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defon
1 year ago
15

how many bonding pairs and nonbonding pairs of electrons does each nitrogen atom contain?bonding pairs of electrons for nitrogen

atoms:6nonbonding pairs of electrons for nitrogen atom
Chemistry
1 answer:
Len [333]1 year ago
7 0

Five valence electrons are present on each nitrogen atom. Each nitrogen atom contributes three electrons to the triple bond when two nitrogen atoms come together to create the dinitrogen molecule, with the final pair of electrons being a non-bonding lone pair.

<h3>Why Can't Nitrogen Form Four Bonds?</h3>

Nitrogen typically has three bonds, but it has the potential for four. If it happens, a positive charge will be produced. If the atom of nitrogen has a negative charge, it may have two bonds. Oxygen normally has two bonds, but it also has the potential for three. If it has three bonds, there will be a positive charge.

<h3>How many nitrogen atom pairs are free of bonds?</h3>

The triple bonds between the nitrogen atoms hold them together. They each possess one pair of electrons as a result. Therefore, there are a total of two single pairings.

<h3>How many bond pairs does nitrogen have?</h3>

In its no-formal charge state, nitrogen is normally surrounded by three covalent bonds and a lone pair.

To know more about Bonding and Non bonding pair of electron visit:

brainly.com/question/18258856

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A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH. Express your
Mazyrski [523]

Answer:

pH= 1.17

Explanation:

The neutralization reaction between HBr (acid) and KOH (base) is given by the following equation:

HBr(aq) + KOH(aq) → KBr(aq) + H₂O(l)

According to this equation, 1 mol of HBr reacts with 1 mol of KOH. Then, the moles can be expressed as the product between the molarity of the acid/base solution (M) and the volume in liters (V). So, we calculate the moles of acid and base:

<u>Acid</u>:

M(HBr) = 0.15 M = 0.15 mol/L

V(HBr) = 50.0 mL x 1 L/1000 mL = 0.05 L

moles of HBr = M(HBr) x V(HBr) = 0.15 mol/L x 0.05 L = 7.5 x 10⁻³ moles HBr

<u>Base</u>:

M(KOH) = 0.25 M = 0.25 mol/L

V(HBr) = 13.0 mL x 1 L/1000 mL = 0.013 L

moles of HBr = M(HBr) x V(HBr) = 0.25 mol/L x 0.013 L = 3.25 x 10⁻³ moles KOH

Now, we have: 7.5 x 10⁻³ moles HBr > 3.25 x 10⁻³ moles KOH

HBr is a strong acid and KOH is a strong base, so they are completely dissociated in water: the acid produces H⁺ ions and the base produces OH⁻ ions. So, the difference between the moles of HBr and the moles of KOH is equal to the moles of remaining H⁺ ions after neutralization:

moles of H⁺ = 7.5 x 10⁻³ moles HBr - 3.25 x 10⁻³ moles KOH = 4.25 x 10⁻³ moles H⁺

From the definition of pH:

pH = -log [H⁺]

The concentration of H⁺ ions is calculated from the moles of H⁺ divided into the total volume:

total volume = V(HBr) + V(KOH) = 0.05 L + 0.013 L = 0.063 L

[H⁺] = (moles of H⁺)/(total volume) = 4.25 x 10⁻³ moles/0.063 L = 0.067 M

Finally, we calculate the pH after neutralization:

pH = -log [H⁺] = -log (0.067) = 1.17

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