Which of the following is INCORRECT regarding | vedclass

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(C) $1$. In $B_2H_6$ (diborane),the bridge bond is a $3c-2e^-$ bond formed by $sp^3$ hybrid orbital of $B$,$1s$ orbital of $H$,and $sp^3$ hybrid orbit

Vedclass · Accepted Answer

$(BeCl_2)_n$: $sp^2-p-sp^2$

Vedclass · Answer

(C) $1$. In $B_2H_6$ (diborane),the bridge bond is a $3c-2e^-$ bond formed by $sp^3$ hybrid orbital of $B$,$1s$ orbital of $H$,and $sp^3$ hybrid orbital of $B$. Thus,$sp^3-s-sp^3$ is correct.$2$. In $Al_2Cl_6$,the bridge bond involves $sp^3$ hybrid orbital of $Al$,$p$ orbital of $Cl$,and $sp^3$ hybrid orbital of $Al$. Thus,$sp^3-p-sp^3$ is correct.$3$. In $(BeCl_2)_n$,$Be$ is $sp$ hybridized. The bridge bond involves $sp$ hybrid orbital of $Be$,$p$ orbital of $Cl$,and $sp$ hybrid orbital of $Be$. Thus,$sp-p-sp$ is the correct overlapping,making $sp^2-p-sp^2$ incorrect.$4$. In $Al_2(CH_3)_6$,the bridge bond is a $3c-2e^-$ bond involving $sp^3$ hybrid orbital of $Al$,$sp^3$ hybrid orbital of $C$ (from $CH_3$),and $sp^3$ hybrid orbital of $Al$. Thus,$sp^3-sp^3-sp^3$ is correct.

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

Which of the following is $INCORRECT$ regarding the molecule and the type of overlapping in the bridge bond?

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