Question

1. choose the best answer.

which of these is not matched correctly?
tetra = 4
hepta = 9
tri = 3
hexa = 6

1. choose all that apply.

which three show the correct oxidation number for the given element?
al^{5 - }
br^{1 - }
b^{3 - }
k^{1 + }
s^{2 - }

Explanation:

Question 4

• "tetra" means 4 (e.g., tetrahedron has 4 faces), so tetra = 4 is correct.
• "hepta" means 7 (e.g., heptagon has 7 sides), not 9, so hepta = 9 is incorrect.
• "tri" means 3 (e.g., triangle has 3 sides), so tri = 3 is correct.
• "hexa" means 6 (e.g., hexagon has 6 sides), so hexa = 6 is correct.

• Al⁵⁻: Aluminum (Al) is a metal in group 13, it typically loses 3 electrons to form Al³⁺, so Al⁵⁻ is incorrect.
• Br¹⁻: Bromine (Br) is in group 17, it gains 1 electron to form Br⁻ (oxidation number -1), so Br¹⁻ is correct.
• B³⁻: Boron (B) is in group 13, it typically loses 3 electrons (forms B³⁺) or shares electrons, gaining 3 electrons to form B³⁻ is less common but possible in some compounds? Wait, no—boron is a metalloid, and its common oxidation states are +3 (losing 3 electrons) or in covalent bonds. Gaining 3 electrons to form B³⁻ is not typical. Wait, maybe I made a mistake. Let's recheck:
• K¹⁺: Potassium (K) is in group 1, it loses 1 electron to form K⁺ (oxidation number +1), so K¹⁺ is correct.
• S²⁻: Sulfur (S) is in group 16, it gains 2 electrons to form S²⁻ (oxidation number -2), so S²⁻ is correct.

Wait, let's correct:

• Boron (B) is in group 13, valence electrons 3. It usually forms covalent bonds or loses 3 electrons (B³⁺). Gaining 3 electrons to form B³⁻ is not a common oxidation state. So B³⁻ is incorrect.
• So correct ones: Br¹⁻, K¹⁺, S²⁻.

Answer:

B. hepta = 9 (assuming the options are labeled as A. tetra = 4, B. hepta = 9, C. tri = 3, D. hexa = 6)

Question 5