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3 years ago

How many nonbonding electron pairs are there in the lewis structure of the peroxide ion, o22−?

2 answers:

8 0

There is 6 non - bonding pairs.
Let me show you one easy method to do this.
o22-, oxygen valence electron = 6 here we have two so total 12, and -2 that means we add electrons so it’s all equal to 14 right.
whenever need to find lone pair, subtract the number you get with the lowest multiple of 8.
here we total 14 valence electron right so lowest multiple of 8 would be 8.
so 14 - 8 = 6 and that is our answer.
Let me know if you have Problem with chemistry.

frosja888 [35]3 years ago

8 0

Answer:

The correct answer is that there are 6 pairs of non-binding electrons.

Explanation:

Hello!

Let's solve this!

In the Lewis structure of hydrogen peroxide we have a simple covalent bond between each of the hydrogen with each of the oxygen. To this is added a simple covalent bond between the oxygen. So we will have:

O22-: this means that we have 6 electrons for each of the oxygen, they are two oxygen so there are 12 electrons and there is a -2 so in total there are 14 electrons.

This must be deducted from the used electrons that are 2 + 2 (with the hydrogen) +2 (of the two oxygen): in total there are six electrons.

14-6 = 8

In other words, there are 4 pairs of non-binding electrons in hydrogen peroxide.

As is the peroxide ion, we must add two freer pairs, which are those that bind with the hydrogen.

The correct answer is that there are 6 pairs of non-binding electrons.

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Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found

Igoryamba

Answer:

The concentrations are :

$[HAsc^-]=0.000702 M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

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Explanation:

$H_2Asc\rightleftharpoons HAs^-+H^+$ $K_{a1}=1.0\times 10^{-5}$

Initial

c 0 0

Equilibrium

c-x x x

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$1.0\times 10^{-5}=\frac{x\times x}{(c-x)}$

$1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}$

Solving for x:

x = 0.000702 M

$[HAsc^-]=0.000702 M$

$HAsc^-\rightleftharpoons As^{2-}+H^+$ $K_{a2}=5\times 10^{-12}$

Initially

x 0 0

At equilibrium ;

(x - y) y y

$K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}$

$5\times 10^{-12}=\frac{y\times y}{(x-y)}$

$5\times 10^{-12}=\frac{y^2}{(x-y)}$

Putting value of x = 0.000702 M

$5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}$

$y=5.92\times 10^{-8} M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

Total concentration of $[H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M$

The pH of the solution :

$pH=-\log[H^+]$

$pH=-\log[7.0206\times 10^{-4} M}=3.15$

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Answer:

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