Boron hydrides are prepared by the action of dil. HCl on \({Mg_3B_2}\) .
All the B-H bonds in \(B_2H_6\) are not equal. The four terminal B-H bonds are shorter (B-H bond length = 1.19 \(\mathring A\)) than the two bridged B-H-B bonds (bond length - 1.33 \(\mathring A\)).
So, the correct option is (B): All the B-H bonds in $B_2H_6$ are equal.
Given below are two statements.
In the light of the above statements, choose the correct answer from the options given below:
Given below are two statements:
Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen.
Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it.
In the light of the above statements, choose the correct answer from the options given below:
Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. $(\mathrm{B}<\mathrm{Al}),(\mathrm{Al}<\mathrm{Ga}),(\mathrm{Ga}<\mathrm{In})$ and $(\mathrm{In}<\mathrm{Tl})$ Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius $\left(\mathrm{M}^{3+}\right)$ than the other one. The atomic number of the element (X) is
Match List-I with List-II: List-I