Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?
Oxygen can form multiple bonds with metals, while fluorine can't form multiple bond with metals. Hence, oxygen has more ability to stabilize higher oxidation state rather than fluorine.
Although $\mathrm{Cr}^{3+}$ and $\mathrm{Co}^{2+}$ ions have same number of unpaired electrons but the magnetic moment of $\mathrm{Cr}^{3+}$ is 3.87 BM and that of $\mathrm{Co}^{2+}$ is 4.87 BM, Why?
Magnetic moment of any metal ion can be decided on the basis of spin as well as orbital contribution of electron. Due to symmetrical electronic configuration, there is no orbital contribution in $\mathrm{Cr}^{3+}$ ion. However, appreciable orbital contribution takes place in $\mathrm{Co}^{2+}$ ion.
Ionisation enthalpies of $\mathrm{Ce}, \mathrm{Pr}$ and Nd are higher than $\mathrm{Th}, \mathrm{Pa}$ and U . Why?
$\mathrm{Ce}, \mathrm{Pr}$ and Nd are lanthanoids and have incomplete $4 f$ shell while $\mathrm{Th}, \mathrm{Pa}$ and U are actinoids and have $5 f$ shell incomplete. In the beginning, when $5 f$-orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The 5 -electrons will therefore, be more effectively shielded from the nuclear charge than $4 f$ electrons of the corresponding lanthanoids.
Therefore, outer electrons are less firmly held and they are available for bonding in the actinoids.
Although Zr belongs to 4 d and Hf belongs to 5 d transition series but it is quite difficult to separate them, Why?
Separation of Zr and Hf are quite difficult because of lanthanoid contraction. Due to lanthanoid contraction, they have almost same size ( $\mathrm{Zr}=160 \mathrm{pm}$ and $\mathrm{Hf}=159 \mathrm{~pm}$ ) and thus, similar chemical properties. That's why it is very difficult to separate them by chemical methods.
Although +3 oxidation states is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?
It is due to the fact that after losing one more electron Ce acquires stable $4 f^{\circ}$ electronic configuration. So, Ce shows +4 oxidation state also alongwith +3 oxidation state.