In an ionic bond, one atom essentially donates an electron to stabilize the other atom. The atoms in an ionic bond have different electronegativity values from each other.
While in a covalent bond, the atoms are bound by shared electrons and the electronegativity values are the same.
Electrons does not have state like matter of Solid, liquid, gas or plasma states. Electrons together with other molecular and atom components makes up matter. Their characteristic and elemental inclination determines the state for which the matter will exist in.
Answer:
B) conducts electricity
Explanation:
This answer can be answered by the process of elimination. Metals are excellent conductors of electricity, which means A is out, and metals are also known to have medium reactivity and have high melting points, meaning C and D are also out.
The element of the group 17 that is most active non metal is fluorine.
The group 17 of the periodic table contains bromine(Br), iodine(I), Chlorine(Cl) and fluorine(F).
Among all the elements of the group 17. Fluorine is the smallest in size.
Because of the small size of fluorine it has the highest electronegativity in group 17.
This high electronegativity makes it a very active non metal. It provides a very high oxidizing power and low dissociation energy to the fluorine atom.
Also because of the very small size the source of attraction between the nucleus and the electrons is very high in floor in atom.
It reacts readily to form oxides and hydroxides.
So, we can conclude here that fluorine is the most active non metal of group 17.
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Answer:
emf generated by cell is 2.32 V
Explanation:
Oxidation: 
Reduction: 
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Overall: 
Nernst equation for this cell reaction at
-
![E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE_%7Bcell%7D%5E%7B0%7D-%5Cfrac%7B0.059%7D%7Bn%7Dlog%7B%5BAl%5E%7B3%2B%7D%5D%5E%7B2%7D%5BI%5E%7B-%7D%5D%5E%7B6%7D%7D)
where n is number of electrons exchanged during cell reaction,
is standard cell emf ,
is cell emf ,
is concentration of
and
is concentration of 
Plug in all the given values in the above equation -
![E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.20-%5Cfrac%7B0.059%7D%7B6%7Dlog%5B%284.5%5Ctimes%2010%5E%7B-3%7D%29%5E%7B2%7D%5Ctimes%20%280.15%29%5E%7B6%7D%5DV)
So, 