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jasenka [17]
3 years ago
15

Dr. Patel and her team have been using GPS to track two plates that are moving toward each

Chemistry
1 answer:
Andru [333]3 years ago
7 0

C: One plate is going underneath the other plate and sinking into the soft rock below.

Explanation:

Where plates are moving towards each other they are said to converging, and are called convergent margins.

The lithosphere is broken into series of slabs called plates. The plates moves on the weak and relatively soft asthenosphere below.

Plates have different motion. At some places, they move apart and they are said to be divergent.

When plates moves towards each other, they are convergent. At a convergent margin, a plate collides with another thereby causing the denser plate usually the oceanic plate to subduct into the asthenosphere. In some other cases, the plates can collide and build upward.

Learn more:

Lithosphere brainly.com/question/9582362

#learnwithBrainly

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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+}.

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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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