Question

$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}$ is the electronic configuration of?
(A) $M{n^{3 + }}$
(B) $F{e^{3 + }}$
(C) $C{r^ + }$
(D) $C{o^{4 + }}$

Answer 1

Hint: The number of electrons in the given electronic configuration can be compared to the atomic number of the elements and their charges in all given options. The answer is multi-correct in this case.

Complete step by step solution:
Let us calculate the number of electrons present in the given ions in all the options.as all the given ions are cations, they lack electrons and the respective charges have to be subtracted from their original atomic numbers to get the number of electrons in them. Let us check all the options-
(A) $M{n^{3 + }}$ (Manganese ion)-
Manganese is a d-block element with an atomic number in neutral atomic form is 25. In the given option Mn is in +3 oxidation state which indicates that the ion lacks three electrons.
$\because $ Mn (Z=25), $M{n^{3 + }}$ (Z=25-3=22). The number of electrons in $M{n^{3 + }}$ are 22.
(B) $F{e^{3 + }}$ (Ferric ion)-
Iron is a d-block element with the atomic number in elemental form is 26. The given ion is in +3 state, so three electrons are lacking from 26 electrons of iron.
$\therefore $ In $F{e^{3 + }}$, Z=26-3 = 23. $\because $ For Fe, Z=26. So the number of electrons in $F{e^{3 + }}$ are 23.
(C) $C{r^ + }$(Chromium ion)-
Chromium atomic number is 24, $C{r^ + }$ lacks only one electron. So, we need to subtract one electron from 25.$\therefore $In $C{r^ + }$(Z = 25-1 = 24).
(D) $C{o^{4 + }}$(Cobalt ion)-
Cobalt atomic number in the free state is 27. In$C{o^{4 + }}$, cobalt lacks four electrons. So let us remove four electrons from 27. We get $C{o^{4 + }}$(Z=27-4 = 23).
Now let us calculate the number of electrons on the given electronic configuration by counting the numbers on the superscripts of the alphabets S, P, D. so, from the given configuration,$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}$, the number of electrons are 23. From the above calculations, 23 electrons are in two ions which are$F{e^{3 + }},{\kern 1pt} C{o^{4 + }}$.

So the answer is correct for two options (B)$F{e^{3 + }}$, (D)$C{o^{4 + }}$.

Note: To give the electronic configuration of any element it is necessary to know the atomic number (Z) of the elements first. But the given elements are in their ionic forms. This means the number of electrons present in these species in their respective states is different from their atomic numbers.