Wishboy's 'O' Level Chem Qn can be upgraded to 'A' Level Qn!

wishboy SG

When chlorine is bubbled through a solution of iodine in hot aqueous sodium hydroxide, the two halogens react in the Cl2 : I2 ratio of 7 : 1, forming a white ppt A and a solution of sodium chloride.

A has the following composition by mass:

Na: 16.9% ; H: 1.1% ; I: 46.7% ; O: 35.3%

Calculate the empirical formula of A and thus deduce the balanced equation for the reaction.

my ans for empirical formula is Na2H3IO6.

for the balanced eqn, i wrote down Cl2 (g) + I2 (aq) + NaOH (aq) ---> NaCl (aq) + Na2H3IO6

but i found out that it was impossible to balance this eqn

am i lacking of something in the eqn? or have i done any part wrongly?

tinuviel07 SG

When chlorine is bubbled through a solution of iodine in hot aqueous sodium hydroxide, the two halogens react in the Cl2 : I2 ratio of 7 : 1, forming a white ppt A and a solution of sodium chloride.

A has the following composition by mass:

Na: 16.9% ; H: 1.1% ; I: 46.7% ; O: 35.3%

Calculate the empirical formula of A and thus deduce the balanced equation for the reaction.

my ans for empirical formula is Na2H3IO6.

for the balanced eqn, i wrote down Cl2 (g) + I2 (aq) + NaOH (aq) ---> NaCl (aq) + Na2H3IO6

but i found out that it was impossible to balance this eqn

am i lacking of something in the eqn? or have i done any part wrongly?

No. of moles of Na = 0.7

" H = 1.1

" I =0.4

" O = 35.3/16 = 2.2

Na : H : I : O

= 1.8 : 2.8 : 1 : 5.5

= 2 : 3 : 1 : 6?

wishboy SG

No. of moles of Na = 0.7

" H = 1.1

" I =0.4

" O = 35.3/16 = 2.2

Na : H : I : O

= 1.8 : 2.8 : 1 : 5.5

= 2 : 3 : 1 : 6?

yea tat's wad i got

wad abt the balancing of the equation?

Chemfreak022 SG

I2 + 7Cl2 + 18NaOH = 2Na2H3IO6 + 14NaCl + 6H2O

wishboy SG

may i ask why do u add H2O to the RHS of the equation, and not on the LHS?

UltimaOnline SG

When chlorine is bubbled through a solution of iodine in hot aqueous sodium hydroxide, the two halogens react in the Cl2 : I2 ratio of 7 : 1, forming a white ppt A and a solution of sodium chloride.

A has the following composition by mass:

Na: 16.9% ; H: 1.1% ; I: 46.7% ; O: 35.3%

>>> When chlorine is bubbled through a solution of iodine in hot aqueous sodium hydroxide, the two halogens react in the Cl2 : I2 ratio of 7 : 1, forming a white ppt A and a solution of sodium chloride.

A has the following composition by mass:

Na: 16.9% ; H: 1.1% ; I: 46.7% ; O: 35.3%

Calculate the empirical formula of A and thus deduce the balanced equation for the reaction. <<<

UltimaOnline's enlightening comments :

For those of you who are intrigued, fascinated and intellectually excited by this wonderfully mysterious chemical reaction, and are eager to more deeply understand the process proper, I'll describe what actually happens :

The iodine undergoes disproportionation with hot aqueous sodium hydroxide, forming iodate(V) ions.

Only participating ions :

6l2 + 12OH- --> 10l- + 2lO3- + 6H2O

Including spectator ions :

6l2 + 12NaOH --> 10Nal + 2NalO3 + 6H2O

The excess iodide ions are oxidized by chlorine.

Only participating ions :

10I- + 5Cl2 --> 10Cl- + 5I2

Including spectator ions :

10NaI + 5Cl2 --> 10NaCl + 5I2

The iodate(V) ions are subsequently and simultaneously hydrolyzed (ie. nucleophilically attacked) by hydroxide ions and oxidized by chlorine (transfer of electrons), into a partially protonated form of the iodate(VII) ion.

Watch carefully what happens, many students may find the following mechanism somewhat tricky. The iodate(V) ion has a central I atom, 2 double bonded O atoms, 1 single bonded O atom with a -1 formal charge. The central I atom has 5 bond pairs and 1 lone pair (no formal charge). When 3 hydroxide nucleophiles attack, the I atom now has 3 additional electrons (hence 3- formal charge). To be precise, it now has 2 double bonds and 4 single bonds. One of the pi bond pairs, becomes a lone pair on the O atom, giving it a -1 formal charge. All in all, we have a 2- formal charge on I (7 bond pairs, 1 lone pair) and 2 O atoms each with a -1 formal charge. Here's where the chlorine comes in. It glady accepts (reduction potential of +1.36V) the lone pair on the I atom, thereby removing its -2 formal charge, and is itself reduced to Cl- ion. Summarily, we obtain the partially protonated form of the iodate(VII) ion.

