Name: Electronic Arrangement of s-block Elements
Uploaded: 2019-07-19
Duration: 24 min 8 s
Description: Do you know how to write the electronic configuration of s-block elements? Is there any similarity in the electronic configuration of all the members of the...

Question

The atomic radius for lithium is

1.23 Å

. When the outermost

2 s

electron is ionized off, the ionic radius of

{Li}^{+}

is

0.76 Å

. Assuming that the difference in radii relates to the space occupied by the

2 s

electron, calculate what percentage of the volume of the lithium atom is occupied by the single valence electron. Is this assumption fair? (Volume of a sphere is

\frac{4}{3} {πr}^{3}

)

Accepted Answer

Electronic configuration of lithium is $1 s^{2} 2 s^{1} .$

Given the atomic radius of lithium to be $1.23 Å$

Ionic radius of ${Li}^{+}$ is $0.76 Å$

Volume of lithium $= \frac{4}{3} {πr}^{3} = \frac{4}{3} \times π \times {(1.23)}^{3} = 7.79 Å^{3}$

Volume of ${Li}^{+}$ $= \frac{4}{3} {πr}^{3} = \frac{4}{3} \times π \times {(0.76)}^{3} = 1.83 Å^{3}$

Based on the assumption $\Rightarrow$ Volume occupied by $2 s$ electron $= 7.79 - 1.83 = 5.96 Å$

Percentage of volume occupied by the $2 s$ electron $= \frac{5.96}{7.79} \times 100 = 76.5 %$

The volume is decreased by $76.5 %$ after the electron is removed. The assumption is fair since, after the removal of the electron the orbit is gone, which means the probability of finding an electron becomes zero.

Important Questions on Group 1 - the Alkali Metals

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