Chemistry 2210: Group 15 Worksheet Chemistry 2210 - Dr. Clase

Worksheet on Group 15.

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Warming up excercises.

Identify the following substances and write balanced equations for any reactions mentioned.

1. An anion that gives two gaseous oxides when its salts are treated with dilute acids.

Nitrite:

2NO2-(aq) + 2H+(aq) ---> H2O(l) + NO(g) + NO2(g)

2. An anion that does not decompose when treated with dilute acids, but which, when some of its salts are strongly heated, gives a brown gas.

Nitrate:

e.g. 2Pb(NO3)2(s) ---> 2PbO(s) + 4NO2(g) + O2(g)

3. Two basic hydrides.

Ammonia, NH3, and hydrazine, N2H4. (PH3 is not really basic.)

4. A substance that is pyrophoric (spontaneously catches fire in air.)

White phosphorus, P4 or phosphine, PH3.

5. The repeating units of two different inorganic polymers with more that one element in the chain. (At least one in group 15.)

(PCl2N) in phosphazines (SN) in S4N4.

6. A group 15 chloride that does NOT hydrolyse with water to give hydrochloric acid and another acid and one which does.

NCl3(l) + 3H2O(l) ---> NH3(aq) + 3HOCl(aq)

(N is more electronegative than Cl so the bond polarity is opposite to most covalent involving chlorine, in this compound the Cl is positive and attracts the negative OH- entity during hydrolysis.)

E.g. PBr3(l) + 3H2O(l) ---> H3PO3(aq) + 3HBr(aq)

(Typical of most non-metal halides.)

Explain the following observations:

7. H3PO4 reacts with three moles of base, but H3PO2 reacts with only one mole.

The structural formulas are: (HO)3P=O and (HO)H2P=O. As in all oxoacids it is the HO- group attached to an electro- negative atom that readily looses a proton and is responsible for the acidity. H3PO4 has three of these groups but H3PO2 has only one, the other two hydrogens are bonded directly to the phosphorus atom and are not acidic.

8. Two of the most common oxides of nitrogen are paramagnetic and have unusual oxidation states for a group 15 element.

They are NO and NO2 in oxidation states 2 and 4 - we would expect a group 15 element to have oxidation states 5, 3, 1 not even numbered ones. Since these compounds have an odd number of electrons they must be paramagnetic (N.B. the reverse is not true, paramagnetic O2 and Fe+2 both have even numbers of electrons.) The best explanation is that these are more stable than, e.g. N2O3 because they have higher entropy without any significant loss in enthalpy.

9. Liquid PCl5 on standing forms a crystalline solid that gives a conducting solution when dissolved in SO2(l). (The SO2 is just a non-aqueous polar solvent, it doesn't react.)

The liquid consists of individual trigonal-bipyramidal PCl5 molecules, which over a period of time transfer Cl- ions among themselves to form PCl4+ and PCl6- ions. In a polar non-aqueous solvent like SO2(l) these ions conduct electricity in the same way that ions carry current in water.

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Identify the lettered compounds in the following and write balanced equations for all the reactions. All reactions are assumed to go to completion, without any reactants remaining.

10. A is a white solid available at garden supply stores. When A is treated with aqueous sodium hydroxide a pungent gas B is evolved which turns moist red litmus blue and a solution of F remains. When B is mixed with air and passed over Rh/Pt gauze the gauze glows red hot and a brown gas C is produced. C dissolves in water to give an acidic solution of D and a colourless paramagnetic gas E which reacts with air to form C.

When F is isolated by evaporating off the water it is found to be a white solid, which on treatment with concentrated sulphuric acid yields pure D as a rather unstable liquid. When aluminium powder and aqueous sodium hydroxide are added to F and the mixture warmed B is produced.

When A is heated it decomposes to give G and H. G condenses at 100ºC to a colourless liquid which is an excellent solvent for A and F.

H was accidentally inhaled by the experimentor, rendering him unconscious for a short while, and when he came round he seemed to find the situation extremely amusing.

Reading through the story the obvious clues are the basic gas, B, which must be ammonia, suggesting that A is an ammonium salt. The brown gas C sounds like NO2. The next reaction is the method used to make nitric acid, D, from ammonia. Since D is also produced from A i via F a nitrate must be involved therefore A is ammonium nitrate, sold as a fertilizer. G sounds like water and H like laughing gas, N2O. 

 NH4NO3(aq) + NaOH(aq) ---> NH3(g) + H2O(l) + NaNO3(aq)
    A                        B                 F

 4NH3(g) + 7O2(g) ---> 4NO2(g) + 6H2O(l)
                         C        (G)
 3NO2(g) + H2O(l) ---> 2HNO3(aq) + NO(g)
            D                      E

 2NO(g) + O2(g) ---> 2NO2(g)

 2NaNO3(s) + H2SO4(l) ---> Na2SO4(s) + 2HNO3(l)

 3NO3-(aq) + 8Al(s) + 5OH-(aq) +18H2O(l) 
                   ---> 3NH3(g) + 8Al(OH)4-(aq)

 NH4NO3(s) ---> 4H2O(l) + N2O(g)
                  G        H

11. P is a yellowish white waxy solid which catches fire spontaneously in warm air. When the supply of air is limited the oxidation product, a white solid is Q, which reacts vigorously with water giving a solution of R, which is acidic and reducing. When R oxidised (e.g. with permanganate) another acid S is formed. When a neutralised solution of S is treated with molydbate reagent a yellow precipitate quickly forms.

P reacts with an excess of chlorine to give a liquid, T which reacts vigorously with water to give a solution which contains S and another acid U. Neutralisation with Ca(OH)2 gives a white precipitate, V and a solution, W. When silver nitrate solution is added to W a white precipitate X forms which is readily soluble in aqueous ammonia.

As well as identifying P to X describe or sketch the structures of P, Q, & R.

The modybdate test clearly indicates that phosphates are present, and P sounds very like white phosphorus itself. 


 P4(s) + 3O2(g) ---> P4O6(s)
 P                   Q

 P4O6(s) + 6H2O(l) ---> 4H3PO3(aq)
                         R

 H3PO4(aq) + [O] ---> H3PO4(aq)
                       S
 Molybdate test confirms phosphate - equation not expected! 

 P4(s) + 10Cl2(g) ---> 4PCl5(l)
                        T

PCl5(l) + 4H2O(l) ---> H3PO4(aq) + 5HCl(aq)
                                     U

 2H3PO4(aq) + 3Ca(OH)2(s) ---> Ca3(PO4)2(s) + 6H2O(l)
                                 V
 
 2HCl(aq) + Ca(OH)2(s) ---> Ca+2(aq) +2Cl-(aq) +2H2O(l)
                             ^^^^W^^^^

 Cl-(aq) + Ag+(aq) ---> AgCl(s)
                         X

 AgCl(s) + 2NH3(aq) ---> Ag(NH3)2+(aq) + Cl-(aq)

P, P4 is a tetrahedron of four P atoms.

Q, P4O6 is also a tetrahedron with a P atom at each apex and an O atom in each edge each linking two of the P atoms.

R, H3PO3 is (HO)2HP=O. I.e. there are two acidic HO groups one hydridic H atom and a doubly bound oxygen on the central phosphorus atom.

(C) H.J. Clase 1998. Updated 1999.03.11

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