Question

Among ${N_2},{\text{ }}C{l_2},{\text{ }}N{H_3},\;$ and ${O_2}$ ​ choose the molecule which contains only triple bond.
A.${O_2}$
B.$C{l_2}$
C.${N_2}$
D.$N{H_3}$

Answer 1

Triple bond: A triple bond is a type of chemical bond between two atoms involving total six bonding electrons instead of the usual two in a covalent single bond. Triple bonds are usually stronger than the equivalent single bonds or double bonds containing a bond order of 3.

Complete step by step answer:
Now we will study each atom and their electronic configuration in order to find their bonding states.
${O_2}$: The Atomic Number of Oxygen is 8 and its atomic mass will be double the value of its atomic number which is 15.999 $g/mo{l^{\; - 1}}$.The electronic configuration of oxygen is $1{s^2}2{s^2}2{p^4}$. Therefore it needs 2 more electrons to complete its octet which is its stable configuration. This is the primary reason by which it can’t form triple bonds.
$C{l_2}$: The Atomic number of chlorine is 17 and its atomic mass will be approximately double the value of its Atomic Number which is 35.453 c.
The electronic configuration of oxygen is Therefore it needs 1 more electron to complete its octet which is its stable configuration. This is the primary reason by which it can’t form double or triple bonds.
${N_2}$: The Atomic number of nitrogen is 7 and its atomic mass will be approximately double the value of its atomic Number which is 14 $g/mo{l^{\; - 1}}$.The electronic configuration of nitrogen is $1{s^2}2{s^2}2{p^3}$. Therefore it needs 3 more electrons to complete its octet which is its stable configuration. This is the primary reason by which it can form triple bonds.
Thus the correct option is D.

Note:
If we see ammonia it has two elements present in it. It has a nitrogen central atom which has 5 electrons in its valence shell and needs three more electrons to complete its octet which will be given by hydrogen. So, it will form three single bonds, not triple bonds.