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Ganezh [65]
4 years ago
12

A graph of gas pressure versus the number of particles in a container is a straight line. Which other relationship will have a s

imilar graph?
Chemistry
1 answer:
Elanso [62]4 years ago
3 0
Volume has the same relationship with the number of particles than the pressure.

You can see it in the ideal gas equation: PV = nRT

As much as V and P are proportional to n.
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Rank these transition metal ions in order of decreasing number of unpaired electrons.
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Answer: The given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

Explanation:

In atomic orbitals, the distribution of electrons of an atom is called electronic configuration.

The electronic configuration in terms of noble gases for the given elements are as follows.

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Fe^{3+} - [Ar] 3d^{5}

So, there is only 1 unpaired electron present in Fe^{3+}.

  • Atomic number of Mn is 25.

Mn^{4+} - [Ar]3d^{3}

So, there are only 3 unpaired electrons present in Mn^{4+}.

  • Atomic number of V is 23.

V^{3+} - [Ar] 3d^{2}

So, there are only 2 unpaired electrons present in V^{3+}.

  • Atomic number of Ni is 28.

Ni^{2+} - [Ar] 3d^{8}

So, there will be 2 unpaired electrons present in Ni^{2+}.

  • Atomic number of Cu is 29.

Cu^{+} - [Ar] 3d^{10}

So, there is no unpaired electron present in Cu^{+}.

Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.

Mn^{4+} > V^{3+} = Ni^{2+} > Fe^{3+} > Cu^{+}

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