Question

Q 37

	No. of 3c-2e bonds	No. of 2c-2e bonds
$Be(BH_4)_2$	x	p
$Al(BH_4)_3$	y	q

Then $[(y + q) – (x + p)]$ is ______.

• A. 4
• B. 8
• C. 12
• D. 2

Answer 1

Answer: (A)

4

Answer 2

In metal borohydrides, the borohydride ion ($BH_4^-$) typically coordinates to the metal center through bridging hydrogen atoms. The most common coordination mode for $BH_4^-$ ligands in compounds like $Be(BH_4)_2$ and $Al(BH_4)_3$ is bidentate bridging, where two hydrogen atoms from each $BH_4$ group bridge to the metal center.

1. For $Be(BH_4)_2$:

   • There are two $BH_4$ ligands.
   • Each $BH_4$ ligand acts as a bidentate bridging ligand, meaning it forms two B-H-Be bridges.
   • Each B-H-Be bridge is a 3c-2e bond. Thus, each $BH_4$ unit contributes 2 such bonds.
   • Total number of 3c-2e bonds in $Be(BH_4)_2$ is $x = 2 \text{ ligands} \times 2 \text{ bonds/ligand} = 4$.
   • In each $BH_4$ unit, there are a total of 4 B-H bonds. If 2 are bridging (3c-2e), the remaining $4 - 2 = 2$ are terminal B-H bonds.
   • Each terminal B-H bond is a 2c-2e bond.
   • Total number of 2c-2e bonds in $Be(BH_4)_2$ is $p = 2 \text{ ligands} \times 2 \text{ bonds/ligand} = 4$.

2. For $Al(BH_4)_3$:

   • There are three $BH_4$ ligands.
   • Each $BH_4$ ligand acts as a bidentate bridging ligand, forming two B-H-Al bridges.
   • Each B-H-Al bridge is a 3c-2e bond. Thus, each $BH_4$ unit contributes 2 such bonds.
   • Total number of 3c-2e bonds in $Al(BH_4)_3$ is $y = 3 \text{ ligands} \times 2 \text{ bonds/ligand} = 6$.
   • The remaining $4 - 2 = 2$ B-H bonds in each $BH_4$ unit are terminal 2c-2e bonds.
   • Total number of 2c-2e bonds in $Al(BH_4)_3$ is $q = 3 \text{ ligands} \times 2 \text{ bonds/ligand} = 6$.

Now, we need to calculate $[(y + q) – (x + p)]$:

• $y + q = 6 + 6 = 12$ (Total number of B-H bonds in $Al(BH_4)_3$)
• $x + p = 4 + 4 = 8$ (Total number of B-H bonds in $Be(BH_4)_2$)

Therefore, $[(y + q) – (x + p)] = 12 - 8 = 4$.

Alternatively, we can note that $x+p$ is the total number of B-H bonds in $Be(BH_4)_2$ ($2 \times 4 = 8$) and $y+q$ is the total number of B-H bonds in $Al(BH_4)_3$ ($3 \times 4 = 12$). The difference is $12 - 8 = 4$.