Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
69 views
in Chemistry by (94.2k points)
closed by
A magnetic moment of 1.73 B.M. will be shown by one among the following:
A. `TiCl_4`
B. `[CoCl_6]^(4-)`
C. `[Cu(NH_3)_4]^(2+)`
D. `[Ni(CN)_4]^(2-)`

1 Answer

0 votes
by (89.5k points)
selected by
 
Best answer
Correct Answer - C
`mu_(MM)=1.73B.M.=sqrt(3)BM`
`impliesforn=1(n=`no. of unpaired electrons).
a. In `TiCl_4:Ti(Z=22)implies3d^24s^2,Ti^(4+)=3d^(0)`
It does not have any unpaired electron.
So `mu_(MM)=0` (dimagnetic)
b.In `[Co^(2+)Cl_6^(-6)]^(4-):Co(Z=27)implies3d^74s^2,Co^(2+)=3d^7`
`Cl^(ɵ)` is a weak firld ligand, so pairing does not occur, it has `n=3`.
So `mu_(MM)=sqrt(n(n+2))BM=sqrt(15)BM`
c. In `[Cu^(2+)(Noverset(0)H_3)_4]^(2+),Cu(Z=29)implies3d^(10)4s^(1)impliesCu^(2+)=3d^(9)` It has one unpaired electron, so its `mu_(MM)=1.73BM`
image

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...