Question

write orbital diagram for each ion and determine if the ion is diamagnetic or paramagnetic.

a. Cd²⁺ b.Au⁺ c.Mo³⁺ d. Zr²⁺

Provide your answer: example :paramagnetic, diamagnetic, etc., accordingly to a, b, c,and d.

Answer 1

a.

The atomic number of Cd is 48 and its outermost electronic configuration is 4d¹⁰ 5s² , since Cd²⁺ is in +2 state, so two electrons are removed from its valence shell therefore, now the outermost electronic configuration of Cd²⁺ becomes 4d¹⁰ 5s⁰ . The orbital diagram is shown as follows:

Since, there is no unpaired electrons in the Cd²⁺ ion hence, it is diamagnetic.

b.

The atomic number of Au is 79 and its outermost electronic configuration is 5d¹⁰ 6s¹ , since Au⁺ is in +1 state, so one electron is removed from its valence shell therefore, now the outermost electronic configuration of Au⁺ becomes 5d¹⁰ 6s⁰ . The orbital diagram is shown as follows:

Since, there is no unpaired electrons in the Au⁺ ion hence, it is diamagnetic.

c.

The atomic number of Mo is 42 and its outermost electronic configuration is 4d⁵ 5s¹ , since Mo³⁺ is in +3 state, so three electrons are removed from its valence shell therefore, now the outermost electronic configuration of Mo³⁺ becomes 4d³ 5s⁰ . The orbital diagram is shown as follows:

Since, there are three unpaired electrons in the Mo³⁺ ion hence, it is paramagnetic.

d.

The atomic number of Zr is 40 and its outermost electronic configuration is 4d² 5s² , since Zr²⁺ is in +2 state, so two electrons are removed from its valence shell therefore, now the outermost electronic configuration of Zr²⁺ becomes 4d² 5s⁰ . The orbital diagram is shown as follows:

Since, there are two unpaired electrons in the Zr²⁺ ion hence, it is paramagnetic.