Calculate the ratio of $\sigma$ and $\pi$ bonds for the following compounds and determine the correct order:
$A$: Tetracyanomethane $(C(CN)_4)$
$B$: Carbon dioxide $(CO_2)$
$C$: Benzene $(C_6H_6)$
$D$: $1, 3$-Butadiene $(CH_2=CH-CH=CH_2)$

Why does the type of overlap shown in the following figure not result in bond formation? Explain.

Match the orbital overlap figures shown in List-$I$ with the description given in List-$II$ and select the correct answer using the code given below the lists.
List-$I$:
$P$. Two $d$-orbitals overlapping axially with same phase.
$Q$. $A$ $p$-orbital and a $d$-orbital overlapping laterally with same phase.
$R$. $A$ $p$-orbital and a $d$-orbital overlapping laterally with opposite phase.
$S$. Two $d$-orbitals overlapping axially with opposite phase.
List-$II$:
$1$. $p-d$ $\pi$ antibonding
$2$. $d-d$ $\sigma$ bonding
$3$. $p-d$ $\pi$ bonding
$4$. $d-d$ $\sigma$ antibonding
Codes: $P \quad Q \quad R \quad S$

The nodal plane in an ethylene $(C_2H_4)$ molecule is:

The $p-p$ orbital overlapping is present in the following molecule:

Explain $\sigma$ and $\pi$ bonds,and mention the orbital overlapping involved in their formation.