In how many isomeric forms does B_3N_3H_4Cl_2 | vedclass

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(B) The compound $B_3N_3H_4Cl_2$ is a derivative of borazine $(B_3N_3H_6)$.In borazine,there are three boron atoms and three nitrogen atoms in the rin

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$4$

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(B) The compound $B_3N_3H_4Cl_2$ is a derivative of borazine $(B_3N_3H_6)$.In borazine,there are three boron atoms and three nitrogen atoms in the ring.When two hydrogen atoms are replaced by chlorine atoms $(Cl)$,the possible isomers are determined by the relative positions of the chlorine atoms on the ring.The isomers are:$1$. $1,2$-dichloro isomer (where $Cl$ atoms are on adjacent $B$ atoms).$2$. $1,3$-dichloro isomer (where $Cl$ atoms are on $B$ atoms separated by one $N$ atom).$3$. $1,4$-dichloro isomer (where $Cl$ atoms are on opposite $B$ atoms).$4$. Another $1,3$-dichloro isomer exists due to the asymmetry of the $B-N$ ring,where the chlorine atoms are attached to different nitrogen-boron environments.Thus,there are $4$ distinct isomeric forms.

Column $I$ (Molecules)	Column $II$ (Number of lone pairs on central atom)
$A. \ NH_3$	$1. \ \text{Two}$
$B. \ H_2O$	$2. \ \text{Three}$
$C. \ XeF_2$	$3. \ \text{Zero}$
$D. \ CH_4$	$4. \ \text{Four}$
	$5. \ \text{One}$

In how many isomeric forms does $B_3N_3H_4Cl_2$ exist?

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