Elements are ionized because they aspire to be stable. The most stable form are the ones with full octet of electrons, the noble gases which consist of the last column in the periodic table. The rest of the elements either accept or readily donate electrons to conform to the electronic configuration that is the same with the nearest noble gas.
1. Potassium's nearest noble gas is Ar which is one electron fewer. So, when ionized, it donates 1 electron. Hence, K⁺.
2. The nearest noble gas for fluorine is Neon which is 1 electron more. Hence, it has to accept one more electron. Hence, F⁻.
See electronegativity is the tendency of an atom to gain an electron and flourine with a valecy of one and a vey small size is the most electronegetive because its orbitals are quite closed to the nucleus and hence the attraction is quite strong so it can attract an electron.the question that arises is that some smaller atoms should be more electronegetive as they are closer to the nucleus but it need more energy for them as compared to flourine to complete their octet. now polarity increases when two atoms of quite different sizes form a compound ... the more electronegetive atom will always attract the bond pair forming a negetive charge on it and positive on the less electroneg. one and polarity increases with electronegetivity of the anion.now as your question says
<span>5=I2.. because both the atoms are same there wont be permanent polarity </span>
<span>4=HI iodine is the least electronegetive of all the halogens due to its large size,electronegetivity decreases down the group </span>
<span>3=HBr bromine is the 2nd largest halogen </span>
<span>2=HCl chlorine is the 3rd largest halogen </span>
<span>1=HF fluorine is the smallest halogen making and hence makes the most polar</span>
We should apply Boyle's Law here given initial pressure, initial volume and final volume.
P1V1= P2V2
(6.5 atm) (13 L) = P2 (3.3 L)
Solve for P2 on your calculator and that should get you to the answer.
Check attached file for the answer.
