Lewis Structures of BeI2

The video tutorial explains how to write the Lewis structure for beryllium iodide (BeI2). It begins by counting the valence electrons: beryllium has 2 and each iodine has 7, totaling 16 valence electrons. Beryllium is placed in the center with iodine atoms on either side. Electrons are paired to form bonds, and octets are completed for iodine. Beryllium is an exception to the octet rule, needing only 4 electrons. Alternative structures with double bonds are considered but found less favorable due to formal charges. The best structure is presented, concluding the tutorial.

How many valence electrons does beryllium contribute to the Lewis structure of BeI2?

• 4

• 7

• 2

• 1

What is the total number of valence electrons used in the Lewis structure of BeI2?

• 16

• 18

• 14

• 8

Where is beryllium placed in the Lewis structure of BeI2?

• On the left

• On the right

• In the center

• Above iodine

What is the role of electron pairs in forming chemical bonds in BeI2?

• To repel atoms

• To form chemical bonds

• To break bonds

• To increase energy

Why is beryllium an exception to the octet rule in BeI2?

• It needs 4 electrons

• It needs 6 electrons

• It needs 8 electrons

• It needs 2 electrons

How many valence electrons does each iodine atom have in the Lewis structure of BeI2?

• 6

• 8

• 5

• 7

What happens if you try to form double bonds in BeI2?

• Electrons are lost

• Beryllium gains an octet

• Structure becomes unstable

• Iodine loses its octet

What is the most likely Lewis structure for BeI2 based on formal charges?

• With double bonds

• With triple bonds

• With single bonds

• With no bonds