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den301095 [7]
3 years ago
10

Give the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each element:_____

___. a. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 b. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2 c. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4 d. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p8e. None of these
Chemistry
1 answer:
Evgesh-ka [11]3 years ago
7 0

The question is incomplete, here is the complete question:

Give the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each element: (a) P, (b)  Mg,  (c) Se

<u>Answer:</u>

<u>For a:</u> The electronic configuration of the monoatomic ion formed is 1s^22s^22p^63s^23p^6

<u>For b:</u> The electronic configuration of the monoatomic ion formed is 1s^22s^22p^6

<u>For c:</u> The electronic configuration of the monoatomic ion formed is 1s^22s^22p^63s^23p^64s^23d^{10}4p^6

<u>Explanation:</u>

An ion is formed when a neutral atom looses or gains electrons.

  • When an atom looses electrons, it results in the formation of positive ion known as cation.
  • When an atom gains electrons, it results in the formation of negative ion known as anion.

For the given elements:

(a) Phosphorus

This is the 15th element of the periodic table having electronic configuration of 1s^22s^22p^63s^23p^3

This element will gain 3 electrons to form P^{3-} ion

The electronic configuration of the monoatomic ion formed is 1s^22s^22p^63s^23p^6

(b) Magnesium

This is the 12th element of the periodic table having electronic configuration of 1s^22s^22p^63s^2

This element will loose 2 electrons to form Mg^{2+} ion

The electronic configuration of the monoatomic ion formed is 1s^22s^22p^6

(c) Selenium

This is the 34th element of the periodic table having electronic configuration of 1s^22s^22p^63s^23p^64s^23d^{10}4p^4

This element will gain 2 electrons to form Se^{2-} ion

The electronic configuration of the monoatomic ion formed is 1s^22s^22p^63s^23p^64s^23d^{10}4p^6

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Now put all the given values in the above expression, we get

\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]\Delta H=34279.8J=34.3kJ        (1 KJ = 1000 J)

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