Question
What is the radius of iodine atom ?
(atomic no. 53, mass no. 126)
A.
$$2.5 \times {10^{ - 11}}m$$
B.
$$2.5 \times {10^{ - 9}}m$$
C.
$$7 \times {10^{ - 9}}m$$
D.
$$7 \times {10^{ - 6}}m$$
Answer :
$$2.5 \times {10^{ - 11}}m$$
Solution :
Electronic configuration of iodine $$\left( {53} \right)$$ is $$= 2, 8, 18, 18, 7$$
∴ Principal quantum number $$n = 5$$
Radius of $$n$$th orbit is given by
$${r_n} = {r_0}\left( {\frac{{{n^2}}}{Z}} \right)\,\,\left( {Z = {\text{atomic number}}} \right)$$
where $${r_0} = 0.53\,\mathop {\text{A}}\limits^ \circ = 0.53 \times {10^{ - 10}}m$$
$$\therefore {r_n} = \left( {0.53 \times {{10}^{ - 10}}} \right) \times \frac{{{5^2}}}{{53}} = 2.5 \times {10^{ - 11}}m$$