MHT-CET Chemistry · Formula sheet

Chemical Bonding and Molecular Structure formulas

6 formulas, 10 reference tables and 41 common traps for MHT-CET Chemistry Chemical Bonding and Molecular Structure, grouped by subtopic.

Full notes with worked examples

Ionic and Covalent Bonding, Lewis Structures and Octet Rule

Learn this subtopic in the notes

Formal charge on an atom in a Lewis structure

Formal charge

\text{FC} = V - L - \tfrac{1}{2}\,B

$V$ valence electrons of the free atom
$L$ lone-pair (non-bonding) electrons on the atom
$B$ bonding electrons around the atom (2 per single bond)

The octet rule and the three ways atoms bond

Bond type	How the octet is reached	Formed between	Example
Ionic (electrovalent)	Electrons transferred (lost / gained)	Metal + non-metal	$\text{NaCl}$ , $\text{MgO}$
Covalent	One pair shared, one electron from each atom	Non-metal + non-metal	$\text{H}_2\text{O}$ , $\text{SCl}_2$
Coordinate (dative)	Shared pair donated by one atom only	Donor with a lone pair	$\text{NH}_4^{+}$ , $\text{H}_3\text{O}^{+}$ Once formed, a coordinate bond is identical to any ordinary covalent bond — the label only records where the pair came from.

Transfer = ionic; share = covalent; one-sided share = coordinate.

Fajans' rules — covalent character of an ionic bond

Factor	Effect on covalent character	Bank example
Smaller cation	More covalent (stronger polariser)	$\text{LiI}$ most covalent among LiCl, LiI, NaCl, NaIQ
Larger anion	More covalent → least ionic	$\text{MI}$ has the lowest ionic character ( $\text{MF}>\text{MCl}>\text{MBr}>\text{MI}$ )Q
Higher cation charge	More covalent	$\text{SnCl}_4$ more covalent than $\text{SnCl}_2$ , $\text{PbCl}_2$ , $\text{SbCl}_3$ Q
Small + highly-charged ions	Highest lattice enthalpy	$\text{BeF}_2$ highest among LiCl, NaCl, $\text{BeF}_2$ , $\text{CaCl}_2$ Q Lattice enthalpy scales with charge density (charge / size), the same driver as polarising power.

Small cation, large anion, high cation charge — all push an ionic bond toward covalent.

Exceptions to the octet rule

Exception type	Electron count on central atom	Examples
Incomplete octet	Fewer than 8	$\text{BF}_3$ , $\text{BeCl}_2$ , $\text{LiCl}$ Q $\text{LiCl}$ is quoted as incomplete because $\text{Li}^{+}$ has a 2-electron duplet, not an octet.
Expanded octet	More than 8 (uses d-orbitals)	$\text{PCl}_5$ , $\text{SF}_6$ , $\text{H}_2\text{SO}_4$
Odd-electron molecule	Odd total → one unpaired electron	$\text{NO}$ , $\text{NO}_2$ Q
Obeys the octet (for contrast)	Exactly 8	$\text{SCl}_2$ , $\text{H}_2\text{O}$ , $\text{CH}_4$ Q

Fewer than 8 = incomplete; more than 8 = expanded; odd total = odd-electron.

Lewis structures, resonance count and electrons around an atom

Species / term	Key count or definition	Answer the bank wants
$\text{NO}_2^{-}$ (nitrite)	Double bond can sit on either O	2 resonance (Lewis) structuresQ
Electrons around S in $\text{H}_2\text{SO}_4$	2 single + 2 double bonds = 4 bonds	12 electronsQ
Lewis acid	Electron-pair acceptor	Accepts an electron pair (not 'donates $\text{H}^{+}$ ')Q A Lewis acid need not contain hydrogen — $\text{BF}_3$ is a Lewis acid because boron accepts a lone pair.
Lewis base	Electron-pair donor	Donates an electron pair (e.g. $\text{NH}_3$ )

1 bond = 2 electrons; resonance = the count of equivalent double-bond placements.

Common traps

A coordinate bond is still a covalent bond

A coordinate (dative) bond shares a pair of electrons exactly like an ordinary covalent bond — the only difference is that one atom supplied both electrons. Don't count it as a separate third kind of bonding force.

