Chemistry Quiz Dreamers Quiz Blog

Chemical Bonding MCQs: 5 Hybridisation and Geometry Questions

By Aniket Bhardwaj · 8 November 2026 · Dreamers Quiz

You will get five hybridisation and geometry MCQs with answers, each solved step by step.

You will also see how to use the Dreamers Quiz portal to revise this chapter against the clock.

In short: To find hybridisation and shape, count the electron pairs around the central atom. Add the sigma bonds and the lone pairs. That total, called the steric number, gives the hybridisation. The lone pairs then decide the final shape.

The method before the questions

Many students lose marks in this topic because they guess the shape from the formula. Do not do that. Follow the same steps every time.

  1. Pick the central atom. It is usually the least electronegative atom (the one that pulls bonding electrons less strongly), other than hydrogen.
  2. Count its valence electrons (the electrons in the outermost shell).
  3. Count the sigma bonds and the lone pairs. A sigma bond is the first bond between two atoms. A double bond has one sigma bond plus one pi bond. A triple bond has one sigma bond plus two pi bonds. Pi bonds do not count toward the steric number.
  4. Add the sigma bonds and the lone pairs. Match the total to the table below.
  5. Look at how many of those pairs are lone pairs. Name the shape from the atoms only.

The idea behind this is simple. Electron pairs repel each other, so they spread out as far as possible around the central atom. A lone pair is a pair of valence electrons not used in bonding.

Steric numberHybridisationArrangement of electron pairsExample
2spstraight lineethyne, C2H2
3sp2flat triangle (trigonal planar)BF3
4sp3tetrahedralCH4
5sp3dtrigonal bipyramidalPCl5
6sp3d2octahedralSF6

Learn this table by heart. It answers most hybridisation MCQs in well under a minute.

Five sample MCQs with answers

Question 1: BF3

What are the hybridisation of boron and the shape of BF3?

Answer: (b). Boron has three valence electrons and forms three sigma bonds with fluorine. It has no lone pair. Steric number = 3 + 0 = 3, so the hybridisation is sp2. With no lone pair, the shape is trigonal planar. All four atoms lie in one flat plane.

Question 2: NH3

What is the shape of NH3?

Answer: (c). Nitrogen has five valence electrons. Three are used in N–H bonds. One lone pair remains. Steric number = 3 + 1 = 4, so nitrogen is sp3 hybridised. The four electron pairs point to the corners of a tetrahedron, but we name the shape by the atoms only. Three hydrogen atoms and one lone pair give a pyramidal shape.

Option (b) is the common trap. The electron-pair arrangement is tetrahedral, but the molecular shape is pyramidal. The lone pair pushes the bond pairs closer, so the H–N–H angle comes out a little smaller than the ideal tetrahedral angle.

Question 3: ethyne (C2H2)

What is the hybridisation of each carbon in ethyne, and how many sigma and pi bonds does the molecule have?

Answer: (c). Ethyne is H–C≡C–H. Each carbon forms one sigma bond with hydrogen and one sigma bond with the other carbon. That is two sigma bonds per carbon, so steric number = 2 and the hybridisation is sp. The triple bond has one sigma and two pi bonds. In total there are two C–H sigma bonds and one C–C sigma bond, so 3 sigma and 2 pi. All four atoms lie on one straight line.

Question 4: SF6

Sulphur in SF6 is hybridised as:

Answer: (c). Sulphur forms six sigma bonds with six fluorine atoms. There is no lone pair. Steric number = 6 + 0 = 6, so the hybridisation is sp3d2. The shape is octahedral. Sulphur sits at the centre and the six fluorine atoms sit at the six corners of the octahedron.

Question 5: XeF2

What is the shape of XeF2?

Answer: (b). Xenon has eight valence electrons. Two are used for the two Xe–F bonds. The remaining six form three lone pairs. Steric number = 2 + 3 = 5, so the hybridisation is sp3d and the electron pairs form a trigonal bipyramid. The three lone pairs sit in the middle plane, where they repel each other least. This leaves the two fluorine atoms on the vertical axis, so the molecule is linear.

Where students go wrong

The same mistakes appear again and again in school tests and entrance practice.

A worked example for your notebook

Find the shape of H2O.

  1. Central atom: oxygen, with six valence electrons.
  2. Two electrons form two O–H sigma bonds. The other four electrons form two lone pairs.
  3. Steric number = 2 + 2 = 4, so oxygen is sp3 hybridised.
  4. Four pairs form a tetrahedral arrangement. Two atoms and two lone pairs give a bent shape.
  5. Two lone pairs repel more than one, so the H–O–H angle is smaller than the H–N–H angle in NH3.

Use the same steps for any molecule. Change only the central atom and the numbers.

How to practise on the Dreamers Quiz portal

Reading solved questions is not enough. You must attempt questions yourself, with the clock running. The Dreamers Quiz portal is built for this.

Here is how it works:

  1. Open the portal from the Chemistry Quiz page on 4ab.in. It opens in one tap.
  2. Pick a class or exam, such as Class 11, Class 12, NEET, JEE or CUET.
  3. Attempt the set against the clock.
  4. See your score and the answer key.
  5. Attempt again. You can attempt as many times as you like.

The portal suits chapter revision before a school test, a quick check after a class, or speed practice for the NEET and CUET MCQ sections. For chemical bonding, try this routine.

Your score on a repeat attempt shows whether you learned the method or only remembered the answers.

Key takeaways

  • Steric number = sigma bonds + lone pairs on the central atom. Pi bonds are not counted.
  • Steric numbers 2, 3, 4, 5 and 6 give sp, sp2, sp3, sp3d and sp3d2.
  • Lone pairs change the molecular shape but not the hybridisation. NH3 is pyramidal and H2O is bent, though both are sp3.
  • In five-pair systems, lone pairs go to the middle-plane positions first.
  • Timed practice with an answer key shows which step you are getting wrong.

Frequently asked questions

Does a double bond change the hybridisation?

Not by itself. Count only sigma bonds and lone pairs. In ethene, each carbon has three sigma bonds, so it is sp2. The double bond has one sigma and one pi bond.

Why is the H–N–H angle in NH3 smaller than the ideal tetrahedral angle?

The lone pair on nitrogen repels the bond pairs more strongly than bond pairs repel each other. This squeezes the N–H bonds closer together.

Can two molecules have the same hybridisation but different shapes?

Yes. CH4, NH3 and H2O are all sp3. Their shapes are tetrahedral, pyramidal and bent, because they have zero, one and two lone pairs.

What can I use the Dreamers Quiz portal for?

You can use it for chapter revision before a school test, a quick check after a class, or speed practice for the NEET and CUET MCQ sections. You see your score and the answer key after each attempt.

Can I attempt a quiz more than once?

Yes. The portal lets you attempt as many times as you like, so you can repeat a chapter set until your method is steady.

What to do next. Try it yourself on the page below.

Chemistry Quiz — Free Timed Chemistry Quizzes →