UO3 is heated in hydrogen to give a solid Y in which U is in the +4 state. Y reacts with HF to give Z, and Z reacts with F2 to give a volatile compound W. (UO3 + H2 -> UO2 + H2O; UO2 + 4HF -> UF4 + 2H2O; UF4 + F2 -> UF6.) The oxidation state of U in W and the number of 5f electrons on the uranium cation of Y (U: 5f^36d^17s^2) are respectively

JEE Advanced levelSingle correct optionDifficulty 3 of 3df213

$UO X_{3}$ is heated in hydrogen to give a solid Y in which U is in the $+ 4$ state. Y reacts with HF to give Z, and Z reacts with $F X_{2}$ to give a volatile compound W. ( $UO X_{3} + H X_{2} UO X_{2} + H X_{2} O$ ; $UO X_{2} + 4 HF UF X_{4} + 2 H X_{2} O$ ; $UF X_{4} + F X_{2} UF X_{6}$ .) The oxidation state of U in W and the number of $5 f$ electrons on the uranium cation of Y (U: $5 f^{3} 6 d^{1} 7 s^{2}$ ) are respectively

The sequence gives Y =

UO X_{2}

, Z =

UF X_{4}

and W =

UF X_{6}

. In

UF X_{6}

x + 6 (- 1) = 0

gives U(+6). In Y the cation is

U X^{4 +}

, which has lost

7 s^{2}

6 d^{1}

and one

5 f

electron, leaving

5 f^{2}