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vekshin1
3 years ago
6

If two atoms have an electronegativity difference of 1.3, what type of bond would form between them?

Chemistry
2 answers:
Debora [2.8K]3 years ago
5 0

Answer:

The correct answer is b polar covalent

Explanation:

When two atoms joined by covalent bond has difference in their electronegativities at that time polarity arise.

    When the electronegativity difference is low such as 1.3 then the polar bond formed by the two atoms are called polar covalent bond.For example H2O

 on the other hand polar bond formed by two atoms having high difference in their electronegativities is called ionic bond.For example NaCl.

zloy xaker [14]3 years ago
3 0

Answer:

a. polar covalent

Explanation:

Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself.

The higher the electronegativity, the more an atom attracts electrons towards itself.

The most commonly used method of calculation of electronegativity is done by Pauling's scale.

Cesium is the least electronegative (0.79), and fluorine is most electronegative (3.98)

Atoms that form bonds with high electronegativity difference among them form ionic bonds (metal and non-metal). If their electronegativity difference (Δ)=0, then they form non polar covalent bond.

  • Δ=0 : Non- polar covalent bond.
  • 0 < Δ < 1.7 : Polar covalent bond.
  • Δ = 1.7 : 50% ionic and 50% covalent bond
  • Δ > 1.7 : Ionic bond

The answer is a. polar covalent

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Answer:

\huge 1 × {10}^{-11} \: \: M

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So we have

11 =  -  log({H}^{+})  \\ {H}^{+} =  {10}^{ - 11}

We have the final answer as

1 × {10}^{-11} \: \: M

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Explanation:

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Thus, for one mole of C₂H₆O₂ (ethylene glycol), 2.5 moles of O₂ is needed.  

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= 4.5/2 = 2.25 mol

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