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Problems · Problem 3.8

Q.Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements;

(a) silicon and bromine
(b) aluminium and sulphur.
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✓ Free question

The key idea is to use the group number (and hence the number of valence electrons) to determine the charge each element forms as an ion. Silicon (Group 14) forms a +4+4 ion, bromine (Group 17) forms a −1-1 ion, so the compound is SiBr4\text{SiBr}_4. Aluminium (Group 13) forms a +3+3 ion, sulphur (Group 16) forms a −2-2 ion, so the compound is Al2S3\text{Al}_2\text{S}_3.

When you’re asked to predict the formula of a compound formed by two elements, you’re essentially being asked: What’s the simplest whole-number ratio of ions that makes the total charge zero? The Periodic Table is your cheat sheet for this — it tells you the charge each element wants to take.

The logic is simple. Metals (on the left) tend to lose electrons and become positive ions (cations). Non-metals (on the right) tend to gain electrons and become negative ions (anions). The group number tells you how many valence electrons an atom has, and that directly gives the charge it will form.

  1. For silicon and bromine:

    Silicon is in Group 14. It has 4 valence electrons. To achieve a stable octet, it’s easier for silicon to lose those 4 electrons than to gain 4 more. So silicon forms a cation with a charge of +4+4: Si4+\text{Si}^{4+}.

    Bromine is in Group 17 (the halogens). It has 7 valence electrons, so it needs just 1 more to complete its octet. It gains one electron, forming an anion with a charge of −1-1: Br−\text{Br}^{-}.

    Now, to make a neutral compound, the total positive charge must balance the total negative charge. If you have one Si4+\text{Si}^{4+}, you need four Br−\text{Br}^{-} ions to cancel the +4+4 charge. The formula is therefore SiBr4\text{SiBr}_4.

    Tip

    A quick way: the subscript for one element is the charge number of the other element (without the sign). So for Si4+\text{Si}^{4+} and Br−\text{Br}^{-}, the 4 from silicon becomes the subscript for bromine, and the 1 from bromine becomes the subscript for silicon — giving Si1Br4\text{Si}_1\text{Br}_4, which we write as SiBr4\text{SiBr}_4.

  2. For aluminium and sulphur:

    Aluminium is in Group 13. It has 3 valence electrons and loses them to form Al3+\text{Al}^{3+}.

    Sulphur is in Group 16. It has 6 valence electrons and needs 2 more to complete its octet, so it forms S2−\text{S}^{2-}.

    To balance charges, we need the smallest whole numbers such that the total positive charge equals the total negative charge. The lowest common multiple of 3 and 2 is 6. So we need two Al3+\text{Al}^{3+} ions (total +6+6) and three S2−\text{S}^{2-} ions (total −6-6). The formula is Al2S3\text{Al}_2\text{S}_3.

    Watch out

    A common mistake is to write AlS\text{AlS} or Al3S2\text{Al}_3\text{S}_2. Always check that the total charge is zero: 2×(+3)+3×(−2)=02 \times (+3) + 3 \times (-2) = 0. If it doesn’t sum to zero, the formula is wrong.

✓Final answer

The predicted formulas are SiBr4\boxed{\text{SiBr}_4} for silicon and bromine, and Al2S3\boxed{\text{Al}_2\text{S}_3} for aluminium and sulphur.

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