Answer:
The answer to the 2nd question is C.giants.
Explanation:
Answer:
Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²
Explanation:
Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²
Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48
Ca. #Substrate e⁻ = 6F = 6(8) = 48
#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs
#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs
BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron
Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²
XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)
For each ionic compound formula, identify the main group to group 17 (VIIA).
Group17 (VIIA) is the primary group to which X belongs in each ionic combination with the formula CaX2. X is undoubtedly an electron acceptor. It is capable of taking an electron. We determine that X must be in the oxidation state -1 by looking at the chemical formula. The neutral ionic compound's total number of oxidation states is zero. Since calcium has an oxidation state of +2, we must identify the element that contains an anion with an oxidation state of -1. The halogens have a strong propensity to pick up an electron and create an anion. X belongs to group17 (VIIA) as a result.
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Answer: SnO2 + 2 H2 = Sn + 2 H2O
Explanation: I used a balance equation website. It's called WebQC if you want to check it out for future help.
Small peak at 3000large peak at 1685F: it contains two benzene rings that is connected by a bunch of carbons and ketone-Explanation: The spectrum shows a stretching absorption consistent with a ketone functional group: carbonyl C=O stretching at ~1685 cm-1. (An aldehyde, by contrast, would also show a ~2700 cm-1 absorption for the carbonyl C-H stretch.) The C=O stretching frequency is consistent with an aromatic ketone, such as in compound F (1,4-diphenyl-1,4-butanedione). In contrast, an aliphatic ketone absorbs at higher energy (~1710 cm-1). The spectrum also shows the typical ~1600 & ~1500 cm-1 absorptions of a phenyl group.
