Answer:
The possible valances can be determined by electron configuration and electron negativity
Good Luck even though this was asked 2 weeks ago
Explanation:
All atoms strive for stability. The optima electron configuration is the electron configuration of the VIII A family or inert gases.
Look at the electron configuration of the nonmetal and how many more electrons the nonmetal needs to achieve the stable electron configuration of the inert gases. Non metals tend to be negative in nature and gain electrons. ( They are oxidizing agents)
For example Florine atomic number 9 needs one more electron to reach a valance number of 8 electrons to equal Neon atomic number 10. Hence Flowrine has a valance of -1
Oxygen atomic number 8 needs two more electrons to reach a valance number of 8 electrons to equal Neon atomic number 10. Hence Oxygen has a valance charge of -2.
Non metals with a low electron negativity will lose electrons when reacting with another non metal that has a higher electron negativity. When the non metal forms an ion it is necessary to look at the electron structure to determine how many electrons the element can lose to gain stability.
For example Chlorine which is normally -1 like Florine when it combines with oxygen can be +1, +3, + 5 or +7. It can lose its one unpaired electron, or combinations of the unpaired electron and sets of the three pairs of electrons.
If you take your mom's advice, a. the room will be cooled by minimizing heat transfer by reflection.
The Sun is heating your room by its <em>radiation</em>.
The aluminium foil is shiny, so it reflects the Sun’s radiation back outside the window.
<em>Convection</em> is the transfer of heat caused by the air moving around in your room.
<em>Conduction</em> occurs only when a hot object is in contact with a cooler object.
Fe + O2 → Fe2O3
After balancing the eq.
4Fe + 3O2 → 2Fe2O3
Hope this will help u mate :)
This reaction is most likely to fall under SN2 because the
thing called carbonication does not occur in SN1. The carbon forms a partial
bond with the nucleophile during the intermediate phase and the leaving group.
So for this question the reaction will fall under SN2.
