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Andrej [43]
3 years ago
9

The central atom in the chlorite anion, ClO2- is surrounded by: Question 5 options: one bonding and three unshared pairs of elec

trons. two bonding and two unshared pairs of electrons. two bonding and one unshared pair of electrons. two double bonds and no unshared pairs of electrons.
Chemistry
1 answer:
PolarNik [594]3 years ago
3 0

Answer:

  • <u>two bonding and two unshared pair of electrons.</u>

Explanation:

<em>Chlorite anion, ClO₂⁻ </em>has one atom of chlorine and two atoms of oxygen.

Chlorine atom has 7 valence electrons (group 17 of the periodic table) and oxygen has 6 valence electrons (group 16 of the periodic table).

Thus, in total the anion has 7 + 6×2 electrons, plus an additional electron indicated by the negative charge of the anion: 7 + 12 + 1 = 20 valence electrons, that must be distributed among the three atoms.

Being chlorine the most electronegative atom, you place it in the center. surrounded by the two oxygen atoms.

Thus, so far:

             O Cl  O (pending to place the valence electrons.

You can put two electrons between each chlorine and oxygen atoms to form the bonds:

     O : Cl : O

Also, complete the octets:

 . .      . .    . .

: O  :  Cl :  O :

 . .     . .     . .

That structure works because:

  • there are 20 electrons
  • every atom has 8 valence electrons

But you need to calculate the formal charges on each atom:

For chlorine:

  • 7 valence electrons - 4 nonbonding valence electrons - 4/2 bonding electrons = 7 - 4 - 2 = + 1.

For each oxygen:

  • 6 valence electrons - 6 nonbonding valence electrons - 2/2 bonding electrons = 6 - 6 - 1 = -1

The total charge of the anion is +1 -1 - 1 = -1.

And the complete structure with the charge is:

   . .      . .    . .

[ : O  :  Cl :  O : ] ⁻

   . .      . .     . .

The square bracketts comprise the entire anion and the negative charge is for the entire structure.

In this structure, <em>the central atom (Cl) is surrounded by two bonding and two unshared pairs of electrons. </em>(this is the answer to the question).

There is another structure which minimizes the formal charge, which will be preferred. It is by making a double bond between chlorine and one of the oxygen atoms, by moving two of the atoms beside one chlorine.

That structure is:

 . .      . .    . .

: O  :  Cl ::  O

 . .     . .     . .

There you have 20 valence electrons and complete octets too, but the formal charges are:

Chlorine:

  • 7valence electrons - 4 nonbonding electrons - 6/2 bonding electrons = 7 - 4 - 3 = 0

The oxygen on the left is equal to the previous structure, thus - 1 forma charge.

The oxygen on the right:

  • 6 valence electrons - 4 nonbonding electrons - 4/2 bonding electrons = 6 - 4 - 2 = 0

Hence, this las structure minimizes the formal charges and is preferred.

This is:

 . .      . .    . .

: O  :  Cl ::  O

 . .     . .     . .  

But you must add the negative charge which belongs to the complete anion:

      . .      . .     . .

[     : O  :  Cl ::  O  ] ⁻

      . .     . .      . .

The square brackets comprise the entire anion and the negative charge is for it.

Hence, in this other structue the central atom, chlorine, is surrounded by six bonding electrons and four nonbonding electrons.

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50.0 g of NaNO3 are dissolved into enough water to make 250 mL of solution. What is the molarity of this solution?
d1i1m1o1n [39]

Answer:

2.35 M

Explanation:

Molarity is mol/L of solution. We have to convert the g to mol and the mL to L. G to mol uses the molar mass of the compound. The molar mass of NaNO₃ is 85.00g/mol.

50.0gNaNO3*\frac{1molNaNO3}{85.00gNaNO3} = 0.588molNaNO3

Then you have to convert mL to L.

250mL*\frac{1L}{1000mL} = 0.250L

Now divide the mol by the L.

\frac{0.588mol NaNO3}{0.250L} = 2.352 M

Round to the smallest number of significant figures = 2.35M

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3 years ago
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Veseljchak [2.6K]
It would be 4.2, hope this helps.
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Is there a mathematical pattern in the number of
aleksandrvk [35]

Answer:

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Explanation:

hope it helps

4 0
3 years ago
what is the concentration of hydroxide ions after 50.0 ml of 0.250 m naoh is added to 120 ml of 0.200 m na2so4? please show all
Galina-37 [17]

The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

What is meant by concentration?

Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.

Concentration of hydroxide ions can be calculated by,

M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.

where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.

Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

To learn more about concentration click on the given link brainly.com/question/17206790

#SPJ4

5 0
10 months ago
When converting an acid salt dissolved in water to its acid form, you are instructed to adjust the ph well into the acidic range
777dan777 [17]
First of all, the problem says that you have to convert the acid salt to its acidic form. If you take it to the neutral pH, that won't be acidic at all. As simple as that, you don't take it to neutral pH because it would lose its definition of being acidic afterall.
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3 years ago
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