SnCl₂ Shape and Bond Angle, Explained with VSEPR

If you guessed that SnCl₂ is linear like CO₂, you have walked straight into the exam trap. SnCl₂ is bent, and the reason is a single lone pair sitting on the tin atom. Here is the VSEPR reasoning, the bond angle, and why SnCl₂ behaves so differently from the linear BeCl₂.

VSEPR in one line

Electron pairs, bonding and lone alike, arrange themselves around the central atom to be as far apart as possible, because electron pairs repel. Count the pairs, arrange them, then describe the shape using only where the atoms are.

Counting SnCl₂’s electron domains

Tin is in Group 14, so it has 4 outer electrons. In SnCl₂ it uses two of them to bond to the two chlorine atoms. That leaves two electrons as a lone pair on tin.

So around the central tin atom:

• 2 bonding pairs (to the two Cl atoms)
• 1 lone pair
• a total of 3 electron domains.

Three electron domains give a trigonal planar arrangement of electron pairs as the base shape, with the pairs about 120° apart.

The lone pair on Sn, and the bent shape

Here is the crucial step. One of those three domains is a lone pair, and a lone pair is invisible when we name the shape, because we name shapes from where the atoms are. Two bonded atoms plus one “missing” position gives a bent (V-shaped, angular) molecule.

A lone pair also repels more strongly than a bonding pair, because it is held by only one nucleus and so spreads out more. It therefore pushes the two Sn–Cl bonds closer together, squeezing the angle down. So the Cl–Sn–Cl angle comes out at a little under 120° (compare the bent SO₂ at about 119°). That “slightly less than 120°” is the answer to give in an A-level exam.

Going further (beyond the syllabus): the measured gas-phase angle in SnCl₂ is actually around 95 to 100°, noticeably less than 120°. Tin is a large, heavy atom that resists mixing its s and p orbitals, so the real angle is smaller than simple VSEPR predicts. For A-level, stick with “just under 120°”. This note is only here so you are not thrown if you see the real figure quoted somewhere.

Why it is not linear like BeCl₂

This is the comparison examiners love. Beryllium is in Group 2, so it has only 2 outer electrons, and in BeCl₂ both are used in bonding. That leaves no lone pair:

• BeCl₂: 2 bonding pairs, 0 lone pairs, 2 domains, so linear, 180°.
• SnCl₂: 2 bonding pairs, 1 lone pair, 3 domains, so bent, just under 120°.

Same formula type, MX₂, completely different shape, and the whole difference is that one extra lone pair on tin. That is why you can never read a shape straight off the formula. You must count the electron pairs.

A quick compare table

Read down the “lone pairs” column and you can watch the shape bend further as lone pairs are added: 0 gives linear, 1 gives a gentle bend, 2 (as in water) bends it right down to 104.5°.

The exam takeaways

• SnCl₂ is bent, not linear, because tin keeps one lone pair.
• Bond angle: slightly less than 120° (the lone pair repels more than the bonding pairs).
• The contrast with BeCl₂ (linear) is all about that lone pair.

Frequently asked questions

What is the shape of SnCl₂? Bent (V-shaped). Tin has 2 bonding pairs and 1 lone pair, giving 3 electron domains, so the molecule is angular rather than linear.

What is the bond angle in SnCl₂? Just under 120° at A-level, because the lone pair repels more strongly than the bonding pairs and squeezes the Cl–Sn–Cl angle below the ideal 120°.

Why is SnCl₂ bent but BeCl₂ linear? Beryllium has only 2 outer electrons, both used in bonding, so BeCl₂ has no lone pair and is linear. Tin keeps a lone pair, giving SnCl₂ a third electron domain and a bent shape.

See it and practise

Rotate these molecules and count the pairs yourself in the free VSEPR simulation (no login needed), then drill the theory in the Chemical Bonding topic on ChemBank.

Try the VSEPR sim, and when you are ready to push your grade, book a trial lesson with Mr Khemistry.