Question

in the space below, write the abbreviated electron configurations of the following elements: 6) cobalt 7) silver 8) tellurium 9) radium 10) lawrencium

Explanation:

Step1: Determine the atomic number

Cobalt (\(Co\)) has an atomic number of \(27\). The noble gas before it is argon (\(Ar\), atomic number \(18\)). So we start with \([Ar]\).

Step2: Fill the remaining electrons

After \([Ar]\), we have \(27 - 18=9\) electrons. The next sub - shells are \(4s\) (holds \(2\) electrons) and \(3d\) (holds \(10\) electrons). But for transition metals, the \(4s\) is filled before \(3d\). So the electron configuration is \([Ar]4s^{2}3d^{7}\).

Step3: For silver (\(Ag\), atomic number \(47\))

The noble gas before it is krypton (\(Kr\), atomic number \(36\)). We have \(47-36 = 11\) electrons left. Due to the stability of the full \(d\) - subshell, the configuration is \([Kr]5s^{1}4d^{10}\) (instead of \([Kr]5s^{2}4d^{9}\) as we might expect from simple filling rules because a full \(d\) - subshell (\(d^{10}\)) is more stable).

Step4: For tellurium (\(Te\), atomic number \(52\))

The noble gas before it is krypton (\(Kr\), atomic number \(36\)). We have \(52 - 36=16\) electrons left. The configuration is \([Kr]5s^{2}4d^{10}5p^{4}\) (filling \(5s\) ( \(2\) electrons), \(4d\) ( \(10\) electrons) and \(5p\) ( \(4\) electrons)).

Step5: For radium (\(Ra\), atomic number \(88\))

The noble gas before it is radon (\(Rn\), atomic number \(86\)). We have \(88 - 86 = 2\) electrons left. The configuration is \([Rn]7s^{2}\).

Step6: For lawrencium (\(Lr\), atomic number \(103\))

The noble gas before it is radon (\(Rn\), atomic number \(86\)). We have \(103-86 = 17\) electrons left. The configuration is \([Rn]7s^{2}5f^{14}6d^{1}\) (filling \(7s\) ( \(2\) electrons), \(5f\) ( \(14\) electrons) and \(6d\) ( \(1\) electron)).

Answer:

1. \([Ar]4s^{2}3d^{7}\)
2. \([Kr]5s^{1}4d^{10}\)
3. \([Kr]5s^{2}4d^{10}5p^{4}\)
4. \([Rn]7s^{2}\)
5. \([Rn]7s^{2}5f^{14}6d^{1}\)