Duplet for H and Li, octet for the rest

Hydrogen and lithium are 'complete' with just 2 outer electrons (a duplet, like helium), not 8. So

\text{Li}^{+}

has a stable duplet even though it has no octet — this is why LiCl is quoted as an 'incomplete octet' example while still being perfectly stable.

Ionic character is the reverse of covalent character

The bank flips the question between 'most covalent' and 'least ionic' — they are the same answer. If MI is the most covalent halide it is automatically the least ionic. Read which one is asked, but the winning compound is identical.

Charge density, not molar mass, sets lattice enthalpy

\text{BeF}_2

beats

\text{CaCl}_2

on lattice enthalpy because

\text{Be}^{2+}

and

\text{F}^{-}

are tiny and highly charged, not because of formula mass. Rank by charge / size (charge density), the same quantity that drives Fajans' rules.

Odd electrons cannot complete an octet

NO has 11 valence electrons and

\text{NO}_2

has 17 — an odd count leaves one electron unpaired, so these can never reach a full octet. Any molecule with an odd valence-electron total is an octet exception by definition.

Expanded octet needs period-3 (or lower) and d-orbitals

Only central atoms from period 3 downward (S, P, Cl...) can expand past 8 using empty d-orbitals. Second-period atoms (C, N, O, F) can never exceed an octet — a tempting distractor to reject.

Lewis acid = electron-pair acceptor, NOT proton donor

A Lewis acid accepts an electron pair; 'gives

\text{H}^{+}

' and 'donates a proton' describe a Brønsted acid.

\text{BF}_3

has no hydrogen yet is a strong Lewis acid — pick 'accepts electron pair'.

A double bond is 4 electrons when you total around an atom

When counting electrons around a central atom, a single bond contributes 2 and a double bond contributes 4. For

\text{H}_2\text{SO}_4

the two S=O double bonds add 8, not 4 — giving 12 total, not 8.

Use half the bonding electrons, not all of them

Formal charge splits each bond evenly, so an atom keeps only half its bonding electrons:

\text{FC} = V - L - \tfrac{1}{2}B

. Forgetting the

\tfrac{1}{2}

doubles the bonding contribution and gives a wrong sign or magnitude.

Count lone-pair electrons, not lone pairs

L

is the number of non-bonding electrons, so one lone pair contributes 2, not 1. For carbon in

\text{CO}_2

(no lone pairs)

L = 0

, giving the clean FC

= 0

the bank expects.

Hybridization

Learn this subtopic in the notes

Determining a central atom's hybridization

Steric number

\text{SN} = (\sigma\text{-bonded atoms}) + (\text{lone pairs on central atom})

$\text{SN}$ steric number, which fixes the hybridization
$\sigma\text{-bonded atoms}$ atoms directly bonded (each multiple bond counts once)
$\text{lone pairs}$ non-bonding electron pairs on the central atom

Valence bond theory: sigma and pi bonds

Bond	Sigma and pi	Example
Single bond	$1\sigma$	$\text{H}-\text{H}$ ; C-C in ethane
Double bond	$1\sigma + 1\pi$	$\text{C}=\text{C}$ in ethene; $\text{O}=\text{O}$
Triple bond	$1\sigma + 2\pi$	$\text{C}\equiv\text{C}$ in ethyne; $\text{N}\equiv\text{N}$

The first bond between two atoms is always a sigma bond; any extra bonds are pi.

The steric-number master table

Steric number	Hybridization	Geometry	Bond angle	Example
2	$sp$	Linear	$180^\circ$	$BeCl_2$ , $C_2H_2$ Q Acetylene $C_2H_2$ has $sp$ carbons (two atoms + one triple bond that counts as one $\sigma$ ) — the bank's classic $sp$ example.
3	$sp^2$	Trigonal planar	$120^\circ$	$BF_3$ , $C_2H_4$ Q Trigonal planar geometry means $sp^2$ — the answer to 'which hybridisation gives trigonal geometry'.
4	$sp^3$	Tetrahedral	$109.5^\circ$	$CH_4$ , $NH_3$ , $H_2O$
5	$sp^3d$	Trigonal bipyramidal	$90^\circ,\,120^\circ$	$PCl_5$ , $SF_4$ Q $SF_4$ is $sp^3d$ (4 bond pairs + 1 lone pair = SN 5); the lone pair distorts it to a see-saw shape but the hybridization stays $sp^3d$ .
6	$sp^3d^2$	Octahedral	$90^\circ$	$SF_6$ , $XeF_4$ Q $XeF_4$ is $sp^3d^2$ (4 bond pairs + 2 lone pairs = SN 6), square planar — NOT $sp^3$ .

