Question

If Hund's rule is not followed and energy of orbitals is governed by "n" only and not $(n+l)$ rule then

• A. $K$ would have been $d-$ block element and paramagnetic
• B. $Cu$ would have been $s-$ block element
• C. $Cr$ would have been diamagnetic
• D. $F{e}^{3+}$ ion would have 5 unpaired electrons

Answer 1

Answer: (A), (B), (D)

The correct options are
A $K$ would have been $d-$ block element and paramagnetic
B $Cu$ would have been $s-$ block element
D $F{e}^{3+}$ ion would have 5 unpaired electrons

If Hund's rule isn't followed and energy of orbitals is governed by 'n' only and not ${}^{\prime }n+l^{\prime }$ rule, then the order of energy of orbitals are in order.
$1s2s2p3s3p3d4s4p4d4f\dots \dots \dots$
a) Electronic Configuration of $K$ is
${}_{19}K=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^1$
So, $K$ would be a $d-$ block element and paramagnetic due to an unpaired electron in d - orbital.

b) ${}_{29}Cu=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^1$
So, $Cu$ would have been $s-$ block element.

c) ${}_{24}Cr=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^6$
(or)
As there are 4 unpaired electron, then $Cr$ would be paramagnetic.

d) In $F{e}^{3+}$, no. of electron = 23
Electronic configuration of $F{e}^{3+}$ is
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$

So, $F{e}^{3+}$ has 5 unpaired electrons.
$\therefore$ a, b, d are correct options.