Question

The number of d-electrons in Ni (Atomic number = 28) is equal to that type of
A. s and p electrons in $N{i}^{2+}$
B. p-electrons in Ar (atomic number=18)
C. d-electrons in $N{i}^{2+}$
D. Total number of electrons in N (atomic number=7)

Answer 1

Hint: The electron in an atom as well as we know reside in orbitals, which are thought of as zones where there is probability of finding electrons. The orbitals are s, p, d, f etc. and based on true shape. To calculate the number of d-electrons in Ni, we have to write the electronic configuration as $[Ar]3d^{8}4s^2$ and configure if any option has eight electrons in d-orbital.

Complete step by step answer:
First let us find out the electrons present in the d-orbital of Ni which has an atomic number 28. We write the electronic configuration which is $[Ar]3d^{8}4s^2$. Thus, from here we can see that Ni has 8 electrons in its d-orbitals.
S and p electrons in $F^{-}$:
$F^{-}$ has total electrons 10 and its configuration is $1s^{2}2s^{2}2p^6$. Thus, total electrons in s and p is 10. So, it does not match with Ni.
p-electrons in Argon:
Argon has an atomic number 18. When we write its configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$ , there are 12 electrons in one p-orbital. Thus, it is not the same as Ni.
d-electrons in $N{i}^{2+}$:
We know that Ni has electronic configuration $[Ar]3d^{8}4s^2$, when two electrons are removed to form Ni cation ($N{i}^{2+}$) we get the configuration of $N{i}^{2+}$ as $[Ar]3d^8$. So, once the e(-s) are removed from the s-orbital, the number of e(-s) in an orbital remains the same. Therefore, the d-electrons in $N{i}^{2+}$ = Ni.
The total number of electrons in N (atomic number=7) is equal to atomic number (z), which is 7. Thus, it is not the same with d-electrons in Ni.

Thus, the correct option is C.

Note:
Students must calculate for each option and then compare. Also remember that for $N{i}^{2+}$ the electrons are removed from s-orbital and not d-orbital. One must also be through with the writing of electronic configuration.