Valence electrons · Bonds & bond order · The 4-step method · Counting electrons · Polyatomic ions · Formal charge & resonance · Octet exceptions — 75 cards
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The electrons in an atom's outermost shell. They are the only electrons that take part in bonding.
Use the last digit of its group number: group 1 → 1, group 2 → 2, groups 13–18 → 3 to 8.
H 1 · C 4 · N 5 · O 6
F 7 · Cl 7 · S 6 · P 5
The element symbol with its valence electrons drawn as dots around it. Place one dot on each of the 4 sides first, then pair them up.
Atoms bond so that each ends up with 8 electrons around it, like a noble gas.
2, called a duet (like helium).
C 4 · N 3 · O 2 · halogen 1. That is the number of electrons each is missing from an octet.
3 bonds + 1 lone pair.
2 bonds + 2 lone pairs.
A drawing of a molecule or ion that shows how the atoms share valence electrons so that each gets an octet (a duet for H).
A shared pair of electrons between two atoms. It counts toward the octet of both atoms.
A pair of electrons that belongs to only one atom and is not involved in bonding.
2 electrons (one shared pair).
Double: 4 (2 pairs). Triple: 6 (3 pairs).
The number of bonds between two atoms: single = 1, double = 2, triple = 3.
Length gets shorter; strength (the energy needed to break it) gets higher.
C≡C (1.203 Å) < C=C (1.339 Å) < C–C (1.535 Å).
C–C (365 kJ/mol) < C=C (598 kJ/mol) < C≡C (962 kJ/mol).
N≡N: 3 · O=O: 2 · F–F: 1.
The C=O bond. A higher bond order means a shorter bond.
Count the total valence electrons of all the atoms (adjusted for the charge if it's an ion).
Connect the atoms with single bonds to make the skeleton, with the central atom in the middle.
Add lone pairs to the outer atoms to complete their octets. Put any leftover electrons on the central atom.
If the central atom still lacks an octet, move lone pairs from outer atoms into double or triple bonds.
The least electronegative atom, which usually forms the most bonds.
H and F. Each can form only one bond.
No. It only moves an existing lone pair so that it becomes shared.
(1) Electrons drawn = electrons counted. (2) Every atom has 8 electrons around it (H: 2).
After step 3, both O atoms are full but C has only 4 electrons. Moving one lone pair from each O gives O=C=O.
After step 3, C has only 4 electrons. Moving two lone pairs from N gives C≡N, so C and N both have 8.
2(1) + 6 = 8
5 + 3(1) = 8
4 + 4(1) = 8
4 + 2(6) = 16
4 + 4(1) + 6 = 14
1 + 4 + 5 = 10
4 + 2(1) + 6 = 12
5 + 3(7) = 26
2(4) + 4(1) = 12
2(5) = 10
A negative charge: add 1 electron per − charge. A positive charge: subtract 1 per + charge.
Put square brackets around the whole structure and write the charge outside, top right.
Covalent bonds (shared pairs). The whole group carries one overall charge.
6 + 1 + 1 = 8
5 + 4 − 1 = 8
6 + 3 − 1 = 8
4 + 5 + 1 = 10
5 + 18 + 1 = 24
4 + 18 + 2 = 24
6 + 24 + 2 = 32
5 + 24 + 3 = 32
5 + 12 + 1 = 18
[O–H]⁻ with 3 lone pairs on O.
[N with 4 single bonds to H]⁺, and no lone pairs.
[:C≡N:]⁻, a triple bond with one lone pair on each atom.
N in the centre; one N=O (O has 2 lone pairs) and two N–O (each O has 3 lone pairs); brackets, − charge.
C in the centre; one C=O and two C–O. Same pattern as nitrate, with a 2− charge.
S in the centre with 4 single bonds to O; each O has 3 lone pairs; S has none; brackets, 2− charge.
FC = V − L − ½B, where V = valence electrons of the free atom, L = electrons in its lone pairs, and B = electrons in its bonds. So ½B is simply the number of bonds.
The overall charge of the molecule or ion (0 for a neutral molecule).
The best structure has formal charges closest to zero, with any negative charge on the more electronegative atom.
6 − 6 − 1 = −1
5 − 0 − 4 = +1
N +1; O in N=O 0; each O in N–O −1. Sum = −1 ✓
S +2; each O −1. Sum = +2 − 4 = −2 ✓
Both give octets, but O=C=O has all formal charges 0; O≡C–O has +1 and −1.
Two or more valid Lewis structures that differ only in where the electrons (double bonds) are placed. They are joined with ↔.
3. The double bond can go to any of the 3 O atoms.
It is a blend: all three N–O bonds are identical, somewhere between a single and a double bond (bond order 1⅓).
2
Incomplete octet · odd number of electrons · expanded octet.
BF₃: B has only 6 electrons. BeCl₂: Be has only 4.
NO or NO₂. The total is odd, so one atom must have 7 electrons.
Atoms in period 3 or below (P, S, Cl…), e.g. PCl₅ (10 e⁻ on P) and SF₆ (12 e⁻ on S). beyond NCERT
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