Question

The shielding of electrons gives rise to an effective nuclear charge,
`Z_(eff)`, which explains why boron is larger than oxygen. Estimate the approximate
`Z_(eff)` felt by a valence electron of boron and oxygen, respectively?a. +5 and +8b. +3 and +6c. +5 and +6d. +3 and +8e. +1 and +4

Answer 1

Option b: +3 and +6.

Step-by-step explanation:

To calculate the
`Z_(eff)` of the valence electron of boron and oxygen, we need to use the next equation:

`Z_(eff) = Z - s`

where Z: is the proton number of the atom and s: is the shielding constant

The electronic configuration of the atom of boron and oxygen is the following:

B (Z=5): 1s²2s²2p¹ → number of valence electrons: 3 (2s²2p¹) → number of nonvalence electrons: 2 (1s²)

O (Z= 8): 1s²2s²2p⁴ → number of valence electrons: 6 (2s²2p⁴) → number of nonvalence electrons: 2 (1s²)

Assuming that the shielding constant is approximately equal to the number of the nonvalence electrons, the
`Z_(eff)` of the valence electron of boron and oxygen is:

B:
`Z_(eff) = 5 - 2 = +3`

O:
`Z_(eff) = 8 - 2 = +6`

So, the correct option is b, +3 for the boron and +6 for the oxygen.

I hope it helps you!