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jolli1 [7]
3 years ago
15

What is the total charge of the nucleus of a nitrogen atom

Chemistry
2 answers:
deff fn [24]3 years ago
8 0
This easy man.its positive charge. i learned this at school.
Soloha48 [4]3 years ago
4 0
+ positive charge
because nucleus always has a positive charge due to proton
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What is this element?<br> 1s22s22p63s2<br> Magnesium<br> Aluminum<br> Potassium<br> Neon
valkas [14]

Answer:

It should be silicon... I even googled it to see if I was correct and that’s what it says so can someone figure out why silicon is not an answer

Explanation:

6 0
3 years ago
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Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the nu
Alja [10]

Answer:

(a) [Noble gas] ns² np⁵: Number of unpaired electron is 1 and belongs to group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²: Number of unpaired electron is 2 and belongs to group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹: Number of unpaired electron is 1 and belongs to group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶ : Number of unpaired electron is 6 and belongs to group 8 of the periodic table.

Explanation:

In the Periodic table, the chemical elements are arranged in 7 rows, called periods and 18 columns, called groups. They are organized in increasing order of atomic numbers.

(a) [Noble gas] ns² np⁵

As the total number of electrons in the p-orbital is 5. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 5 electrons in np orbital. <u>So there are 7 valence electrons.</u>

Therefore, this element belongs to the group 17 i.e. halogen group of the periodic table.

(b) [noble gas] ns² (n-1)d²

As the total number of electrons in the d-orbital is 2. Therefore, the number of unpaired electrons is 2.

This element has 2 electrons in ns orbital and 2 electrons in (n-1)d orbital. So there are <u>4 valence electrons.</u>

Therefore, this element belongs to the group 4 of the periodic table.

(c) [noble gas] ns² (n-1)d¹⁰ np¹

As the total number of electrons in the p-orbital is 1. Therefore, the number of unpaired electron is 1.

This element has 2 electrons in ns orbital and 1 electron in np orbital. So there are <u>3 valence electrons</u>.

Therefore, this element belongs to the group 13 of the periodic table.

(d)[noble gas] ns² (n-2)f⁶

As the total number of electrons in the f-orbital is 6. Therefore, the number of unpaired electron is 6.

This element has 2 electrons in ns orbital and 6 electrons in (n-2)f orbital. So there are <u>8 valence electrons.</u>

Therefore, this element belongs to the group 8 of the periodic table.

3 0
3 years ago
Draw a Lewis structure for HCCl3 . Include all hydrogen atoms and show all unshared pairs and the formal charges, if any. Assume
andrey2020 [161]

Answer:

See answer below

Explanation:

First, let's explain what a Lewis structure is.

A lewis structure in short words is a draw of an atom or molecule showing how the electrons bond with another electrons of atoms, and also shows the unshared pair of electrons, which are the electrons that do not bond in the molecule. In the case of an atom, it shows all the available electrons it has to be shared and bonded with another atom to form a molecule.

With this said, in order to draw the lewis structure we need to know how many electrons the atoms involved have. To know this, we need to write the electronic configuration of the atoms, based on it's atomic number.

In the case of Hydrogen (Z = 1), Carbon (Z = 6) and Cl (Z = 17):

[H] = 1s¹     1 electron available.

[C] = 1s² 2s² 2p²   In this case, we have 4 electrons.

[Cl] = 1s² 2s² 2p⁶ 3s² 3p⁵  In this case, 7 electrons.

Now that we have the configuration, and the available electrons, we need to draw the atoms. The Carbon is the more electronegative atom of them, so, the bonding will be formed based on this atom as the central. So the other atoms will just bond and shared a pair of electron with the carbon. The HCCl₃ can be treated as CH₄, with a tetrahedrical form.

The picture below shows the lewis structure.

Hope this helps

3 0
3 years ago
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Do you know the formula? I can’t find it.
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Taking a test please help. Thank you.
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It’s the 3rd one obviously bro
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