A polar molecule<span> has a net dipole as a result of the opposing charges (i.e. having partial positive and partial negative charges) from </span>polar<span> bonds arranged asymmetrically. Water (H</span>2<span>O) is an example of a </span>polar molecule<span> since it has a slight positive charge on one side and a slight negative charge on the other.</span>
Option (i) would have the highest 2nd Ionization Energy.
Option (i) is Sodium.
Can be Written as 2, 8 , 1
For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.
Now After Removing that Electron...
Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.
Neon has a very stable Octet.
It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.
So
Option (i) [Sodium] has the highest 2nd Ionization Energy
Answer:
CFC’s illustrate that technical advances can be a “double-edged” sword as CFC's were made for the advantages but soon or later it came as a disadvantage for environment.
CFC's was discovered with many advantages such as they are stable, non-corrosive, and replacement of harmful ammonia used in refrigerators and air conditioning systems. But soon it is also discovered that CFC's are destroying the upper atmosphere layer called ozone, which is a very big disadvantage for us. CFCs are major contributor to global warming after carbon dioxide.
Hence, it proves that CFC’s shows how technological advancement can be “double-edged” sword.
Lewis Structure is drawn in following steps,
1) Calculate Number of Valence Electrons: # of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
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Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
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12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,