Question

Match the column I with column II and mark the appropriate choice.

	Column I		Column II
(A)	An element which can show +8 oxidation state	(i)	$Ce$
(B)	An element with 7+ as the most stable oxidation state in its oxides	(ii)	$Pm$
(C)	Radioactive lanthanoid	(iii)	$Os$
(D)	Lanthanoid which shows +4 oxidation state	(iv)	$Mn$

• A. $\left(A\right)\to \left(i\right),\left(B\right)\to \left(ii\right),\left(C\right)\to \left(iii\right),\left(D\right)\to \left(iv\right)$
• B. $\left(A\right)\to \left(ii\right),\left(B\right)\to \left(iii\right),\left(C\right)\to \left(iv\right),\left(D\right)\to \left(i\right)$
• C. $\left(A\right)\to \left(iv\right),\left(B\right)\to \left(i\right),\left(C\right)\to \left(ii\right),\left(D\right)\to \left(iii\right)$
• D. $\left(A\right)\to \left(iii\right),\left(B\right)\to \left(iv\right),\left(C\right)\to \left(ii\right),\left(D\right)\to \left(i\right)$

Answer 1

The correct option is D $\left(A\right)\to \left(iii\right),\left(B\right)\to \left(iv\right),\left(C\right)\to \left(ii\right),\left(D\right)\to \left(i\right)$

A${⟶}_{76}Os⟶$ Outer electronic configuration is $Xe4f^{14}5d^{6}6s^2$

$\downarrow$

Losing these $8e^{-}s$ it will attain fully filled configuration. Hence, can show an oxidation of $+8$.

B${⟶}_{25}Mn⟶$ Shows $+7$ as the most stable oxidation state in its' oxides, configuration$⟶4s^{2}3d^5$: losing these $7e^{-}$ attains $+7$ oxidation state.

C$⟶Pm⟶$ Radioactive Lanthanoid as it emits $\beta -$ radiations.

D$⟶$Lanthanoid showing $+4$ oxidation state$⟶$$Ce$ losing $4e^{-}s$ attains stable fulfilled $e^{-}$ and hence exhibits $+4$ oxidation state.