'O' & 'A' Level Qns.
Qn 1.
When 1.01g sample of potassium nitrate(V) is heated above its
melting point, oxygen gas is evolved. The mass of the sample
decreases by 0.16g. Construct a balanced equation for this
decomposition reaction, name the residue and (for 'A' level
students) draw its structural formula.
Q2.
Sulfur and chlorine can react together to form disulfur
dichloride. When 1.00g of disulfur dichloride was reacted with
water, 0.36g of a yellow precipitate of sulfur was formed together
with a solution containing a mixture of sulfuric(IV) acid (a.k.a.
sulfurous acid), and hydrochloric acid. Given that the most common
isotope of sulfur is S8, use the above data to deduce the balanced
equation for the reaction between disulfur dichloride and water.
(For 'A' level students) draw the structural formulae of disulfur
dichloride and sulfuric(IV) acid.
Solutions (partial).
Qn1. Since 1 mole of potassium nitrate(V) decomposes to give 0.5
mole of oxygen gas, the residue is potassium nitrate(III) {stock
name} or potassium nitrite {latin name}.
For the nitrate(III) ion (aka nitrite ion), uninegative formal
charge on a singly bonded O atom; central N atom has 1 lone pair, 3
bond pairs; electron geometry tetrahedral, ionic geometry trigonal
pyramidal.
Qn2. Since 1 mole of disulfur dichloride undergoes hydrolysis to
give 1.5 moles of S(s), or 1.5/8 moles of S8(s), the coefficients
of the balanced equation are : 2, 3, 1, 4, 3/8
For disulfur dichloride, the 2 central S atoms have 2 lone pairs
and 2 bond pairs; electron geometry tetrahedral, molecular geometry
v-shape.
For sulfuric(IV) acid {stock name} or sulfurous acid {latin
name}, the central S atom has 1 lone pair, 4 bond pairs (S can
expand its octet using its vacant and energetically accessible 3d
orbitals); electron geometry tetrahedral, molecular geometry
trigonal pyramidal.