Class 11 Chemistry · Original EZT practice
Chemical Bonding and Molecular Structure (Class 11 Chapter 4): summary and practice
Chemical Bonding explains why atoms stick and what shapes they make: the octet rule and Lewis dots, ionic and covalent bonds, bond parameters, VSEPR geometries, hybridisation, and molecular orbital theory with bond order. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Kossel-Lewis approach
Atoms bond to reach eight valence electrons - the octet rule. Lewis dot symbols track the valence electrons, formal charge assigns each atom its share (valence - non-bonding - half the bonding electrons), and ionic bonds form when electrons transfer outright, held by lattice enthalpy.
2. Bond parameters
Every bond carries measurable parameters: length, angle, enthalpy and order. Resonance spreads double-bond character over several positions - ozone's two bonds come out equal, each effectively one and a half.
3. VSEPR theory
Electron pairs repel and settle as far apart as possible: two pairs linear, three trigonal planar, four tetrahedral, five trigonal bipyramidal, six octahedral. Lone pairs squeeze harder than bond pairs, so NH3 compresses to 107 degrees and H2O to 104.5.
4. Valence bond theory and hybridisation
Valence bond theory builds bonds from orbital overlap - sigma head-on, pi sideways. Hybridisation mixes s and p orbitals into equivalent sp, sp2 or sp3 sets, matching the geometries VSEPR predicts: sp linear, sp2 trigonal planar, sp3 tetrahedral.
5. Molecular orbital theory and hydrogen bonding
MO theory combines atomic orbitals into bonding and antibonding orbitals across the whole molecule; bond order = (Nb - Na)/2 explains why N2 is triple-bonded and O2 paramagnetic. Section 4.9 closes with hydrogen bonding, the weak link that shapes water and ice.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
MO bond order of O2
Choose the method: Bond order = (Nb - Na) / 2 from the MO configuration.
EZT practice question
The MO configuration of O2 places 10 electrons in bonding and 6 in antibonding orbitals. Find its bond order.
Show the worked solution
- Bond order = (Nb - Na) / 2.
- = (10 - 6) / 2.
Answer: 2 - a double bond, matching the O = O Lewis picture.
Check: The two unpaired antibonding electrons also explain why O2 is paramagnetic, something valence bond theory cannot show.
Watch for this: Count only the electrons in the MO levels the chapter lists - the 1s cores cancel out.
Now try it yourself
Bond order of N2 (10 bonding, 4 antibonding)?
Check your answer
3.
(10 - 4) / 2 = 3, a triple bond.
Removing an electron changes bond order
Choose the method: Track which orbital loses the electron, then recompute (Nb - Na) / 2.
EZT practice question
O2+ loses one electron from an antibonding orbital, giving 10 bonding and 5 antibonding. Find its bond order.
Show the worked solution
- Bond order = (10 - 5) / 2.
- = 5 / 2.
Answer: 2.5 - removing an antibonding electron STRENGTHENS the bond.
Check: O2+ has a shorter, stronger bond than O2 - the opposite of what 'losing something weakens' intuition says.
Watch for this: Check WHERE the electron leaves from - removing a bonding electron would lower the order instead.
Now try it yourself
CO has the same MO count as N2 (10 bonding, 4 antibonding). Bond order?
Check your answer
3.
Isoelectronic with N2, so (10 - 4) / 2 = 3.
Assign a formal charge
Choose the method: FC = valence electrons - non-bonding electrons - half the bonding electrons.
EZT practice question
In ozone, the central oxygen holds 2 non-bonding electrons and makes 3 bonds. Find its formal charge.
Show the worked solution
- Oxygen's valence is 6.
- FC = 6 - 2 - (6 / 2).
- = 6 - 2 - 3.
Answer: +1.
Check: The singly bonded terminal oxygen carries -1, so the molecule stays neutral overall - charges must sum to the molecule's charge.
Watch for this: Halve the BONDING electrons only - the lone-pair electrons count whole.
Now try it yourself
An oxygen with 4 non-bonding electrons and 2 bonds. Formal charge?
Check your answer
0.