Count the steric number first; the row it lands in gives the hybridization, geometry and angle.

Common traps

The first bond is always sigma

In any single, double or triple bond the FIRST bond is a sigma bond; only the additional bonds are pi. So a double bond is

1\sigma + 1\pi

(not

2\pi

), and a triple bond is

1\sigma + 2\pi

Sigma is stronger than pi; pi locks rotation

Axial overlap in a sigma bond is more effective than the sideways overlap of a pi bond, so a sigma bond is stronger. A pi bond also fixes the geometry — there is no free rotation about a double or triple bond.

Hybrids formed = orbitals mixed, not bonds made

The number of hybrid orbitals equals the number of atomic orbitals that intermix, not the number of bonds. Water's oxygen is

sp^3

(four hybrids) even though it makes only two bonds — the other two hybrids hold lone pairs.

Geometry name reports atoms only; SN includes lone pairs

SF_4

has steric number 5, so it is

sp^3d

— but its shape is called see-saw, not trigonal bipyramidal, because one of the five positions is a lone pair. The hybridization follows the steric number; the shape name follows only the bonded atoms.

d-orbitals only appear from steric number 5

sp^3d

and

sp^3d^2

need d-orbitals and only occur for central atoms with an expanded octet (period 3 and below). If the steric number is 4 or less, the answer is

sp

sp^2

sp^3

— never a d-form.

Count lone pairs on the central atom, not just the atoms

H_2O

and

NH_3

each have only 2 or 3 bonded atoms, yet both are

sp^3

— because oxygen carries 2 lone pairs and nitrogen 1. If you count only the bonded atoms you would wrongly call water

sp

and ammonia

sp^2

. Always add the lone pairs.

A multiple bond is one atom, not two, in the steric count

In acetylene

C_2H_2

each carbon is bonded to one H and one C (a triple bond). The triple bond counts as one

\sigma

bond, so the steric number is 2, giving

sp

— not

sp^3

. Never count the extra

\pi

bonds toward the steric number.

VSEPR Theory and Molecular Geometry

Learn this subtopic in the notes

Counting bond pairs and lone pairs on the central atom

Lone pairs on the central atom

\text{lp} = \frac{V - \text{bp}}{2}

$V$ valence electrons of the central atom
$\text{bp}$ bond pairs = number of atoms bonded to it (single bonds)
$\text{lp}$ lone pairs left on the central atom

The master shape table (AXnEm to geometry)

Type (AXnEm)	Bond pairs / Lone pairs	Shape	Ideal bond angle	Example
$\text{AX}_2$	2 / 0	Linear	$180^\circ$	$\text{BeCl}_2$ , $\text{C}_2\text{H}_2$
$\text{AX}_3$	3 / 0	Trigonal planar	$120^\circ$	$\text{BF}_3$
$\text{AX}_2\text{E}$	2 / 1	Bent (angular)	about $119.5^\circ$	$\text{SO}_2$
$\text{AX}_4$	4 / 0	Tetrahedral	$109.5^\circ$	$\text{CH}_4$ , $\text{SiCl}_4$ , $\text{NH}_4^{+}$ Q
$\text{AX}_3\text{E}$	3 / 1	Trigonal pyramidal	$107^\circ$	$\text{NH}_3$
$\text{AX}_2\text{E}_2$	2 / 2	Bent (angular)	$104.5^\circ$	$\text{H}_2\text{O}$ , $\text{SCl}_2$ Q
$\text{AX}_5$	5 / 0	Trigonal bipyramidal	$120^\circ$ and $90^\circ$	$\text{PCl}_5$ Q
$\text{AX}_4\text{E}$	4 / 1	See-saw	$90^\circ$ , $120^\circ$	$\text{SF}_4$ , $\text{TeF}_4$ Q $\text{AB}_4\text{E}$ has a trigonal-bipyramidal parent geometry but a see-saw shape — the bank tests both the type-to-shape and the parent-geometry versions.
$\text{AX}_3\text{E}_2$	3 / 2	T-shaped	about $90^\circ$	$\text{ClF}_3$ , $\text{BrF}_3$ , $\text{ICl}_3$
$\text{AX}_2\text{E}_3$	2 / 3	Linear	$180^\circ$	$\text{XeF}_2$
$\text{AX}_6$	6 / 0	Octahedral	$90^\circ$	$\text{SF}_6$
$\text{AX}_5\text{E}$	5 / 1	Square pyramidal	about $90^\circ$	$\text{BrF}_5$ , $\text{IF}_5$ Q
$\text{AX}_4\text{E}_2$	4 / 2	Square planar	$90^\circ$	$\text{XeF}_4$ Q

Read off the shape from the AXnEm type: count X (bonded atoms) and E (lone pairs), then look up the row.

