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प्रश्न
Write down the electronic configuration of the following elements from the given atomic numbers. Answer the following question with explanation.
19K, 3Li, 11Na, 4Be Which of these atoms has smallest atomic radius?
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उत्तर
19K, 3Li, 11Na, 4Be
Electronic configuration of the following elements is:
19K = 2,8,8,1
3Li = 2,1
11Na = 2,8,1
4Be = 2,2
4Be has smallest atomic radius because 19K, 3Li, 11Na are present in same group 1 but Be is present in group 2. According to the trend, as we move from left to right atomic size of an atoms decreases. Due to large positive charge on the nucleus, the electrons are pulled closer to the nucleus and the size of atom decreases.
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संबंधित प्रश्न
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| 1) Alkaline earth metals | a) Group 18 |
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