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Over [174]
3 years ago
6

How many electrons are transferred between the cation and anion to form the ionic bond in one formula unit of each compound? (1

electrons, 2 electrons, 3 electrons, 4 electrons) ?
NaCI
CaS
BaO
KBr
LiF
Chemistry
1 answer:
rosijanka [135]3 years ago
3 0
Here are the solutions:
For NaCl, there would be one electron transferred from the sodium atom, now producing a cation to the chlorine atom resulting in a chlorine anion. Forming ionic bond.
For CaS, there would be 2 electrons transferred from an atom of Ca to S, this can be obtained by simply looking at the ionic charges and or combining capacities of Ca.
For BaO, likewise, 2 electrons that are valence electrons are transferred from Ba to Oxygen.
In the ionic compound of KBr, the atom of K, potassium donated its one valence electron to Br, in need of it to achieve a stable octet, and produce the chemical bond.
Finally, LiF, a single valence electron is transferred from a metal atom of Li to F, for both of the atoms that would form the ionic bond to achieve a stable octet, and or full electron shell.
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which pair shares the same empirical formula. c2h2 and c6h6, c2h2 and c2h4, ch2 and c6h6, ch and c2h4
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Explanation:

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Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3
vichka [17]

Answer:

0.13 m of Mn(NO_3)_2 → Highest boiling point

0.19 m of AgNO_3 → Second  Highest boiling point

0.17 m of CrSO_4 → Third highest boiling point

0.31 m Sucrose (nonelectrolyte)  → Lowest boiling point

Explanation:

Elevation in boiling is given by :

\Delta T_b=i\times k_b\times m

Where :

i = van't Hoff factor

k_b= Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of AgNO_3

AgNO_3\rightarrow Ag^++NO_3^{-}

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

\Delta T_b=2\times k_b\times 0.19 m

\Delta T_b=0.38 m\times k_b

2) 0.17 m of CrSO_4

CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

\Delta T_b=2\times k_b\times 0.17 m

\Delta T_b=0.34 m\times k_b

3) 0.13 m of Mn(NO_3)_2

Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

\Delta T_b=3\times k_b\times 0.13 m

\Delta T_b=0.39 m\times k_b

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

\Delta T_b=1\times k_b\times 0.31 m

\Delta T_b=0.31 m\times k_b

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b

0.13 m of Mn(NO_3)_2 → Highest boiling point

0.19 m of AgNO_3 → Second  Highest boiling point

0.17 m of CrSO_4 → Third highest boiling point

0.31 m Sucrose (nonelectrolyte)  → Lowest boiling point

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