Question

1. based on the periodic table shown, elements from columns b and e will combine in which ratio to form an ionic compound? 2b:3e no likely combination 3b:2e 3b:1e

Explanation:

Step1: Determine the charge of element B

Assume column B is group 2 (alkaline earth metals). Elements in group 2 lose 2 electrons and have a charge of \(+2\).

Step2: Determine the charge of element E

Assume column E is group 16. Elements in group 16 gain 2 electrons and have a charge of \(-2\).

Step3: Use the criss - cross method

For an ionic compound, the charges must balance. Let the formula be \(B_xE_y\). Using the criss - cross rule (the magnitude of the charge of one ion becomes the subscript of the other ion), \(x = 2\) (charge of E) and \(y = 2\) (charge of B). But simplifying \(B_2E_2\) gives \(BE\) (ratio \(1:1\)). Wait, no, re - check.
Wait, another approach: assume column B is group 3 (if we consider the general periodic table structure in some basic models for this problem - maybe a simplified table). If B is in a group where it loses 3 electrons (\(+3\) charge) and E is in group 16 (gains 2 electrons, \(-2\) charge). Using the criss - cross rule: for \(B^{3 +}\) and \(E^{2-}\), the formula is \(B_2E_3\) (ratio \(2B:3E\)).

Answer:

\(2B:3E\)