6 - 4 - 2 = 0, like the oxygen in water.
Predict a shape with VSEPR
Choose the method: Count electron pairs, place them, then name the shape by atoms only.
EZT practice question
NH3 has 4 electron pairs around nitrogen, one of them a lone pair. Name its molecular shape and approximate bond angle.
Show the worked solution
- 4 pairs point tetrahedrally.
- One corner is a lone pair, invisible in the shape.
- The lone pair squeezes the bonds below 109.5 degrees.
Answer: Trigonal pyramidal, about 107 degrees.
Check: The chapter's own value is 107 degrees - the electron GEOMETRY is tetrahedral but the molecular SHAPE is pyramidal.
Watch for this: Do not call NH3 tetrahedral - that describes the electron pairs, not the atom positions.
Now try it yourself
H2O has 4 pairs, two of them lone pairs. Shape and angle?
Check your answer
Bent, about 104.5 degrees.
Two lone pairs squeeze harder, closing the angle further to 104.5 degrees.
Identify the hybridisation
Choose the method: Match the electron-pair count to sp, sp2 or sp3.
EZT practice question
BF3 has three bonding pairs and no lone pair on boron, trigonal planar at 120 degrees. What is boron's hybridisation?
Show the worked solution
- Three electron domains need three equivalent orbitals.
- One s mixes with two p orbitals.
Answer: sp2.
Check: sp3 covers four domains (CH4), sp covers two (BeCl2) - the count alone fixes the letter.
Watch for this: Count DOMAINS, not atoms - a double bond still counts as one domain.
Now try it yourself
Hybridisation of carbon in CH4?
Check your answer
sp3.
Four bonding domains mix one s with all three p orbitals.
Count sigma and pi bonds
Choose the method: First bond between two atoms is sigma; extras are pi.
EZT practice question
How many sigma and pi bonds does a molecule of N2 contain?
Show the worked solution
- N2 has a triple bond.
- One head-on overlap plus two sideways overlaps.
Answer: 1 sigma and 2 pi bonds.
Check: The same rule gives ethene's double bond as 1 sigma + 1 pi between the carbons.
Watch for this: Pi bonds come from SIDEWAYS p-orbital overlap - a second sigma bond between the same atoms does not exist.
Now try it yourself
Sigma and pi count in ethene's C = C double bond?
Check your answer
1 sigma, 1 pi.
The first overlap is sigma; the second is pi.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Bond order?
Show answer and reason
(Nb - Na) / 2.
From the MO configuration.
2. NH3 shape?
Show answer and reason
Trigonal pyramidal, about 107 degrees.
4 pairs, one a lone pair.
3. H2O shape?
Show answer and reason
Bent, about 104.5 degrees.
Two lone pairs squeeze more.
4. Sigma vs pi?
Show answer and reason
Sigma head-on, pi sideways.
First bond is always sigma.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Find the bond order of O2+ (10 bonding, 5 antibonding).
- Name the shape and angle of H2O.
- How many sigma and pi bonds does the C = C bond in ethene carry?
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Questions about this practice page
- Why do noble gases rarely form bonds?
- Their outer shells already hold eight electrons - a complete octet - so there is no energetic gain in sharing or transferring electrons. Bonding exists to finish the octet; theirs is finished.
- What is the difference between sigma and pi bonds?
- A sigma bond forms by head-on orbital overlap along the line joining the nuclei - every single bond is one. A pi bond forms by sideways overlap of p orbitals above and below that line, and appears only as the second or third bond in double and triple bonds.
- How does VSEPR predict molecular shape?
- Count the electron pairs around the central atom - they repel into the geometry that keeps them farthest apart. Lone pairs repel harder than bond pairs, so they distort the ideal angles: NH3 drops from 109.5 to about 107 degrees, H2O to 104.5.
- What does bond order tell you?
- Roughly how many bonds link two atoms: order 1, 2, 3 match single, double, triple. A higher bond order means a shorter, stronger bond, and an order of zero in MO theory means the molecule does not hold together at all.
- Are these the NCERT exercise questions for Chemical Bonding (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.