The paramagnetism of an element is determined by the number of unpaired electrons in its electron configuration. The more unpaired electrons, the greater the paramagnetism.

Let's look at the electron configurations of the different oxidation states of manganese (Mn):

a. Mn(II): [Ar] 3d5 4s0 - Mn(II) means it has lost 2 electrons, so its configuration becomes [Ar] 3d3 4s0. This has 3 unpaired electrons.

b. Mn(III): [Ar] 3d5 4s0 - Mn(III) means it has lost 3 electrons, so its configuration becomes [Ar] 3d2 4s0. This has 2 unpaired electrons.

c. Mn(IV): [Ar] 3d5 4s0 - Mn(IV) means it has lost 4 electrons, so its configuration becomes [Ar] 3d1 4s0. This has 1 unpaired electron.

d. Mn(VII): [Ar] 3d5 4s0 - Mn(VII) means it has lost 7 electrons, so its configuration becomes [Ar] 3d0 4s0. This has 0 unpaired electrons.

Therefore, Mn(II) with 3 unpaired electrons would exhibit the greatest paramagnetism.

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The paramagnetism of an element is determined by the number of unpaired electrons in its electron configuration. The more unpaired electrons, the greater the paramagnetism.

Let's look at the electron configurations of the different oxidation states of manganese (Mn):

a. Mn(II): [Ar] 3d5 4s0 -> Mn(II) means it has lost 2 electrons, so its configuration becomes [Ar] 3d3 4s0. This has 3 unpaired electrons.

b. Mn(III): [Ar] 3d5 4s0 -> Mn(III) means it has lost 3 electrons, so its configuration becomes [Ar] 3d2 4s0. This has 2 unpaired electrons.

c. Mn(IV): [Ar] 3d5 4s0 -> Mn(IV) means it has lost 4 electrons, so its configuration becomes [Ar] 3d1 4s0. This has 1 unpaired electron.

d. Mn(VII): [Ar] 3d5 4s0 -> Mn(VII) means it has lost 7 electrons, so its configuration becomes [Ar] 3d0 4s0. This has 0 unpaired electrons.

Therefore, Mn(II) with 3 unpaired electrons would exhibit the greatest paramagnetism.

Knowee AI · Accepted Answer

The paramagnetism of an element is determined by the number of unpaired electrons in its electron configuration. The more unpaired electrons, the greater the paramagnetism.

Let's look at the electron configurations of the different oxidation states of manganese (Mn):

a. Mn(II): [Ar] 3d5 4s0 -> Mn(II) means it has lost 2 electrons, so its configuration becomes [Ar] 3d3 4s0. This has 3 unpaired electrons.

b. Mn(III): [Ar] 3d5 4s0 -> Mn(III) means it has lost 3 electrons, so its configuration becomes [Ar] 3d2 4s0. This has 2 unpaired electrons.

c. Mn(IV): [Ar] 3d5 4s0 -> Mn(IV) means it has lost 4 electrons, so its configuration becomes [Ar] 3d1 4s0. This has 1 unpaired electron.

d. Mn(VII): [Ar] 3d5 4s0 -> Mn(VII) means it has lost 7 electrons, so its configuration becomes [Ar] 3d0 4s0. This has 0 unpaired electrons.

Therefore, Mn(II) with 3 unpaired electrons would exhibit the greatest paramagnetism.

Which of the following oxidation states of manganese could exhibit the greatest paramagnetism?Select one:a.Mn(II)b.Mn(III)c.Mn(IV)d.Mn(VII)

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