Assertion (A): Among the lanthanoids, Ce and Tb are the two whose +4 state is most commonly met in solid compounds, such as CeO2 and TbO2. Reason (R): Ce^4+ and Tb^4+ attain the 4f^0 and 4f^7 configurations respectively. Choose the correct option.

Both A and R are true and R is the correct explanation of A

Both A and R are true but R is not the correct explanation of A

JEE Advanced levelSingle correct optionDifficulty 3 of 3df203

Assertion (A): Among the lanthanoids, Ce and Tb are the two whose $+ 4$ state is most commonly met in solid compounds, such as $CeO X_{2}$ and $TbO X_{2}$ . Reason (R): $Ce X^{4 +}$ and $Tb X^{4 +}$ attain the $4 f^{0}$ and $4 f^{7}$ configurations respectively. Choose the correct option.

Ce (

4 f^{1} 5 d^{1} 6 s^{2}

) loses four electrons to give

4 f^{0}

. Tb (

4 f^{9} 6 s^{2}

) loses

6 s^{2}

and two

4 f

electrons to give

4 f^{7}

. The empty and half-filled

4 f

subshells are the extra stabilisation behind the +4 states; R explains A (in NCERT-level terms).