Question

1. an ionic bond would form between which pairs of elements? (1 point) *

atomic number 7 and atomic number 8
atomic number 8 and atomic number 12
atomic number 4 and atomic number 18
atomic number 3 and atomic number 4

1. which compound does not have a bent molecular shape? (1 point) *

beh₂
h₂s
h₂o
seh₂

Explanation:

Question 3

Step1: Identify elements by atomic number

Atomic number 7 is N (non - metal), 8 is O (non - metal), 12 is Mg (metal), 4 is Be (metal), 18 is Ar (noble gas), 3 is Li (metal), 4 is Be (metal).

Step2: Recall ionic bond formation

Ionic bonds form between metals and non - metals (or polyatomic ions).

• For atomic number 7 (N) and 8 (O): both non - metals, covalent bond likely.
• For atomic number 8 (O, non - metal) and 12 (Mg, metal): metal - non - metal, ionic bond likely.
• For atomic number 4 (Be, metal) and 18 (Ar, noble gas): noble gas is inert, no bond.
• For atomic number 3 (Li, metal) and 4 (Be, metal): both metals, metallic bond.

Step1: Recall VSEPR theory for molecular shape

• \(BeH_2\): Be has 2 valence electrons, forms 2 bonds with H. The central atom Be has no lone pairs. According to VSEPR, the electron - pair geometry and molecular geometry is linear (\(AX_2\) type, where A is central atom, X is bonded atom).
• \(H_2S\): S is central atom, has 6 valence electrons. It forms 2 bonds with H, so it has 2 lone pairs. The molecular shape is bent (\(AX_2E_2\) type, E is lone pair).
• \(H_2O\): O is central atom, has 6 valence electrons. Forms 2 bonds with H, has 2 lone pairs. Molecular shape is bent (\(AX_2E_2\) type).
• \(SeH_2\): Se is central atom, has 6 valence electrons. Forms 2 bonds with H, has 2 lone pairs. Molecular shape is bent (\(AX_2E_2\) type).

Step2: Determine the non - bent shape

From the above analysis, \(BeH_2\) has a linear shape, not bent.

Answer:

atomic number 8 and atomic number 12

Question 4