SATHEE: Co-Ordination Compounds Question 297

Co-Ordination Compounds Question 297

Question: Among the following complexes (K-P)

$K_{3} [F e {(C N)}_{6}]$ (K), $[C o {(N H_{3})}_{6}] C l_{3} (L),$ $N a_{3} [C o {(o x a l a t e)}_{3}]$ (M), $[N i {(H_{2} O)}_{6}] C l_{2}) (N),$ $K_{2} [P t {(C N)}_{4} (O)$ and $[Z n {(H_{2} O)}_{6}] {(N O_{3})}_{2} (P)$ the diamagnetic complexes are

Options:

A) K, L, M, N

B) K, M, O, P

C) L, M, O, P

D) L, M, N, O

Show Answer

Answer:

Correct Answer: C

Solution:

Complex	No.of electrons in outer d orbital	No.of unpaired electrons (s)
${[F e {(C N)}_{6}]}^{3 -}$	$3 d^{5}$	1 ( $C N^{-}$ causes pairing of electrons)
${[C o {(N H_{3})}_{6}]}^{3 -}$	$3 d^{6}$	0
${[C o {(O X)}_{3}]}^{3 -}$	$3 d^{6}$	0
${[N i {(H_{2} O)}_{6}]}^{2 +}$	$3 d^{8}$	2
${[P t {(C N)}_{4}]}^{2 -}$	$5 d^{8}$	0 ( $C N^{-}$ cause pairing of electrons)
${[Z n {(H_{2} O)}_{6}]}^{2 +}$	$3 d^{10}$	0

Thus L, M, O and P are diamagnetic.

Co-Ordination Compounds Question 297

Question: Among the following complexes (K-P)

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Co-Ordination Compounds Question 297

Question: Among the following complexes (K-P)

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu