Answer:
Diagram B shows the correct arrangement of electrons in the product.
The molar concentration of the original HF solution : 0.342 M
Further explanation
Given
31.2 ml of 0.200 M NaOH
18.2 ml of HF
Required
The molar concentration of HF
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)
Titrant = NaOH(1)
Titrate = HF(2)
Input the value :

Volume= Mass/ Density= 20/184
= 0.108 cubic cm
N oxidation number in N₂O = 1
S oxidation number in SO₂ = 4
S oxidation number in SO₃ = 6
P oxidation number in P₄O₆ = 3
Therefore, the Sulfur in SO₃ will not react with molecular oxygen as Sulfur is already using all of its valence electrons in bonding.
Explanation:
a single bond such a (C-H) has one sigma bond whereas a double ( C=C) and triple (C=C) bond has one sigma bond with remaining being pi bond