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s344n2d4d5 [400]
2 years ago
12

Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbol

s for the monatomic ions formed from the following elements: (a) Cl (b) Na (c) Mg (d) Ca (e) K (f) Br (g) Sr (h) F

Chemistry
1 answer:
77julia77 [94]2 years ago
5 0

Answer:

The Lewis structures are in image attached.

Explanation:

Lewis symbol is a representation of an element symbol along with its valence electrons around it in the form of dot(s).

Mono-atomic ions formed from the following :

(a) Cl

Chlorine's atomic number is 17 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

Cl=1s^22s^22p^63s^23p^5

Cl^-=1s^22s^22p^63s^23p^6

(b) Na

Sodium's atomic number is 11 in which only 1 electrons are present in its valence shell .So in order to gain noble gas stability it will loose 1 electron to completes its octet. In the Lewis symbol no dot shown as sodium has lost its 1 electron.

Na=1s^22s^23p^63s^1

Na^+=1s^22s^23p^63s^0

(c) Mg

Magnesium's atomic number is 12 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electrons to completes its octet.

In the Lewis symbol no dot shown as magnesium has lost its 2 electrons.

Mg=1s^22s^23p^63s^2

Mg^{2+}=1s^22s^23p^63s^0

(d)Ca

Calcium's atomic number is 20 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

Ca= 1s^22s^22p^63s^23p^64s^2

Ca^{2+}=1s^22s^23p^6^23p^64s^0

(e) K

Potassium's atomic number is 19 in which only 1 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 1 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 1 electron.

K= 1s^22s^22p^63s^23p^64s^1

K^{+}=1s^22s^23p^6^23p^64s^0

(f) Br

Bromine's atomic number is 35 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

Br=1s^22s^22p^63s^23p^63d^{10}4s^24p^5

Br^-= 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

(g) Sr

Strontium's atomic number is 38 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

Sr=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2

Sr^{2+}=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^0

(h) F

Florine's atomic number is 7 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet.

F=1s^22s^22p^5

F^-=1s^22s^22p^6

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Explanation:

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Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH   →  Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

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Number of moles = 0.08 mol

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Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂.      NaOH             :      Cu(OH)₂

                               2                   :          1

                               0.32              :           1/2×0.32 = 0.16 mol

                            Cu(NO₃)₂         :           Cu(OH)₂

                                  1                  :               1

                             0.08                :              0.08

The number of moles produced by  Cu(NO₃)₂  are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield ×  100

percentage yield = 5.23 g/  7.808 g ×  100

percentage yield = 0.67 ×  100

percentage yield = 67%

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