Answer:
d. Adding 5% HCl solution to the crude reaction mixture will protonate aniline increasing its solubility in the aqueous solution.
Explanation:
On this case, we have to check the <u>structures of each compound</u> (figure 1). For naphthalene we dont have <u>any functional groups</u> therefore, the addition of HCl or NaOH it will not affect naphthalene so <u>we can discard "B" and"C".</u>
When we add HCl solution we will have the production 
the presence of this <u>hydronium ion will protonate the acid</u>, so we can <u>discard a.</u>
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Finally, for d when we add the <u>hydronium ion will react with aniline</u> (a base) and will produce an <u>ammonium ion</u>. This ammonium ion have a <u>positive charge</u>, therefore the <u>polarity will increase</u> and the molecule would be more soluble on water (figure 2).
I hope it helps!
Let's check Electronic configuration of N in ground state
In hydrogen case
Hence Hybridization
Structure is tetrahedral .
Spare way:-
- Bond pairs==3=>Bonding electrons=6
- Lone pair=1=>Anti bonding electrons=2
Hybridization
![\\ \sf\longmapsto \dfrac{1}{2}[\sigma+\sigma *]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cdfrac%7B1%7D%7B2%7D%5B%5Csigma%2B%5Csigma%20%2A%5D)
![\\ \sf\longmapsto \dfrac{1}{2}[6+2]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cdfrac%7B1%7D%7B2%7D%5B6%2B2%5D)
![\\ \sf\longmapsto \dfrac{1}{2}[8]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cdfrac%7B1%7D%7B2%7D%5B8%5D)
- sp3 Hybridization
- Shape-Tetrahedral .
Answer:
Using the drop down Menu
Proton : Possessing a Charge of +1. It is found in the Nucleus and has a Mass of "A"
Neutron; Possesses a charge of 0. It is situated in the Nucleus and Has a Mass of "B"
Electron: Has a charge of -1. Found in the Orbitals of the atom and has a mass of "C"
Not helping or supporting either side of the argument
