- A$A$
- B$C$
- C$D$
- ✓$B$
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$(A)$ In $B _{2} H _{6}$, all $B - H$ bonds are equivalent.
$(B)$ In $B _{2} H _{6}$ there are four $3-$centre$-2-$electron bonds.
$(C)$ $B _{2} H _{6}$ is a Lewis acid.
$(D)$ $B _{2} H _{6}$ can be synthesized form both $BF _{3}$ and $NaBH _{4}$.
$(E)$ $B _{2} H _{6}$ is a planar molecule.
Choose the most appropriate answer from the options given below..... .
$C{H_3} - C \equiv C - C{H_3}\xrightarrow{{{H_2}/Pd - BaS{O_4}}}$ $A\xrightarrow{{B{r_2}/CC{l_4}}}C$
$C{H_3} - C \equiv C - C{H_3}\xrightarrow{{Na/Liq.\,N{H_3}}}$ $B\xrightarrow{{B{r_2}/{H_2}O}}D$

Assertion $(A)$: Cis form of alkene is found to be more polar than the trans form
Reason $(R)$: Dipole moment of trans isomer of $2$-butene is zero.
In the light of the above statements, choose the correct answer from the options given below :
Reason : $NaOH$ is strong alkali.