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:
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2007-07-19 06:05:13
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answer #1
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answered by Anonymous
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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
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answer #2
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answered by s0u1 reaver 5
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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
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answer #3
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answered by niuchemist 6
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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
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answer #4
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answered by ♣DreamDancer♣ 5
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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
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answer #5
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answered by ag_iitkgp 7
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It's called an 'ALKYNE' as in 'ETHYNE'
Where Ethane C2H6 (CH3-CH3) has been de-hydogenated to C2H2 (CHâ¡CH).
2007-07-19 14:28:01
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answer #6
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answered by Norrie 7
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