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Lunna [17]
2 years ago
15

For the atoms that do not follow the octet rule state how many electrons surround these atoms. Express your answers as integers

separated by commas. NNON N O, NBF3, NICl2−, NOPBr3, NXeF4 = nothing electrons
Chemistry
1 answer:
wolverine [178]2 years ago
7 0

Answer:

The octet rule isn't all its cracked up to be.  It works fine for some of the elements in the second period, but begins to come unglued for many others.

NNO does follow the octet rule, and has at least three resonance structures

N=N=O <−> N≡N−O <−> N−N≡O  

Formal charges suggest that the one in the middle is most likely.

NBF3 --  Really?  Show me, because I don't think there is such a compound.

NICl2^- or NICl^2- ..... which is it?

NICl2 (neutral) is an analogue of NCl3 where iodine has replaced one chlorine.  It follows the octet rule.

Explanation:

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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
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Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

The value of x = C\alpha = 0.28 M

Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

Concentration of N_2O_4 = 1.0 - x  = 1.0 - 0.28 = 0.72 M

Concentration of NO_2 = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Now put all the values in this expression, we get :

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

8 0
2 years ago
Write the electron configurations for the following ions:
Ket [755]

Answer:

Co²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Sn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zr⁴⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Ag⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

S²⁻ : 1s² 2s² 2p⁶ 3s² 3p⁶

Explanation:

Cobalt (Co): atomic number 27

<u>The electronic configuration of Co in ground state: </u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷

<u>The electronic configuration of Co in +2 oxidation state (Co²⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Tin (Sn): atomic number 50

<u>The electronic configuration of Sn in ground state: </u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p²

<u>The electronic configuration of Sn in +2 oxidation state (Sn²⁺) </u>:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zirconium (Zr): atomic number 40

<u>The electronic configuration of Zr in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d²

<u>The electronic configuration of Zr in +4 oxidation state (Zr⁴⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Silver (Ag): atomic number 47

<u>The electronic configuration of Ag in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰

<u>The electronic configuration of Ag in +1 oxidation state (Ag⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

Sulphur (S): atomic number 16

<u>The electronic configuration of S in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁴

<u>The electronic configuration of S in -2 oxidation state (S²⁻) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶

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Hafnium has six naturally occurring isotopes: 0.16% of 174Hf, with an atomic weight of 173.940 amu; 5.26% of 176Hf, with an atom
ser-zykov [4K]

Answer:

177.277amu

Explanation:

the total occuring isotopes for Hafnium is =6.

First isotope had an atomic weight of 173.940amu

Second isotope =175.941amu

Third isotope =176.943amu

Fourth isotope=177.944amu

Fifth isotope. =178.946amu

sixth isotope .179.947amu

<em>Avera</em><em>ge</em><em> </em><em>ato</em><em>mic</em><em> </em><em>wei</em><em>ght</em><em> </em><em>of</em><em> </em><em>Haf</em><em>nium</em><em>=</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>the </em><em>atomi</em><em>c</em><em> </em><em>weights</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>iso</em><em>topes</em><em>/</em><em> </em><em>Tota</em><em>l</em><em> </em><em>occu</em><em>ring</em><em> </em><em>isotopes</em>

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