give the complete electron configuration for yttrium. do not use the noble gas abbreviation. y has __1__…

give the complete electron configuration for yttrium. do not use the noble gas abbreviation. y 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. 5p⁶

give the complete electron configuration for yttrium. do not use the noble gas abbreviation. y 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. 5p⁶

Answer

Explanation:

Step1: Determine atomic number

Yttrium (Y) has an atomic number of 39. In a neutral atom, the number of electrons is equal to the atomic number. So Y has 39 electrons.

Step2: Fill electron - orbitals

We fill the orbitals according to the Aufbau principle, Pauli - exclusion principle, and Hund's rule. After filling up to $3p^6$, the next orbitals are filled as follows: $4s^2$, then $3d^{10}$, then $4p^6$, then $5s^2$, and finally $4d^1$.

Answer:

  1. H. 39
  2. O. $4s^2$
  3. M. $3d^{10}$
  4. P. $4p^6$
  5. Q. $5s^2$
  6. N. $4d^1$