the condensed electron configuration of $Ni$ is $$Ni: [Ar]4s^23d^8$$
Work Step by Step
*Strategy: 1) Find the nearest noble gas element of the lower atomic number. 2) Find out which shell is the outer shell and how many electrons there are in the outer shell (by looking at the periodic table). 3) Put the outer-shell electron in the orbitals and subshells according to Hund's rule. 1) The nearest noble gas element of the lower atomic number of nickel ($Ni$) is argon ($Ar$). Therefore, we would use $Ar$ in the condensed electron configuration of $Ni$. 2) Looking at the periodic table, - $Ni$ is on the 4th row. The outer shell is the 4th shell. - $Ni$ is on the VIIIB column, with number 10 on top. The number of outer-shell electrons is 10. (With the transition metals, which are those fall in the B-columns, you have to be careful when calculating the number of outer-shell electrons. Find out which column the element falls, or look at the number above the column to see how many outer-shell electrons it has) 3) $Ni$ is a transition metal (those in the B-columns), so its electron arrangments are a bit out of the ordinary. After the electrons fill the s- and p- subshells of the 3rd shell, it does not fill the d-subshell as usual. Instead, it jumps to fill the s-subshell of the 4th shell, which is lower in energy than the 3rd d-subshell. That means the first 2 outer-shell electrons of $Ni$ occupy the s-subshell of the 4th shell ($4s^2$). The rest 8 electrons continue to fill the d-subshell of the 3rd shell ($3d^8$). In conclusion, the condensed electron configuration of $Ni$ is $$Ni: [Ar]4s^23d^8$$ (Writing $4s$ before or after $3d$ is up to you. In this case, I write $4s$ before $3d$ to show that the energy level of $4s$ is lower than that of $3d$).