Among the following , the number of elements show only one non-zero 
oxidation state : $\mathrm{O},\hspace{0.17em}\hspace{0.17em}\mathrm{Cl},\hspace{0.17em}\mathrm{F},\hspace{0.17em}\mathrm{N},\hspace{0.17em}\mathrm{P},\hspace{0.17em}\mathrm{Sn},\hspace{0.17em}\mathrm{Tl},\hspace{0.17em}\mathrm{Na},Ti$

Na, F two elements show only one non-zero oxidation state 
$\text{F}\left({\mathrm{ns}}^2{\mathrm{np}}^5\right)=-1;\text{\hspace{0.17em}Na}\left({\mathrm{ns}}^1\right)=+1$ F is highly electro negative Na is electro positive.  
Remaining all elements exhibit more than one oxidations state 
$\text{Oxygen,O\hspace{0.17em}}\left(\mathrm{Z}=8\right)=-2,-1,-1/2,+1 \text{For\hspace{0.17em}C}\mathrm{l}\left({\mathrm{ns}}^2{\mathrm{np}}^5\right)\to -1,+1,+3,+5,+7$
$$\begin{gathered} \text{For\hspace{0.17em}Nitrogen,N\hspace{0.17em}}\to -3,-2,-1/3,+1,+2,+3,+4,+5 \\ \text{For P}\left({\mathrm{ns}}^2{\mathrm{np}}^3\right)\to -3,+3,+5 \text{Sn}\left(\mathrm{Z}=50\right){\mathrm{ns}}^2{\mathrm{np}}^2\to +2,+4 \\ \mathrm{for} \text{T}\mathrm{l}\left(\mathrm{Z}=81\right){\mathrm{ns}}^2{\mathrm{np}}^1\to +1,+3 \text{For Ti}\left(\mathrm{Z}=22\right)\to +2,+3,+4 \end{gathered}$$