent chain and methyl groups at 2, 2 and 4, so its name is 2,2,4-trimethylpentane. Commas separate numbers and hyphens separate numbers from words.

Compare bonds and functional groups in reactants and products. Classify the actual transformation, not the reagent name: converting C=C to C–C by adding atoms is addition, while replacing C–H by C–Cl is substitution.

Under ultraviolet light, Cl–Cl can split homolytically: $\mathrm{Cl_2 \rightarrow 2Cl\boldsymbol{\cdot}}$. H–Cl can be represented as breaking heterolytically to H$^+$ and Cl$^-$ when both electrons go to chlorine.

After drawing arrows, count electrons and charges in every product. A radical has an unpaired electron; heterolysis produces opposite charges whose total equals the reactant charge.

The unpaired electron makes many radicals highly reactive. An electrophile is electron-deficient and is attracted to an electron-rich region such as a lone pair or a $\pi$ bond.

A hydrocarbon contains carbon and hydrogen only. Saturated means that it contains no carbon–carbon multiple bond, so each carbon has the maximum number of hydrogen atoms allowed by its C–C connectivity.

Closing an alkane chain to make one ring removes two hydrogen atoms, which explains the change from C$_n$H$_{2n+2}$ to C$_n$H$_{2n}$. For example, propane is C$_3$H$_8$ and cyclopropane is C$_3$H$_6$.

C$_4$H$_{10}$ can be CH$_3$CH$_2$CH$_2$CH$_3$ (butane) or CH$_3$CH(CH$_3$)CH$_3$ (2-methylpropane). Both contain four carbons and ten hydrogens, but the carbon skeletons differ.

To find isomers, vary the carbon skeleton, functional-group position or functional group where the formula permits it. After each drawing, recount atoms, check valencies, assign an IUPAC name and reject any structure that is just a rotated or renumbered duplicate.

For a cycloalkane, choose a ring size from three up to the total carbon count, then distribute the remaining carbons as substituents. Number substituted rings to give the lowest set of locants and reject rotations, reflections and alternative numbering of the same connectivity.

C$_4$H$_8$ gives cyclobutane and methylcyclopropane. For C$_5$H$_{10}$, valid connectivities include cyclopentane, methylcyclobutane, ethylcyclopropane, 1,1-dimethylcyclopropane and 1,2-dimethylcyclopropane. This same ring-size method extends to six carbons.

Use displayed, structural or skeletal formulae consistently. In a skeletal ring, every unlabelled vertex is carbon and enough hydrogens are implied to give carbon four bonds.

Alkanes burn exothermically, so fractions containing them are used as fuels. Supply and demand rarely match crude-oil composition, so cracking increases smaller fuels and alkene feedstocks, while reforming improves combustion quality.

For every cracking or reforming equation, conserve the number of carbon and hydrogen atoms. Fractional distillation has no reaction equation because no covalent bonds change.

Pollutant formation depends on fuel composition and combustion conditions. Improving oxygen mixing can reduce CO and soot, but high combustion temperature can favour nitrogen-oxide formation.

Acid deposition lowers the pH of soils and surface waters and can damage carbonate stone and living systems. In an enclosed space, incomplete combustion makes CO especially dangerous because it is colourless and toxic.

Alternative fuels are developed to reduce dependence on finite crude oil, improve long-term security of supply and reduce harmful or greenhouse-gas emissions over the fuel's life cycle.

Combustion of fossil alkanes transfers geologically stored carbon to atmospheric CO$_2$. CO$_2$ absorbs outgoing infrared radiation, so increasing its concentration strengthens the greenhouse effect and contributes to climate change.

State the system boundary, identify every carbon or energy input, compare atmospheric uptake with emissions, and give a conditional conclusion rather than relying on a label such as 'bio' or 'hydrogen'.

To balance complete combustion of C$_x$H$_y$, form $x$CO$_2$ and $y/2$H$_2$O, then balance O$_2$. Include state symbols when requested.

The carbon skeleton remains while a C–H bond and X–X bond are replaced by C–X and H–X bonds. Ultraviolet radiation initiates radical formation.

A curly half-arrow moves one electron. In each propagation step, use half-arrows so bond breaking and bond formation account for the unpaired electron without inventing charge.

The product can undergo further substitution because it still contains C–H bonds. Longer alkanes can also substitute at different carbon positions. The resulting mixture of products and isomers makes the reaction poorly selective for synthesis.

The C=C consists of one $\sigma$ bond and one $\pi$ bond. Addition reactions break the weaker $\pi$ component and form two new $\sigma$ bonds, converting the two carbon centres from double-bonded to single-bonded.