Only participating ions :

2IO3- + 6OH- + 2Cl2 -> 2(H3IO6)2- + 4Cl-

Including spectator ions :

2NaIO3 + 6NaOH + 2Cl2 -> 2Na2H3IO6 + 4NaCl

Additional information on the fully protonated iodate(VII) ion, that may also be obtained from iodic(VII) acid :

H5IO6 = HIO4.2H2O <---- iodic(VII) acid (central I atom double bonded to 3 O atoms, single bonded to OH) hydrated with 2 water molecules that nucleophilically attack and dative bond to the partial positively charged I atom; 2 proton transfers ensue, from the 2 water molecules to 2 double bonded O atoms; end product molecule (with central I atom) has 5 acidic protons at the end of 5 single bonded O atoms, and 1 double bonded O atom without acidic proton. Summarily, HIO4.2H2O becomes H5IO6, known as ortho-per-iodic acid, or hexa-oxo-iodic(VII) acid, or penta-hydro-oxido-iodine.

Therefore, the correct balanced equation is obtained by combining the following equations that occur stepwise :

6l2 + 12NaOH --> 10Nal + 2NalO3 + 6H2O

10NaI + 5Cl2 --> 10NaCl + 5I2

2NaIO3 + 6NaOH + 2Cl2 -> 2Na2H3IO6 + 4NaCl

We obtain a summary balanced equation as follows :

(6-5)I2 + (12+6)NaOH + (5+2)Cl2 --> 2Na2H3IO6 + (10+4)NaCl + 6H2O

I2 + 18NaOH + 7Cl2 --> 2Na2H3IO6 + 14NaCl + 6H2O

UltimaOnline SG

may i ask why do u add H2O to the RHS of the equation, and not on the LHS?

Water is produced in the disproportionation reaction of the halogen which occurs in the presence of the hydroxide nucleophile. But if you're at 'O' level, and not expected to understand this, then if the question is to be fair, it should indicate that water is also a product.

So wishboy, if you're at 'O' levels, this is not your fault. It is the fault of the question for not specifying this clearly. Don't worry about it. The actual 'O' level question should be fair and let you know water is a product.

UltimaOnline SG

Since Wishboy asked about why water is produced, here's why. (Below, I used chlorine as the halogen. You can substitute chlorine with iodine.)

Qn. Draw the mechanisms to explain the following disproportionation reactions :

a) cold aqueous NaOH at 15 deg C

Cl2 + 2NaOH --> NaCl + NaClO + H2O

3Cl2 + 6NaOH --> 3NaCl + 3NaClO + 3H2O

b) hot aqueous NaOH at 70 deg C

Substitute 3ClO- --> 2Cl- + ClO3- into

3Cl2 + 6NaOH --> 3NaCl + 3NaClO + 3H2O to obtain

3Cl2 + 6NaOH --> 5NaCl + NaClO3 + 3H2O

Solution :

The mechanisms are as follows :

a) cold aqueous NaOH at 15 deg C

Cl2 + 2NaOH --> NaCl + NaClO + H2O

3Cl2 + 6NaOH --> 3NaCl + 3NaClO + 3H2O

1) A hydroxide ion nucleophile attacks Cl-Cl

2) Heterolytic cleavage or fission occurs - bond pair between O and Cl, becomes lone pair on Cl leaving group to produce Cl- ion

3) Proton transfer (acid-base reaction) from Cl-O-H to hydroxide ion, forming the chlorate(I) ion (aka hypochlorite ion) and water.

b) hot aqueous NaOH at 70 deg C

Substitute 3ClO- --> 2Cl- + ClO3- into

3Cl2 + 6NaOH --> 3NaCl + 3NaClO + 3H2O to obtain

3Cl2 + 6NaOH --> 5NaCl + NaClO3 + 3H2O

The following steps occur simultaneously (not consecutively; oxygen has no empty d orbitals and cannot violate its octet) when in close proximity and with sufficient activation energy (hence high temperature needed) to overcome repulsion between the anions :

1) Double (and non-dative) covalent bonds form between the O atoms of 2 chlorate(I) ions and the Cl atom of the 3rd chlorate(I) ion, to achieve stable octets for all atoms. (To represent formation of non-dative double bonds, draw 2 pairs of 2 single-hook arrows coming from opposite directions meeting in the middle; similar to 2 free radicals bonding in a termination step)

2) To avoid violating oxygens' octets, heterolytic cleavage or fission occurs on the 2 chlorate(I) ions - the bond pairs between O and Cl, become lone pairs on the 2 Cl leaving groups, to produce 2Cl- ions.

(Side note : Cl- ions are stable (hence excellent leaving groups; stability of conjugate base explains why HCl is a strong acid) because chlorine is relatively highly electronegative, and Cl- ion has a relatively large ionic radius to stablize the negative charge.)

wishboy SG

wow... thanks UltimaOnline for the detailed explaination, and anyone else who had helped ! this is an interesting reaction indeed

btw, im at JC1 (or 'A' levels). the teacher did not really teach us abt this reaction though