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CBSE · Class 11 · Chemistry

Chemical Bonding and Molecular Structure

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The octet rule says atoms combine so that each has:
  1. eight neutrons
  2. eight bonds
  3. eight protons
  4. eight electrons in its outer shell
2.
According to VSEPR theory, the shape of PH3 is:
  1. trigonal pyramidal
  2. square planar
  3. square pyramidal
  4. see-saw
3.
The hybridisation of the central atom in PH3 is:
  1. sp3d
  2. sp3d2
  3. sp
  4. sp3
4.
The bond order of O2− is:
  1. 1.5
  2. 2
  3. 1
  4. 0
5.
Which of these is paramagnetic?
  1. H2
  2. N2
  3. N2−
  4. O22−
6.
Which of these molecules has a dipole moment (is polar)?
  1. CCl4
  2. CO2
  3. XeF4
  4. H2O
7.
How many σ and π bonds are there in CH3–C≡C–CH3?
  1. 9σ and 3π
  2. 9σ and 2π
  3. 11σ and 0π
  4. 8σ and 3π
8.
Assertion (A): SF4 has zero dipole moment.
Reason (R): A molecule’s dipole moment is the vector sum of its bond dipoles, so it is zero only when the shape is symmetrical and they cancel.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Define bond order. How is it related to bond length and bond stability?
10.
Use VSEPR theory to predict the shape of CH4. State the numbers of bond pairs and lone pairs on the central atom.
11.
Write the molecular orbital configuration of O2+. Find its bond order and say whether it is paramagnetic or diamagnetic.
12.
Arrange N2+, O2−, O22− in order of increasing stability.
13.
Give the hybridisation of each carbon atom in CH3–CH3, from left to right.
14.
Arrange NH3, H2O, CH4 in order of decreasing bond angle.
15.
Show how calcium (Z = 20) and sulphur (Z = 16) form an ionic compound by transfer of electrons, and give its formula.
16.
Which bond is more polar, Al–H or P–F? Which atom carries the partial negative charge in it? (Pauling electronegativities: Al 1.5, H 2.1, P 2.1, F 4.0)
Section C — Word problems4 × 4 = 16 marks
17.
Explain sp, sp2 and sp3 hybridisation with one example each, giving the shape and bond angle.
18.
Using molecular orbital theory, compare He2 and He2+: (a) write their MO configurations, (b) find their bond orders, (c) state their magnetic behaviour, (d) say which is more stable.
19.
For XeF4 and CH4, give (a) the hybridisation of the central atom, (b) the numbers of bond pairs and lone pairs on it, (c) the shape and bond angle, (d) whether the molecule is polar.
20.
For CH2=CH2: (a) how many σ and π bonds does it have? (b) give the hybridisation of each carbon atom, from left to right; (c) state the bond angle around each kind of carbon present.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 11 · Chemistry

Answer key

Chemical Bonding and Molecular Structure · AW-11C-CBM-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) eight electrons in its outer shellKössel and Lewis
2(a) trigonal pyramidal3 bond pairs and 1 lone pair on the central atom
3(d) sp34 electron pairs around it
4(a) 1.5½ (bonding − antibonding electrons); σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 π2py2 π*2px2 π*2py1
5(c) N2−N2−: 1 unpaired electron
6(d) H2OH2O is bent (V-shaped): its bond dipoles do not cancel
7(b) 9σ and 2πevery single bond is σ; a double bond is σ + π; a triple bond is σ + 2π
8(d) A is false, but R is true.A is false: it is see-saw. R is true.
9Bond order = ½ (bonding electrons − antibonding electrons). The higher the bond order, the shorter and stronger (more stable) the bond. A bond order of zero means the molecule does not exist.—
104 bond pairs, 0 lone pairs; tetrahedral4 electron pairs: sp3
11σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 π2py2 π*2px1; bond order 2.5; paramagnetic (1 unpaired)15 electrons
12O22− < O2− < N2+bond orders: N2+ 2.5, O2− 1.5, O22− 1; higher bond order means shorter, stronger bond
13sp3, sp3four single bonds: sp3; one double bond: sp2; a triple bond or two double bonds: sp
14CH4 > NH3 > H2ONH3 107°, H2O 104.5°, CH4 109.5°
15Each Ca (2, 8, 8, 2) loses 2 electrons to give Ca2+ (2, 8, 8); each S (2, 8, 6) gains 2 to give S2− (2, 8, 8). One of each balances the charges: CaS.both ions reach a noble-gas configuration
16P–F (difference 1.9 against 0.6); δ− on Fthe larger the electronegativity difference, the more polar the bond; the more electronegative atom is δ−
17sp: one s + one p orbital give two hybrids at 180°: BeCl2 is linear. sp2: one s + two p give three hybrids at 120°: BF3 is trigonal planar. sp3: one s + three p give four hybrids at 109.5°: CH4 is tetrahedral.—
18(a) He2: σ1s2 σ*1s2; He2+: σ1s2 σ*1s1. (b) 0 and 0.5. (c) He2 diamagnetic; He2+ paramagnetic (1 unpaired). (d) He2+ (higher bond order).—
19XeF4: sp3d2, 4 bond pairs and 2 lone pairs, square planar, 90°, non-polar. CH4: sp3, 4 bond pairs and 0 lone pairs, tetrahedral, 109.5°, non-polar.—
20(a) 5σ and 1π; (b) sp2, sp2; (c) sp2: 120°single bond σ; double σ + π; triple σ + 2π