Since $O _{2}$ and $Cl _{2}$ are taken at NTP, so $T_{O_{2}}=T_{C L_{2}}=20^{\circ} C =293.15 K$ and $PO _{2}= PCr _{2}=$ $1$ atm.

Also, given $V_{ Cl _{2}}= VO _{2}$

So,

$\frac{ P_{O _{2}} V_{ O _{2}}}{P_{ Cl _{2}} V_{ Cl _{2}}}=\frac{n_{ O _{2}}}{n_{ Cl_{2}}} \times \frac{ R }{ R } \times \frac{ T_{O _{2}}}{ T_{Cl _{2}}}$

$\Rightarrow \frac{n_{c e_{2}}}{n_{ O _{2}}}=\frac{1}{1}$

Thus, option $(A)$ is corrrect.