$(a)$ Electropositive nature of the element$(s)$ increases down the group and decreases across the period.
$(b)$ Electronegativity of the element decreases down the group and increases across the period.
$(c)$ Atomic size increases down the group and decreases across a period (left to right).
$(d)$ Metallic character increases down the group and decreases across a period.
On the basis of the above trends of the Periodic Table,answer the following about the elements with atomic numbers $3$ to $9$:
$(a)$ Name the most electropositive element among them.
$(b)$ Name the most electronegative element.
$(c)$ Name the element with the smallest atomic size.
$(d)$ Name the element which is a metalloid.
$(e)$ Name the element which shows maximum valency.

(A) The elements with atomic numbers $3$ to $9$ are: Lithium ($Li$,$Z=3$),Beryllium ($Be$,$Z=4$),Boron ($B$,$Z=5$),Carbon ($C$,$Z=6$),Nitrogen ($N$,$Z=7$),Oxygen ($O$,$Z=8$),and Fluorine ($F$,$Z=9$). These elements belong to the second period.
$(a)$ Electropositive nature decreases across a period. Thus,the first element,Lithium $(Li)$,is the most electropositive.
$(b)$ Electronegativity increases across a period. Thus,the last element,Fluorine $(F)$,is the most electronegative.
$(c)$ Atomic size decreases across a period from left to right. Thus,Fluorine $(F)$ has the smallest atomic size.
$(d)$ Boron $(B)$ is a metalloid located between metals and non-metals.
$(e)$ Valency increases from $1$ to $4$ and then decreases to $1$. Carbon $(C)$ has a valency of $4$,which is the maximum in this series.