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faust18 [17]
1 year ago
13

what is the maximum number of covalent bonds that an oxygen atom with atomic number 8 can make with hydrogen? question 4 options

: 1 6 4 2
Chemistry
1 answer:
Oliga [24]1 year ago
3 0

Maximum number of covalent bonds that an oxygen atom can make with hydrogen is 2.

  • the ground state electronic configuration of oxygen is 2s² 2p⁴ that means it has 6 electrons in its valence shell and require two electrons are required to complete its octate.
  • Two bonds are created when an electron donor atom shares the two needed electrons with oxygen. The ability of two oxygen atoms to share valence electrons results in the creation of a double bond between the two atoms.
  • There are no longer any empty orbitals in the octet of oxygen after it is complete. As a result, it is unable to accept more electrons or create more bonds.

Therefore, Oxygen can only generate two bonds because it needs two additional electrons to complete its octet, after which it will run out of empty orbitals in which to receive additional electrons and create additional bonds.

learn more about octate here:

https://brainly.in/question/24161245

#SPJ4

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How many electrons does nitrogen (N) need to gain to have a stable outer electron shell
earnstyle [38]
It need "3 electrons" to have a stable electronic configuration. 

'cause it has 5 electrons in it's outer shell & every atom needs 8 electrons. So, it requires 3 more!

Hope this helps!
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2 years ago
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Na + CI2 =2NaC is It balanced
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Answer:

no, the correct answer is NaCI

Explanation:

you're welcome

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2. Why do you think electroplated jewelleries are in demand?<br> Chemistry plsss
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Explanation:

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8 0
2 years ago
From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
Nina [5.8K]

<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

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The answer is temperature (may be wrong)
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