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katrin2010 [14]
3 years ago
6

An element belonging to the halogen family would be expected to have a ________ ionization energy and a ________ electron affini

ty.
Chemistry
1 answer:
nikklg [1K]3 years ago
8 0
<span>An element belonging to the halogen family would be expected to have a large ionization energy and a large electron affinity.
Flourine, Chlorine, Bromine, Iodine and astatine are the elements that belongs to the halogen family and mostly they have high values of ionization energy. 
The amount of energy released when an electron is added to an atom or molecule to form a negative ion or anion is electron affinity.</span>Chlorine from this family has highest electron affinity.

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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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