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Allushta [10]
3 years ago
6

For the following pair, indicate which element has the lower first ionization energy: Match the words in the left column to the

appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
1. Given the elements Cl and Ge ,_________ has the smaller first ionization energy.
2. Given the elements Te and Se , __________ has the smaller first ionization energy.
3. Given the elements Ba and Ti , __________ has the smaller first ionization energy.
4. Given the elements Cu and Ag ,___________ has the smaller first ionization energy.

to fill in the blank:

(A) Ag
(B) Ge
(C) Cl
(D) Ti
(E) Cu
(F) Ba
(G) Se
(H) Te
Chemistry
1 answer:
emmainna [20.7K]3 years ago
5 0

<u>Answer:</u>

<u>For 1:</u> The correct answer is Ge.

<u>For 2:</u> The correct answer is Te.

<u>For 3:</u> The correct answer is Ba.

<u>For 4:</u> The correct answer is Ag.

<u>Explanation:</u>

Ionization energy is defined as the energy required to remove an electron from the outermost shell of an isolated gaseous atom. It is represented as E_i

X(g)\rightarrow X^+(g)+1e^-;E_i

Ionization energy increases as we move from left to right in a period. This happens because the atomic radius of an element decreases moving across a period, which increases the effective attraction between the negatively charged electrons and positively-charged nucleus. Hence, the removal of electron from the outermost shell becomes difficult and requires more energy.

Ionization energy decreases on moving from top to bottom in a group. This happens because the number of shells increases as we move down the group. The electrons get added in the new shell. So, the shielding of outermost electrons from the inner ones is more which decreases the attraction between the electrons and the nucleus. Hence, the removal of electron from the outermost shell becomes easy and requires less energy.

For the given options:

  • <u>Option 1:</u>

Chlorine is the 17th element of the periodic table belonging to Period 3 and Group 17.

Germanium is the 32nd element of the periodic table belonging to Period 4 and Group 14.

Hence, germanium will have smaller first ionization energy.

  • <u>Option 2:</u>

Tellurium is the 52nd element of the periodic table belonging to Period 5 and Group 16.

Selenium is the 34th element of the periodic table belonging to Period 4 and Group 16.

Hence, tellurium will have smaller first ionization energy.

  • <u>Option 3:</u>

Barium is the 56th element of the periodic table belonging to Period 6 and Group 2.

Titanium is the 22nd element of the periodic table belonging to Period 4 and Group 4.

Hence, barium will have smaller first ionization energy.

  • <u>Option 4:</u>

Copper is the 29th element of the periodic table belonging to Period 4 and Group 11.

Silver is the 47th element of the periodic table belonging to Period 5 and Group 11.

Hence, silver will have smaller first ionization energy.

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Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

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And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

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(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

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Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

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              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

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