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One element has atomic weight 39. Its electronic configuration is 1 s2, 2s2, 2p6, 3s2, 3p6, 4s1. The true statement for that element is : [JEE 1995]
- a)High value of IE
- b)Transition element
- c)Isotone with 18Ar38
- d)None
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
One element has atomic weight 39. Its electronic configuration is 1 s2...
Explanation:
Atomic weight = 39
Electronic configuration = 1 s2, 2s2, 2p6, 3s2, 3p6, 4s1
The atomic number can be calculated as follows:
1s2 = 2 electrons
2s2, 2p6 = 8 electrons
3s2, 3p6 = 18 electrons
4s1 = 1 electron
Total number of electrons = 29
Therefore, the element must have an atomic number of 29.
Option A: High value of IE
The first ionization energy (IE) is the energy required to remove one electron from an atom in its gaseous state. The IE of an element depends on the number of protons in its nucleus, the distance between the nucleus and the valence electrons, and the shielding effect of the inner electrons.
The element with the given electronic configuration has a valence electron in the 4s orbital. The distance between the nucleus and the valence electron is relatively large, and there are 18 inner electrons that shield the valence electron from the nucleus. Therefore, the IE of this element is relatively low.
Option B: Transition element
A transition element is an element that has a partially filled d subshell in one or more of its common oxidation states. The element with the given electronic configuration does not have any electrons in its d subshell, so it is not a transition element.
Option C: Isotone with 18Ar38
Isotones are atoms that have the same number of neutrons but different numbers of protons.
The atomic number of the element with the given electronic configuration is 29. To find an isotone of this element, we need to look for an element with 29 neutrons.
The atomic weight of the element is 39, which means it has 39 - 29 = 10 neutrons.
The only element with 29 neutrons and an atomic weight of 38 is Argon-38 (18Ar38).
Therefore, the given element is an isotone of Argon-38.
Option D: None
Option C is the correct answer.
Atomic weight = 39
Electronic configuration = 1 s2, 2s2, 2p6, 3s2, 3p6, 4s1
The atomic number can be calculated as follows:
1s2 = 2 electrons
2s2, 2p6 = 8 electrons
3s2, 3p6 = 18 electrons
4s1 = 1 electron
Total number of electrons = 29
Therefore, the element must have an atomic number of 29.
Option A: High value of IE
The first ionization energy (IE) is the energy required to remove one electron from an atom in its gaseous state. The IE of an element depends on the number of protons in its nucleus, the distance between the nucleus and the valence electrons, and the shielding effect of the inner electrons.
The element with the given electronic configuration has a valence electron in the 4s orbital. The distance between the nucleus and the valence electron is relatively large, and there are 18 inner electrons that shield the valence electron from the nucleus. Therefore, the IE of this element is relatively low.
Option B: Transition element
A transition element is an element that has a partially filled d subshell in one or more of its common oxidation states. The element with the given electronic configuration does not have any electrons in its d subshell, so it is not a transition element.
Option C: Isotone with 18Ar38
Isotones are atoms that have the same number of neutrons but different numbers of protons.
The atomic number of the element with the given electronic configuration is 29. To find an isotone of this element, we need to look for an element with 29 neutrons.
The atomic weight of the element is 39, which means it has 39 - 29 = 10 neutrons.
The only element with 29 neutrons and an atomic weight of 38 is Argon-38 (18Ar38).
Therefore, the given element is an isotone of Argon-38.
Option D: None
Option C is the correct answer.
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Community Answer
One element has atomic weight 39. Its electronic configuration is 1 s2...
Isotones are compound with same nutron number In the provided config and atomic mass and mass number difference is 20 same as in.
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One element has atomic weight 39. Its electronic configuration is 1 s2, 2s2, 2p6, 3s2, 3p6, 4s1. The true statement for that element is : [JEE 1995]a)High value of IEb)Transition elementc)Isotone with18Ar38d)NoneCorrect answer is option 'C'. Can you explain this answer?
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One element has atomic weight 39. Its electronic configuration is 1 s2, 2s2, 2p6, 3s2, 3p6, 4s1. The true statement for that element is : [JEE 1995]a)High value of IEb)Transition elementc)Isotone with18Ar38d)NoneCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One element has atomic weight 39. Its electronic configuration is 1 s2, 2s2, 2p6, 3s2, 3p6, 4s1. The true statement for that element is : [JEE 1995]a)High value of IEb)Transition elementc)Isotone with18Ar38d)NoneCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One element has atomic weight 39. Its electronic configuration is 1 s2, 2s2, 2p6, 3s2, 3p6, 4s1. The true statement for that element is : [JEE 1995]a)High value of IEb)Transition elementc)Isotone with18Ar38d)NoneCorrect answer is option 'C'. Can you explain this answer?.
One element has atomic weight 39. Its electronic configuration is 1 s2, 2s2, 2p6, 3s2, 3p6, 4s1. The true statement for that element is : [JEE 1995]a)High value of IEb)Transition elementc)Isotone with18Ar38d)NoneCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One element has atomic weight 39. Its electronic configuration is 1 s2, 2s2, 2p6, 3s2, 3p6, 4s1. The true statement for that element is : [JEE 1995]a)High value of IEb)Transition elementc)Isotone with18Ar38d)NoneCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One element has atomic weight 39. Its electronic configuration is 1 s2, 2s2, 2p6, 3s2, 3p6, 4s1. The true statement for that element is : [JEE 1995]a)High value of IEb)Transition elementc)Isotone with18Ar38d)NoneCorrect answer is option 'C'. Can you explain this answer?.
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