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Shehab

  • 3 years ago

Without looking at the periodic table , identify the period block and group of the element that has the electron configuration [Ar]3d^34s^2.

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  1. aaronq
    • 3 years ago
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    to do this question you count the number of electrons present and since it's neutral it will correspond to the atomic number (which is the number of protons) which identifies the element. you NEED to memorize the periodic table in order to do this question or at least know where 3d^3 is. Hint: transition metals .. Scandium, Titanium..V

  2. Carl_Pham
    • 3 years ago
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    Since the 3d subshell is not filled, it must lie in the d block. That is confirmed by the fact that the 4s subshell is filled, because you know the s block is to the left of the d block, so it gets filled first. Now, take note of the fact that you have 3 eletrons in the 3d subshell, and 2 in the 4, for a total of 5 electrons outside the noble-gas core. That tells you this element is in the 5th column of the Periodic Table, reading left to right. If you are using the international labeling of groups, then you now know this element is in Group 5. If you are using the typical American group labeling, it's a little trickier. You have to remember that the d block starts with Group 3B, so that the 3rd element of any d block period is in Group 5B. It will help to remember that the American group labels are strongly related to the number of valence electrons. In this case, you have in principle 5 valence electrons (three 3d and two 4s) so it's in Group 5B.

  3. Preetha
    • 3 years ago
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    I really like Carl's strategy. Look at the electron config.

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