Ncert Exercises & Solutions

Identify the outer orbital octahedral complexes among the following that have the same number of unpaired electrons:
(Atomic numbers of Mn, Fe, Co, and Ni are 25, 26, 27, and 28 respectively)

(a)	${[{MnCl}_{6}]}^{3 -}$	(b)	${[{FeF}_{6}]}^{3 -}$
(c)	${[{CoF}_{6}]}^{3 -}$	(d)	${[Ni {({NH}_{3})}_{6}]}^{2 +}$

1. (a, b)
2. (b, c)
3. (c, d)
4. (a, c)

Molecular orbital electronic configuration of Mn $^{3 +} i n {[M n C l_{6}]}^{3 -} i s$

Number of unpaired electrons =4

Magnetic property = Paramagnetic

Molecular orbital electronic configuration of $C o^{3 +} i n {[C o F_{6}]}^{3 -} i s$

Number of unpaired electrons =4

Magnetic property = Paramagnetic

Molecular orbital electronic configuration of $F e^{3 +} i n {[F e F_{6}]}^{3 -}$ is

Number of unpaired electrons =4

Magnetic property = Paramagnetic

Molecular orbital electronic configuration of ${Ni}^{2 +} in {[Ni {({NH}_{3})}_{6}]}^{2 +}$ is

Number of unpaired electrons =4

Magnetic property = Paramagnetic

Thus, ${[M n C l_{6}]}^{3 -} a n d {[C o F_{6}]}^{3 -}$ are paramagnetic having four electrons each.

Hence, correct choices are (a) (c).

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