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Al(3+)<Mg(2+)<Na+<F−<O(2−)<N(3−)Al(3+)<Mg(2+)<Na+<F−<O(2−)<N(3−) The above can be aptly described as
  • a)
    the same number of electrons (9 electrons).
  • b)
    isoelectronic species
  • c)
    the same number of electrons (8 electrons).
  • d)
    the same number of electrons (7 electrons).
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Al(3+)<Mg(2+)<Na+<F−<O(2−)<N(3−)...
They have same number of electrons.
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Most Upvoted Answer
Al(3+)<Mg(2+)<Na+<F−<O(2−)<N(3−)...
Explanation:

1. Isoelectronic Species:
Isoelectronic species are those which have the same number of electrons. In the given sequence of elements, Al, Mg, Na, FO, and N are present. To determine if they are isoelectronic, we need to compare the number of electrons in each species.

2. Electron Configuration:
To find the number of electrons in each species, we need to determine their electron configurations. The electron configurations of the given elements are as follows:

- Al: 1s² 2s² 2p⁶ 3s² 3p¹
- Mg: 1s² 2s² 2p⁶ 3s²
- Na: 1s² 2s² 2p⁶ 3s¹
- FO: 1s² 2s² 2p⁶
- N: 1s² 2s² 2p³

3. Counting Electrons:
By counting the number of electrons in each species, we get:
- Al: 13 electrons
- Mg: 12 electrons
- Na: 11 electrons
- FO: 10 electrons
- N: 7 electrons

4. Conclusion:
From the above count, we can see that Al, Mg, and Na do not have the same number of electrons. However, FO and N have the same number of electrons, i.e., 10 electrons. Therefore, the given sequence can be described as isoelectronic species.

5. Answer:
Hence, the correct answer is option 'B' - isoelectronic species.
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Al(3+)<Mg(2+)<Na+<F−<O(2−)<N(3−)Al(3+)<Mg(2+)<Na+<F−<O(2−)<N(3−)The above can be aptly described asa)the same number of electrons (9 electrons).b)isoelectronic speciesc)the same number of electrons (8 electrons).d)the same number of electrons (7 electrons).Correct answer is option 'B'. Can you explain this answer?
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