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2 years ago

Assign formal charges to each atom in the resonance forms of n2o.

Chemistry

1 answer:

inn [45]2 years ago

8 0

Answer:

Here's what I get

Explanation:

The three resonance forms of N₂O are shown in the first diagram below (you can also use horizontal dashes to represent the bonding pairs).

To get the formal charge (FC) on the atoms, cut each bond in half, as in the second diagram.

Each atom gets the electrons on its side of the cut.

Formal charge = valence electrons in isolated atom - electrons on bonded atom

FC = VE - BE

(a) In Structure A

Left-hand N:

VE = 5

BE = 1 lone pair (2)+ 3 bonding electrons = 2 + 3 = 5

FC = 5 - 5 = 0.

Central N:

VE = 5

BE = 4

FC = 5 - 4 = +1

On O:

VE = 6

BE = 3 lone pairs(6) + 1 bonding electron = 7

FC = 5 - 6 = -1

(b) In Structure B

Left-hand N:

VE = 5

BE = 2 lone pairs (4)+ 2 bonding electrons = 4 + 2 = 6

FC = 5 - 6 = -1.

Central N:

VE = 5

BE = 4

FC = 5 - 4 = +1

On O:

VE = 6

BE = 2 lone pairs(4) + 2 bonding electrons = 6

FC = 6 - 6 = 0

(c) In Structure C

Left-hand N:

VE = 5

BE = 3 lone pairs (6)+ 1 bonding electrons = 6 + 1 = 7

FC = 5 - 7 = -2.

Central N:

VE = 5

BE = 4

FC = 5 - 4 = +1

On O:

VE = 6

BE = 1 lone pair(2) + 3 bonding electrons = 5

FC = 6 - 5 = +1

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If 200.0 g of copper(II) sulfate react with an excess of zinc metal, what is the theoretical yield of copper? 1.253 g 50.72 g 79

Helen [10]

1)we need a balanced equation: CuSO₄ + Zn ---> ZnSO₄ + Cu

2) we need to convert the grams of CuSO₄ to moles using the molar mass.

molar mass CuSO₄= 63.5 + 32.0 + (4 x 16.0)= 160 g/mol

$200.0 g CuSO_4 ( \frac{1 mol}{160 grams} )= 1.25 mol CuSO_4$

3) convert moles of CuSO₄ to moles of Cu

$1.25 mol CuSO_4 ( \frac{1 mol Cu}{1 mol CuSO_4} )= 1.25 mol Cu$

4) convert moles of Cu to grams using it's molar mass.

molar mass Cu= 63.5 g/mol

$1.25 mol (\frac{63.5 grams}{1 mol} )= 79.4 grams Cu$

I did it step-by-step as the explanation but you can do all of this in one step.

$200.0 g CuSO_4 ( \frac{1 mol CuSO_4}{160 g} ) ( \frac{1 mol Cu}{1 mol CuSO_4} ) ( \frac{63.5 grams}{1 mol Cu} )= 79.4 grams Cu$

4 0

2 years ago

A rod, X has a positive charge of 8. An otherwise identical rod, Y has a negative charge of 4. The rods are touched together, an

gtnhenbr [62]

Answer:

1. electrons

2. From "Y" to "X"

Explanation:

1. Electrons move between the rod since the electrons are the only charge carriers which are free to move.

2. The particles move from from "Y" to "X" since the electrons are the only charge carriers which are free to move. The positive charge on rod x is due to a deficit of electrons while the negative charge on rod Y is due to the excess of electrons. When the rods come together, the electrons move from "Y" to "X" since the electrons are the only charge carriers which are free to move.

4 0

1 year ago

Exactly 17.0 mL of a H2SO4 solution was required to neutralize 45.0 mL of 0.235 M NaOH. What was the concentration of the H2SO4

aleksandr82 [10.1K]

Answer:

Molarity for the sulfuric acid is 0.622 M

Explanation:

When we neutralize an acid with a base, molarity of both . both volume are the same. The formula is:

M acid . volume of acid = M base . volume of base

M acid = unknown

Volume of acid = 17 mL

Volume of base = 45 mL

M base = 0.235 M

Therefore, we replace: M acid . 17 mL = 0.235 M . 45 mL

M acid = (0.235 M . 45 mL) / 17 mL

M acid = 0.622 M

6 0

2 years ago

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

2 years ago

How many moles are there in 8.50 X 1024 molecules of sodium sulfate, Na2SO3?

DiKsa [7]

Hello!

To solve this question, we need to use the Avogadro's Number, which is a constant first discovered by Amadeo Avogadro, an Italian scientist. He discovered that in a mole of a substance, there are 6,02*10²³ molecules. Using this relationship, we apply the following conversion factor:

$8,50* 10^{24}molecules* \frac{1 mol Na_2SO_3}{6,02* 10^{23}molecules}=14,12 molesNa_2SO_3$

So, 8,50 * 10²⁴ molecules of Na₂SO₃ represent 14,12 moles of Na₂SO₃

Have a nice day!

7 0

2 years ago