Question

In which of the following element, both ultimate and penultimate shells are incompletely filled?
(A) Ni
(B) Cd
(C) Hg
(D) Zn

Answer 1

Hint: To solve this question, we need to determine the electronic configurations of Ni (Nickel), Cd (Cadmium), Hg (Mercury), and Zn (Zinc). The electronic configuration of the elements can be determined using their atomic number.

Complete solution:
We know that the outermost shell of an atom is known as the valence shell or the ultimate shell. And the shell that is just before the ultimate shell or the second last shell of an atom is known as the penultimate shell.
Now, let us determine the electronic configurations of the elements.
The atomic number of Ni (Nickel) is 28. Its electronic configuration is as follows
\[\begin{align}
  & _{28}Ni=[Ar]3{{d}^{8}}4{{s}^{2}} \\
 & _{28}Ni=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{8}} \\
\end{align}\]
We can see that its electronic configuration is of the type
$(1{{s}^{2}})(2{{s}^{2}}{{p}^{6}})(3{{s}^{2}}3{{p}^{6}}3{{d}^{8}})(4{{s}^{2}})$
So, it has an incomplete ultimate shell as well as an incomplete penultimate shell.
The atomic number of Cd (Cadmium) is 48. Its electronic configuration is as follows
\[\begin{align}
  & _{48}Cd=[Kr]4{{d}^{10}}5{{s}^{2}} \\
 & _{48}Cd=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{6}}5{{s}^{2}}4{{d}^{10}} \\
\end{align}\]
We can see that its electronic configuration is of the type
$(1{{s}^{2}})(2{{s}^{2}}{{p}^{6}})(3{{s}^{2}}3{{p}^{6}}3{{d}^{10}})(4{{s}^{2}}{{p}^{6}}{{d}^{10}})(5{{s}^{2}})$
So, it has an incomplete ultimate shell but a complete penultimate shell.
The atomic number of Hg (Mercury) is 80. Its electronic configuration is as follows
\[\begin{align}
  & _{80}Hg=[Xe]4{{f}^{14}}5{{d}^{10}}6{{s}^{2}} \\
 & _{80}Hg=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{6}}5{{s}^{2}}4{{d}^{10}}5{{p}^{6}}4{{f}^{14}}5{{d}^{10}}6{{s}^{2}} \\
\end{align}\]
We can see that its electronic configuration is of the type
$(1{{s}^{2}})(2{{s}^{2}}{{p}^{6}})(3{{s}^{2}}3{{p}^{6}}3{{d}^{10}})(4{{s}^{2}}{{p}^{6}}{{d}^{10}}4{{f}^{14}})(5{{s}^{2}}5{{p}^{6}}5{{d}^{10}})(6{{s}^{2}})$
So, it has an incomplete ultimate shell but a complete penultimate shell.
The atomic number of and Zn (Zinc) is 30. Its electronic configuration is as follows
\[\begin{align}
  & _{30}Zn=[Ar]3{{d}^{10}}4{{s}^{2}} \\
 & _{30}Zn=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}} \\
\end{align}\]
We can see that its electronic configuration is of the type
$(1{{s}^{2}})(2{{s}^{2}}{{p}^{6}})(3{{s}^{2}}3{{p}^{6}}3{{d}^{10}})(4{{s}^{2}})$
So, it has an incomplete ultimate shell but a complete penultimate shell.
Hence both ultimate and penultimate shells are incompletely filled in Nickel (Ni).

Note:
It should be noted that the electrons in the valence shell of an atom are the ones that participate in chemical bond formation.
Also, sometimes when a central non-metal atom is bonded to highly electronegative atoms and has empty d-orbitals in its outermost shell and may expand its octet to accommodate more than 8 electrons and covalent bonds.