Suppose MgSe(a=545 pm) and SrSe (a=623 pm) both crystallize with a NaCl (rock salt) structure. Calculate the ionic radius of Sr2+ (Hint:Assume that the Se2- ions actually touch in MgSe.)
To calculate the ionic radius of Sr2+ in SrSe, we can use the concept of ionic radii and assuming that the Se2- ions actually touch in MgSe.
In a NaCl structure, the cations (Sr2+ in this case) occupy the face-centered cubic (FCC) sites, while the anions (Se2-) occupy the octahedral sites.
In an FCC unit cell, there are 4 atomic spheres per unit cell. The edge length, a, of the unit cell can be calculated using the given lattice parameter:
a = 2 * r + d, where r is the radius of the atomic spheres and d is the distance between the centers of neighboring spheres.
In a NaCl structure, d = 2^(1/2) * r, because the centers of the neighboring spheres touch each other.
For MgSe, we assume that the Se2- ions touch, so the distance between the centers of neighboring spheres is equal to twice the radius of the Se2- ion:
d = 2 * r(Se)
Therefore, the equation becomes:
a(MgSe) = 2 * r(Mg) + 2 * r(Se)
Since we know the lattice parameter, a(MgSe), for MgSe and the lattice parameter, a(SrSe), for SrSe, we can compare the radii:
For MgSe:
a(MgSe) = 545 pm (given)
a(MgSe) = 2 * r(Mg) + 2 * r(Se)
For SrSe:
a(SrSe) = 623 pm (given)
a(SrSe) = 2 * r(Sr) + 2 * r(Se)
Comparing the two equations, we can see that:
r(Sr) = r(Mg) + (a(SrSe) - a(MgSe)) / 2
Plugging in the values:
r(Sr) = r(Mg) + (623 pm - 545 pm) / 2
Solving the equation gives:
r(Sr) = r(Mg) + 39 pm
Therefore, the ionic radius of Sr2+ (r(Sr)) in SrSe is equal to the ionic radius of Mg2+ (r(Mg)), plus the half of the difference between the lattice parameters of SrSe and MgSe.
To calculate the ionic radius of Sr2+, we need to make use of the ionic radius of Se2- in MgSe and the assumption that the Se2- ions actually touch in MgSe.
In a NaCl (rock salt) structure, the cation (sodium or strontium) is surrounded by six anions (chloride or selenide), and vice versa. The cation-anion distance can be calculated by summing up their respective ionic radii.
For MgSe, the anion-anion distance is equal to the distance between the centers of two Se2- ions, which can be calculated as:
Anion-anion distance (MgSe) = a - 2 × ionic radius of Se2-
= 545 pm - 2 × ionic radius of Se2-
Since MgSe crystallizes with a NaCl structure, the anion-anion distance must match that of SrSe. Therefore, we can equate the anion-anion distances of both compounds:
Anion-anion distance (MgSe) = Anion-anion distance (SrSe)
Substituting the values:
545 pm - 2 × ionic radius of Se2- = 623 pm - 2 × ionic radius of Se2-
Now, we can solve for the ionic radius of Se2-:
Ionic radius of Se2- = (545 pm - 623 pm) / 2
To find the ionic radius of Sr2+, we can use the determined ionic radius of Se2- and substitute it into the equation:
Anion-cation distance (SrSe) = ionic radius of Sr2+ + ionic radius of Se2-
Since we have previously found the anion-anion distance (SrSe) as 623 pm and the ionic radius of Se2-, we can rearrange the equation:
ionic radius of Sr2+ = Anion-cation distance (SrSe) - ionic radius of Se2-
By plugging in the values:
ionic radius of Sr2+ = 623 pm - ionic radius of Se2- (previously calculated)
By solving this equation, we can determine the ionic radius of Sr2+ in nm.