Chemistry, 10.10.2019 05:30, Queenbee2304

Calculate the radius of tantalum (ta) atom, given that ta has a bcc crystal structure, a density of 16.6 g/cm, and an atomic weight of 180.9 g/mol. (avogadro number, 6.023 x 103 atoms/mol).

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answered: carlohillyard4633

The radius of tantalum (Ta) atom is $R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

Explanation:

From the Body-centered cubic (BBC) crystal structure we know that a unit cell length a and atomic radius R are related through

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

So the volume of the unit cell $V_{c}$ is

$V_{c}= a^3=(\frac{4R}{\sqrt{3} } )^3=\frac{64\sqrt{3}R^3}{9}$

We can compute the theoretical density ρ through the following relationship

$\rho=\frac{nA}{V_{c}N_{a}}$

where

n = number of atoms associated with each unit cell

A = atomic weight

$V_{c}$ = volume of the unit cell

$N_{a}$ = Avogadro’s number ( $6.023 \times 10^{23}$ atoms/mol)

From the information given:

A = 180.9 g/mol

ρ = 16.6 g/cm^3

Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.

We can use the theoretical density ρ to find the radio R as follows:

$\rho=\frac{nA}{V_{c}N_{a}}\\\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}$

Solving for R

$\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}$

Substitution for the various parameters into above equation yields

$R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

answered: Guest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.