Question

question 37 (4 points) give the electron configuration for hafnium, hf, element 72, using the noble gas abbreviated method. use the aufbau principle. 1 2 3 4 a. he b. ne c. ar d. kr e. xe f. rn g. uuo h. 6s^0 i. 6s^1 j. 6s^2 k. 6p^0 l. 6p^1 m. 6p^2 n. 6p^3 o. 6p^4 p. 6p^5 q. 6p^6 r. 5d^0 s. 5d^1 t. 5d^2 u. 5d^3 v. 5d^4 w. 5d^5 x. 5d^6 y. 5d^7 z. 5d^8 aa. 5d^9 bb. 5d^10 cc. 5s^0 dd. 5s^1 ee. 5s^2 ff. 4f^0 gg. 4f^14 hh. 4p^2 ii. 4p^3

Explanation:

Step1: Identify the nearest noble gas

The nearest noble gas to hafnium (atomic number 72) with a lower atomic number is xenon (Xe), atomic number 54.

Step2: Determine remaining electrons

Hafnium has 72 electrons. After accounting for the 54 electrons of xenon, there are \(72 - 54=18\) remaining electrons.

Step3: Apply Aufbau Principle

According to the Aufbau Principle, the next sub - shells are filled in the order \(6s\), \(4f\), \(5d\). The \(6s\) sub - shell fills first and can hold 2 electrons (\(6s^{2}\)). Then the \(4f\) sub - shell fills and can hold 14 electrons (\(4f^{14}\)). After that, the remaining \(18-(2 + 14)=2\) electrons go into the \(5d\) sub - shell (\(5d^{2}\)).

Answer:

E. Xe, J. \(6s^{2}\), GG. \(4f^{14}\), T. \(5d^{2}\)