is the number of electrons in orbitals of the most stable complex ion among , and . The nature of oxide of vanadium of the type is: [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
$' X'$ is the number of electrons in $t_{2 g}$ orbitals of the most stable complex ion among ${[Fe {({NH}_{3})}_{6}]}^{3 +}$ , ${[Fe {(Cl)}_{6}]}^{3 -}, {[Fe {(C_{2} O_{4})}_{3}]}^{3 -}$ and ${[Fe {(H_{2} O)}_{6}]}^{3 +}$ . The nature of oxide of vanadium of the type $V_{2} O_{x}$ is: [2025]

1 Acidic

2 Neutral

3 Basic

4 Amphoteric

Ans.
(4)

Because of chelation effect, ${[Fe {(C_{2} O_{4})}_{3}]}^{3 -}$ is most stable. As $C_{2} O_{4}^{2 -}$ is a weak field ligand, ${Fe}^{3 +}$ $(3 d^{5})$ in ${[Fe {(C_{2} O_{4})}_{3}]}^{3 -}$ splits as $t_{2 g}^{3} e_{g}^{2}$ . Thus, ${[Fe {(C_{2} O_{4})}_{3}]}^{3 -}$ has five unpaired electrons. $V_{2} O_{5}$ is amphoteric.

Want Us to Email you About Special Offers & Updates?