Question

In ${N_2}$ molecule ,the atoms are bonded by-
(A) $1\sigma$ and $2\pi$ bonds , 2 L.P.
(B) $1\sigma$ and $1\pi$ bonds, 1L.P.
(C) $2\sigma$ and $1\pi$ bonds ,No L.P.
(D) $1\sigma$and $1\pi$bonds ,No L.P.

Answer 1

Draw the Lewis Dot Structure of the nitrogen molecule . Observe the number of electrons being shared between both the nitrogen atoms . According to the number of electrons being shared , calculate the number of bonds( bonding electrons) and lone pairs.

Complete answer:
The atomic number of Nitrogen is 7. The electronic distribution of N is 2,5. It means it has 5 valence electrons . The Lewis symbol of nitrogen will look like $:N \vdots$
Consider the nitrogen molecule and draw its Lewis Dot Structure. Before drawing Lewis Dot Structure ,we need to keep in mind a few points in mind.
While drawing the structure ,consider only the valence shell electron i.e. only 5 electrons. The octet of each atom should be complete.
Keeping all this in mind we draw the structure of ${N_2}$ molecule .
$:N \vdots \vdots N:$
Here three pairs of electrons are being shared between two nitrogen atoms, so a triple bond is formed.
$:N \equiv N:$
The octet of both the nitrogen atom is complete. There are three bonds between two nitrogen atoms. We know that only one $\sigma$ (sigma) bond can be present between any two atoms . So ,one out of the three is a $\sigma$ bond and the rest two are $\pi$ bonds.
From the structure it is clear that 3 electrons of each atom are in bonding and each atom has a pair of nonbonding or lone pairs of electrons.
Hence, option (A) is correct.

Note:
Sigma bonds are stronger than pi bonds. Because of the presence of three bonds in nitrogen molecules, it has a very high bond enthalpy . Nitrogen gas is inert at room temperatures( at lower temperatures).