All categories

4 years ago

How do you figure out the electron configuration of an element?

Chemistry

1 answer:

Rus_ich [418]4 years ago

6 0

Writing the Configuration

Starting at the top of the periodic table and moving from left to right across the rows, write the row number, the block letter and how many squares are in each block's section until you get to the desired element. To calculate the electron configuration for phosphorus (P), which is in the third row, p-block, third element in that block, write: 1s2 2s2 2p6 3s2 3p3. Check your work by adding the electron numbers to see if they equal the atomic number of the element; for this example, you would write: 2+2+6+2+3=15, which is the atomic number of phosphorus.

Special Instructions

Because an atom's orbitals sometimes overlap in real life, you must account for that in your configurations. For the d-block, change the row number to one number less than it actually is. For example, the electron configuration for germanium (Ge) is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2. Notice even though you are on row four, it is called "3d" to account for overlap.

Hope i helped you :) Brainliest?

You might be interested in

What are some examples of some monotonic ions found in the body and how do they affect the body? Please help, anything will help

Luba_88 [7]

Atoms are electrically neutral because the number of protons, which carry a 1+ charge, in the nucleus of an atom is equal to the number of electrons, which carry a 1- charge, in the atom. The result is that the total positive charge of the protons cancels out the total negative charge of the electrons so that the net charge of the atom is zero. Most atoms, however, can either gain or lose electrons; when they do so, the number of electrons becomes different from the number of protons in the nucleus. The resulting charged species is called an ion.

4 0

3 years ago

A certain atom has a nucleus containing six protons and eight neutrons and has six electrons orbiting the nucleus

zavuch27 [327]

Answer:

The answer is a carbon isotope.

Explanation:

Carbon has 6 protons and electrons so the atom is neutral.

However it is a carbon isotope because it has more neutrons than protons so be aware of that

5 0

4 years ago

What is the frequency of radiation whose wavelength is 2.51 x 10 -7m?

hodyreva [135]

The frequency of radiation : 1.2 x 10¹⁵ Hz

<h3>Further explanation </h3>

Radiation energy is absorbed by photons

The energy in one photon can be formulated as

$\large{\boxed{\bold{E\:=\:h\:.\:f}}}$

Where

h = Planck's constant (6,626.10⁻³⁴ Js)

f = Frequency of electromagnetic waves

c = speed of light

= 3.10⁸ m/s

λ = wavelength

Wavelength-λ is 2.51 x 10⁻⁷m

The frequency :

$\tt f=\dfrac{3.10^8}{2.51\times 10^{-7}}=1.2\times 10^{15}$

8 0

3 years ago

Give the formulas of the compounds in each set (a) lead(l|) oxide and lead(lV) oxide; (b) lithium nitride, lithium nitrite, and

Ivanshal [37]

Answer:

For a: The formula of lead (II) oxide and lead (IV) oxide is $PbO\text{ and }PbO_2$ respectively.

For b: The formula of lithium nitride, lithium nitrite and lithium nitrate is $Li_3N,LiNO_2\text{ and }LiNO_3$ respectively.

For c: The formula of Strontium hydride and strontium hydroxide is $SrH_2\text{ and }Sr(OH)_2$ respectively.

For d: The formula of magnesium oxide and manganese (II) oxide is $MgO\text{ and }MnO$ respectively.

Explanation:

All the given compounds are ionic compounds. This means that between the atoms complete sharing of electrons takes place.

For a:

Lead is the 82th element of periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^2$ .

To form $Pb^{2+}$ ion, this element will loose 2 electrons and to form $Pb^{4+}$ ion, this element will loose 4 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for lead (II) oxide is PbO and for lead (IV) oxide is $PbO_2$

Thus, the formula of lead (II) oxide and lead (IV) oxide is $PbO\text{ and }PbO_2$ respectively.

For b:

Lithium is the 3rd element of periodic table having electronic configuration of $[He]2s^1$ .

To form $Li^{+}$ ion, this element will loose 1 electron.

Nitrogen is the 7th element of periodic table having electronic configuration of $[He]2s^22p^3$ .

To form $N^{3-}$ ion, this element will gain 3 electrons.

Nitrite ion is a polyatomic ion having chemical formula of $NO_2^{-}$

Nitrate ion is a polyatomic ion having chemical formula of $NO_3^{-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for lithium nitride is $Li_3N$ , for lithium nitrite is $LiNO_2$ and for lithium nitrate is $LiNO_3$

Thus, the formula of lithium nitride, lithium nitrite and lithium nitrate is $Li_3N,LiNO_2\text{ and }LiNO_3$ respectively.

For c:

Strontium is the 38th element of periodic table having electronic configuration of $[Kr]5s^2$ .

To form $Sr^{2+}$ ion, this element will loose 2 electrons.

Hydrogen is the 1st element of periodic table having electronic configuration of $1s^1$ .

To form $H^{-}$ ion, this element will gain 1 electron and is named as hydride ion.

Hydroxide ion is a polyatomic ion having chemical formula of $OH^{-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for strontium hydride is $SrH_2$ and for strontium hydroxide is $Sr(OH)_2$

Thus, the formula of Strontium hydride and strontium hydroxide is $SrH_2\text{ and }Sr(OH)_2$ respectively.

For d:

Magnesium is the 12th element of periodic table having electronic configuration of $[Ne]3s^2$ .

To form $Mg^{2+}$ ion, this element will loose 2 electrons.

Manganese is the 25th element of the periodic table having electronic configuration of $[Ar]3d^54s^2$

To form $Mn^{2+}$ ion, this element will loose 2 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for magnesium oxide is MgO and for manganese (II) oxide is MnO.

Thus, the formula of magnesium oxide and manganese (II) oxide is $MgO\text{ and }MnO$ respectively.

7 0

3 years ago

In the equation for the formation of magnesium sulfide shown, which atom gains electrons?

Zarrin [17]

S + 2e⁻ → S²⁻ (sulfur will gain electrons)

Mg - 2e⁻ → Mg²⁺ (magnesium will lose electrons)

Magnesium is a metal which tends to give electrons which are received by the sulfur, which is a nonmetal and tends to gain electrons.

7 0

3 years ago

Read 2 more answers