Question

question 11 0.75 pts what is the complete electronic configuration for ru^+4 based on the aufbau order? this is the sentence form starting with 1s2. answer format: input the answer as a \text\ typed sentence without superscripts or subscripts. it must be entered in the aufbau order. for example, nitrogen (n) is input as 1s22s22p3. this problem can be solved by dimensional analysis as the primary method, formula where permitted, or simple math. using ratio and proportion methods is not allowed. be sure to show the formula on homework and exams. show work here as a file upload in the last question. be sure to report the units for all values when showing your work. keys will be posted for exemplary practice problems. you must show work on the exams. practice showing the calculator answer and the sig figs answer.

Explanation:

Step1: Determine the atomic number of Ru

Ruthenium (Ru) has an atomic number \( Z = 44\). The \(Ru^{+4}\) ion has \(44 - 4=40\) electrons.

Step2: Fill the orbitals according to Aufbau principle

Start filling orbitals:

• \(1s\) can hold \(2\) electrons: \(1s2\)
• \(2s\) can hold \(2\) electrons: \(2s2\)
• \(2p\) can hold \(6\) electrons: \(2p6\)
• \(3s\) can hold \(2\) electrons: \(3s2\)
• \(3p\) can hold \(6\) electrons: \(3p6\)
• \(4s\) can hold \(2\) electrons: \(4s2\)
• \(3d\) can hold \(10\) electrons: \(3d10\)
• \(4p\) can hold \(6\) electrons: \(4p6\)
• \(5s\) can hold \(2\) electrons: \(5s2\)
• \(4d\) (after filling \(5s\) for transition - metal ions, remove electrons from \(5s\) first). Since we have \(40\) electrons in total, after filling \(1s22s22p63s23p64s23d104p65s2\) (\(2 + 2+6 + 2+6 + 2+10 + 6+2=38\) electrons), we need \(40-38 = 2\) more electrons in \(4d\). So the configuration is \(1s22s22p63s23p64s23d104p64d2\)

Answer:

1s22s22p63s23p64s23d104p64d2