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Luba_88 [7]
3 years ago
15

An atom's configuration based on its number of electrons ends at 3p3. Another atom has nine more electrons. Starting at 3p, what

is the remaining configuration?
3p64s23d4
3p34s23d7
3p63d44s2
3p33d74s2
Chemistry
2 answers:
77julia77 [94]3 years ago
8 0
<h3><u>Answer;</u></h3>

3p34s23d7

<h3><u>Explanation</u>;</h3>
  • Electrons in an atom are contained in specific energy levels that are different distances from the nucleus.
  • Within each energy level is a volume of space where specific electrons are likely to be located, called orbitals. Orbitals are of different shapes, denoted by a letter (s, p, d, f, g).
  • S-orbital takes a maximum of two electrons, p-orbital take a maximum of six electrons, d-orbital takes a maximum of 10 electrons, and so fourth.
  • The electron filling pattern takes; 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p.........
  • Therefore; in this case after 3p3, we then go to 4s, with 2 electrons, then 3d which takes 7 electrons.

Maslowich3 years ago
3 0

Answer:

A) 3p64s23d4

Explanation:

First off, it's important to understand the following concepts;

An atom's electron configuration is a numeric representation of its electron orbitals. The S-orbital takes a maximum of two electrons, p-orbital take a maximum of six electrons, d-orbital takes a maximum of 10 electrons, and so fourth.

The electron filling follows this pattern; 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p.........

An electron configuration can quickly and simply tell a reader how many electron orbitals an atom has as well as the number of electrons populating each of its orbitals.

Now back to the question, since the atom's configuration ends at 3p3 and we need to add nine more electrons,

1) P orbital can only accommodate 6 electrons, so we add three electrons to the 3p3 to make it 3p6 leaving us off with (9-3 = 6 electrons left).

2) The next orbital is the 4s. The s orbital can only take in 2 electrons so we add two electrons leaving us with 4 (6-2) electrons. we have 4s2 here.

3) The next orbital on the order is 3d. Since the d orbital can take in op to 10 electrons, it means we can add the remaining 4 electrons to give us 3d4.

The final electronic configuration from 3p is now given as;

3p64s23d4

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(a) Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of t
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Answer:

a

The radius of an impurity atom occupying FCC octahedral site is 0.414{\rm{R}}

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The radius of an impurity atom occupying FCC tetrahedral site is 0.225{\rm{R}} .

Explanation:

In order to get a better understanding of the solution we need to understand that the concept used to solve this question is based on the voids present in a unit cell. Looking at the fundamentals

An impurity atom in a unit cell occupies the void spaces. In FCC type of structure, there are two types of voids present. First, an octahedral void is a hole created when six spheres touch each other usually placed at the body center. On the other hand, a tetrahedral void is generated when four spheres touch each other and is placed along the body diagonal.

Step 1 of 2

(1)

The position of an atom that fits in the octahedral site with radius \left( r \right)is as shown in the first uploaded image.

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The radius of the impurity is as follows:

2r=a-2R------(A)

The relation between radius of atom and edge length is calculated using Pythagoras Theorem is shown as follows:

Consider \Delta {\rm{XYZ}} as follows:

(XY)^ 2 =(YZ) ^2 +(XZ)^2

Substitute XY as{\rm{R}} + 2{\rm{R + R}} and {\rm{YZ}} as a and {\rm{ZX}} as a in above equation as follows:

(R+2R+R) ^2 =a ^2 +a^ 2\\16R ^2 =2a^ 2\\ a =2\sqrt{2R}

Substitute value of aa in equation (A) as follows:

r= \frac{2\sqrt{2}R -2R }{2} \\ =\sqrt{2} -1R\\ = 0.414R

The radius of an impurity atom occupying FCC octahedral site is 0.414{\rm{R}}

Note

An impure atom occupies the octahedral site, the relation between the radius of atom, edge length of unit cell and impure atom is calculated. The relation between the edge length and radius of atom is calculated using Pythagoras Theorem. This further enables in finding the radius of an impure atom.  

Step 2 of 2

(2)

The impure atom in FCC tetrahedral site is present at the body diagonal.

The position of an atom that fits in the octahedral site with radius rr is shown on the second uploaded image :

In the above diagram, R is the radius of atom and a is the edge length of the unit cell.

The body diagonal is represented by AD.

The relation between the radius of impurity, radius of atom and body diagonal is shown as follows:

AD=2R+2r----(B)

   In    \Delta {\rm{ABC}},

(AB) ^2 =(AC) ^2 +(BC) ^2

For calculation of AD, AB is determined using Pythagoras theorem.

Substitute {\rm{AC}} as a and {\rm{BC}} as a in above equation as follows:

(AB) ^2 =a ^2 +a ^2

AB= \sqrt{2a} ----(1)

Also,

AB=2R

Substitute value of 2{\rm{R}} for {\rm{AB}} in equation (1) as follows:

2R= \sqrt{2} aa = \sqrt{2} R

Therefore, the length of body diagonal is calculated using Pythagoras Theorem in \Delta {\rm{ABD}} as follows:

(AD) ^2 =(AB) ^2 +(BD)^2

Substitute {\rm{AB}} as \sqrt 2a   and {\rm{BD}} as a in above equation as follows:

(AD) ^2 =( \sqrt 2a) ^2 +(a) ^2 AD= \sqrt3a

For calculation of radius of an impure atom in FCC tetrahedral site,

Substitute value of AD in equation (B) as follows:

\sqrt 3a=2R+2r

Substitute a as \sqrt 2{\rm{R}} in above equation as follows:

( \sqrt3 )( \sqrt2 )R=2R+2r\\\\

r = \frac{2.4494R-2R}{2}\\

=0.2247R

\approx 0.225R

The radius of an impurity atom occupying FCC tetrahedral site is 0.225{\rm{R}} .

Note

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