DABCO reacts with one equivalent of BH3 to generate a white powder with composit

Question

DABCO reacts with one equivalent of BH3 to generate a white powder with composition C 57.23 %, H 11.92 %, B 8.59 %, N 22.26 %. The X-ray diffraction pattern ( = 1.54056 A) of this product displays peaks at the following values of 2 angular range 10-30 °, a peak at 9.627" was not recorded) 13.630°, 16.714, 19.322, 21.629°, 23.722, 27.459 29.160 Lattice type is: Reason: Calculate a (b) Lattigetpareaker ar a cubic basis. Determine the lattice type and lattice parameter. [10 marks] Peak position h, k, I (c) Use the elemental analysis data to determine the molecular formula and explain the bonding present in this molecule [3 marks]

Explanation / Answer

For the given reaction of DABCO with BH3,

X-ray pattern is given,

(a) to get line spacing (d), we use,

Bragg's equation,

nl = 2dsin(theta)

with,

theta = 13.630/2 = 6.815 degree

n = 1

l = 1.54056 Angstrom

we get,

1 x 1.54056 = 2dsin(6.815) = 6.49 Angstrom

For a cubic cell, the second peak in the XRD is for 110 plane

Lattice type : cubic

so,

using,

1/d*2 = (h^2 + k^2 + l^2)/a^2

with,

h = 1, k = 1, l = 0

we get,

1/(6.49)^2 = (1^2 + 1^2 + o^2)/a^2

so,

lattice parameter (a) = 9.18 Angstrom

(b) elemental analysis

moles C = 57.23/12 = 4.77 mol

moles H = 11.92/1 = 11.92 mol

moles B = 8.59/11 = 0.781 mol

moles N = 22.26/14 = 1.59 mol

divide by smallest number

C = 4.77/0.781 = 6

H = 11.92/0.781 = 15

B = 0.781/0.781 = 1

N = 1.59/0.781 = 2

So molecular formula becomes = C6H15BN

Reaction of DABCO with BH3 generates a ionic compound, wherein N is +vely charged and B is negatively charged. This is an ionic bond.

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