Answer: Option C. -> 4f7 5d1 6s2.
:
C
Gadolinium $\left(X_{6}4Gd\right)=[Xe{]}^{54}4f^{7}5d^{1}6s^2$.

Answer: Option C. -> Availability of 4f electrons results in the formation of compounds in +4 state for all the members of the series.
:
C
Availibilty of 4f electrons does not result in the formation of compounds in +4 oxidation state for all the members of the lanthanoid series.

Answer: Option B. -> V2+ 
:
B
Option a) – Number of oxidation states increases as we move from group 3 to group 7 in same period.
Option b) – V${}^{2+}$(3 unpaired electrons) %undefined Cr${}^{2+}$ ( 4 unpaired electrons) = Fe${}^{2+}$ (4 unpaired electrons) %undefined Mn${}^{2+}$ ( 5 unpaired electrons)
Option c)- Ionic size decreases from left to right in the same period.
Option d)- As the standard reduction potential value for the couple M${}^{3+}$/M${}^{2+}$ decreases, the stability of M${}^{3+}$ ion in aqueous solution increases. Thus, correct increasing order of stability of M${}^{3+}$ ions in aqueous solution is: Co${}^{3+}$ < Fe${}^{3+}$<Cr${}^{3+}$ <Sc${}^{3+}$.