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2007-07-19 05:32:35 · 6 answers · asked by Anonymous in Science & Mathematics Chemistry

6 answers

You have to draw it out. This would be easier if you gave a compound but if you have CO (Carbon Monoxide) then:

C = 4 (4th from left)
O = 6 (6th from left)
= 10

So you have the octet rule which means an ion can't have more than 8 bonding pairs, so CO means 8+8=16

So you have 16 bonding ions and only need 10, so the you have 6 move inbetween the C and the O to make it a triple bond (2 = bond pair)

Hard to show here, but:
.. ..
: C:O: becomes :C:::O:
·· ··

2007-07-19 06:05:13 · answer #1 · answered by Anonymous · 0 0

In addition to the methods mentioned above , in case of diatomic molecules u may also calculate the bond order which is = 0.5 x (no of bonding electrons - no of antibonding electrons)

if bond order = 3 , then the molecule contains a triple bond.

2007-07-19 12:41:29 · answer #2 · answered by s0u1 reaver 5 · 0 0

Take a sample of the compound and run an IR spectrum.

Triple Bonds will appear in the 1900-2300 region, depending on the exact kind of triple bond you have.

2007-07-20 03:40:23 · answer #3 · answered by niuchemist 6 · 0 0

one easy way is to do a lewis dot structure with the compound. You can see the various bonds and lone pairs

2007-07-19 13:07:19 · answer #4 · answered by ♣DreamDancer♣ 5 · 0 0

By conducting some experiments as how many H or Br atoms it takes up, does it respond to Ag-NH3 test etc.

2007-07-19 12:37:29 · answer #5 · answered by ag_iitkgp 7 · 1 0

It's called an 'ALKYNE' as in 'ETHYNE'
Where Ethane C2H6 (CH3-CH3) has been de-hydogenated to C2H2 (CH≡CH).

2007-07-19 14:28:01 · answer #6 · answered by Norrie 7 · 0 0

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