An atom of calcium has 2 valence electrons and an atom of sulfur has 6 valence electrons how many lone pairs of electrons are present in the lewis structure of calcium sulfur

I believe the answer is Two

Total valence electrons is 8

Ca - - S ( - - is single bond) this needs 6 more electrons as a single bond is only 2 electrons

So you put 3 lone pairs on each atom but this makes it so there are 14 electrons

Take 2 lone pairs off of both atoms so each Atom only has 1 Lone pair

The 4 lone pairs you took off of the atoms become a triple bond between Ca and S.

This leaves you with 10 Electrons which is 2 too many but Octet should be filled over valency

Now you have Ca = - S ( = - is triple bond) and you have one lone pair on each atom which gives you two total lone pairs