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3 years ago

What’s the distance from the nucleus of an atom to its theoretical outer edge of electron orbits

Chemistry

1 answer:

ollegr [7]3 years ago

4 0

Answer:

The range of atoms = (30-300 pm) depending upon the element

Explanation:

The Atomic radii of the atom is the distance from the center of the circle to the outermost orbital.

The center of the circle is the nucleus and the radii is the outermost boundary.

The actual size of the atom is decided on the basis of the Zeff . Also known as effective nuclear charge.

Zeff: It is the net positive charge felt by the outermost electron by the nucleus.

The value of Zeff depends upon the shielding constant. More the shielding less will be the Zeff . Hence the size of the atom increases.

Due to shielding the outermost electrons feel less pull of nucleus.

The greater the Zeff , the smaller the radius of the atom.

The formula used to calculate the atomic mass is :

$r_{n}=\frac{52.9n^{2}}{Z}$ pm

Here "pm"= picometers

$1 pm = 10^{-12}m$

The size of the smallest atom H-atom = 120 pm

The range of atoms = (30-300 pm)

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Rank these transition metal ions in order of decreasing number of unpaired electrons.

lesya692 [45]

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.

Atomic number of Fe is 26.

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

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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What data should be plotted to show that experimental concentration data fits a zeroth-order reaction?

lorasvet [3.4K]

Download organic chemistry 2

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Describe how u would separate a mixture of salt , sand and iodine into different components. please help i will give a brainlies

almond37 [142]

<h2>answer.</h2>

to remove iodine sublime

to remove sand filter

to remove water evaporate

Explanation:

1. put the mixture in a beaker

2.Heat the mixture using a watch glass filled with cold water, the iodine will sublime at the base of the watch glass.

3.Add water to sand and salt. Stir for salt to diffuse in the water

4.Filter sand from the solution using a filter paper

5.evaporate the remaining water to remain with Salt crystals.

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4 years ago

What happens to an atom in a liquid as it freezes?

siniylev [52]

When a liquid is frozen, the atoms in the liquid slow their movement due to lack of energy so the substance becomes hard.

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3 years ago

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Talja [164]

Answer:

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Explanation:

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