Question

(a) calculate %ic of the interatomic bonds for the intermetallic compound al6mn. (b) on the basis of this result what type of interatomic bonding would you expect to be found in al6mn?

Answer 1

The answer is:

a)0.25%

b) metallic bond

The explanation:

A) The percentage of ionic character in a compound having some covalent character can be calculated by the following equation. The percent ionic character = Observed dipole moment/Calculated dipole moment assuming 100% ionic bond × 100.

-The percent ionic character is a function of the electron negativities of the ions XA and XB .  The electronegativities for Al and Mn  are 1.5 and 1.6, respectively :

when %IC = [1-exp(-1/4) (XB-XA)^2].100

so, %IC =  [1  -  exp(- 0.25)(1.6- 1.5)^2] .  100 = 0.25%

 

(b) Because the percent ionic character is so small (0.25%) and this intermetallic compound is composed of two metals Al and Mn, the bonding is completely metallic.

Metallic bond:

• Metallic bonding can be either weak (68 kJ/mole or 0.7 eV/atom for Hg) or strong (850 kJ/mole or 8.8 eV/atom for W)

• Metallic bonding gives rise to high electrical and thermal conductivity.

-The electrons are loosely held since each atom has several unoccupied valence orbitals; it is relatively easy for the electrons to move about. In this manner the electrons allow atoms to slide past each other.

Answer 2

Interatomic bonds are the bonds found in the atomic or molecule structures of the compounds. The percentage is 0.25% and the type of interatomic bonding is metallic bond.

What are ionic characters and metallic bonds?

Ionic characters are the difference in the electronegativity of the atoms and their percentage.  Whereas, metallic bonding is the bond between the electron and the positive metal ions.

The percentage ionic character can be estimated by the formula:

$\begin{aligned}\text{Percent ionic character} = \dfrac{\text{Observed dipole moment}}{\text{Calculated dipole moment}} \times 100\%\end{aligned}$

The ions $\rm X_{A}$ and $\rm X_{B}$ have electronegativities of 1.5 and 1.6.

Substituting values in the above equation:

$\begin{aligned}\%\;\rm IC& =\rm [1-exp(-\dfrac{1}{4}) (X_{B}-X_{A})^{2}] \times 100\\\\&= \rm [1 - exp(- 0.25)(1.6- 1.5)^{2}] \times 100\\\\&= 0.25\%\end{aligned}$

The bond present is metallic as the percentage of the ionic character is very small and have two metal aluminium and manganese in it. The metallic bonds can be either weak or strong and have thermal and electrical conductivity.

Therefore, the per cent ionic character is 0.25% and the bond is a metallic bond.

Learn more about ionic character and metallic bonds here:

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