Bond angles and how lone pairs shrink them

Molecule	Bond pairs / Lone pairs	Bond angle	Note
$\text{CH}_4$	4 / 0	$109.5^\circ$	Ideal tetrahedral — no lone pair to distort.
$\text{NH}_3$	3 / 1	$107^\circ$	One lone pair shrinks $109.5^\circ$ a little.
$\text{H}_2\text{O}$	2 / 2	$104.5^\circ$	Two lone pairs shrink it further.
$\text{BF}_3$	3 / 0	$120^\circ$	Trigonal planar, no lone pair — full angle.Q
$\text{SO}_2$	2 / 1	about $119.5^\circ$	Bent; one lone pair barely dents the $120^\circ$ parent.Q SO $_2$ is the O–S–O $119.5^\circ$ the bank tests — not $109.5^\circ$ or $180^\circ$ ; its parent is trigonal, not tetrahedral.

Take the ideal angle for the parent geometry, then subtract for each lone pair.

Common traps

Lone pairs count toward the electron geometry but not the described shape

The parent (electron-pair) geometry counts every pair, but the reported shape names only where the atoms sit.

\text{XeF}_4

has an octahedral electron geometry, yet its shape is square planar — the 2 lone pairs occupy positions but aren't drawn as part of the shape. Always answer with the atom-only shape unless the question asks for the parent geometry.

'Regular geometry as expected' means zero lone pairs

When the bank asks which molecule has its regular or expected geometry, it wants the one with no lone pairs on the central atom —

\text{SiCl}_4

, not

\text{SF}_4

\text{BrF}_5

\text{XeF}_4

. A lone pair always distorts, so any lone-pair molecule is disqualified.

Count lone pairs on the central atom only

For IF, the question asks for lone pairs on the central iodine: I has

V=7

, one electron goes into the I–F bond, leaving

6/2 = 3

lone pairs on I. Don't add the 3 lone pairs sitting on F — the central-atom count is 3.

BF₃ has zero lone pairs — boron is electron-deficient

Boron has only 3 valence electrons and forms 3 bonds, so nothing is left over — 0 lone pairs (an incomplete octet with 6 electrons). Students often assume every central atom carries a lone pair;

\text{BF}_3

\text{SF}_6

and

\text{PCl}_5

are common zero-lone-pair molecules.

H₂O is bent, not linear

Water is

\text{AX}_2\text{E}_2

: the 2 lone pairs on oxygen push the two O–H bonds down to about

104.5^\circ

, giving a bent shape — not the

180^\circ

linear shape you might expect from just 'two bonds'. Its shape-twin in the bank is

\text{SCl}_2

, also bent.

SF₄ is not tetrahedral — it has a lone pair

\text{SF}_4

has 4 bonded atoms but S carries 1 lone pair (

\text{AX}_4\text{E}

), so 5 electron domains give a see-saw shape, not tetrahedral. Only the zero-lone-pair

\text{AX}_4

molecules (

\text{CH}_4

\text{SiCl}_4

\text{NH}_4^{+}

) are tetrahedral.

Parent geometry versus molecular shape

For

\text{TeF}_4

(

\text{AX}_4\text{E}

) the parent geometry is trigonal bipyramidal (5 domains) but the molecular shape is see-saw. If the question says 'geometry', answer the parent trigonal bipyramidal; if it says 'shape', answer see-saw. Read the wording.

SO₂ is 119.5°, not 109.5°

\text{SO}_2

has a trigonal (not tetrahedral) parent — 2 bond pairs and 1 lone pair around S — so its O–S–O angle is about

119.5^\circ

, close to the

120^\circ

trigonal ideal. The

107.5^\circ

109^\circ

distractors are tetrahedral-family angles that don't apply here.

