Question

Part of one row of the periodic table is shown.

4 blue boxes in a row. The first has S n at the center and 50 above with tin and 118.71 below. The second has S b at the center and 51 above with antimony and 121.76 below. The third has T e at the center and 52 above with tellurium and 127.60 below. The fourth has I at the center and 53 above with iodine and 126.90 below.

Which element’s atoms have the greatest average number of neutrons?

tin (Sn)
antimony (Sb)
tellurium (Te)
iodine (I)

Answer 1

Tellurium (Te).

Reason: average neutrons ≈ atomic mass − atomic number.
- Sn: 118.71 − 50 ≈ 68.7
- Sb: 121.76 − 51 ≈ 70.8
- Te: 127.60 − 52 ≈ 75.6
- I: 126.90 − 53 ≈ 73.9

Te has the largest value.