Number of $s p^{3}$ hybridised atom in borax $N a_{2} B_{4} O_{7}, 10 H_{2} O$ is $x$ . Find the value of $\frac{x}{19}$

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Solution

$(N a_{2} B_{4} O_{7} . 10 H_{2} O)$
is actually $N a_{2} [B_{4} O_{5} (O H)_{4}] . 8 H_{2} O$
Anionic part of borax $\Rightarrow [B_{4} O_{5} (O H)_{4}]^{2 -}$

In the anionic part,
Number of $s p^{3}$ hybridized atom = 11 [2 boron + 9 oxygen]
Number of $s p^{2}$ hybridized atom = 2[2 boron]
Also, $8 H_{2} O$ has $8 s p^{3}$ hybridized atoms.
So, total $s p^{3}$ hybridised atoms $= 11 + 8 = 19$
$x = 19$
$∴ \frac{x}{19} = \frac{19}{19} = 1$

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