Question

The number of $d$-electrons in $Ni$ (Atomic number $=28$) is equal to that of the:

• A. $s$ and $p$-electrons in $F^{-}$
• B. $p$-electrons in $Ar$ (Atomic number $=18$)
• C. $d$-electrons in ${Ni}^{2+}$
• D. Total number of electrons in $N$ (Atomic number $=7$)

Answer 1

The correct option is C $d$-electrons in ${Ni}^{2+}$

The number of d-electrons in Ni (Atomic number =28) is equal to that of the d-electrons in $N{i}^{2+}$
The electronic configuration of Ni is $\left[Ar\right]3d^{8}4s^2$. It has eight d electrons.
The electronic configuration of $N{i}^{2+}$ is $\left[Ar\right]3d^{8}4s^0$. It has eight d electrons.
The electronic configuration of $F^{-}$ is $1s^{2}2s^{2}2p^6$. It has 10 s and p electrons.
The electronic configuration of Ar is $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$. It has 12 p electrons.
The electronic configuration of N is $1s^{2}2s^{2}2p^3$. It has 7 electrons.