Question

question 34 (2 points) give the complete electron configuration of the titanium (iv) ion. ti is element 22. if a subshell or orbital is empty you still need to include it. 1s² 2s² 2p⁶ 3s² 1 2 3 a. 3s⁰ b. 3s¹ c. 3s² d. 3p⁰ e. 3p¹ f. 3p² g. 3p³ h. 3p⁴ i. 3p⁵ j. 3p⁶ k. 3d⁰ l. 3d¹ m. 3d² n. 3d³ o. 3d⁴ p. 3d⁵ q. 3d⁶ r. 3d⁷ s. 3d⁸ t. 3d⁹ u. 3d¹⁰ v. 4s⁰ w. 4s¹ x. 4s² y. 4p⁰ z. 4p¹ aa. 4p² bb. 4p³

Explanation:

Step1: Determine titanium's neutral electron config

The neutral titanium (Ti) atom with atomic number 22 has an electron - configuration of $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{2}$.

Step2: Consider the charge of titanium(IV) ion

Titanium(IV) ion ($Ti^{4 + }$) has lost 4 electrons. First, 2 electrons are removed from the 4s orbital and then 2 electrons from the 3d orbital.

Step3: Write the electron - configuration of $Ti^{4+}$

The electron - configuration of $Ti^{4+}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$. So the blanks should be filled with $3p^{6}$, $3d^{0}$, $4s^{0}$.

Answer:

J. $3p^{6}$, K. $3d^{0}$, V. $4s^{0}$