Ionic bonds are formed when there is complete transfer of valence electrons between two atoms.
Electronegativity tells the trend of an atom to atract electrons.
You should search for the complete set of rules that indicate whether an ionic or covalent bond happens.
There are two relevant rules to state if whether an ionic bond will happen:
- When the difference of electronegativities between the two atoms is greater than 2.0, then the bond is ionic.
- When the difference is between 1.6 and 2.0, the bond is ionic if one of the elements is a metal.
You need to list the electronegativities of the five elements (there are tables with this information)
Element electronegativity
Cu: 1.9
H: 2.2
Cl 3.16
I: 2.66
S: 2.58
Differences:
Cu / S: 2.58 - 1.9 = 0.68
H / S: 2.58 - 2.2 = 0.38
Cl / S: 3.16 - 2.58 =0.58
I / S: 2.66 - 2.58 = 0.08
Those differences are too low to consider that the bond is ionic.
Then the answer is that none of those atoms forms an ionic bond with sulfur.
Answer:
there is some difference
Explanation:
hydrogen ion --
symbol-H
Charge- 1+ or 1-
it is atom
hydroxide ion --
symbol- OH
charge-. -1
contains.- one oxygen atom and one hydrogen atom.
they are bonded.
it is a molecule
Answer:
See explanation
Explanation:
Let us recall that a negative ion is formed by addition of electrons to an atom. When electrons are added to the atom, greater interelectronic repulsion increases the size of the Te^2− hence it is greater in size than Te atom. Therefore, the ionic radius of Te^2− is greater than the atomic radius of Te.
In the second question, oxygen is positioned so far to the right because it has a far smaller nuclear charge compared to Te. Hence in the PES spectrum, the 1s sublevel of oxygen lies far to the right of that of Te.
Answer:
Yes, Pb3(PO4)2.
Explanation:
Hello there!
In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.
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