Question

Select the atoms whose ions have an electron configuration of nd5(n = 3, 4, 5 ...). For example, the 2+ ion for V ends the electron configuration with 3d 3.
Ir
Rn
Fe
Pa
Y
Re

Answer 1

Answer: The atoms whose ions have electronic configuration as $nd^5$ are Iridium (Ir), Iron (Fe) and Rhenium (Re)

Explanation:

An ion is formed when a neutral atom looses or gains electrons.

• When an atom looses electrons, it results in the formation of positive ion known as cation.
• When an atom gains electrons, it results in the formation of negative ion known as anion.

For the given elements:

Iridium:  The electronic configuration of this element is $[Xe]6s^25d^7$. The stable oxidation state of iridium is +4.

The electronic configuration of $Ir^{4+}$ is $[Xe]5d^5$

Radon:  The electronic configuration of this element is $[Xe]6s^25d^{10}6p^6$. This is a stable element.

Iron:  The electronic configuration of this element is $[Ar]4s^23d^6$. The stable oxidation state of iron is +3.

The electronic configuration of $Fe^{3+}$ is $[Ar]3d^5$

Protactinium:  The electronic configuration of this element is $[Rn]7s^26d^15f^2$. The stable oxidation state of protactinium is +3.

The electronic configuration of $Pa^{3+}$ is $[Rn]7s^2$

Ytterium:  The electronic configuration of this element is $[Kr]5s^24d^1$. The stable oxidation state of ytterium is +3.

The electronic configuration of $Y^{3+}$ is $[Ar]4s^23d^{10}4p^6$

Rhenium:  The electronic configuration of this element is $[Xe]6s^25d^5$. The stable oxidation state of rhenium is +2.

The electronic configuration of $Re^{2+}$ is $[Xe]5d^5$

Hence, the atoms whose ions have electronic configuration as $nd^5$ are Iridium (Ir), Iron (Fe) and Rhenium (Re)

Answer 2

Ir = [Xe] 4f14 5d7 6s2 
Rn = [Xe] 4f14 5d10 6s2 6p6 
Fe = [Ar] 3d6 4s2 
Pa = [Rn] 5f2 6d1 7s2 
Y = [Kr] 4d1 5s2 
Re = [Xe] 4f14 5d5 6s2 
So only Rhenium (Re) has an electronic configuration where the d orbital has 5 electrons in it.