More lone pairs, smaller angle

Because a lone pair repels harder than a bond pair, adding lone pairs to the same parent geometry always shrinks the bond angle:

\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}

. Don't quote

109.5^\circ

for all three — only the zero-lone-pair member keeps the ideal.

Molecular Orbital Theory and Bond Order

Learn this subtopic in the notes

Bond order from the MO configuration

Bond order

\text{Bond order} = \tfrac{1}{2}\left(N_b - N_a\right)

$N_b$ number of electrons in bonding molecular orbitals
$N_a$ number of electrons in antibonding molecular orbitals

Magnetic behaviour, bond length and stability

Bond order controls length and strength

\text{Bond order} \uparrow \;\Rightarrow\; \text{bond length} \downarrow,\quad \text{bond strength} \uparrow

Bond order and magnetic nature of common species

Species	Total electrons	Bond order	Magnetic nature
$\text{H}_2$	2	1	Diamagnetic
$\text{Li}_2$	6	1	Diamagnetic
$\text{N}_2$	14	3	Diamagnetic
$\text{N}_2^+$	13	2.5	Paramagnetic One electron removed from a bonding orbital, so bond order drops to 2.5.
$\text{O}_2$	16	2	Paramagnetic Two unpaired electrons in $\pi^*2p$ — the classic paramagnetic diatomic.
$\text{O}_2^+$	15	2.5	Paramagnetic
$\text{O}_2^-$	17	1.5	Paramagnetic
$\text{F}_2$	18	1	Diamagnetic
$\text{CO}$	14	3	Diamagnetic Isoelectronic with $\text{N}_2$ ; MOT gives bond order 3, not the Lewis double bond.
$\text{NO}$	15	2.5	Paramagnetic Odd-electron molecule: one unpaired electron in a $\pi^*2p$ orbital.

Bond order rises to a maximum of 3 at

\text{N}_2

/CO; paramagnetic species are the ones with an unpaired electron.

Common traps

Count TOTAL electrons, and adjust for an ion's charge

Always start from the total electron count.

\text{N}_2

has 14,

\text{O}_2

has 16 — but an ion shifts this:

\text{N}_2^+

has 13 (one removed),

\text{O}_2^-

has 17 (one added). Filling the wrong number of electrons is the single biggest source of wrong bond orders here.

Only σ* and π* orbitals count as antibonding

When asked for antibonding electrons in

\text{N}_2

, count only

\sigma^*1s

and

\sigma^*2s

(= 4). The

\pi^*2p

orbitals are empty in

\text{N}_2

, so do not add them; and never count the bonding

\sigma

\pi

electrons here.

Ions can have a fractional bond order

Removing or adding one electron changes

N_b - N_a

by 1, so the bond order shifts by

\tfrac12

\text{N}_2^+

and

\text{O}_2^+

are both 2.5,

\text{O}_2^-

is 1.5 — a half-integer answer is correct, not an arithmetic slip. Do not round it.

MOT bond order can differ from the Lewis picture

The Lewis structure of CO looks like a double bond, but MOT gives bond order 3 (CO is isoelectronic with

\text{N}_2

). Trust the MO count

\tfrac12(N_b - N_a)

over a quick Lewis guess.

O2 is paramagnetic — the two unpaired electrons

The Lewis structure of

\text{O}_2

shows all electrons paired, but MOT places two unpaired electrons in the

\pi^*2p_x

and

\pi^*2p_y

orbitals (one each, by Hund's rule). So

\text{O}_2

is paramagnetic — a classic exam favourite. Never call it diamagnetic.

Higher bond order = shorter bond, not longer

Bond length runs opposite to bond order.

\text{N}_2

(BO 3) has the shortest, strongest bond;

\text{Cl}_2

(BO 1) the longest, weakest. In a 'decreasing bond length' question the order is the reverse of the bond-order order.

O2 is paramagnetic even though its bond order is a whole number

Bond order being an integer (2) does not make

\text{O}_2

diamagnetic — magnetic nature depends on unpaired electrons, not on the bond order.

\text{O}_2

has two unpaired

\pi^*2p

electrons, so it is paramagnetic. Read the two properties independently.

