Question

Consider following compounds:

$I:K_4\left[Fe\right(CN{)}_6],II:N{H}_{4}Cl,III:H_{2}S{O}_4$

Ionic, covalent and coordinate bonds are present in :

• A. I, II and III
• B. I and III
• C. II and III
• D. I and II

Answer 1

The correct option is A I, II and III

$H_{2}S{O}_4$ has $3$ types of bonds- ionic, covalent and coordinate. Sulphur is the central atom,surrounded by $4$ oxygen atoms and $2$ hydrogen atoms places on left of any $2$ oxygen atoms. Now, oxygen shares one electron each with the sulphur and hydrogen atom. But by this only $2$ oxygen atoms will be able to complete octet. The remaining $2$ will complete by coordinate bonding. Both the oxygen atoms will take electrons from sulphur but sulphur won't take theirs ie sharing from only one end. This is called coordinate bond. Hydronium ions are attached with sulphate ions using ionic bonding
$K_4\left[{Fe\left(CN\right)}_6\right]$: has ionic bond between $K^{+}and{\left[{Fe\left(CN\right)}_6\right]}^{4-}$, coordinate bonding in $CN$ and covalent bond in ${\left[{Fe\left(CN\right)}_6\right]}^{4-}$
${NH}_{4}Cl$: has ionic bond between ${{NH}_4}^{+}and{Cl}^{-}$, coordinate bonding in ${{NH}_4}^{+}$ and covalent bond in ${{NH}_3.}^{}$