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3 years ago

What is the total number of orbitals found in the second energy level? openstudy?

Chemistry

2 answers:

kkurt [141]3 years ago

4 0

Explanation:

Second energy level consists of one "s" sub-shell and a "p" sub-shell. So, in a "s" sub-shell there is only one orbital. Whereas in a "p" sub-shell there are 3 orbitals which are $p_{x}$ , $p_{y}$ , and $p_{z}$ .

Also, each orbital can contain a maximum of two electrons.

Therefore, we can conclude that in total there are four orbitals (one from s and three from p) found in the second energy level.

Setler79 [48]3 years ago

3 0

On the second shell there are two individual subshells:
The "s" subshell has only 1 orbital with max. two electrons spinning around; and the so-called "p" subshell has 3 orbitals with max. 6 electrons (2 on each!)
In total, there are four orbitals with 8 revolving electrons on the second shell.
Hope could help :)

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ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

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