Dipole Moment, Polarity and Intermolecular Forces

Learn this subtopic in the notes

Dipole moment: definition and comparison

Dipole moment

\mu = q \times d

$\mu$ dipole moment (debye, D)
$q$ magnitude of the separated charge
$d$ distance between the positive and negative charge centres

Types of intermolecular force

Force	Acts between	Strength	Example pair
London / dispersion	Any molecules (even non-polar)	Weakest (grows with size)	CH4 + C2H6 Present in every substance; the ONLY force in non-polar molecules. Largest among HX in HI (biggest, most polarisable).
Dipole–induced dipole (Debye)	One polar + one non-polar molecule	Weak	NH3 + C6H6Q
Dipole–dipole	Two polar molecules	Moderate (bigger dipole → stronger)	HF, HCl (polar HX) Strongest dipole–dipole among the hydrogen halides is HF, because F gives the largest bond dipole.
Hydrogen bonding	H on N/O/F, near a lone pair on N/O/F	Strongest of these	H2O, NH3, HF, alcohols

Polar–polar → dipole–dipole; polar–non-polar → dipole-induced dipole; non-polar only → dispersion; H on N/O/F → hydrogen bond.

Common traps

Dipole moment is a vector — add directions, not magnitudes

NF3 has very polar N–F bonds yet a tiny dipole moment (~0.23 D) because its lone-pair dipole points opposite to the resultant of the N–F bond dipoles and nearly cancels it. In NH3 the two point the same way and add. Always sum the bond dipoles as vectors, and remember the lone pair contributes too.

Bigger electronegativity difference → bigger bond dipole

Down a group electronegativity falls, so the bond dipole falls:

\mu(\text{CH}_3\text{F}) > \mu(\text{CH}_3\text{Cl}) > \mu(\text{CH}_3\text{Br}) > \mu(\text{CH}_3\text{I})

. Don't rank by molecular mass — the heaviest (CH3I) has the smallest dipole.

Polar bonds do NOT guarantee a polar molecule

BF3, CCl4, CH4 and CO2 are built from polar bonds yet have zero net dipole because their symmetric shapes make the bond dipoles cancel. Never answer 'it has a net dipole' just because you see electronegative atoms — check the shape first.

CHCl3 is polar; CCl4 is not

The single most-tested pair here.

\text{CCl}_4

is perfectly tetrahedral, so its four C–Cl dipoles cancel →

\mu = 0

\text{CHCl}_3

swaps one Cl for H, breaking that symmetry, so it has a net dipole. Same atoms, opposite answer — symmetry decides.

Dipole–dipole vs dispersion point to different HX

For the hydrogen halides, the strongest dipole–dipole force is in HF (largest bond dipole), but the largest dispersion force is in HI (biggest, most polarisable). The question wording decides — read whether it asks for dipole–dipole or for total van der Waals / dispersion.

Match the force to the pair's polarity

Two polar molecules → dipole–dipole. One polar + one non-polar → dipole-induced dipole. Both non-polar → dispersion only. Check the polarity of both species before naming the force.

No H on N/O/F means no hydrogen-bond donor

A molecule can contain N, O or F and still not hydrogen-bond as a donor if there is no H directly on that atom. Ethers (CH3–O–CH3) and tertiary amines (no N–H) cannot donate a hydrogen bond, so they boil lower than alcohols and primary/secondary amines of similar size.

H2S does not hydrogen-bond like H2O

Students extend H-bonding to

\text{H}_2\text{S}

by analogy with water, but sulphur is large and not electronegative enough —

\text{H}_2\text{S}

shows only weak dipole–dipole/dispersion forces, which is exactly why it is a gas while water is a liquid.

Chemical Bonding and Molecular Structure formulas

Ionic and Covalent Bonding, Lewis Structures and Octet Rule

Formal charge on an atom in a Lewis structure

The octet rule and the three ways atoms bond

Fajans' rules — covalent character of an ionic bond

Exceptions to the octet rule

Lewis structures, resonance count and electrons around an atom

Common traps

Hybridization

Determining a central atom's hybridization

Valence bond theory: sigma and pi bonds

The steric-number master table

Common traps

VSEPR Theory and Molecular Geometry

Counting bond pairs and lone pairs on the central atom

The master shape table (AXnEm to geometry)

Bond angles and how lone pairs shrink them

Common traps

Molecular Orbital Theory and Bond Order

Bond order from the MO configuration

Magnetic behaviour, bond length and stability

Bond order and magnetic nature of common species

Common traps

Dipole Moment, Polarity and Intermolecular Forces

Dipole moment: definition and comparison

Types of intermolecular force

Common traps

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