Answer:
Option E!
Explanation:
If we were to draw the lewis dot structure for IBr2 -, we would first count the total number of valence electrons ( " available electrons " ). Iodine has 7 valence electrons, and so does Bromine, but as Bromine exists in 2, the total number of valence electrons would be demonstrated below;

Don't forget the negative on the Bromine!
Now go through the procedure below;
1 ) Place Iodine in the middle and draw single bonds to each of the bromine.
2 ) Add three lone pairs on each of the Bromine's
3 ) Now we have 6 electrons left, if we were to exclude the electrons shared in the " single bonds. " This can be placed as three lone pairs on Iodine ( central atom )!
The molecular geometry can't be linear, as there are lone pairs on the atoms. This makes it bent.
The answer is B=1
-2(3b-2)=8b-10
-6b+4=8b-10
-14b+4=-10
-14b=-14
B=1
Answer: IUPAC name of
will be hept-1-yne.
Explanation:
The IUPAC nomenclature of organic compounds are given as follows:
1. Select the longest possible carbon chain.
2. For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'hept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
3. A suffix 'yne' is added at the end of the name if triple bond is present.

The IUPAC name will be hept-1-yne.
Increasing the pressure of the gas increases the intermolecuar forces of attraction between the molecules because the distance between the molecules are being reduced by the action of the piston.