The Student Room Group
Reply 1
FoOtYdUdE
Liquid ammonia, NH3, and water, H20, both show H-Bonding.


So basically, i know how to draw it.. but erm could someone please clarify whether it's a or b:

a) only link two NH3 via h-bonding...h20 molecule not relevant - so don't include.

or

b) link two NH3 using a water molecule.

Two NH3.
Reply 2
Kinkerz
Two NH3.

so basically...a??

aah, cheers.

bdway,

how do i suggest the formula for sodium chlorate (vii) ?? :confused:
Reply 3
FoOtYdUdE
so basically...a??

:yep:

FoOtYdUdE
bdway,

how do i suggest the formula for sodium chlorate (vii) ?? :confused:

Not sure.
Reply 4
Kinkerz
:yep:


I'd go with NaClO3, but I may be wrong.

it's NaCl04 butt i dunno y. oh nvm. thnxxx.
Reply 5
FoOtYdUdE
it's NaCl04 butt i dunno y. oh nvm. thnxxx.

At this point, do you need to know why?
Reply 6
If you're using oxidation states:

O = -2
Na = +1
---
4(-2) + 1 = -7

So the Cl must have an oxidation number of +7, hence your sodium chlorate (vii) ion.
Reply 7
Kinkerz
If you're using oxidation states:

O = -2
Na = +1
---
4(-2) + 1 = -7

So the Cl must have an oxidation number of +7, hence your sodium chlorate (vii) ion.

genius!

i get it! :biggrin:
Reply 8
so basically if it was sodium chlorate (III) then the Cl would have an ox. no. of +3 ?? aaah and then the formula would be NaCl02 ??:biggrin:
Reply 9
FoOtYdUdE
so basically if it was sodium chlorate (III) then the Cl would have an ox. no. of +3 ?? aaah and then the formula would be NaCl02 ??:biggrin:

That'd be my approach, yes.