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Gennadij [26K]
3 years ago
8

For which of the following elements (in their normal, stable, forms) would it be correct to describe the bonding as involving a

"sea of electrons"? a. hydrogen b. helium c. sulfur d. iodine e. lithium
Chemistry
1 answer:
Dvinal [7]3 years ago
4 0

Answer: The correct option is e.

Explanation: In the given question it is asked to tell the element which has 'Sea of electrons' while bonding.

This means that the element should have extra electrons so that it can bond easily or this also means which element can easily donate its electron to form a bond.

  • Electronic configuration of Hydrogen = 1s^1

Hydrogen cannot loose electrons easily because the electron is present in 1s sub-shell which is the closest to the nucleus and hence will be tightly bonded to it.

  • Electronic configuration of Helium = 1s^2

This element will not loose electrons easily because the 1s sub-shell is fully filled and hence will not be available for bonding.

  • Electronic configuration of Sulfur = 3s^23p^4

This element lacks 2 electrons from attaining stable configurations. Hence, it will gain 2 electrons rather than donating for bonding.

  • Electronic configuration of Iodine = 5s^25p^5

This element lacks 1 electron to attain stable configuration. Hence, it will gain 1 electron rather than donating for bonding.

  • Electronic configuration of Lithium = 1s^22s^1

This element has an extra electron in its valence shell and can be easily lost in order to form bond.

Hence, the correct option is e.

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Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
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Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

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Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

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