Question

Match the column I with column II and mark the appropriate choice.

	column I			column II
A	NaH		i	Interstitial hydride
B	$C{H}_4$		ii	Molecular hydride
C	$V{H}_{0.56}$		iii	Ionic hydride
D	$B_2{H}_6$		iv	Electron-deficient hydride

• A. $\left(A\right)\to \left(iii\right),\left(B\right)\to \left(iv\right),\left(C\right)\to \left(ii\right)\left(D\right)\to \left(i\right)$
• B. $\left(A\right)\to \left(ii\right),\left(B\right)\to \left(iv\right),\left(C\right)\to \left(iii\right)\left(D\right)\to \left(i\right)$
• C. $\left(A\right)\to \left(i\right),\left(B\right)\to \left(ii\right),\left(C\right)\to \left(iv\right)\left(D\right)\to \left(iii\right)$
• D. $\left(A\right)\to \left(iii\right),\left(B\right)\to \left(ii\right),\left(C\right)\to \left(i\right)\left(D\right)\to \left(iv\right)$

Answer 1

The correct option is D $\left(A\right)\to \left(iii\right),\left(B\right)\to \left(ii\right),\left(C\right)\to \left(i\right)\left(D\right)\to \left(iv\right)$

NaH is an ionic compound and is made of sodium cations $\left(N{a}^{+}\right)$ and hydride anions $\left(H^{-}\right)$. It has the octahedral crystal structure with each sodium ion surrounded by six hydride ions.

$C{H}_4$ forms molecular or covalent hydrides. These are mainly formed by p - block elements and some s - block elements.

The third compound has an oxidation number of hydrogen which is zero. So it belongs to Interstitial hydride.

In $B_2{H}_6$ the Boron atom is surrounded by 6 electrons, so it is electron deficient due to its incomplete octet.

The correct option is D.