Chemistry, 23.07.2019 17:00, jwbri

Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank items as equivalent, overlap them. hardesteasiest

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The rate of Formation of Carbocation mainly depends on two factors'

1) Stability of Carbocation:
The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.

2) Ease of detaching of Leaving Group:
The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

R-I > R-Cl > R-F

B > C > A
Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from

Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from

answered: Guest

screening

explanation:

answered: Guest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.