give the complete electron configuration for the yttrium ion in problem 9. do not use the noble gas…

give the complete electron configuration for the yttrium ion in problem 9. do not use the noble gas abbreviation. the y ion has __1__ electrons 1s² 2s² 2p⁶ 3s² 3p⁶ __2__ __3__ __4__ __5__ __6__ a. 10 b. 12 c. 14 d. 16 e. 18 f. 36 g. 38 h. 39 i. 40 j. 3p⁶ k. 3p⁴ l. 3s² m. 3d¹⁰ n. 4d¹ o. 4s² p. 4p⁶ q. 5s² r. 5s⁰ s. 4d⁰ t. 4d³ u. 5s¹ v. 3p⁵

give the complete electron configuration for the yttrium ion in problem 9. do not use the noble gas abbreviation. the y ion has __1__ electrons 1s² 2s² 2p⁶ 3s² 3p⁶ __2__ __3__ __4__ __5__ __6__ a. 10 b. 12 c. 14 d. 16 e. 18 f. 36 g. 38 h. 39 i. 40 j. 3p⁶ k. 3p⁴ l. 3s² m. 3d¹⁰ n. 4d¹ o. 4s² p. 4p⁶ q. 5s² r. 5s⁰ s. 4d⁰ t. 4d³ u. 5s¹ v. 3p⁵

Answer

Explanation:

Step1: Determine atomic number of yttrium

Yttrium (Y) has an atomic number of 39, which means a neutral Y atom has 39 electrons.

Step2: Consider common yttrium ion charge

Yttrium commonly forms a +3 ion ($Y^{3 +}$). When it forms a +3 ion, it loses 3 electrons. So the number of electrons in $Y^{3+}$ is $39 - 3=36$.

Step3: Write electron - configuration

The electron - configuration of an atom/ion with 36 electrons follows the Aufbau principle. After $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$, the next orbitals are filled as $4s^{2}3d^{10}4p^{6}$.

Answer:

  1. F. 36
  2. O. $4s^{2}$
  3. M. $3d^{10}$
  4. P. $4p^{6}$
  5. R. $5s^{0}$
  6. S. $4d^{0}$